Solutions to exercises

Appendix I Solutions to exercises

Here we collect all solutions to the exercises which have been posted so far. In the html version these are already accessible via links, and so this is mostly useful for the pdf version.

1 Maximum principles for ordinary differential equations
1.1 Maximum principles

Exercises

1.1.1. (PS1) Some basic counterexamples.

1.1.1.a

Solution.

Consider the function \(u(x) = \cos x\) on the interval \([-\pi/2,\pi/2]\text{.}\) Then \(u \in C^2((-\pi/2,\pi/2))\cap C^0([\pi/2,\pi/2])\) and \(Lu \equiv 0\text{,}\) but \(u\) achieves its maximum over \([-\pi/2,\pi/2]\) at the interior point \(0\text{.}\)

Comment 1.

Remember that

\begin{gather*} Lu = \left(\frac{d^2}{dx^2} + 1\right) u = u'' + u\text{.} \end{gather*}

In particular, \(Lu\) is not \(u''+1\text{.}\)

Comment 2.

Of course this counterexample is in no way unique. Some other counterexamples that appeared on the sheets in previous years include

\(u=2-\frac 12 x^2\) on \([-1,1]\text{,}\)
\(u=9-4(x-1/2)^2\) on \([0,1]\text{,}\)
\(u=\sin x\) on \([0,\pi]\text{.}\)

Note that the first two are solutions of differential inequality \(Lu \ge 0\) rather than the ODE \(Lu=0\text{.}\)

1.1.1.b

Solution.

Consider the function \(u(x) = -\cosh x\) on the interval \([-1,1]\text{.}\) Then \(u \in C^2((-1,1))\cap C^0([-1,1])\) and \(Lu \equiv 0\text{,}\) but \(u\) achieves its maximum over \([-1,1]\) at the interior point \(0\text{.}\)

Comment 1.

Remember that

\begin{gather*} Lu = \left(\frac{d^2}{dx^2} - 1\right) u = u'' - u\text{.} \end{gather*}

In particular, \(Lu\) is not \(u''-1\text{.}\)

Comment 2.

Again this counterexample is in no way unique. Some other counterexamples that appeared on the sheets in previous years include

\(u=-2-\frac 12 x^2\) on \([-1,1]\text{,}\)
\(u=-9-4(x-1/2)^2\) on \([0,1]\text{,}\)
\(u=-e^x-e^{-x}\) on \([-1,1]\text{,}\)
\(u=-(e^x+e^{1-x})\) on \([0,1]\text{.}\)

If you came up with a counterexample involving exponentials, it can be an interesting exercise to try and come up with one using polynomials instead, and conversely.

A solution from a few years ago which I particularly liked, from a pedagogical point of view, went as follows. Consider functions of the form \(u=-x^2+a\) on \([-1,1]\) where \(a \in \R\) is a constant ‘to be determined’. The \(-x^2\) guarantees that the maximum of \(u\) occurs at the interior point \(0\text{,}\) and also that \(u\) is non-constant, which is a good start. It remains to show that by picking \(a\) appropriately we can ensure the inequality \(Lu \ge 0\) holds on \((-1,1)\text{.}\)

1.1.2. (PS1) A concrete application.

1.1.2.a

Solution.

The operator \(L\) is of the form (1.1) with \(g \equiv 0\) and \(h \equiv -1\text{.}\) Both of these are clearly bounded functions, and moreover \(h \le 0\) so that we can apply Theorem 1.4. If the desired inequality does not hold, then we must have \(u \gt 0\) at some point in \([0,\log 3]\text{.}\) Thus the maximum \(M\) of \(u\) must also be strictly positive. Since \(u\) vanishes at the endpoints of the interval, we deduce that it achieves \(M\) at some point \(c \in (0,\log 3)\text{.}\) But then Theorem 1.4 implies that \(u \equiv M \gt 0\) is constant, contradicting the fact that \(u\) vanishes at the endpoints.

Comment.

Do make sure that you are checking the hypotheses of results before you apply them. The official solution is quite verbose about this, and on an exam you could probably get away with less. But on an exam the question could also have been engineered so that one of these hypotheses failed in a subtle way!

1.1.2.b

Solution.

Following the hint, we observe that the particular solution \(u_p=-1\) satisfies \(Lu_p=1\text{.}\) Thus \(Lu_1 = 1\) if and only if \(u_h = u_1 - u_p\) solves \(Lu_h = 0\text{.}\) Recalling that the general solution to this latter ODE is \(u_h = Ae^x + Be^{-x}\) for some constants \(A,B \in \R\text{,}\) we conclude that \(u_1\) must be of the form

\begin{equation*} u_1 = Ae^x + Be^{-x} - 1 \text{.} \end{equation*}

Inserting into the boundary conditions at \(0\) and \(\log 3\text{,}\) we find the linear system

\begin{align*} A+B \amp= 1 \\ 3A + \frac 13 B \amp = 1 \end{align*}

which has the unique solution \(A=1/4\text{,}\) \(B=3/4\text{.}\) We conclude that

\begin{equation*} u_1 = \tfrac 14e^x + \tfrac 34e^{-x} - 1 \end{equation*}

is the unique solution.

Comment.

Sometimes students write something to effect that the solution has to be unique because there were two boundary conditions. This is not true in general. For a counterexample, consider the problem \(u''+u=0\) on \((-\pi/2,\pi/2)\) with the boundary conditions \(u(-\pi/2)=u(\pi/2)=0\text{.}\) Then \(u = \alpha \cos x\) is a solution for any constant \(\alpha \in \R\text{,}\) and so there are an infinite number of solutions.

We could also have gotten uniqueness – crucially without having to think about how to solve (✶) – simply by citing Theorem 1.6.

1.1.2.c

Solution.

Let \(v=u_1-u\text{.}\) Then \(v(0)=v(\log 3)=0\text{,}\) and moreover

\begin{equation*} Lv = Lu_1 - Lu = 1-f \ge 0 \ina (0,\log 3). \end{equation*}

Certainly \(v \in C^2((0,\log3) \cap C^0([0,\log 3])\text{,}\) and so applying Theorem 1.4 we find as in Part a that \(v \le 0\) on \([0,\log 3]\text{.}\) Rearranging yields the desired inequality.

1.1.2.d

Solution.

We have shown that \(u_1(x) \le u(x) \le 0\) for all \(x \in [0,\log 3]\text{.}\) Plugging in \(x=\log 2\) we obtain

\begin{equation*} -\frac 18 = u_1(\log 2) \le u(\log 2) \le 0\text{.} \end{equation*}

Suppose that \(u(\log 2)=0\text{.}\) Then by Part a \(u\) achieves its maximum at the interior point \(\log 2 \in (0,\log 3)\text{,}\) and so Theorem 1.4 implies that \(u \equiv 0\text{.}\) Similarly, if \(u(\log 2) = -1/8\) then by Part c \(v=u_1-u\) achieves its maximum at the interior point \(\log 2\text{,}\) and so Theorem 1.4 implies that \(v \equiv 0\text{,}\) i.e. \(u \equiv u_1\text{.}\)

1.1.2.e

Solution.

We have shown that \(u\) achieves it maximum \(M=0\) at the left endpoint \(0\text{.}\) Applying Theorem 1.5, we deduce that either \(u'(0) \lt 0\) or else \(u \equiv 0\text{.}\) In either case \(u'(0) \le 0\text{.}\) Arguing similarly at the right endpoint \(\log 3\) gives \(u'(\log 3) \ge 0\text{.}\)

We have also shown that \(v=u_1-u\) achieves its maximum at the left and right endpoints, and hence that \(v'(0) \le 0\) and \(v'(\log 3) \ge 0\text{.}\) Since \(u_1'(0)=-1/2\) and \(u_1'(\log 3)=1/2\text{,}\) the desired inequalities follow from substituting the definition of \(v\) and rearranging.

If \(u'(0)=0\) or \(u'(\log 3)=0\text{,}\) then our above argument involving Theorem 1.5 implies that \(u \equiv 0\text{.}\) Similarly, if \(u'(0)=-1/2\) or \(u_1'(\log 3)=1/2\) then \(u \equiv u_1\text{.}\)

1.1.3. (PS1) Basic lemma for \(h \le 0\).

Solution.

Suppose that \(u(c)=M \ge 0\) for some \(c \in (a,b)\text{.}\) Then from basic calculus we have

\begin{equation*} u'(c) = 0, \qquad u''(c) \le 0. \end{equation*}

Moreover, by assumption we have \(h(c) M \le 0\text{.}\) Thus

\begin{gather*} Lu(c) = u''(c) + h(c) M \le 0, \end{gather*}

which contradicts the assumption that \(Lu \gt 0\text{.}\)

1.1.4. Theorem 1.4.

Solution 1.

Using Exercise 1.1.3 as a replacement for Lemma 1.1 in the proof of Theorem 1.2, the only thing we need to check is that we can still choose \(\alpha \gt 0\) large enough that \(Lz \gt 0\) on \((a,b)\text{,}\) or indeed on the smaller interval \((a,d)\text{.}\) Differentiating, we find

\begin{gather*} Lz = z'' + gz' + hz= (\alpha^2+\alpha g + h)e^{\alpha(x-c)} - h. \end{gather*}

Since \(h \le 0\text{,}\) for \(Lz \gt 0\) it is enough to have \(\alpha^2 + \alpha g + h \gt 0\text{.}\) Since \(g\) and \(h\) are bounded, we can easily choose \(\alpha \gt 0\) large enough so that this is the case on \((a,b)\) and hence \((a,d)\text{.}\)

Solution 2.

Here is an alternative self-contained version which repeats the relevant parts of the proof of Theorem 1.2. Suppose that \(u(c) = M\) for some \(c \in (a,b)\) and that \(u\) is non-constant. Then there exists \(d \in (a,b)\) such that \(u(d) \lt M\text{;}\) we assume without loss of generality that \(d \gt c\text{.}\) Consider the function

\begin{gather*} z(x) = e^{\alpha(x-c)} - 1 \end{gather*}

where \(\alpha \gt 0\) is a constant to be determined. Note that \(z\) is negative on \((a,c)\) and positive on \((c,b)\text{.}\) Differentiating we find

\begin{gather*} Lz = z'' + gz' + hz= (\alpha^2+\alpha g + h)e^{\alpha(x-c)} - h. \end{gather*}

Now consider

\begin{gather*} w = u + \varepsilon z \end{gather*}

where \(\varepsilon \gt 0\) is a small parameter to be determined. Since \(z(a) \lt 0\text{,}\) we have \(w(a) \lt u(a) \le M\text{.}\) Similarly, since \(u(d) \lt M\text{,}\) we can pick \(\varepsilon \gt 0\) small enough that

\begin{gather*} w(d) = u(d) + \varepsilon z(d) \lt M \end{gather*}

as well. Finally, since \(z(c) = 0\) we have \(w(c) = u(c) = M\text{.}\) But this means that \(w\) achieves its maximum \(\ge M\) over \([a,d]\) at some interior point in \((a,d)\text{,}\) contradicting Exercise 1.1.3.

1.1.6. (PS1) Alternative proof of Theorem 1.2.

Solution.

Following the hint, assume for the sake of contradiction that \(u(c)=M\) for some \(c \in (a,b)\text{.}\) By calculus, we then have that \(u'(c)=0\text{.}\) Applying Theorem 1.3 to \(u\) on \((a,c)\) we have that either \(u'(c) \gt 0\) or else \(u \equiv M\) on \((a,c)\text{.}\) Since \(u'(c)=0\text{,}\) the only possibility is that \(u \equiv M\) on \((a,c)\text{.}\) Similarly, applying Theorem 1.3 to \(u\) on \((c,b)\) we conclude that \(u \equiv M\) on \((c,b)\text{.}\) Thus \(u \equiv M\) on all of \((a,b)\) as desired.

Comment.

When we apply Theorem 1.3 (or indeed our other results from this section) to \(u\) on a subinterval such as \([a,c]\text{,}\) we are really applying the theorem to the restriction \(\tilde u = u|_{[a,c]}\text{.}\) The possible conclusion is therefore \(\tilde u \equiv M\text{,}\) or in other words that \(u \equiv M\) on \([a,c]\text{.}\) In particular, we cannot conclude anything whatsoever about the behaviour of \(u\) outside of \([a,c]\text{.}\) If this tripped you up, it may be worth making another attempt before peeking at the official solution.

1.2 Applications

Exercises

1.2.2. (PS1) A ‘generalised’ maximum principle.

1.2.2.a

Solution.

First we observe that \(\cos x \gt 0\) for \(\abs x \le \frac \pi 4\text{.}\) Thus we can obtain (✶) simply by calculating

\begin{align*} 0 \amp \le u'' + u \\ \amp = (w\cos x)'' + w\cos x\\ \amp = (\cos x) w'' - 2(\sin x) w' \end{align*}

and then dividing by \(\cos x\text{.}\)

Comment.

It is crucial here that \(\cos x \gt 0\) on the interval \((-\frac \pi 4, \frac \pi 4)\) in question. If we had the reverse inequality \(\cos x \lt 0\text{,}\) then in the end we would get the reverse inequality

\begin{gather*} w'' - 2\frac{\sin x}{\cos x} w' \le 0 \end{gather*}

and so would be in a position to apply minimum principles rather than maximum principles.

1.2.2.b

Solution.

The inequality (✶) can be written as \(L w \ge 0\text{,}\) where here the coefficients are \(h \equiv 0\) and \(g = -2\tan x\text{,}\) which are bounded on the interval \((a,b)=(-\frac\pi 4,\frac \pi 4)\text{.}\) By Theorem 1.2, \(w\) therefore achieves its maximum \(M\) at one of the endpoints \(a,b\) and not in the interior, unless \(w \equiv M\) is constant. If \(w \equiv M\text{,}\) then \(u \equiv M \cos x\text{,}\) and so assume that \(w \not \equiv M\text{.}\) Then we have

\begin{equation*} w \lt M = \max(w(a),w(b)) \text{ for } x \in (a,b) \text{,} \end{equation*}

or in other words

\begin{equation*} \frac u{\cos x} \lt \max\Big( \frac{u(\frac \pi 4)}{\cos(\frac \pi 4)}, \frac{u(-\frac \pi 4)}{\cos(-\frac \pi 4)} \Big) \text{ for } x \in (-\tfrac\pi 4,\tfrac \pi 4) \text{.} \end{equation*}

Evaluating the cosines on the right hand side and multiplying through by \(\cos x\) yields the desired inequality.

2 Preliminaries
2.1 Index notation

Exercises

2.1.1. (PS2) Summation convention.

2.1.1.a

Solution.

\(\trace A = A_{ii}\text{.}\)

2.1.1.b

Solution.

\((Ax)_i = A_{ij} x_j\text{.}\)

2.1.1.c

Solution.

