Section 1.1 Maximum principles
Let \([a,b]\) be a closed interval and let \(u \maps [a,b] \to \R\text{.}\) We suppose that \(u\) is continuous, in which case its maximum
\begin{equation*}
M = \max_{[a,b]} u
\end{equation*}
exists, and is attained at least one point in \([a,b]\text{.}\) We further assume that \(u\) is twice continuously differentiable on \((a,b)\text{,}\) so that in symbols we now have \(u \in C^2((a,b)) \cap C^0([a,b])\text{.}\) This allows us to do basic calculus with \(u\text{.}\) In particular, we can apply the differential operator
\begin{equation}
L = \frac{d^2}{dx^2} + g \frac d{dx} + h\tag{1.1}
\end{equation}
to \(u\text{,}\) where here the coefficients \(g,h\) are bounded functions on \((a,b)\text{.}\) The result is the well-defined function \(Lu \maps (a,b) \to \R\) given by
\begin{equation*}
Lu(x) = u''(x) + g(x)u'(x) + h(x)u(x)\text{.}
\end{equation*}
The goal of this chapter is to show how inequalities for \(Lu\) can lead to conclusions about \(M\text{.}\)
Lemma 1.1. Basic lemma for \(h\equiv 0\).
Suppose that \(h \equiv 0\) and that \(Lu \gt 0\) on \((a,b)\text{.}\) Then \(u\) can equal \(M\) only at the endpoints \(a\) and \(b\text{.}\)
1
By \(Lu \gt 0\) on \((a,b)\) we mean \(Lu(x) \gt 0\) for all \(x \in (a,b)\text{.}\)
Proof.
Suppose that \(u(c)=M\) 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*}
and hence
\begin{equation*}
Lu(c) = u''(c) + 0 \le 0,
\end{equation*}
which contradicts the assumption that \(Lu \gt 0\) on \((a,b)\text{.}\)
By applying Lemma 1.1 to \(-u\text{,}\) we obtain a related minimum principle.
While already quite useful, a deficiency in Lemma 1.1 is that we cannot weaken the assumptions to \(Lu \ge 0\text{.}\) In particular, we cannot apply it to solutions of the differential equation \(Lu = 0\text{.}\) Indeed, the function \(u \equiv M\) is an easy counter example. Fortunately, however, this is the only counter example.
Theorem 1.2. One-dimensional strong maximum principle for \(h\equiv 0\).
Suppose that \(h\equiv 0\) and \(Lu \ge 0\) in \((a,b)\text{.}\) If \(u\) attains its maximum \(M\) at some \(c \in (a,b)\text{,}\) then \(u \equiv M\text{.}\)
Proof.
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 \((c,b)\text{;}\) see Figure 1.2. Differentiating we find
\begin{gather*}
Lz = z'' + gz' = \alpha(\alpha+g)e^{\alpha(x-c)}.
\end{gather*}
Since \(g\) is bounded, we can choose \(\alpha \gt 0\) large enough that \(Lz \gt 0\) on \((a,b)\text{.}\)
Now consider
\begin{gather*}
w = u + \varepsilon z
\end{gather*}
where \(\varepsilon \gt 0\) is a small parameter to be determined. By linearity, we have
\begin{equation*}
Lw = Lu + \varepsilon L z \gt 0
\end{equation*}
on \((a,b)\text{.}\) Moreover, 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 Lemma 1.1.
The above proof ‘upgrades’ Lemma 1.1 into Theorem 1.2 by applying the lemma to a well-chosen function \(w=u+\varepsilon z\text{.}\) This will be a recurring theme throughout the course. Note that the choice of \(z\) highly non-unique! It is a useful exercise to list the properties of \(z\) needed in the proof.
Theorem 1.3. One-dimensional Hopf lemma for \(h \equiv 0\).
Suppose that \(h \equiv 0\) and \(Lu \ge 0\) in \((a,b)\text{.}\) If \(u(a)=M\) and \(u\) has a one-sided derivative at \(a\text{,}\) then either \(u'(a) \lt 0\) or \(u \equiv M\text{.}\) Similarly if \(u(b) = M\) and \(u\) has a one-sided derivative at \(b\text{,}\) then either \(u'(b) \gt 0\) or \(u \equiv M\text{.}\)
The non-strict inequalities are clear from the definition of the (one-sided) derivative. For instance, if \(u(a)=M\) and \(u \le M\) on \([a,b]\text{,}\) then clearly
\begin{equation*}
u'(a) = \lim_{x \searrow a} \frac{u(x)-u(a)}{x-a} \le 0
\end{equation*}
provided the limit exists. The content of the theorem is that this inequality is in fact strict.
Proof.
Suppose without loss of generality that \(u(a) = M\) and that \(u(d) \lt M\) for some \(d \in (a,b)\text{.}\) As in the proof of Theorem 1.2, we set
\begin{gather*}
w = u + \varepsilon z,
\qquad
z = e^{\alpha(x-a)}-1,
\end{gather*}
where \(\alpha \gt 0\) is first chosen sufficiently large so that \(Lz \gt 0\) on \((a,d)\) and \(\varepsilon \gt 0\) is then chosen sufficiently small so that
\begin{gather}
w(a) = M \gt w(d). \tag{1.2}
\end{gather}
Applying Lemma 1.1 to \(w\) on \((a,d)\text{,}\) we deduce that \(w\) achieves its maximum over \([a,d]\) at either \(a\) or \(d\) and not in the interior. By (1.2), \(w\) cannot achieve its maximum at \(d\) and must achieve it at \(a\text{.}\) Thus the one-sided derivative
\begin{gather*}
w'(a) = u'(a) + \varepsilon z'(a) \le 0,
\end{gather*}
which implies
\begin{gather}
u'(a) \le -\varepsilon z'(a) = -\varepsilon\alpha \lt 0\tag{1.3}
\end{gather}
as desired.