\((ABC)_{ij} = A_{ik} B_{k\ell} C_{\ell j}\text{.}\)

2.1.1.d

Solution.

\(y \cdot Ax = y_i A_{ij} x_j\text{.}\)

2.1.1.e

Solution.

\(\langle Ax,By\rangle = A_{ij} x_j B_{ik} y_k\text{.}\)

Comment.

Suppose we are multiplying together two finite sums

\begin{gather*} \left(\sum_{i=1}^N a_i\right) \left(\sum_{i=1}^N b_i\right) = \left(a_1 + \cdots + a_N\right) \left(b_1 + \cdots + b_N\right)\text{.} \end{gather*}

If we expand this out completely, all possible products \(a_i b_j\) will appear, and not just the ‘diagonal’ products \(a_i b_i\) (no sum). One way to write this out is to say

\begin{gather*} \left(\sum_{i=1}^N a_i\right) \left(\sum_{j=1}^N b_j\right) = \sum_{i=1}^N \sum_{j=1}^N a_i b_j\text{.} \end{gather*}

Applying the same logic here, we have

\begin{align*} \langle Ax,By\rangle \amp= \sum_{i=1}^N (A x)_i (By)_i\\ \amp= \sum_{i=1}^N \left(\sum_{j=1}^N A_{ij} x_j\right) \left(\sum_{k=1}^N B_{ik} x_k\right)\\ \amp = \sum_{i=1}^N \sum_{j=1}^N \sum_{k=1}^N A_{ij} x_j B_{ik} x_k\text{,} \end{align*}

where the last equality only works because we use different indices (\(j\) and \(k\)) in the two inner sums.

2.1.1.f

Solution.

\(\trace(A^\top B) = A_{ij} B_{ij}\)

2.2 Partial derivatives

Exercises

2.2.1. (PS2) Derivatives of linear and quadratic functions.

2.2.1.a

Solution.

Using the definition of the Jacobian matrix, we have

\begin{gather*} (Dx)_{ij} = \partial_j x_i = \delta_{ji}\text{,} \end{gather*}

i.e. that \(Dx\) the identity matrix.

2.2.1.b

Solution.

We calculate

\begin{align*} (D(Ax))_{ij} \amp= \partial_j (Ax)_i\\ \amp= \partial_j (A_{ik} x_k)\\ \amp= A_{ik} \partial_j x_k\\ \amp= A_{ik} \delta_{jk}\\ \amp= A_{ij} \text{.} \end{align*}

In other words, \(D(Ax)=A\text{.}\)

2.2.1.c

Solution.

Using the previous part, we calculate

\begin{gather*} \nabla \cdot (Ax) = \trace D(Ax) = \trace A = A_{ii}\text{.} \end{gather*}

2.2.1.d

Solution.

We calculate

\begin{align*} (\nabla (x \cdot Ax))_i \amp= \partial_i ( x_j A_{jk} x_k)\\ \amp= \delta_{ij}A_{jk} x_k + x_j A_{jk} \delta_{ki}\\ \amp= A_{ik} x_k + x_j A_{ji}\\ \amp= A_{ij} x_j + x_j A_{ji}\\ \amp= (A_{ij}+A_{ji}) x_j\text{.} \end{align*}

In other words, \(\nabla (x \cdot Ax) = (A+A^\top)x\text{.}\)

Comment.

It is easy to check that, for any \(x \in \R^N\text{,}\) \(x \cdot Ax = x \cdot A^\top x\text{.}\) In particular, the function \(x \mapsto x \cdot Ax\) in this question is exactly the same function as \(x \mapsto \tfrac 12 x \cdot (A+A^\top)x\text{.}\) This strongly suggests that the expression \(A+A^\top\) should somehow appear in the formula for the gradient, either explicitly or implicitly.

2.2.1.e

Solution.

Starting from the formula from the previous part, we have

\begin{align*} (D^2 (x \cdot Ax))_{ik} \amp = \partial_k \partial_i (x \cdot Ax)\\ \amp = \partial_k [(A_{ij}+A_{ji}) x_j]\\ \amp = (A_{ij}+A_{ji}) \delta_{jk}\\ \amp = A_{ik}+A_{ki}\text{.} \end{align*}

In other words, \(D^2 (x \cdot Ax) = A + A^\top\text{.}\)

Comment.

Since \(x \mapsto x \cdot Ax\) is a smooth function, its Hessian \(D^2(x \cdot Ax)\) has to be a symmetric matrix. So it can’t possibly be \(2A\text{,}\) which isn’t necessarily symmetric, but could plausibly be \(A+A^\top\text{,}\) which is always symmetric.

2.2.1.f

Solution.

Starting from the formula from the previous part, we have

\begin{align*} \Delta (x \cdot Ax) \amp= \partial_i \partial_i (x \cdot Ax)\\ \amp = A_{ii}+A_{ii}\\ \amp = 2 \trace A\text{.} \end{align*}

2.2.1.g

Solution 1. Direct argument

We calculate

\begin{align*} \Delta (\abs{Ax}^2) \amp= \partial_i \partial_i (A_{jk} x_k A_{j\ell} x_\ell)\\ \amp= \partial_i (A_{jk} \delta_{ki} A_{j\ell} x_\ell + A_{jk} x_k A_{j\ell} \delta_{\ell i})\\ \amp= \partial_i (A_{ji} A_{j\ell} x_\ell + A_{jk} x_k A_{ji} )\\ \amp= A_{ji} A_{j\ell} \delta_{\ell i} + A_{jk} \delta_{ki} A_{ji} \\ \amp= A_{ji} A_{ji} + A_{ji} A_{ji} \\ \amp= 2 \abs A^2, \end{align*}

where in the last step we have recognised the matrix norm from Definition 2.5.

Solution 2. Slicker argument

Perhaps unsurprisingly, we can save a bit of work by reusing one of the earlier parts. Note that

\begin{equation*} \abs{Ax}^2 = \langle Ax,Ax\rangle = \langle x,A^\top Ax\rangle = x \cdot Bx \end{equation*}

where \(B=A^\top A\text{.}\) By an earlier part we therefore have

\begin{equation*} D^2 (\abs{Ax}^2) = B + B^\top = 2B\text{,} \end{equation*}

where in the last step we have used that \(B\) is symmetric. In particular,

\begin{equation*} \Delta(\abs{Ax}^2) = \trace(2B) = 2A_{ij}A_{ij} = 2 \abs{A}^2\text{.} \end{equation*}

2.2.2. (PS2) Derivative of the modulus.

Solution 1.

We have

\begin{align*} \partial_i \abs x \amp= \partial_i \sqrt{x_1^2+\cdots+x_N^2}\\ \amp= \tfrac 12 (x_1^2+\cdots+x_N^2)^{-1/2} 2x_i\\ \amp= \frac{x_i}{\abs x}\text{.} \end{align*}

In other words,

\begin{equation*} \nabla \abs x = \frac x{\abs x} \text{.} \end{equation*}

Solution 2.

We can actually use the summation convention here if we square things first and then work backwards. Let \(f \maps x \mapsto \abs x\) and \(g \maps x \mapsto \abs x^2 = (f(x))^2\text{.}\) Then using the chain rule we have

\begin{gather} \nabla g(x) = 2f(x) \nabla f(x) = 2 \abs x \nabla f(x)\text{.}\tag{✶} \end{gather}

Calculating the left hand side using the summation convention and the product rule we get

\begin{equation*} (\nabla g(x))_i = \partial_i (x_j x_j) = 2 x_j \partial_i x_j = 2 x_j \delta_{ij} = 2 x_i\text{,} \end{equation*}

i.e. \(\nabla g(x) = 2x\text{.}\) Rearranging (✶) (recall that \(x \ne 0\) in \(\Omega\)) we get

\begin{equation*} \nabla f(x) = \frac 1{2 \abs x} \nabla g(x) = \frac 1{2 \abs x} 2x = \frac x{\abs x}\text{.} \end{equation*}

Comment.

As we talked about in class and in the notes, there are many different ways to write down and think about the chain rule. Some of these involve introducing ‘intermediate variables’. For instance, in this example we could set \(y=\abs x^2\) and then write

\begin{equation*} \partial_i \abs x = \frac{\partial \sqrt y}{\partial x_i} = \frac{d\sqrt y}{dy} \frac{\partial y}{\partial x_i} = \frac 12 y^{-1/2} 2x_i = \frac{x_i}{\abs x}\text{.} \end{equation*}

When writing things this way you must use different letters for the different variables \(x \in \R^N\) and \(y=\abs x^2 \in \R\text{.}\) Using the same letter for both of these leads to writing things like

\begin{equation*} \partial_i \abs x = \frac{\partial \sqrt x}{\partial x_i} = \frac{d\sqrt x}{dx} \frac{\partial x}{\partial x_i} = \frac 12 x^{-1/2} 2x_i = \frac{x_i}{\abs x} \end{equation*}

which are extremely confusing if taken literally.

2.3 The chain rule

Exercises

2.3.1. (PS2) Derivatives after shifting and scaling.

2.3.1.a

Solution.

\(x \mapsto r^{-1}(x-y)\text{.}\)

2.3.1.b

Solution.

To simplify the notation we introduce an intermediate variable \(z = r^{-1}(x-y)\text{.}\) Using the chain rule and the previous part we have

\begin{align*} \partial_i u(x) \amp = \partial_i [v(z)]\\ \amp = \partial_j v(z) \partial_i z_j\\ \amp = \partial_j v(z) r^{-1}\delta_{ij}\\ \amp = r^{-1} \partial_i v(z)\text{,} \end{align*}

or in other words \(\nabla u(x) = r^{-1} \nabla v(r^{-1}(x-y))\text{.}\)

2.3.1.c

Solution.

We repeatedly use the previous part, with \(z\) defined in the same way,

\begin{align*} \partial_{ij} u(x) \amp = \partial_i \partial_j [v(z)]\\ \amp = r^{-1}\partial_i [\partial_j u(z)]\\ \amp = r^{-2} \partial_i \partial_j u(z)\\ \amp = r^{-2} \partial_{ij} u(z),\text{,} \end{align*}

or in other words \(D^2 u(x) = r^{-2} D^2 v(r^{-1}(x-y))\text{.}\)

The question asked for \(D^2 v\) in terms of derivatives of \(u\text{,}\) and so we rewrite this as \(D^2 v(x) = r^2 D^2 u(y+rx)\text{.}\)

2.3.1.d

Solution.

We have now spotted the pattern: each time we differentiate \(u\) we simply get the same derivative applied to \(v\) but with an additional factor of \(r^{-1}\text{.}\) So \(\partial^\alpha u(x) = r^{-\abs \alpha} \partial^\alpha v(r^{-1}(x-y))\text{.}\)

2.3.2. (PS2) Derivatives of symmetric functions.

2.3.2.a

Solution.

We calculate

\begin{align*} \partial_i f(x) \amp= \partial_i [\phi(\abs x)]\\ \amp= \phi'(\abs x) \partial_i \abs x\\ \amp= \phi'(\abs x) \frac{x_i}{\abs x}, \end{align*}

where in the last step we have used Exercise 2.2.2.

Comment.

See this comment for Exercise 2.2.2.

2.3.2.b

Solution.

By the previous part we have

\begin{equation*} \nabla \abs x^p = p\abs x^{p-2} x\text{.} \end{equation*}

2.3.2.c

Solution.

Using the previous part and the product rule, we have

\begin{align*} (D(\abs x^p x))_{ij} \amp= \partial_j (\abs x^p x_i )\\ \amp= p\abs x^{p-2} x_j x_i + \abs x^p \delta_{ij}\text{.} \end{align*}

Comment.

Some authors write the matrix with entries \(x_i x_j\) as \(x \otimes x\text{,}\) in which case we can write the above formula as

\begin{equation*} D(\abs x^p x) = p\abs x^{p-2} (x \otimes x) + \abs x^p I, \end{equation*}

where here \(I\) is the identity matrix.

2.3.2.d

Solution.

We calculate

\begin{align*} \nabla \log \abs x \amp = \frac 1{\abs x} \nabla \abs x\\ \amp = \frac x{\abs x^2}\text{.} \end{align*}

2.3.2.e

Solution.

Using our formula for \(\nabla \abs x^p\text{,}\) we calculate

\begin{align*} \Delta \abs x^p \amp = \partial_i (p\abs x^{p-2} x_i)\\ \amp = p(p-2)\abs x^{p-4} x_i x_i + p\abs x^{p-2} \delta_{ii}\\ \amp = p (p-2+N) \abs x^{p-2}\text{.} \end{align*}

2.3.2.f

Solution.

Using our formula for \(\nabla \log \abs x\text{,}\) we calculate

\begin{align*} \Delta \log \abs x \amp = \partial_i (\abs x^{-2} x_i)\\ \amp = -2\abs x^{-4}x_i x_i +\abs x^{-2}\delta_{ii}\\ \amp = (N-2)\abs x^{-2}\text{.} \end{align*}

2.3.2.g

Solution.

This follows immediately from the previous two parts.

2.5 Regularity of domains

Exercises

2.5.1. (PS3) A domain with a corner.

2.5.1.a

Solution.

To see that \(\Omega\) has the interior ball property at the origin, let \(B=B_1(-e_2)\text{.}\) Then we easily check that \(0 \in \partial B\text{.}\) Moreover, all points \((x,y) \in B\) have \(y \lt 0\text{,}\) which in particular implies that \((x,y) \in \Omega\text{.}\)

2.5.1.b

Solution.

The following is adapted from a student’s solution in a previous year. Let \(B=B_r(x_0,y_0)\) with \(0 \in \partial B_r(x_0,y_0)\) and hence \(\abs{(x_0,y_0)}=r\text{.}\) Supposing for contradiction that \(B \subseteq \R^2 \without \overline \Omega\text{,}\) we can assume without loss of generality that \(x_0 \ge 0\text{,}\) in which case \(y_0 \gt \abs{x_0} = x_0 \ge 0\text{.}\) We will show that points \((x,x)\) with \(0 \lt x \ll 1\) lie in \(\overline \Omega \cap B\text{,}\) which is the desired contradiction. Clearly all such points lie in \(\partial \Omega \subset \overline \Omega\text{.}\) To see that they lie in \(B\text{,}\) we simply expand

\begin{align} \abs{(x,x)-(x_0,y_0)}^2 \amp = (x-x_0)^2+(x-y_0)^2 \notag\\ \amp = r^2 - 2(x_0+y_0)x + 2x^2\text{.}\tag{✶} \end{align}

Since \(x_0 \ge 0\) and \(y_0 \gt 0\text{,}\) the coefficient of \(x\) in (✶) above is strictly negative. Thus the right hand side of (✶) is \(\lt r^2\) for \(x \gt 0\) sufficiently small, implying that \((x,x) \in B\text{.}\)

Comment 1.

On the exam, you may need to determine whether a given domain \(\Omega\) satisfies the interior or exterior ball properties, as these are hypotheses of some of our main results. But doing this ‘by eye’ without giving a detailed rigorous proof will be completely acceptable.

Comment 2.

There were several other nice solutions this year, several of which did not involve any calculus.

2.5.2. (PS3) Locally constant functions.

2.5.2.a

Solution.

Fix a value \(z \in u(\Omega)\text{,}\) and suppose for the sake of contradiction that \(u \not \equiv z\) on \(\Omega\text{.}\) Then \(\Omega = A \cup B\) where the sets \(A = \{ x \in \Omega : u(x) = z \}\) and \(B = \{x \in \Omega : u(x) \ne z\}\) are disjoint and nonempty. Moreover, since \(u\) is continuous and \(\R \without \{z\}\) is open, a standard argument shows that \(B\) is open. Since \(u\) is locally constant, we easily check that \(A\) is open as well. But then we have written \(\Omega\) as the disjoint union of two nonempty open sets, contradicting the assumption that \(\Omega\) was connected.

Comment 1.

If \(u \maps \Omega \to \R\) is locally constant, then it is very easy to check that it is automatically continuous. So the requirement \(u \in C^0(\Omega)\) is somewhat redundant.

Comment 2.

Often students attempt a more hands-on argument involving chaining together balls on which \(u\) is constant. While this can be made to work, it requires some serious effort because the definition of ‘locally constant’ gives us no control (from below) on the radii of these balls, and so if we are not careful they could rapidly shrink to zero.

2.5.2.b

Solution.

Let \(A,B \subset \R^N\) be any two nonempty disjoint open sets, \(\Omega = A \cup B\text{,}\) and define a continuous function \(u\) by \(u \equiv 0\) in \(A\) and \(u \equiv 1\) in \(B\text{.}\) Then \(u\) is locally constant in the sense above, but clearly it is not globally constant.

2.6 Symmetric matrices

Exercises

2.6.2. (PS3) Invariance of the Laplacian.

Solution.

As always, we use the convention in Notation 2.14. Repeatedly using both the chain rule and the second part of Exercise 2.2.1, we calculate

\begin{align*} \Delta v \amp = \partial_i \partial_i [u(Ax)]\\ \amp = \partial_i[ \partial_j u(Ax)\partial_i (Ax)_j ]\\ \amp = \partial_i[ \partial_j u(Ax) A_{ji} ]\\ \amp = \partial_{kj} u(Ax) A_{ji} \partial_i(Ax)_k\\ \amp = \partial_{kj} u(Ax) A_{ji} A_{ki}\\ \amp = \partial_{kj}u(Ax) \delta_{jk}\\ \amp = \partial_{jj}u(Ax)\\ \amp = \Delta u(Ax)\text{,} \end{align*}

where towards the end we have used the identity

\begin{equation*} A_{ji}A_{ki} = (AA^\top)_{jk} = \delta_{jk}\text{,} \end{equation*}

which holds since \(A\) is orthogonal.

2.7 Maxima and minima

Exercises

2.7.1. (PS3) Minimising a quadratic polynomial.

2.7.1.a

Solution.

We calculate

\begin{align*} \partial_1 f(x) \amp = 2x_1 + \alpha (x_2 + 1)\\ \partial_2 f(x) \amp = 2x_2 + \alpha x_1\text{.} \end{align*}

Setting these both equal to zero, we get a linear system of equations for \((x_1,x_2)\text{.}\) When \(\alpha = \pm 2\text{,}\) this system has no solutions. Otherwise the unique solution is

\begin{gather*} x = \Big( \frac{2\alpha}{\alpha^2-4}, -\frac{\alpha^2}{\alpha^2-4} \Big)\text{.} \end{gather*}

Comment.

It’s important here for us to realise that there are no solutions when \(\alpha = \pm 2\text{.}\) Certainly our formula for the critical point involves \(\alpha^2-4\) in a denominator and so doesn’t make sense for \(\alpha = \pm 2\text{.}\)

2.7.1.b

Solution.

The Hessian is

\begin{align*} D^2 f(x) = \begin{pmatrix} 2 \amp \alpha \\ \alpha \amp 2 \end{pmatrix}\text{.} \end{align*}

The eigenvalues of this matrix are \(2 - \alpha\) and \(2+\alpha\text{,}\) and so it is positive definite only if both \(2 - \alpha \ge 0\) and \(2 + \alpha \ge 0\text{,}\) i.e. if \(\abs \alpha \le 2\text{.}\) Similarly \(D^2 f\) is positive definite when \(\abs \alpha \lt 2\text{.}\)

2.7.1.c

Solution.

By Proposition 2.34, a local minimum of \(f\) must be a point \(y \in \R^2\) where \(\nabla f(y) = 0\) and \(D^2 f(y)\) is positive semi-definite. We have shown that \(\nabla f(y) = 0\) is possible only when \(\alpha \ne 2\text{,}\) and that \(D^2 f(y)\) is positive semi-definite only when \(\abs \alpha \le 2\text{,}\) and so to have a local minimum we must have \(\abs \alpha \lt 2\text{.}\)

2.7.1.d

Solution.

From previous parts we know that \(f\) can only have a minimum if \(\abs \alpha \lt 2\text{,}\) and that in this case the Hessian matrix \(D^2 f\) has smallest eigenvalue \(\min(2+\alpha,2-\alpha) \gt 0\text{.}\) The lower bound in Corollary 2.29 therefore gives

\begin{gather*} f(y+h) \ge \tfrac 12 \min(2+\alpha,2-\alpha) \abs h^2 + f(y) \ge f(y) \end{gather*}

for all \(h \in \R^2\text{,}\) which implies

\begin{gather*} \inf_{\R^2} f = \min_{\R^2} f = f(y) = \cdots = \frac{\alpha^2}{\alpha^2 - 4}\text{.} \end{gather*}

Comment.

In general, \(D^2 f(y)\) positive definite and \(\nabla f(y) =0\) at an interior point \(y\) are sufficient conditions for \(f\) to have a local minimum at \(y\text{.}\) (This can be proved using Proposition 2.35.) They are far from sufficient, however, for \(f\) to have a global minimum at \(y\text{.}\) What saves us in this example is the fact that \(f\) is a quadratic polynomial, which makes Corollary 2.29 much more powerful.

2.7.1.e Optional.

Solution.

The eigenvectors of \(D^2 f\) are \((1,1)\) and \((1,-1)\text{,}\) which leads us to look at

\begin{align*} f(t,t) \amp= (2+\alpha) t^2 + \alpha t,\\ f(t,-t) \amp= (2-\alpha) t^2 + \alpha t \text{.} \end{align*}

We consider four cases:

If \(\alpha \lt -2\) then \(f(t,t) \to -\infty\) as \(t \to \pm\infty\text{.}\)
If \(\alpha \gt 2\) then \(f(t,-t) \to -\infty\) as \(t \to \pm\infty\text{.}\)
If \(\alpha = -2\) then \(f(t,t) \to -\infty\) as \(t \to +\infty\text{.}\)
If \(\alpha = 2\) then \(f(t,-t) \to -\infty\) as \(t \to -\infty\text{.}\)

Thus, whenever \(\abs \alpha \ge 2\text{,}\) we have \(\inf_{\R^2} f = -\infty\text{.}\)

2.8 Compactness and diagonal subsequences

Exercises

2.8.1. (PS3) Suprema, maxima, and closures.

Solution.

Since \(\overline\Omega \supset \Omega\text{,}\) we clearly have \(\sup_{\overline\Omega} f \ge \sup_\Omega f\text{,}\) and so it suffices to prove the reverse inequality. For any \(x \in \overline\Omega\text{,}\) there exists a sequence of points \(x_n \in \Omega\) with \(x_n \to x\text{.}\) Since \(x_n \in \Omega\text{,}\) we have \(f(x_n) \le \sup_\Omega f\) for each \(n\text{.}\) Passing to the limit using the sequential continuity of \(f\text{,}\) we find \(f(x) \le \sup_\Omega f\text{.}\) Since \(x \in \overline\Omega\) was arbitrary, we conclude that \(\sup_{\overline\Omega} f \le \sup_\Omega f\) as desired.

If \(\Omega\) is bounded, then \(\overline\Omega\) is compact, and so Theorem 2.40 guarantees that \(\sup_{\overline\Omega} f\) is achieved, i.e. \(\sup_{\overline\Omega} f = \max_{\overline \Omega} f\text{.}\) Combining with the previous part we have \(\sup_\Omega f = \max_{\overline\Omega} f\text{.}\)

2.8.2. (PS4) Sufficient condition for equicontinuity.

Solution 1. Using the mean value theorem

Fix \(x \in B\) and \(n \in \N\text{,}\) and define \(g(t) = f(tx_0+(1-t)x)\text{.}\) Then \(g \in C^1([-1,1])\text{,}\) and hence the mean value theorem implies that there exists \(t \in (0,1)\) such that

\begin{align*} f_n(x)-f_n(x_0) \amp = \frac{g(1)-g(0)}{1-0} \\ \amp = g'(t)\\ \amp = \nabla f_n(tx_0+(1-t) x) \cdot (x_0-x)\text{.} \end{align*}

Estimating the right hand side we find

\begin{equation*} \abs{f_n(x)-f_n(x_0)} \le M \abs{x-x_0}\text{.} \end{equation*}

Since the constant \(M\) on the right hand side is independent of both \(x \in B\) and \(n \in \N\text{,}\) the desired equicontinuity follows.

Solution 2. Using the fundamental theorem of calculus

For any \(x \in B\) and \(n \in \N\text{,}\) the fundamental theorem of calculus followed by the chain rule give

\begin{align*} f_n(x)-f_n(x_0) \amp = \int_0^1 \frac d{dt} f_n(tx+(1-t)x_0)\, dt\\ \amp = \int_0^1 \nabla f_n (tx+(1-t)x_0) \cdot (x-x_0)\, dt\\ \amp = \left(\int_0^1 \nabla f_n(tx+(1-t)x_0)\, dt \right) \cdot (x-x_0)\text{.} \end{align*}

Taking the norm of both sides, we then estimate

\begin{align*} \abs{f_n(x)-f_n(x_0)} \amp \le \left(\int_0^1 \abs{\nabla f_n(tx+(1-t)x_0)}\, dt \right) \abs{x-x_0}\\ \amp \le M\abs{x-x_0}\text{.} \end{align*}

Since the constant \(M\) on the right hand side is independent of both \(x \in B\) and \(n \in \N\text{,}\) the desired equicontinuity follows.

Comment 1.

The solutions above establish, for any \(x \in B_r(x_0)\) and any \(n \in \N\text{,}\) the estimate \(\abs{f_n(x)-f_n(x_0)} \le M\abs{x-x_0}\text{.}\) To see that this implies equicontinuity, let \(\varepsilon \gt 0\) and choose \(\delta = \varepsilon/M\text{.}\) Then for any \(x \in B_r(x_0)\) with \(\abs{x-x_0} \lt \delta\text{,}\) and for any \(n \in \N\text{,}\) our above estimate gives \(\abs{f_n(x)-f(x_0)} \lt M (\varepsilon/M) = \varepsilon\text{.}\)

Note that the definition of equicontinuity, like the definition of continuity, hands us an arbitrary \(\varepsilon \gt 0\) and asks us to produce an appropriate \(\delta \gt 0\text{.}\) We cannot choose our favourite \(\varepsilon \gt 0\) and only consider this value.

Comment 2.

Note that the domain of the function \(g\) in the first solution above is some appropriate closed interval \([-1,1]\text{,}\) and not the same ball \(B\) that \(f\) is defined on. Indeed, \(g\) is a function of a single real variable, and \(B \subset \R^N\) is a set in \(N\) dimensions.

2.9 Integration and vector calculus

Exercises

2.9.4. Gradients and spheres.

2.9.4.a

Solution.

Fix \(p \in \partial B\text{.}\) To show that (✶) holds at \(p\text{,}\) it suffices to show that

\begin{gather} e \cdot \nabla u (p) = \frac{\partial u}{\partial n} e \cdot n\tag{†} \end{gather}

for all unit vectors \(e \in \R^N\text{.}\) Clearly (†) holds when \(e=n\text{,}\) and so, by basic linear algebra, it is enough to show that (†) holds when \(e\) is perpendicular to \(n\text{,}\) in which case the right hand side is zero.

So let \(e\) be a unit vector perpendicular to \(n\text{.}\) An easy calculation shows that

\begin{gather*} x_0 + \cos(\theta) rn + \sin(\theta) re \end{gather*}

lies on \(\partial B\) for any \(\theta \in \R\text{;}\) this is a parametrization of a circle. By the assumption on \(u\text{,}\) we therefore have that

\begin{equation*} \theta \mapsto u(x_0 + r(\cos(\theta) n + \sin(\theta) e)) \end{equation*}

is a constant function. Differentiating using the chain rule, we deduce that

\begin{align*} 0 \amp = \frac d{d\theta}u(x_0 + r(\cos(\theta) n + \sin(\theta) e))\\ \amp = \nabla u(x_0 + r(\cos(\theta) n + \sin(\theta) e)) \cdot r(-\sin(\theta) n + \cos(\theta) e)\text{.} \end{align*}

In particular, at \(\theta = 0\) we find

\begin{gather*} 0 = \nabla u(x_0 + rn) \cdot re = r (\nabla u(p) \cdot e)\text{,} \end{gather*}

and hence \(\nabla u(p) \cdot e = 0\) as desired.

2.9.4.b

Solution.

We argue as in the previous part. As \(u\) is maximised at \(p = x_0 + rn\text{,}\) we know that the function

\begin{equation*} \theta \mapsto u(x_0 + r(\cos(\theta) n + \sin(\theta) e)) \end{equation*}

is maximised at \(\theta = 0\text{.}\) In particular, its derivative at \(\theta = 0\) must vanish, which is all we need for our argument above to work.

3 Maximum principles for elliptic equations
3.1 Ellipticity and uniform ellipticity

Exercises

3.1.1. (PS4) Checking uniform ellipticity.

3.1.1.a

Solution.

We have \(a_{ij} = \delta_{ij}\text{,}\) and hence

\begin{gather*} a_{ij} \xi_i \xi_j = \delta_{ij} \xi_i \xi_j = \abs \xi^2 \ge 1 \cdot \xi^2 \end{gather*}

for any \(\xi \in \R^N\text{.}\) Thus \(\Delta\) is uniformly elliptic with ellipticity constant \(\lambda_0 = 1\text{.}\)

3.1.1.b

Solution.