If we relax the restriction \(h \equiv 0\) in Theorem 1.2 then the conclusions are no longer valid; see Exercise 1.1.1. We can still make progress, however, if we assume both that \(h \le 0\) and that the maximum \(M\) of \(u\) is nonnegative.
Theorem 1.4. One-dimensional strong maximum principle for \(h\le 0\).
Suppose that \(h \le 0\text{.}\) If \(Lu \ge 0\) in \((a,b)\) and \(M \ge
0\text{,}\) then \(u\) can attain \(M\) at some \(c \in (a,b)\) only if \(u \equiv M\text{.}\)
Proof.
Theorem 1.5. One-dimensional Hopf lemma for \(h\le 0\).
Suppose that \(h \le 0\text{,}\) \(Lu \ge 0\) in \((a,b)\) and \(M \ge 0\text{.}\) If \(u(a)=M\) and \(u\) has a one-sided derivative at \(a\text{,}\) then either \(u'(a) \lt 0\) or \(u \equiv M\text{.}\) Similarly if \(u(b) = M\) and \(u\) has a one-sided derivative at \(b\text{,}\) then either \(u'(b) \gt
0\) or \(u \equiv M\text{.}\)
Proof.
Exercises Exercises
1. (PS1) Some basic counterexamples.
Show that the conclusion of Theorem 1.2 does not hold if the operator \(L\)
(a)
is replaced by \(L = \dfrac{d^2}{dx^2} + 1\text{;}\) or
Hint.
It is enough to consider solutions of \(Lu=0\text{,}\) which are explicit. (But not just any solution will do!)
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 \(Lu=0\text{.}\)
(b)
\(L = \dfrac{d^2}{dx^2} - 1\text{.}\)
Hint 1.
Same as for the previous part.
Hint 2.
While \(L\) satisfies the hypotheses of Theorem 1.4, this theorem also requires \(M \ge 0\text{.}\) So your counterexample will have \(M
\lt 0\text{.}\)
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 \((-1,1)\text{.}\)
2. (PS1) A concrete application.
Let \(L = d^2/dx^2 - 1\) and suppose that \(u \in C^2((0,\log 3)) \cap
C^0([0,\log 3])\) solves the boundary value problem
\begin{align}
\left\{
\begin{aligned}
\amp Lu = f \quad \text{ in } (0,\log 3), \\
\amp u(0) = u(\log 3) = 0
\end{aligned}
\right.\text{,}\tag{✶}
\end{align}
but that all we know about the function \(f\) on the right hand side is that it satisfies the inequalities \(0 \le f \le 1\text{.}\)
(a)
Show that \(u \le 0\) on \([0,\log 3]\text{.}\)
Hint.
Use Theorem 1.4.
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!
(b)
Hint.
The constant function \(-1\) is a ‘particular solution’ which solves the ODE but not the boundary conditions. Conclude that \(u_1\) must have the form
\begin{equation*}
u_1 = Ae^x + Be^{-x} - 1
\end{equation*}
for some constants \(A,B\text{,}\) and then solve for these constants using the boundary conditions.
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.
(c)
Show that \(u \ge u_1\) on \([0,\log 3]\text{.}\)
Hint.
Apply Theorem 1.4 to \(v=u_1-u\text{.}\)
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.
(d)
Conclude that \(-1/8 \le u(\log 2) \le 0\text{.}\) What more can you say if one of these inequalities is an equality?
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. \(u \equiv u_1\text{.}\)
(e)
Suppose that the one-sided derivatives \(u'(0),u'(\log 3)\) exist. Explain why \(-1/2 \le u'(0) \le 0\) and \(0 \le u'(\log 3) \le
1/2\text{.}\) What more can you say if one of these inequalities is an equality?
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 \(u \equiv u_1\text{.}\)
(f)
(Optional) Since the coefficients in (✶) are constants, one can in principle find an explicit formula for \(u\) in terms of integrals provided \(f\) is reasonably nice, say continuous on \([0,\log 3]\text{.}\) If you remember how this goes, give it a try and then see if you can use your explicit formula to reproduce the estimates above.
3. (PS1) Basic lemma for \(h \le 0\).
Suppose that \(Lu \gt 0\) on \((a,b)\text{.}\) If \(h \le 0\) and \(M \ge
0\text{,}\) show that \(u\) can equal \(M\) only at the endpoints \(a\) and \(b\text{.}\)
Hint.
Follow the proof of Lemma 1.1.
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 \(Lu \gt 0\text{.}\)
4. Theorem 1.4.
Use Exercise 1.1.3 to prove Theorem 1.4.
Hint.
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*}
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 \((a,d)\text{.}\)
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.
5. Theorem 1.5.
Prove Theorem 1.5.
Hint.
6. (PS1) Alternative proof of Theorem 1.2.
Give an alternative proof of Theorem 1.2 by using Theorem 1.3.
Hint.
Assume for the sake of contradiction that \(u(c)=M\) for some \(c \in
(a,b)\text{,}\) and then apply Theorem 1.3 to the restrictions of \(u\) to \((a,c)\) and \((c,b)\text{.}\)
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.