We have \(a_{ij} = 1\text{,}\) and hence

\begin{align*} a_{ij} \xi_i \xi_j \amp = \xi_1 \xi_1 + \xi_1 \xi_2 + \xi_2 \xi_1 + \xi_2 \xi_2\\ \amp = \xi_1^2 + 2\xi_1 \xi_2 + \xi_2^2\\ \amp = (\xi_1+\xi_2)^2\text{,} \end{align*}

which is \(\ge 0\) but vanishes for instance when \(\xi = (1,-1)\text{.}\) Thus \(L\) is not elliptic.

Alternatively, we could check that one of the eigenvalues of the matrix

\begin{equation*} a = \begin{pmatrix} 1 \amp 1 \\ 1 \amp 1 \end{pmatrix} \end{equation*}

is \(0\text{,}\) which is not positive. A slick way to do this is to just take the determinant and recall how this is related to the eigenvalues.

Comment.

Since \(\partial_{12}=\partial_{21}\) when acting on \(C^2\) functions, many authors will abbreviate the operator \(L\) in this problem as

\begin{equation*} L = \partial_{11} + 2\partial_{12} + \partial_{22}\text{.} \end{equation*}

This does not mean that they consider the non-symmetric matrix

\begin{equation*} a = \begin{pmatrix} 1\amp 2 \\ 0 \amp 1 \end{pmatrix}, \end{equation*}

as this would violate our assumption \(a_{ij}=a_{ji}\text{.}\) Perhaps more importantly, calculating the eigenvalues of this non-symmetric matrix will no longer provide you with any information about the ellipticity of \(L\text{.}\)

3.1.1.c

Solution.

In this case we have, for any \(\xi\in \R^2\text{,}\)

\begin{align*} a_{ij} \xi_i \xi_j \amp = (1+\alpha^2) \xi_1^2 - 2 \alpha\beta \xi_1 \xi_2 + (1+\beta^2) \xi_2^2\\ \amp = \xi_1^2 + \xi_2^2 + (\alpha \xi_1 - \beta \xi_2)^2\\ \amp \ge \abs \xi^2\text{.} \end{align*}

Thus \(L\) is uniformly elliptic with ellipticity constant \(\lambda_0 = 1\text{.}\)

Alternatively, one can simply calculate the eigenvalues of the relevant matrix

\begin{equation*} a = \begin{pmatrix} 1+\alpha^2 \amp -\alpha\beta \\ -\alpha\beta \amp 1+\beta^2 \end{pmatrix} \end{equation*}

by brute force. They are \(1\) and \(1+\alpha^2+\beta^2 \ge 1\text{,}\) which again implies uniform ellipticity with \(\lambda_0 = 1\text{.}\)

3.1.1.d

Solution.

The eigenvalues of the matrix

\begin{equation*} a(x) = \begin{pmatrix} 1 \amp 0 \\ 0 \amp e^{-x_1} \end{pmatrix} \end{equation*}

are clearly \(1\) and \(e^{-x_1}\text{,}\) both of which are strictly positive for any fixed \(x \in \Omega\text{.}\) Thus \(L\) is elliptic. On the other hand, as \(x_1 \to \infty\text{,}\) the second eigenvalue \(e^{-x_1} \to 0\text{,}\) and so this operator cannot be uniformly elliptic.

Comment 1.

To show that \(L\) is not uniformly elliptic it is not enough to show that the smallest eigenvalue of \(a\) is \(\ge e^{-x_1}\text{,}\) or equivalently that \(a_{ij} \xi_i \xi_j \ge e^{-x_1} \abs \xi^2\text{.}\) Indeed, these properties also hold for the Laplacian \(\Delta\text{,}\) which is uniformly elliptic on \(\Omega\text{.}\) To disprove uniform ellipticity need the reverse inequality that the smallest eigenvalue of \(a\) is \(\le e^{-x_1}\text{,}\) or alternatively that for \(\xi=e_2\) we get \(a_{ij} \xi_i \xi_j \le e^{-x_1} \abs \xi^2\text{.}\)

Comment 2.

Note that, because of our choice of \(\Omega\text{,}\) the coefficients of \(L\) are indeed bounded functions.

3.1.2. (PS4) Ellipticity after shifting and scaling.

3.1.2.a

Solution.

By Exercise 2.3.1 and keeping in mind Notation 2.14, we have

\begin{align*} u(y+rx) \amp = v(x) \\ \partial_i u(y+rx) \amp = r^{-1} \partial_i v(x) \\ \partial_{ij} u(y+rx) \amp = r^{-2} \partial_{ij} v(x)\text{.} \end{align*}

Thus

\begin{align*} Lu(y+rx) \amp = a_{ij}(y+rx) \partial_{ij} u(y+rx) \\ \amp \qquad + b_i(y+rx) \partial_i u(y+rx) \\ \amp \qquad + c(y+rx) u(y+rx)\\ \amp = r^{-2} a_{ij}(y+rx) \partial_{ij} v(x) \\ \amp \qquad + r^{-1} b_i(y+rx) \partial_i u(x) \\ \amp \qquad + c(y+rx) v(x)\text{,} \end{align*}

and so we take

\begin{equation*} \tilde L = \tilde a_{ij}(x) \partial_{ij} + \tilde b_i(x) \partial_i + \tilde c(x) \end{equation*}

where

\begin{align*} \tilde a_{ij} (x) \amp= r^{-2} a_{ij}(y+rx),\\ \tilde b_i (x) \amp= r^{-1} b_i(y+rx).\\ \tilde c (x) \amp= c(y+rx)\text{.} \end{align*}

Comment.

In problems like this it is important to keep track of where the functions \(u,v\) and coefficients \(a_{ij},b_i,c\) are being evaluated. This year, several students took

\begin{gather*} \tilde L = \frac 1{r^2} a_{ij} \partial_{ij} + \frac 1r b_i \partial_i + c\text{,} \end{gather*}

but this operator has

\begin{gather*} \tilde L v(x) = \frac 1{r^2} a_{ij}(x) \partial_{ij}v(x) + \frac 1r b_i(x) \partial_iv(x) + c(x) v(x)\text{,} \end{gather*}

which is not quite what we want.

3.1.2.b

Solution.

Let \(x \in B_1(0)\) and \(\xi \in \R^N\text{.}\) Then \(y+rx \in B_r(y)\) and so by our above formulas and the ellipticity of \(L\) we have

\begin{gather*} \tilde a_{ij}(x)\xi_i \xi_j = r^{-2} a_{ij}(y+rx)\xi_i \xi_j \ge r^{-2} \lambda_0|\xi|^2 \text{.} \end{gather*}

Since \(x\) and \(\xi\) were arbitrary, we conclude that \(\tilde L\) is uniformly elliptic with ellipticity constant \(\tilde \lambda_0 = r^{-2} \lambda_0\text{.}\) (The boundedness and symmetry of the coefficients of \(\tilde L\) easily follow from the corresponding properties of the coefficients of \(L\text{.}\))

3.2 The weak maximum principle

Exercises

3.2.1. Non-uniqueness.

3.2.1.a

Solution.

Certainly \(u \in C^2(\Omega)\cap C^0(\overline\Omega)\text{.}\) Moreover, on \(\partial\Omega\) we have either \(x_1 \in \{0,\pi\}\) or \(x_2 \in \{0,\pi\}\text{,}\) and hence \(u=0\text{.}\) To calculate \(\Delta u\) the summation convention is not particularly helpful; we find

\begin{align*} \Delta u \amp = \partial_{11} (\sin x_1 \sin x_2) + \partial_{22} (\sin x_1 \sin x_2)\\ \amp = \sin x_2 \partial_{11} \sin x_1 + \sin x_1 \partial_{22} \sin x_2\\ \amp = \sin x_2 \cdot (-\sin x_1) + \sin x_1 \cdot (-\sin x_2)\\ \amp = -2u\text{.} \end{align*}

Thus the desired PDE holds.

To see that this problem does not have a unique solution, we must produce another one. The obvious choice is \(u \equiv 0\text{.}\) More generally, by linearity we have that \(\alpha u\) is a solution for any \(\alpha \in \R\text{.}\) Thinking back to linear algebra, we could call this a one-dimensional subspace of solutions.

Since the relevant elliptic operator \(L=\Delta + 2\) has zeroth-order coefficient \(c = 2\text{,}\) Corollary 3.5 does not apply; that result required \(c \le 0\) in \(\Omega\text{.}\)

Comment.

When calculating boundaries, it is often helpful to draw a picture. Some students in previous years have assumed there was a rule like

\begin{equation*} \partial (A \times B) = (\partial A) \times (\partial B)\text{,} \end{equation*}

which in this case told them that the boundary of the square \(\Omega\) was only its four corners. In this case the rule is instead

\begin{equation*} \partial (A \times B) = (\partial A \times \overline B) \cup (\overline A \times \partial B)\text{,} \end{equation*}

which will give you both the sides and the corners.

3.2.1.b

Solution.

Since \(u \equiv 0\) is always a solution, it suffices to find another solution \(u \not \equiv 0\text{.}\) One example is \(\Omega = (0,\infty) \times \R^{N-1}\) and \(u = x_1\text{.}\) My personal favourite is to look at regions \(\Omega = (0,\pi) \times \R^{N-1}\) bounded by two parallel hyperplanes and then take a solution like \(u = \sin(x_1)e^{x_2}\) coming from separation of variables. Note that since the equation is linear, if \(u\) is a solution then so is \(\alpha u\) for any \(\alpha \in \R\) — as soon as we have two solutions we necessarily have uncountably many.

3.2.3. (PS4) A two-sided estimate.

3.2.3.a

Solution.

We will use the comparison principle with the comparison functions

\begin{align*} u_1 \amp= \frac m {2N} (r^2 -\abs{x-p}^2),\\ u_2 \amp= \frac M{2N} (R^2 -\abs{x-q}^2), \end{align*}

which are of course defined and smooth on all of \(\R^N\) and hence in particular on \(\Omega\text{.}\) We calculate \(\Delta u_1 = -m\) and \(\Delta u_2 = -M\) (again on all of \(\R^N\) and hence on \(\Omega)\text{,}\) and so

\begin{gather*} \Delta u_2 \le \Delta u \le \Delta u_1 \ina \Omega. \end{gather*}

In order to conclude the desired inequalities

\begin{gather} u_1 \le u \le u_2 \ina \Omega\tag{✶} \end{gather}

from the comparison principle, we need to show

\begin{gather} u_1 \le u \le u_2 \ona \partial\Omega.\tag{✶✶} \end{gather}

Since \(B_r(p) \subseteq \Omega\) we have \(\abs{x-p} \ge r\) on \(\partial\Omega\text{,}\) and hence \(u_1 \le 0 = u\text{.}\) Similarly \(\Omega \subseteq B_R(q)\) implies \(\abs{x-q} \le R\) on \(\partial\Omega\) and hence \(u_2 \ge 0 = u\text{.}\) Thus (✶✶) holds and we can indeed deduce (✶) as desired.

Comment 1.

In the notation of the official solution, many students correctly observed that \(u_1 = 0\) on \(\partial B_r(p)\) and \(u_2 = 0\) on \(\partial B_R(q)\text{.}\) Note, though, that these equalities are not what the comparison principle is asking for. It wants inequalities \(u_1 \le u \le u_2\) on \(\partial\Omega\text{,}\) which are a bit more subtle.

It may be clarifying to sketch a cartoon of the set \(\Omega\) and the inclusions \(B_r(p) \subseteq \Omega \subseteq B_R(q)\)

Comment 2.

While \(\Delta\) is elliptic, \(-\Delta\) is not. If you set \(L=-\Delta\text{,}\) you will have to be quite careful with your inequalities when referencing results from lectures (some of them will flip).

3.2.3.b

Solution.

Let \(u_1,u_2\) be as in the previous part. A cheeky way to respond is to say yes, consider the case where \(m=M=0\) so that \(f,u,u_1,u_2\equiv 0\text{.}\) A slightly less cheeky example would be to take \(m=M \ne 0\) (so \(f\) is constant) and \(p=q\) and \(r=R\) so that \(\Omega\) is a ball. Then we can easily check that \(u_1 = u = u_2\) in all of \(\Omega\text{.}\)

3.2.4. (PS5) Domains contained in a slab.

3.2.4.a

Solution.

Since \(z \ge 0\) and \(c \le 0\text{,}\) arguing as in the proof of Theorem 3.2 we get that

\begin{align*} L z \amp = -k (k a_{11} + b_1) e^{kx_1} + cz\\ \amp \le -k (k a_{11} + b_1) e^{kx_1} \end{align*}

in \(\Omega\text{.}\) By uniform ellipticity the first two factors above satisfy

\begin{gather*} k (k a_{11} + b_1) \ge k \Big(k\lambda_0 - \sup_\Omega \abs{b_1}\Big) \ge 1 \end{gather*}

in \(\Omega\text{,}\) provided we choose \(k \gt 0\) sufficiently large depending on \(\lambda_0\) and \(\sup_\Omega\abs{b_1}\text{,}\) for instance

\begin{equation*} k = \max\Big(1, \frac 1{\lambda_0} \sup_\Omega\abs{b_1}\Big)\text{.} \end{equation*}

Since \(x_1 \ge 0\) in \(\Omega\text{,}\) inserting this into our earlier estimate for \(Lz\) yields

\begin{align*} L z \amp \le -e^{kx_1} \le -1 \end{align*}

in \(\Omega\) as desired.

Comment 1.

The above solution gives an explicit formula for \(k\text{,}\) but I want to emphasise that this is not strictly necessary. What we are using here is the following elementary fact: If

\begin{equation*} p(k)=p_0+p_1 k + \cdots + p_n k^n \end{equation*}

is a polynomial with coefficients \(p_0,\ldots,p_{n-1} \in \R\) and \(p_n \gt 0\text{,}\) then \(\lim_{k \to \infty} p(k) = \infty\text{.}\) Here our polynomial is

\begin{equation*} p(k) = k^2 \lambda_0 - \sup_\Omega \abs{b_1} k \end{equation*}

with \(n=2\) and \(p_2 = \lambda_0 \gt 0\text{.}\) Without even thinking about the quadratic formula, we know that \(p(k) \to \infty\) as \(k \to \infty\text{.}\) In particular, there exists \(K \gt 0\) depending only on \(\lambda_0,\sup_\Omega \abs{b_1}\) such that \(k \ge K\) implies \(p(k) \ge 1\text{.}\)

Comment 2.

Note that we need \(Lz \le -1\) here, and not just \(Lz \le 0\text{.}\) There are many ways to accomplish this, but in the official solutions the strategy is to

use the inequality \(e^{kx_1} \ge 1\text{,}\) which holds in \(\Omega\) as \(x_1,k \ge 0\text{;}\) and
arrange for the remaining factor (up to a sign) to satisfy \(k (k a_{11} + b_1) \ge 1\) in \(\Omega\text{.}\)

If either of these \(\ge 1\) bounds is replaced by \(\ge 0\text{,}\) then the argument fails.

3.2.4.b

Solution.

Since \(c \le 0\) and \(\max_{\partial\Omega} \abs u \ge 0\text{,}\) the previous part gives

\begin{align*} L v \amp = \Big(\sup_\Omega f\Big) Lz + c \max_{\partial\Omega} \abs u\\ \amp \le -\sup_\Omega \abs f \\ \amp \le f = Lu \end{align*}

in \(\Omega\text{.}\) On \(\partial\Omega\) we have \(z \ge 0\) and hence \(v \ge \max_{\partial\Omega} \abs u \ge u\text{.}\) Applying the comparison principle, we deduce that \(v \ge u\) in \(\Omega\text{.}\) Arguing similarly with \(v\) replaced by \(-v\) we get \(-v \le u\) in \(\Omega\text{,}\) and hence \(\abs u \le v\) as desired.

Comment.

Be careful not to immediately discard constant terms when applying differential operators. Here, for instance, we have

\begin{align*} Lv \amp= L\left( \Big(\sup_\Omega f\Big) z + \max_{\partial\Omega} \abs u \right)\\ \amp= \Big(\sup_\Omega f\Big) Lz + \Big(\max_{\partial\Omega} \abs u\Big) L1\\ \amp= \Big(\sup_\Omega f\Big) Lz + \Big(\max_{\partial\Omega} \abs u\Big) c\text{.} \end{align*}

3.2.4.c

Solution.

By the previous part we have

\begin{align*} \max_{\overline\Omega} \abs u \amp \le \max_{\overline\Omega} v \\ \amp = \max_{\partial\Omega} \abs u + \Big(\max_{\overline \Omega} z\Big)\Big(\sup_\Omega \abs f\Big)\\ \amp \le \max_{\partial\Omega} \abs u + (e^{kd}-1)\sup_\Omega \abs f, \end{align*}

which is an inequality of the desired form with \(C = e^{kd}-1\text{.}\) Finally, we observe that \(C\) depends only on \(d\) and on \(k\text{,}\) and that \(k\) in turn depends only on \(\lambda_0\) and \(\sup_\Omega \abs{b_1}\text{.}\)

Comment.

Note that we cannot quite set \(C = \max_{\overline\Omega} z\text{,}\) as that would depend not only on \(\lambda_0,d,\sup_\Omega \abs{b_1}\) but also on the domain \(\Omega\) itself.

3.2.4.d

Solution.

Applying the previous part to \(u_n-u\text{,}\) which has \(L(u_n-u)=f_n-f\text{,}\) we have

\begin{align*} \n{u_n-u}_{C^0(\overline\Omega)} \amp = \max_{\overline\Omega} \abs{u_n-u}\\ \amp \le \max_{\partial\Omega} \abs{u_n-u} + C \sup_\Omega \abs{f_n-f}\\ \amp = \n{u_n-u}_{C^0(\partial\Omega)} + C \n{f_n-f}_{C^0(\overline\Omega)}\\ \amp \to 0 \end{align*}

as \(n \to \infty\text{.}\)

Comment.

Let \(\Omega\) be bounded and \(f_n,f \in C^0(\overline \Omega)\text{.}\) In part thanks to Exercise 2.8.1, the following statements are equivalent

\(f_n \to f\) uniformly on \(\Omega\)
\(f_n \to f\) uniformly on \(\overline \Omega\)
\(\displaystyle \sup_\Omega \abs{f_n - f} \to 0\)
\(\displaystyle \sup_{\overline \Omega} \abs{f_n - f} \to 0\)
\(\displaystyle \max_{\overline \Omega} \abs{f_n - f} \to 0\)

You are free to use this fact without comment, and certainly without looking up the number of Exercise 2.8.1.

3.3 The Hopf lemma and the strong maximum principle

Exercises

3.3.1. (PS5) Normal derivatives of radial functions.

Solution.

Writing \(u = U \circ \rho\) where \(\rho(x)=\abs{x-x_0}\text{,}\) the chain rule gives (for \(x \in B_r(x_0) \without \{x_0\}\))

\begin{equation*} \nabla u(x) = U'(\rho(x)) \nabla \rho(x) = U'(\abs{x-x_0}) \frac{x-x_0}{\abs{x-x_0}}. \end{equation*}

In particular, for \(p \in \partial B_r(x_0)\) we have

\begin{equation*} \nabla u(p) = U'(r) n, \end{equation*}

where \(n\) is the unit normal vector from Definition 2.47. Thus

\begin{equation*} \frac{\partial u}{\partial n}(p) = n \cdot \nabla u(p) = U'(r) \abs n^2 = U'(r), \end{equation*}

where in the last step we have used that \(n\) is a unit vector.

3.3.3. A strong comparison principle.

Solution.

Applying Proposition 3.4, we have immediately that \(u \ge v\) in \(\Omega\text{.}\) Thus the difference \(w = v-u \le 0\) in \(\Omega\text{,}\) and if \(u=v\) at some point in \(\Omega\) then \(w\) achieves \(\max_{\overline\Omega} w = 0\) at this point. But \(Lw = Lv-Lu \ge 0\text{,}\) and so by the Strong maximum principle this is only possible if \(w \equiv 0\text{,}\) i.e. \(u \equiv v\text{.}\)

3.3.4. (PS5) Uniqueness for other boundary conditions.

3.3.4.a

Solution.

Consider the difference \(w=u-v\text{.}\) Since \(w\) is continuous and \(\Omega\) is bounded, either \(\max_{\overline\Omega} w \ge 0\) or \(\min_{\overline\Omega} w \le 0\) or both. Replacing \(w\) with \(-w\) if necessary, we can assume that \(M = \max_{\overline\Omega} w \ge 0\text{.}\) Suppose that \(w\) is not constant. Then, since \(Lw = Lu-Lv = 0\) in \(\Omega\text{,}\) Theorem 3.8 implies that \(w \lt M\) in \(\Omega\text{,}\) and hence that \(w(p)=M\) for some \(p \in \partial \Omega\text{.}\) But then Theorem 3.9 implies that \(\partial w/\partial n \gt 0\) at \(p\text{,}\) which is a contradiction.

3.3.4.b

Solution.

As \(w=u-v\) is constant, we have \(Lw = cw = 0\) in \(\Omega\text{.}\) If \(c \not \equiv 0\text{,}\) then the only way for this to happen is if \(w \equiv 0\text{.}\)

Comment.

In this and other problems students sometimes claim, implicitly or explicitly, that \(c \not \equiv 0\text{,}\) plus the overall assumption that \(c \le 0\text{,}\) implies \(c \lt 0\text{.}\) While it ends up being a very minor detail in this problem, it is important to understand that this is not true. To understand why, let’s unpack what the inequalities for the coefficient function \(c\) mean, precisely, in the context of this unit:

\(c \equiv 0\) is shorthand for “\(c(x) = 0\) for all \(x \in \Omega\)”. The negation of this, which we write as \(c \not \equiv 0\text{,}\) is therefore “there exists \(x \in \Omega\) such that \(c(x) \ne 0\)”.
\(c \le 0\) is by default interpreted “\(c \le 0\) in \(\Omega\)”, which in turn is shorthand for “\(c(x) \le 0\) for all \(x \in \Omega\)”. Similarly for \(c \lt 0\text{.}\)

Thus \(c \not \equiv 0\) combined with \(c \le 0\) in \(\Omega\) implies that there exists \(x \in \Omega\) such that \(c(x) \lt 0\text{.}\) It does not imply that \(c(x) \lt 0\) for all \(x \in \Omega\text{.}\) For an explicit example, take for instance \(\Omega = (-1,1) \subset \R\) and \(c(x)=-x^2\text{.}\) Then \(c \le 0\) in \(\Omega\) and \(c \not \equiv 0\text{,}\) but \(c(0)=0\) and so it is not true that \(c \lt 0\) in \(\Omega\text{.}\)

3.3.6. (PS5) Maximum principles in the reverse order.

3.3.6.a

Solution.

Here and in what follows, let \(L\) be uniformly elliptic and \(u \in C^2(\Omega) \cap C^0(\overline \Omega)\) satisfy \(L u \ge 0\) in \(\Omega\text{.}\) Note that we never have to get our hands dirty by, e.g., introducing a complicated comparison function.

Assuming that \(c \equiv 0\) and that \(\sup_\Omega u\) is achieved at some point \(x \in \Omega\text{,}\) we need to show that \(u\) is constant. Consider the function \(\tilde u = u - \sup_\Omega u\text{.}\) Then \(\tilde u\) achieves \(\sup_\Omega \tilde u = 0\) at \(x\text{.}\) Moreover, since \(c \equiv 0\) we have \(L\tilde u = Lu \ge 0\text{.}\) Thus (6) implies that \(\tilde u\) is constant, and hence that \(u\) is constant as desired.

3.3.6.b

Solution.

Assuming now that \(c \le 0\) and that \(\sup_\Omega u \ge 0\) is achieved at some point \(x \in \Omega\text{,}\) we need to show that \(u\) is constant. Again consider the function \(\tilde u = u - \sup_\Omega u\text{.}\) Then \(\tilde u\) achieves \(\sup_\Omega \tilde u = 0\) at \(x\text{.}\) Moreover, since \(c \le 0\) we have

\begin{gather*} L\tilde u = Lu - c \sup_\Omega u \ge Lu \ge 0. \end{gather*}

Thus (6) implies that \(\tilde u\) is constant, and hence that \(u\) is constant as desired.

3.3.6.c

Solution.

Since \(\Omega\) is bounded, we know that \(\sup_\Omega = \max_{\overline \Omega} u\) is achieved at some point \(x^* \in \overline \Omega\text{.}\) If \(x^* \in \partial\Omega\) then

\begin{gather} \max_{\overline \Omega} u = \max_{\partial\Omega} u\tag{✶} \end{gather}

obviously holds, so it remains to consider the possibility that \(x^* \in \Omega\text{.}\) Whichever case (1)–(3) of the weak maximum principle we are are in, the parallel case (4)–(6) of the strong maximum principle now implies that \(u\) is constant, and so (✶) certainly holds. (Indeed, since \(x^* \in \partial\Omega \subseteq \overline\Omega\) we have \(\max_{\partial\Omega} u \le \max_{\overline \Omega} u = u(x^*) \le \max_{\partial\Omega} u\text{.}\))

3.4 An extended problem

3.4.1. Weak maximum principle.

Solution.

As \(\Omega\) is bounded, \(\Delta\) is uniformly elliptic with no zeroth-order terms, and \(\Delta u = x_2 \ge 0\) in \(\Omega\text{,}\) we can indeed apply the weak maximum principle to get that

\begin{equation*} \max_{\overline \Omega} u = \max_{\partial\Omega} u = 0 \text{.} \end{equation*}

3.4.2. Uniqueness.

Solution.

As \(\Omega\) is bounded and \(\Delta\) is uniformly elliptic with no zeroth-order terms, Corollary 3.5 indeed implies that \(u\) has at most one solution \(u \in C^2(\Omega) \cap C^0(\overline\Omega)\text{.}\)

3.4.3. Strong maximum principle.

Solution.

As \(\Omega\) is bounded and \(\Delta\) is uniformly elliptic with no zeroth-order terms, we can indeed apply Strong maximum principle. By our earlier application of the weak maximum principle, \(\max_{\overline\Omega} u = 0\text{,}\) and so the conclusion is that either \(u \lt 0\) in \(\Omega\) or else \(u \equiv 0\text{.}\) The latter is not possible, since it would imply that \(\Delta u \equiv 0\text{,}\) contradicting (✶). We therefore deduce that \(u \lt 0\) in \(\Omega\text{.}\)

3.4.4. Hopf lemma.

Solution.

As in the previous problem, we know that \(\Omega\) is connected, \(\Delta\) is uniformly elliptic with no zeroth order terms, and \(\Delta u = x_2 \ge 0\) in \(\Omega\text{.}\) Moreover, we are given that \(u \in C^2(\Omega) \cap C^1(\overline\Omega)\text{.}\) By the above problems, we know that \(u\) achieves \(\sup_\Omega u = 0\) at every point \(p \in \partial\Omega\text{,}\) and moreover that \(u\) is not constant. Therefore we can apply the Hopf lemma at every point of \(p \in \partial\Omega\) where \(\Omega\) satisfies the interior ball property, obtaining \(\partial u/\partial n \gt 0\) there. Drawing a picture of \(\Omega\text{,}\) we see that this is every point \(p \in \partial\Omega\) aside from the corner points \((\pm1,0)\text{.}\)

As an example, consider the origin \((0,0)\text{.}\) There the outward pointing normal is clearly \((0,-1)\text{,}\) and so we get

\begin{equation*} 0 \lt \frac{\partial u}{\partial n}(0,0) = (0,-1) \cdot \nabla u(0,0) = -\partial_2 u(0,0), \end{equation*}

i.e. that \(\partial_2 u(0,0) \lt 0\text{.}\)

3.4.5. Comparison.

Solution.

Integrating twice with respect to \(x_2\text{,}\) we see that \(v=x_2^3/6\) is such a function. As \(x_2 \le 1\) on \(\partial \Omega\text{,}\) we see that \(v \le 1/6\) there. Thus we have \(u = 0 \ge v - 1/6\) on \(\partial\Omega\) and \(\Delta u = \Delta (v-1/6)\) in \(\Omega\text{.}\) Again, \(\Omega\) is bounded and \(L\) is uniformly elliptic with no zeroth-order coefficient, and so we can apply the comparison principle to get that \(u \ge v - 1/6\) in \(\Omega\text{.}\) In particular, \(u(0,\frac 12) \ge v(0,\frac 12) - 1/6 = -7/48\text{.}\)

This is just one example; there are many other comparison functions that could have been chosen.

3.4.6. Symmetry.

Solution.

Using the chain rule we can check that if \(u\) solves (✶) then so does \(\tilde u\text{.}\) Uniqueness then implies \(\tilde u \equiv u\text{,}\) i.e. that \(u\) is even in \(x_1\text{.}\)

4 The Dirichlet problem
4.1 Interior estimates

Exercises

4.1.1. (PS6) A maximum principle for the gradient.

Solution.

The assumption \(u \in C^3(\Omega) \cap C^1(\overline\Omega)\) guarantees that \(v \in C^2(\Omega) \cap C^0(\overline\Omega)\text{.}\) Differentiating repeatedly using the product rule, we have

\begin{align*} \Delta v \amp = \partial_i \partial_i [\partial_j u \partial_j u]\\ \amp = 2 \partial_i [\partial_{ij} u \partial_j u]\\ \amp = 2 [\partial_{iij} u \partial_j u + \partial_{ij} u \partial_{ij} u]. \end{align*}

Since \(u\) is harmonic, \(\partial_j \Delta u = 0\) and so the first term vanishes, leaving us with

\begin{equation*} \Delta v = 2 \partial_{ij} u \partial_{ij} u = 2\abs{D^2 u}^2 \ge 0 \end{equation*}

as desired.

Applying the weak maximum principle to \(v\) and the uniformly elliptic operator \(L=\Delta\text{,}\) we conclude that

\begin{equation*} \max_{\overline\Omega} \abs{\nabla u}^2 = \max_{\overline\Omega} v = \max_{\partial \Omega} v = \max_{\overline\Omega} \abs{\nabla u}^2\text{,} \end{equation*}

which yields the desired statement about \(\abs{\nabla u}\) after taking square roots.

4.1.2. (PS6) Interior estimate for second derivatives.

Solution.

Throughout this solution, \(C\) will denote the constant from Lemma 4.2, and not the constant in the desired inequality. Consider the function \(v = \partial_i u\text{.}\) Since \(u\) is \(C^4\) and harmonic, \(v\) is \(C^3\) and harmonic, and so we can apply Lemma 4.2 to \(v\) on any ball \(B \subseteq B_r(x_0)\text{.}\) Our goal is to estimate \(\nabla v\) on \(B_{r/2}(x_0)\text{,}\) and so following the hint we fix \(x \in B_{r/2}(x_0)\text{.}\) By the triangle inequality, we have \(B_{r/2}(x) \subset B_r(x_0)\text{.}\) Thus we can certainly apply Lemma 4.2 on the strictly smaller ball \(B_{r/4}(x)\) to obtain

\begin{equation*} \sup_{B_{r/8}(x)} \abs{\nabla v} \le \frac {4C}r \sup_{B_{r/4}(x)} \abs v\text{,} \end{equation*}

or, in terms of \(u\text{,}\)

\begin{equation*} \sup_{B_{r/8}(x)} \abs{\nabla\partial_i u} \le \frac {4C}r \sup_{B_{r/4}(x)} \abs {\partial_i u}\text{.} \end{equation*}

In particular, for any \(j\) we have

\begin{equation*} \abs{\partial_{ij} u(x)} \le \frac {4C}r \sup_{B_{r/4}(x)} \abs {\partial_i u}. \end{equation*}

To estimate the right hand side, we apply Lemma 4.2 to \(u\) on \(B_{r/2}(x)\) to get

\begin{gather*} \sup_{B_{r/4}(x)} \abs {\partial_i u} \le \sup_{B_{r/4}(x)} \abs {\nabla u} \le \frac {2C}r \sup_{B_{r/2}(x)} \abs {u}. \end{gather*}

Putting things together, we conclude that

\begin{gather*} \abs{\partial_{ij} u(x)} \le \frac {8C^2}{r^2} \sup_{B_{r/2}(x)} \abs u \le \frac {8C^2}{r^2} \sup_{B_r(x_0)} \abs u\text{.} \end{gather*}

Since the above inequality is true for any \(i,j\) and any \(x \in B_{r/2}(x_0)\text{,}\) the desired inequality follows by taking a supremum. Indeed – and this is perhaps more detail than is really needed – we have

\begin{gather*} \sup_{B_{r/2}(x_0)} \abs{D^2 u} \le N \max_{i,j} \sup_{B_{r/2}(x_0)} \abs{\partial_{ij} u} \le \frac {8NC^2}{r^2} \sup_{B_r(x_0)} \abs u\text{.} \end{gather*}

Comment 1.

With these sorts of arguments one can replace the ball \(B_r(x_0)\) on the left hand sides of the inequalities in Lemma 4.2 and Corollary 4.3 with \(B_{\theta r}(x_0)\) for any \(\theta \in (0,1)\text{,}\) with the caveat that the constant \(C=C(N,\theta)\) now depends on \(\theta\) as well as the dimension \(N\text{.}\) One can also prove such a result directly by choosing the ‘cutoff function’ \(\eta\) in the proof of Lemma 4.2 differently.

Comment 2.

In this problem we need to pass back and forth between estimates for Vector and matrix norms and estimates for components. In this problem, since we are not particularly fussed about getting the best possible constants, it is enough to use the basic inequalities

\begin{alignat*}{2} \max_i \abs{a_i} \le \abs a \amp \le \sqrt N \max_i \abs{a_i} \amp \qquad \amp\fora a \in \R^N,\\ \max_{i,j} \abs{A_{ij}} \le \abs A \amp \le N \max_{i,j} \abs{A_{ij}} \amp \qquad \amp\fora A \in \R^{N \times N}. \end{alignat*}

Let me know if you have any questions about these inequalities or how to prove them.

Comment 3.

In the factor \(C/r\) on the right hand side of (4.2), \(r\) refers to the radius of the ball \(B_r(x_0)\) where Lemma 4.2 is being applied. So when we apply this result on balls with radius \(r/2\) and \(r/4\) in the argument above, the factors that appear are \(2C/r\) and \(4C/r\text{,}\) respectively, rather than just \(C/r\) each time.

Comment 4.

Note that Lemma 4.2 as written applies to a single harmonic function, not a vector or matrix whose components are each harmonic functions. In terms of symbols, Lemma 4.2 requires \(u \in C^3(\overline B)\) and does not allow, say, for \(u\) to instead lie in \(C^3(\overline B,\R^N)\text{;}\) see Notation 2.19. But we can get around this quite easily by, e.g., separately considering each of the components \(\partial_i u\) of the gradient \(\nabla u\) in the official solution.

4.2 The Dirichlet problem in a ball

Exercises

4.2.1. (PS6) Explicitly solving a Dirichlet problem.

Solution.

We calculate \(T1 = -4\) and \(Tx_i=-8x_i\text{.}\) Thus with respect to the basis \(\{1,x_1,x_2\}\text{,}\) the operator \(T\) is represented by the diagonal matrix

\begin{equation*} T = \begin{pmatrix} -4 \amp 0 \amp 0 \\ 0 \amp -8 \amp 0 \\ 0 \amp 0 \amp -8 \end{pmatrix}\text{,} \end{equation*}

which is simple to invert. The data \(g = x_1^2 x_2\) satisfies \(\Delta g = 2x_2\text{.}\) Thus, the solution \(u\) of the associated Dirichlet problem (4.6) is given by (4.9) as

\begin{align*} u \amp = (1-\abs x^2) T^{-1}(-\Delta g) + g \\ \amp = (1-\abs x^2) T^{-1}(-2x_2) + x_1^2 x_2\\ \amp = \tfrac 14(1-\abs x^2) x_2 + x_1^2 x_2\\ \amp = \tfrac 14(1+3x_1^2-x_2^2)x_2\text{.} \end{align*}

Once this formula has been obtained, we can of course easily verify that it solves (4.6).

4.2.2. (PS6) Harmonic functions are \(C^\infty\).

4.2.2.a

Solution.

Consider the setting of Lemma 4.5, but assume now that the functions \(f_n\) are \(C^{\ell+2}\text{.}\) Using the bounds in Exercise 4.1.3 for \(0 \le k \le \ell\text{,}\) we find that

\begin{equation*} \n{f_n-f_m}_{C^\ell(B_{r/2}(x))} \le C \sup_{B_r(x)} \abs{f_n-f_m} \to 0 \end{equation*}

as \(n,m \to \infty\text{,}\) where here the constant \(C\) depends on \(N\text{,}\) \(r\text{,}\) and \(\ell\text{.}\) Appealing to Theorem 2.20, we deduce that \(f_n \to f\) in \(C^\ell(\overline{B_{r/2}(x)})\text{.}\) Since \(x\) was arbitrary, this in particular implies that \(f \in C^\ell(\Omega)\text{.}\)

4.2.2.b

Solution.

Using our new and upgraded version of Lemma 4.5 in the proof of Theorem 4.8, we see that the sequence \(u_n\) in fact converges uniformly to some limit \(\tilde u \in C^\ell(\Omega)\text{,}\) which by the uniqueness of limits must be the same as \(u\text{.}\) Note that the functions \(u_n\) are polynomials, and so lie in \(C^\infty(\Omega)\text{.}\)

4.2.2.c

Solution.

If \(u \in C^0(\Omega)\text{,}\) then clearly \(g = u|_{\partial B} \in C^0(\partial B)\) as \(\partial B \subset \Omega\text{.}\) Thus, for any \(\ell\text{,}\) our improved version of Theorem 4.8 there exists a harmonic function \(\tilde u = C^\ell(B) \cap C^0(\overline B)\) which equals \(g\) on the boundary. By uniqueness we must have \(\tilde u = u|_{\overline B}\text{.}\) Since \(B\) and \(\ell\) were arbitrary, we must have \(u \in C^\infty(\Omega)\text{.}\)

4.2.3. Smooth but unbounded harmonic functions.

Solution.

Let \(\Phi\) be the function from Exercise 2.3.2, and note that while \(\Phi\) is \(C^\infty\) and harmonic away from \(x = 0\text{,}\) it is unbounded in the limit as \(x \to 0\text{.}\) So let \(B=B_1(0)\) and \(u(x)=\Phi(x-e_1)\text{.}\) Then \(u \in C^\infty(B)\) and harmonic in \(B\text{,}\) but does not lie in \(C^0(\overline B)\text{.}\) Indeed, as \(x \in B\) tends to \(e_1 \in \partial B\text{,}\) we have \(\abs{u(x)} \to +\infty\text{.}\)

4.3 The mean value property

Exercises

4.3.1. (PS7) Reflecting harmonic functions.

Solution.

Let \(x \in B\text{.}\) If \(x_N \gt 0\text{,}\) then we can pick \(\varepsilon \gt 0\) small enough that \(B_\varepsilon(x) \subseteq B^+\text{.}\) Since \(v|_{B^+} = u\) is harmonic, we have

\begin{equation*} v(x) = \frac 1{\abs{\partial B_r}}\int_{\partial B_r(x)} v\, dS \end{equation*}

for all \(0 \lt r \lt \varepsilon\) by the mean value property.

If \(x_N \lt 0\text{,}\) then instead we pick \(\varepsilon \gt 0\) small enough that \(B_\varepsilon(x) \subseteq B^-\text{,}\) the bottom half of the ball. One can check that

\begin{gather*} v|_{B^-}(x) = -u(x_1,\ldots,x_{N-1},-x_n) \end{gather*}

is harmonic on this set in several ways. We could simply calculate \(\Delta v\) using the chain rule (or recall Exercise 2.6.2), or we could change variables inside the integral to show that \(v|_{B^-}\) satisfies the mean value property.

It remains to consider the case where \(x_N = 0\text{.}\) Pick \(\varepsilon \gt 0\) small enough that \(B_\varepsilon(x) \subseteq B\text{.}\) Using a change of variables we find that, for any \(0 \lt r \lt \varepsilon\text{,}\)

\begin{align*} \int_{\partial B_r(x)} v\, dS \amp= \int_{\partial B_r(x) \cap \{x_N \gt 0\}} v\, dS +\int_{\partial B_r(x) \cap \{x_N \lt 0\}} v\, dS\\ \amp= \int_{\partial B_r(x) \cap \{x_N \gt 0\}} u\, dS -\int_{\partial B_r(x) \cap \{x_N \gt 0\}} u\, dS\\ \amp= 0 = u(x) = v(x). \end{align*}

We can now can now conclude by applying Theorem 4.11.

Comment.

Note that the “for each \(x \in \Omega\)” in Theorem 4.11 is doing a lot of work. If we only know

\begin{gather*} w(x)= \frac 1{\abs{\partial B_r}}\int_{\partial B_r(x)} w\, dS \end{gather*}

for balls centred at a single \(x \in \Omega\) (and all sufficiently small \(r \gt 0\)), then we can conclude nothing about the function \(w\) whatsoever. On past exams students have sometimes gotten confused about this, especially when \(\Omega\) is itself a ball (or related to a ball).

4.4 Subharmonic functions

Exercises

4.4.1. (PS7) Directly checking \(\abs x\) is subharmonic.

4.4.1.a

Solution.

The function \(h\) being harmonic means that \(h''=0\text{,}\) i.e. that \(h\) is an affine function whose graph is a straight line. The identity (✶) is one of the many ‘standard forms’ for affine functions, and can easily be derived from your preferred form by algebraic manipulations, or by appealing to basic facts about lines in the plane.

4.4.1.b

Solution.

For any \(t \in (0,1)\) we have \(t,1-t \ge 0\text{,}\) and so by the triangle inequality

\begin{align*} u(ta+(1-t)b) \amp = \abs{ta+(1-t)b}\\ \amp \le \abs{ta}+\abs{(1-t)b}\\ \amp = t\abs{a}+(1-t)\abs{b}\\ \amp = tu(a)+(1-t)u(b) \end{align*}

as desired.

4.4.1.c

Solution.

Combining the previous two parts, we have, for \(t \in (0,1)\)

\begin{align*} u(ta+(1-t)b) \amp \le tu(a)+(1-t)u(b)\\ \amp \le th(a)+(1-t)h(b)\\ \amp = h(ta+(1-t)b)\text{,} \end{align*}

where in the second step we have used that \(u \le h\) on \(\partial B = \{a,b\}\text{.}\) Thus \(u \le h\) on \(B\text{.}\) By the arbitrariness of \(h\) and \(B\text{,}\) we conclude that \(u\) is subharmonic in the sense of Definition 4.12.

4.4.2. (PS7) Pointwise maximum of subharmonic functions.

Solution.

Let \(B\) be a ball whose closure is contained in \(\Omega\text{,}\) and let \(h \in C^2(B) \cap C^0(\overline B)\) be a harmonic function with

\begin{gather*} u = \max(u_1,u_2) \le h \ona \partial B. \end{gather*}

Then in particular \(u_1 \le h\) on \(\partial B\text{,}\) and so since \(u_1\) is subharmonic we have \(u_1 \le h\) in \(B\text{.}\) Similarly we get \(u_2 \le h\text{.}\) Thus \(u = \max(u_1,u_2) \le h\) as desired.

This can be applied to the function \(u(x)=\abs x\) in Exercise 4.4.1. Indeed, we have \(u(x) = \max(x,-x)\text{,}\) where the functions \(x \mapsto \pm x\) are harmonic and hence in particular subharmonic, and so \(u\) is subharmonic by the above argument.

Comment.

Note that when verifying Definition 4.12 for \(\max(u_1,u_2)\text{,}\) the harmonic function \(h\) that is handed to us satisfies, by assumption, the inequality \(\max(u_1,u_2) \le h\) on \(\partial B\text{.}\) Several students instead assumed that \(u_1 \le h\) and \(u_2 \le h\) on this set and noted that this implies \(\max(u_1,u_2) \le h\text{.}\) But that has the logic exactly backwards, which could easily have cost marks on an exam.

4.4.3. (PS7) Harmonic liftings.

Solution.

By construction, \(U\) is harmonic on \(B\) and agrees with \(u\) on \(\partial B\text{,}\) and so \(u \le U\) on \(B\) by the definition of \(u\) being subharmonic. Moreover, \(u \equiv U\) on \(\Omega\without B\text{,}\) and so we deduce that \(u \le U\) on all of \(\Omega\) as desired.

It remains to show that \(U\) is subharmonic. Let \(B'\) be a ball whose closure is contained in \(\Omega\text{,}\) and let \(h \in C^2(B') \cap C^0(\overline{B'})\) be a harmonic function in \(B'\) with \(U \le h\) on \(\partial B'\text{.}\) We need to show that \(U \le h\) in \(B'\text{.}\) Since \(u \le U\) on \(\Omega\text{,}\) we have \(u \le U \le h\) on \(\partial B'\text{,}\) and since \(u\) is subharmonic this implies \(u \le h\) on \(B'\text{.}\) Thus on \(B' \without B\) (which may of course be empty) we have \(u = U \le h\text{,}\) and so it remains to show that \(U \le h\) on the intersection \(B' \cap B\text{.}\)

But \(U\) and \(h\) are both harmonic on \(B' \cap B\text{,}\) and so by the comparison principle it suffices to show that \(U \le h\) on the boundary (draw a picture!!)

\begin{gather*} \partial(B' \cap B) = (\overline{\partial B' \cap B}) \cup (\overline{B' \cap \partial B}). \end{gather*}

On \(\partial B' \cap B \subseteq \partial B'\) we have \(U \le h\) by assumption. On \(B' \cap \partial B \subseteq \partial B\) we have \(u=U\text{,}\) and since \(B' \cap \partial B \subseteq B'\) this gives \(U=u\le h\) by the argument in the previous paragraph. Since \(U\) and \(h\) are continuous, we don’t have to worry about the closures, and so we conclude that \(U \le h\) on \(\partial(B' \cap B)\) and hence in \(B' \cap B\) by the comparison principle.

4.6 An extended problem

4.6.1. Getting a harmonic function.

Solution.

We’ve shown on Exercises 3.4 that \(v=x_2^3/6\) is such a function. The difference \(w=u-v\) then solves

\begin{align} \left\{ \begin{alignedat}{2} \Delta w \amp= 0 \amp\quad\amp \ina \Omega, \\ w \amp= -v \amp\quad\amp \ona \partial \Omega. \end{alignedat} \right.\tag{✶✶} \end{align}

4.6.2. Existence.

Solution.

Drawing a picture, we visually confirm that \(\Omega\) satisfies the exterior ball property. As it is also bounded and connected, and \(-v \in C^0(\partial\Omega)\text{,}\) Theorem 4.17 guarantees that (✶✶) has a unique solution \(w \in C^2(\Omega) \cap C^0(\overline \Omega)\text{.}\) Indeed, by Exercise 4.2.2 we have \(w \in C^\infty(\Omega)\text{.}\) Thus (✶) has a unique solution \(u = w + v \in C^\infty(\Omega) \cap C^0(\overline\Omega)\text{.}\)

4.6.3. Extension.

Solution.

Our domain \(\Omega\) is precisely the type of domain considered in Exercise 4.3.1. Moreover, \(w\) is a harmonic function with \(w=-v = -x_2^3/6 = 0\) on \(\partial \Omega \cap \{x_2 = 0\}\text{.}\) Thus we can apply the result from this exercise, and conclude that the function

\begin{align*} W(x_1,x_2) = \begin{cases} w(x_1,x_2) \amp x_2 \ge 0, \\ -w(x_1,-x_2) \amp x_2 \lt 0 \end{cases} \end{align*}

is harmonic in \(B_1(0)\text{.}\) Indeed, we have that \(W \in C^2(B_1(0)) \cap C_0(\overline{B_1(0)})\) satisfies

\begin{align} \left\{ \begin{alignedat}{2} \Delta W \amp= 0 \amp\quad\amp \ina B_1(0), \\ W \amp= -x_2^3/6 \amp\quad\amp \ona \partial B_1(0). \end{alignedat} \right.\tag{✶✶✶} \end{align}

4.6.4. Explicit formula.

Solution.

Looking at (✶✶✶) we see that it is precisely the type of problem covered by Lemma 4.7. Thus the unique solution \(W\) is a polynomial, which we can for instance calculate using the technique from the proof of Lemma 4.7. We ultimately find that the corresponding \(u\) is

\begin{equation*} u = - \tfrac 18 (1-\abs x^2)x_2 \text{.} \end{equation*}

Using this explicit formula we can check, for instance, that \(u(0,\frac 12) = -3/64 \in (-7/48,0)\) and that \(\partial_2 u(0,0) = -1/8 \lt 0\text{.}\)

5 Radial symmetry
5.1 Statement of the theorem

Exercises

5.1.1. (PS8) Necessity of \(u\gt0\) in Theorem 5.1.

Solution 1.

Consider \(N=1\text{,}\) \(f(u)=u\text{,}\) and \(R=\pi\text{.}\) Then \(u(x) = \sin(x)\) satisfies all of the hypotheses of the theorem, but is clearly not an even function of \(x\text{.}\)

Solution 2.

Another nice counterexample, which I saw in a solution from a few years ago, is to consider \(f\) with \(f(0)=0\) and the function \(u \equiv 0\text{.}\) Clearly \(u\) satisfies \(\Delta u + f(u) = 0\) in \(B\) and \(u=0\) on \(\partial B\text{.}\) While it is radially symmetric, it fails to satisfy the second conclusion of Gidas–Ni–Nirenberg about monotonicity.

5.1.2. (PS8) Proving \(u\gt0\).

Solution.

Note that \(\Delta u = -f(u) \le 0\text{.}\) If we do not have \(u \gt 0\) in \(\Omega\text{,}\) then \(u\) must achieve \(\inf_\Omega u \le 0\) at some point \(x \in \Omega\text{.}\) But then the strong maximum principle implies that \(u\) is identically constant, and since \(u = 0\) on \(\partial\Omega\) the only possibility is \(u \equiv 0\text{.}\)

Alternatively we could have appealed to Exercise 3.3.3 with comparison function \(v \equiv 0\text{.}\) Note that in either case we need both the boundedness and the connectedness of \(\Omega\) to make the argument work.

5.1.3. (PS8) Very old exam question.

Solution 1. Restricting \(u\) to a subdomain

The hint was to consider the restriction of \(u\) to the open set \(\Omega' := \{x \in \Omega : u \gt 1\}\text{.}\) Assume for the sake of contradiction that \(\Omega'\) is nonempty. Since \(u=0\lt1\) on \(\partial\Omega\text{,}\) we can check that \(\partial\Omega' \subset \Omega\text{,}\) which in particular implies that \(u=1\) on this set. Thus we have

\begin{alignat*}{2} \Delta u \amp= u^3 - u \ge 0 \amp\qquad\amp \ina \Omega',\\ u \amp= 1 \amp\qquad\amp \ona \partial\Omega'. \end{alignat*}

The weak maximum then implies \(u \le 1\) on all of \(\Omega'\text{.}\) But this contradicts the definition of \(\Omega'\text{.}\) The argument for \(u \ge -1\) is similar.

Solution 2. Thinking about \(\Delta u\) at a maximum or minimum

Suppose for the sake of contradiction that \(M =\sup_\Omega u \gt 1\text{.}\) Since \(\Omega\) is bounded, this supremum must be achieved at some \(x \in \overline \Omega\text{,}\) and since \(u=0\) on \(\partial \Omega\) we must have \(x \in \Omega\text{.}\) Thus \(x\) is a local maximum of \(u\) and so the Hessian \(D^2 u(x)\) is negative semi-definite. In particular its trace \(\Delta u(x) = \delta_{ij} (D^2 u(x))_{ij} \le 0\) by Lemma 2.31. But from the equation we have

\begin{gather*} \Delta u(x) = u^3-u = M^3 - M = M(M^2-1) \gt 0, \end{gather*}

a contradiction. Thus \(\sup_\Omega u \le 1\text{.}\) The argument that \(\inf_\Omega u \ge 1\) is similar.

Solution 3. Using Lemma 5.4

It is natural to wonder if Lemma 5.4 can be used here, and indeed it can. Suppose for simplicity that \(\Omega\) is connected, although the argument can be generalised to work when \(\Omega\) is disconnected. (Of course on an exam it would be dangerous to make additional assumptions beyond what is given to us in the problem!) First we write the given PDE as \(\Delta u + f(u) = 0\) where \(f \in C^1(\R)\) is given by \(f(z)=z-z^3\text{.}\) Let \(M=\max_{\overline \Omega} u\text{,}\) and suppose for the sake of contradiction that \(M \gt 1\text{.}\) As \(u=0 \lt 1\) on \(\partial\Omega\text{,}\) \(u\) must achieve \(M\) at an interior point of \(\Omega\text{.}\) As \(f(M) \lt 0\) the constant function \(M\) also satisfies

\begin{align*} \Delta u +f(u) = 0 \amp\gt f(M) = \Delta M + f(M), \end{align*}

in \(\Omega\text{,}\) as well as \(u \le M\text{.}\) The first part of Lemma 5.4 then implies \(u \equiv M\text{,}\) contradicting the fact that \(u \lt M\) on \(\partial\Omega\text{.}\)

5.1.4. (PS8) Laplacian of radially symmetric functions.

Solution.

As in Exercise 2.3.2 we have

\begin{gather*} \partial_i u(x) = \partial_i U(\abs x) = U'(\abs x)\abs x^{-1} x_i \text{.} \end{gather*}

Differentiating a second time using the product rule, we find

\begin{align*} \Delta u(x) \amp = \partial_i \partial_i u(x)\\ \amp = \partial_i \big[ U'(\abs x)\abs x^{-1} x_i \big]\\ \amp = U''(\abs x)\abs x^{-2} x_i x_i - U'(\abs x)\abs x^{-3} x_i x_i + U'(\abs x) \abs x^{-1} \delta_{ii}\\ \amp = U''(\abs x) - U'(\abs x)\abs x^{-3} x_i x_i + U'(\abs x) \abs x^{-1} \delta_{ii}\\ \amp = U''(\abs x) + \frac{N-1}{\abs x} U'(\abs x) \end{align*}

as desired.

5.2 Hyperplanes and reflections

Exercises

5.2.1. (PS9) Reflections are idempotent.

Solution.

We calculate

\begin{align*} (x^{\lambda,e})^{\lambda,e} \amp= x^{\lambda,e} + 2(\lambda - x^{\lambda,e}\cdot e)e\\ \amp= x + 2(\lambda - x\cdot e)e + 2\Big(\lambda - \big(x + 2(\lambda - x\cdot e)e\big) \cdot e\Big)e\\ \amp= x + 2(\lambda - x\cdot e)e + 2\Big(\lambda - x \cdot e - 2(\lambda - x\cdot e)\Big)e\\ \amp= x + \Big[2(\lambda - x\cdot e) - 2(\lambda - x \cdot e)\Big]e\\ \amp= x\text{.} \end{align*}

5.3 A semilinear comparison principle

Exercises

5.3.1. (PS9) Necessity of an assumption in Lemma 5.4.

Solution.

Take \(N=1\text{,}\) \(\Omega=(0,2\pi)\text{,}\) and \(f(s)=s\text{.}\) Then \(u(x) = \sin x\) and \(v \equiv 0\) satisfy

\begin{equation*} \Delta u + f(u) = 0 = \Delta v + f(v) \end{equation*}

in \(\Omega\text{,}\) and \(u=v\) on \(\partial\Omega\) as well as at \(0 \in \Omega\text{.}\) But clearly we do not have \(u \ge v\) in \(\Omega\text{,}\) let alone \(u \equiv v\text{.}\) Moreover, at \(2\pi \in \partial\Omega\) we have

\begin{equation*} \frac{\partial u}{\partial n}(2\pi) = u'(2\pi) = 1 \gt 0 = \frac{\partial v}{\partial n}(2\pi)\text{.} \end{equation*}

5.4 Proof of the theorem

Exercises

5.4.1. A calculation needed in Lemma 5.6.

5.4.1.a

Solution.

The composition \(f \circ \phi\) has a local maximum at \(x_1 = 0\text{.}\) Thus

\begin{equation} 0 = \partial_1 (f \circ \phi) = \partial_1 f(\phi) \partial_1 \phi_1 + \partial_2 f(\phi) \partial_1 \phi_2 \text{.}\tag{5.10} \end{equation}

Now at \(x_1 = 0\) we have

\begin{equation*} \phi = x^*, \quad \partial_1 \phi_1 = 1, \quad \partial_1 \phi_2 = 0, \end{equation*}

and so (5.10) simplifies to \(\partial_1 f(x^*)=0\) as desired.

5.4.1.b

Solution.

Since \(u = 0\) on \(\partial B\text{,}\) we indeed have \(u \circ \phi = 0\) for \(x_1\) near \(0\text{.}\) Differentiating this identity once we have as above that

\begin{equation*} 0 = \partial_1 u(\phi) \partial_1 \phi_1 + \partial_1 u(\phi) \partial_1 \phi_2, \end{equation*}

which yields \(\partial_1 u(x^*)=0\) upon setting \(x_1=0\text{.}\) Differentiating a second time, we find

\begin{align*} 0 \amp = \partial_1(\partial_1 u(\phi) \partial_1 \phi_1 + \partial_1 u(\phi) \partial_1 \phi_2)\\ \amp = (\partial_{11} u(\phi)\partial_1 \phi_1 + \partial_{12} u(\phi) \partial_1 \phi_2) \partial_1 \phi_1 \\ \amp \qquad + \partial_1 u(\phi) \partial_{11} \phi_1 \\ \amp \qquad + (\partial_{21} u(\phi)\partial_1 \phi_1 + \partial_{22} u(\phi) \partial_1 \phi_2) \partial_1 \phi_2 \\ \amp \qquad + \partial_2 u(\phi) \partial_{11} \phi_2 \text{.} \end{align*}

Now at \(x_1=0\text{,}\) we have

\begin{equation*} \phi = x^*, \quad \partial_1 \phi_1 = 1, \quad \partial_{11} \phi_2 = -R^{-1}, \quad \partial_1 \phi_2 = \partial_{11} \phi_1 = \partial_1 u = 0\text{,} \end{equation*}

and so the above formula simplifies dramatically to

\begin{align*} 0 \amp = \partial_{11} u(x^*) - R^{-1} \partial_2 u(x^*)\text{,} \end{align*}

which is the desired identity.

5.4.1.c

Solution.

For \(x \in \partial B\text{,}\) the outward pointing normal vector is \(n=x/R\text{.}\) Since \(u\) achieves its minimum over \(\overline B\) at \(x\text{,}\) we therefore have

\begin{equation*} x \cdot \nabla u = Rn \cdot \nabla u = R\frac{\partial u}{\partial n} \le 0\text{.} \end{equation*}

Since \(x \cdot \nabla u = \partial_2 u = 0\) at \(x^*\text{,}\) we conclude that \(f = x \cdot \nabla u\) achieves \(\max_{\partial B} f\) at \(x^*\text{.}\) Thus by Part a we have \(\partial_1 f = 0\) at \(x^*\text{.}\) Using the product rule we have

\begin{align*} \partial_1 f \amp = \partial_1 (x_1 \partial_1 u + x_2 \partial_2 u) \\ \amp = \partial_1 u + x_1 \partial_{11} u + x_2 \partial_{12} u \text{.} \end{align*}

Plugging in \(x=x^*\) and simplifying, we conclude that

\begin{equation*} 0 = \partial_1 f(x^*) = R \partial_{12} u(x^*)\text{,} \end{equation*}

and hence \(\partial_{12} u(x^*)=0\) as desired.

5.4.1.d

Solution.

First consider the case where \(B_R(0) \subset \R^N\) and \(x^* = Re_N\text{.}\) We can now parametrise \(\partial B\) near \(x^*\) using

\begin{equation*} x = \phi(x_1,\ldots,x_{N-1}) = \Big(x_1,\ldots,x_{N-1},\sqrt{R^2-x_1^2-\cdots-x_{N-1}^2}\Big)\text{.} \end{equation*}

At \(x_1=\cdots=x_{N-1}=0\text{,}\) we calculate

\begin{align*} \partial_i \phi_k \amp= \begin{cases} 1 \amp i=k \lt N \\ 0 \amp \text{otherwise} \end{cases}, \\ \partial_{ij} \phi_k \amp= \begin{cases} -R^{-1} \amp i=j \lt N \text{ and } k=N \\ 0 \amp \text{otherwise} \end{cases}\text{.} \end{align*}

If \(u \in C^2(\overline B)\) satisfies \(u=0\) on \(\partial B\text{,}\) \(u \gt 0\) in \(B\text{,}\) and \(\partial_N u = 0\) at \(x^*\text{,}\) calculations similar to those in the previous two parts then yield \(\nabla u(x^*) = 0\) and \(\partial_{ij} u(x^*) = 0\) unless \(i=j=N\text{,}\) i.e.

\begin{equation*} \partial_{ij} u(x^*) = \delta_{iN} \delta_{jN} \partial_{NN} u(x^*) \text{.} \end{equation*}

Now consider a general \(x^* \in \partial B\text{,}\) let \(A \in \R^{N \times N}\) be an orthogonal matrix with \(x^* = A(Re_N)\text{,}\) and define a function \(v \in C^2(\overline B)\) by \(v(x)=u(Ax)\text{.}\) Assuming that \(u=0\) on \(\partial B\text{,}\) \(u \gt 0\) in \(B\text{,}\) and \(\nabla u = 0\) at \(x^*\text{,}\) we can apply the above argument to \(v\) to get

\begin{equation*} \partial_{ij} v(Re_N) = \delta_{iN} \delta_{jN} \partial_{NN} v(Re_N) \text{.} \end{equation*}

Differentiating \(u(x)=v(A^{-1}x)\) twice as in Exercise 2.6.2, we find that

\begin{align*} \partial_{ij} u(x^*) \amp = A_{ki}^{-1} A_{\ell j}^{-1} \partial_{k\ell} v(Re_N)\\ \amp = A_{ki}^{-1} A_{\ell j}^{-1} \delta_{kN} \delta_{\ell N} \partial_{NN} v(Re_N) \\ \amp = A_{Ni}^{-1} A_{Nj}^{-1} \partial_{NN} v(Re_N) \\ \amp = A_{iN} A_{jN} \partial_{NN} v(Re_N) \\ \amp = \frac{x^*_i x^*_j}{R^2} \partial_{NN} v(Re_N) \text{,} \end{align*}

which is of the desired form in (5.8).

5.4.2. (PS9) Monotonicity in an annulus.

Solution.

The proof of Lemma 5.6 only involves a neighborhood of the boundary \(\abs x = R\text{,}\) and so is unchanged for the outer boundary \(\abs x = R_2\) of our annular set. The proofs of Lemmas 5.7–5.8 also continue to work provided \(\lambda \gt (R_1+R_2)/2\text{,}\) since for these values they cannot ‘see’ the hole in the domain. Thus we conclude that (5.9) holds for any \(\lambda \gt (R_1+R_2)/2\text{.}\) This implies \(\partial_1 u \lt 0\) for \(x\) in our annulus with \(x_1 \gt (R_1+R_2)/2\text{,}\) and hence in particular that \(\partial_r u = \partial_1 u \lt 0\) along the line segment \((x_1,0,\ldots,0)\text{,}\) \((R_1+R_2)/2 \lt x_1 \lt R\text{.}\) Rotating coordinates, we obtain \(\partial_r u \lt 0\) for \((R_1+R_2)/2 \lt \abs x \lt R\) as desired.

5.4.3. (PS9) Gidas–Ni–Nirenberg for an ellipse.

Solution.

Going through the proofs of Lemma 5.6, Lemma 5.7 and Lemma 5.8, everything continues to work. We conclude that \(u\) is even in \(x_1\) with \(\partial_1 u \lt 0\) in \(E \cap \{x_1 \gt 0\}\text{.}\) Rotating coordinates, we get a similar statement with \(x_1\) replaced by \(x_2\text{.}\)

Here we have used the fact that the ellipse is symmetric under reflections in either \(x_1\) or \(x_2\text{.}\) Unlike the ball, it is not symmetric under other reflections, and for this reason we can no longer argue that \(u\) is radially symmetric.

6 Maximum principles for parabolic equations
6.2 The weak maximum principle

Exercises

6.2.1. (PS10) Weak maximum principle for \(c\le0\).

6.2.1.a

Solution.

We have

\begin{equation*} Le^{-kt} = (k+c)e^{-kt} \gt 0 \end{equation*}

provided \(k \gt \sup_D (-c)\text{.}\)

6.2.1.b

Solution.

This follows at once from the fact that \(v \gt u\) in \(\overline D\text{.}\)

6.2.1.c

Solution.

If \(v\) were to achieve this positive maximum at a point \((\bar x,\bar t) \in Q \cup (\Omega \times \{T\})\text{,}\) then arguing exactly as in the proof of Theorem 3.2 we get

\begin{equation*} Lv(\bar x,\bar t) \le c(\bar x,\bar t) \max_{\overline D} v \le 0, \end{equation*}

where in the last step we have used that \(c \le 0\text{.}\) But this contradicts Part a.

6.2.1.d

Solution.

Thus, exactly as in the proof of Theorem 3.2, we have \(\max_{\overline D} v = \max_\Sigma v\text{,}\) and so sending \(\varepsilon \to 0\) we get

\begin{equation*} \max_{\overline D} u = \max_\Sigma u \end{equation*}

as desired.

6.2.2. Asymptotics for the heat equation.

6.2.2.a

Solution.

We calculate

\begin{align*} Lv \amp = -\partial_t v + \Delta v \\ \amp = N(2-\abs x^2) e^{-Nt} - 2N e^{-Nt} \\ \amp = -N\abs x^2 e^{-Nt}\\ \amp \le 0\text{.} \end{align*}

6.2.2.b

Solution.

First we observe that \(Lv \le 0 = Lu\) in \(\Omega \times (0,T]\text{.}\) Since \(\abs x = 1\) on \(\partial \Omega \times [0,T]\text{,}\) we also have \(v=(2-1)e^{-Nt} \ge 0 = u\) on this set. Finally, on \(\Omega \times \{0\}\) we have

\begin{equation*} v = 2-\abs x^2 \ge 2-1 = 1 \ge u\text{.} \end{equation*}

Applying Corollary 6.4, we conclude that \(u \le v\) in \(\overline \Omega \times [0,T]\) as desired.

6.2.2.c

Solution.

Since \(T\) in the previous part was arbitrary, we have \(u \le v\) on \(\Omega \times (0,\infty)\text{.}\) Arguing similarly with \(-u\) we find \(-u \le v\) and hence \(\abs u \le v\) on \(\Omega \times (0,\infty)\text{.}\) Since

\begin{equation*} v(x,t) \le 2e^{-Nt} \to 0 \end{equation*}

as \(t \to \infty\text{,}\) uniformly for \(x \in \Omega\text{,}\) we conclude that \(u(x,t) \to 0\) uniformly as desired.

Comment.

In this context of this problem and this chapter, this is not a particularly important point, but the above convergence is uniform because our estimate \(\abs{u(x,t)-0} \le 2e^{-Nt}\) is independent of \(x\) (‘uniform in \(x\)’). Thus

\begin{equation*} \lim_{t \to \infty} \sup_{x \in \Omega} \abs{u(x,t)-0} = 0\text{.} \end{equation*}

6.2.3. (PS10) Comparison principle.

Solution.

Following the hint, we first reduce to the case where \(c \le 0\) by making the change of variable \(\bar u = e^{\gamma t} u\) and \(\bar v = e^{\gamma t} v\) where \(\gamma \in \R\) is a constant to be determined. The product rule gives

\begin{align*} L u \amp= L (e^{-\gamma t}\bar u) \\ \amp= e^{-\gamma t} L\bar u + \gamma e^{-\gamma t} \bar u\\ \amp= e^{-\gamma t} (L\bar u + \gamma \bar u) \\ \amp= e^{-\gamma t} \bar L \bar u, \end{align*}

where \(\bar L\) is the operator

\begin{gather*} \bar L = -\partial_t + a_{ij} \partial_{ij} + b_i \partial_i + (c+\gamma), \end{gather*}

and similarly for \(v\) and \(\bar v\text{.}\) Since the factor \(e^{\gamma t} \gt 0\text{,}\) our assumptions on \(u\) and \(v\) are equivalent to \(\bar L \bar u \le \bar L \bar v\) in \(D\) and \(\bar u \ge \bar v\) on \(\Sigma\text{,}\) while the desired conclusion is similarly equivalent to \(\bar u \ge \bar v\) in \(D\text{.}\) Moreover, since \(c\) is bounded, we can pick \(\gamma\) large and negative enough that the zeroth order coefficient of \(\bar L\) is \(\bar c = c+\gamma \le 0\text{.}\)

Thus we can assume without loss of generality that \(c \le 0\text{.}\) Defining \(w = v-u\) we have \(Lw \ge 0\) in \(D\) and \(w \le 0\) on the parabolic boundary \(\Sigma\text{.}\) We now conclude as in the proof of Proposition 3.4: if \(M = \max_{\overline D} w \le 0\) then we are already done, while if \(M \gt 0\) then we can apply (the second case of) Theorem 6.3 to get a contradiction.

6.3 Semilinear comparison principles

Exercises

6.3.1. (PS10) Uniqueness for semilinear problems.

Solution.

Let \(\mathcal S(u) = -\partial_t u + \Delta u + f(u)\) be the semilinear parabolic operator appearing in the statement of the problem. Clearly it is uniformly parabolic. Suppose that \(u,v \in C^2(D) \cap C^1(\overline D)\) are two solutions. Since they agree on the parabolic boundary \(\Sigma\) and satisfy \(\mathcal S(u) = \mathcal S(v) = 0\text{,}\) two applications of Theorem 6.7 yield \(u \le v\) and \(v \ge u\) in \(D\text{,}\) or in other words \(u \equiv v\) in \(D\text{.}\)

6.3.2. (PS10) Bounding a solution to a semilinear problem.

Solution.

The semilinear operator \(\mathcal S(u) = -\partial_t u + \Delta u + u\partial_1 u + \sin u \) is clearly uniformly parabolic because of the Laplacian, and \(F(x,t,z,p)=zp_1+\sin z\) is certainly \(C^1\text{.}\) Moreover, a quick calculation shows that the constant functions \(0\) and \(\pi\) satisfy

\begin{gather*} 0 \le u \le \pi \ona \Sigma, \qquad \mathcal S(0) = \mathcal S(\pi) = \mathcal S(u) = 0. \end{gather*}

Thus Theorem 6.7 implies that \(0 \le u \le \pi\) on all of \(D\) as desired.

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Appendix I Solutions to exercises

1 Maximum principles for ordinary differential equations1.1 Maximum principles

Exercises

1.1.1. (PS1) Some basic counterexamples.

1.1.1.a

1.1.1.b

1.1.2. (PS1) A concrete application.

1.1.2.a

1.1.2.b

1.1.2.c

1.1.2.d

1.1.2.e

1.1.3. (PS1) Basic lemma for \(h \le 0\).

1.1.4. Theorem 1.4.

1.1.6. (PS1) Alternative proof of Theorem 1.2.

1.2 Applications

Exercises

1.2.2. (PS1) A ‘generalised’ maximum principle.

1.2.2.a

1.2.2.b

2 Preliminaries2.1 Index notation

Exercises

2.1.1. (PS2) Summation convention.

2.1.1.a

2.1.1.b

2.1.1.c

2.1.1.d

2.1.1.e

2.1.1.f

2.2 Partial derivatives

Exercises

2.2.1. (PS2) Derivatives of linear and quadratic functions.

2.2.1.a

2.2.1.b

2.2.1.c

2.2.1.d

2.2.1.e

2.2.1.f

2.2.1.g

2.2.2. (PS2) Derivative of the modulus.

2.3 The chain rule

Exercises

2.3.1. (PS2) Derivatives after shifting and scaling.

2.3.1.a

2.3.1.b

2.3.1.c

2.3.1.d

2.3.2. (PS2) Derivatives of symmetric functions.

2.3.2.a

2.3.2.b

2.3.2.c

2.3.2.d

2.3.2.e

2.3.2.f

2.3.2.g

2.5 Regularity of domains

Exercises

2.5.1. (PS3) A domain with a corner.

2.5.1.a

2.5.1.b

2.5.2. (PS3) Locally constant functions.

2.5.2.a

2.5.2.b

2.6 Symmetric matrices

Exercises

2.6.2. (PS3) Invariance of the Laplacian.

2.7 Maxima and minima

Exercises

2.7.1. (PS3) Minimising a quadratic polynomial.

2.7.1.a

2.7.1.b

2.7.1.c

2.7.1.d

2.7.1.e Optional.

2.8 Compactness and diagonal subsequences

Exercises

2.8.1. (PS3) Suprema, maxima, and closures.

2.8.2. (PS4) Sufficient condition for equicontinuity.

2.9 Integration and vector calculus

Exercises

1 Maximum principles for ordinary differential equations
1.1 Maximum principles

2 Preliminaries
2.1 Index notation

3 Maximum principles for elliptic equations
3.1 Ellipticity and uniform ellipticity

4 The Dirichlet problem
4.1 Interior estimates