Section 1.2 Applications
Despite their humble origins in basic calculus, the results from Section 1.1 have many non-trivial applications, some of which have analogues for the partial differential equations we will study in later chapters. For instance, solutions to boundary value problems which obey a maximum principle are unique.
Theorem 1.6. Uniqueness for boundary value problems.
Let \(f,g\) and \(h \le 0\) be bounded functions on \((a,b)\) and let \(\alpha,\beta \in \R\text{.}\) Then the boundary value problem
\begin{equation}
\left\{
\begin{aligned}
\amp u'' + gu' + hu = f \quad \text{ in } (a,b), \\
\amp u(a) =\alpha, \quad u(b)=\beta
\end{aligned}
\right.\tag{1.4}
\end{equation}
has at most one solution \(u \in C^2((a,b)) \cap C^0([a,b])\text{.}\)
Proof.
Suppose that \(u,v \in C^2((a,b)) \cap C^0([a,b])\) are both solutions to (1.4). We will show that \(w=u-v \equiv 0\) and hence \(u
\equiv v\text{.}\) Defining the linear operator \(L\) as in (1.1), we have
\begin{equation*}
Lw = Lu-Lv = f - f = 0 \text{ in } (a,b),
\end{equation*}
and similarly on the boundary we find
\begin{equation*}
w(a)=w(b)=0.
\end{equation*}
In particular, if \(w\) is constant, then this constant is zero and we are done. So assume that \(w\) is not constant. Then \(w\) must either achieve a maximum \(M \gt 0\) or a minimum \(m \lt 0\) at some point \(c
\in (a,b)\text{.}\) Applying Theorem 1.4 to \(w\text{,}\) we see that the first of these cannot happen, and applying the same theorem to \(-w\) we deduce that the second cannot happen either. Thus we must have \(w \equiv 0\) as desired.
Maximum principles can also be useful when comparing two solutions of the same problem (or even different problems!). Here is one famous example. Note that unlike in the previous results, there is no sign condition on the coefficient \(h\text{.}\)
Theorem 1.7. Sturm separation theorem.
Let \(g,h\) be bounded functions on \((c,d)\text{,}\) and suppose that \(u,v \in C^2((c,d))\) satisfy the ODE
\begin{equation*}
u'' + gu' + hu = v'' + gv' + hv = 0
\end{equation*}
in \((c,d)\text{.}\) If \(u\not \equiv 0\) and \(v \not \equiv 0\text{,}\) then, between any two consecutive zeros of \(v\text{,}\) \(u\) can have at most one root.
Proof.
Let \(a,b \in (c,d)\) be two consecutive zeros of \(v\) so that \(v(a)=v(b)=0\) and \(v \ne 0\) on \((a,b)\text{.}\) Then the ratio
\begin{equation*}
w = \frac uv
\end{equation*}
is \(C^2\) on the interval \((a,b)\text{,}\) where a calculation shows that it solves the ODE
\begin{equation}
w'' + \Big( 2 \frac{v'}v + g \Big) w' = 0\text{.}\tag{1.5}
\end{equation}
\begin{equation}
\tilde L = \frac {d^2}{dx^2} + \Big( 2 \frac{v'}v + g \Big) \frac
d{dx}\text{.}\tag{1.6}
\end{equation}
For sufficiently small \(\varepsilon \gt 0\) we have \(w \in C^0([a+\varepsilon,b-\varepsilon])\text{,}\) and hence that the coefficient \(2v'/v+g\) in (1.6) is a bounded function on \((a+\varepsilon,b-\varepsilon)\text{.}\) Applying Theorem 1.2 to the function \(w\) and the operator \(\tilde L\) on the interval \([a+\varepsilon,b-\varepsilon]\text{,}\) we therefore conclude that \(w\) cannot have any maxima or minima on \((a+\varepsilon,b-\varepsilon)\) unless it is constant. It follows that \(w\) and hence \(u\) can change sign at most once over the interval \((a,b)\text{.}\)
From the existence theory for the initial value problem (not examinable) we can check that roots of \(u\) are necessarily simple, and the result follows.
Note that, in the above argument, we not only needed to consider a new function \(w=u/v\text{,}\) but also a new differential operator \(\tilde L\text{!}\)
Exercises Exercises
1. Non-uniqueness.
Give a counterexample showing that the conclusion of Theorem 1.6 fails to hold for the operator \(L=d^2/dx^2+1\) on \((0,2\pi)\text{.}\)
2. (PS1) A ‘generalised’ maximum principle.
Suppose that \(u \in C^2((-\frac \pi4,\frac \pi4)) \cap C^0([\frac \pi4,\frac \pi4])\) satisfies \(u'' + u \ge 0\) on \((-\frac \pi4,\frac \pi4)\text{.}\)
(a)
Taking inspiration from the proof of Theorem 1.7, consider the ratio \(w=u/\cos x\text{,}\) and show that it satisfies
\begin{gather}
w'' - 2\frac{\sin x}{\cos x} w' \ge 0\tag{✶}
\end{gather}
on \((-\frac\pi 4,\frac \pi 4)\text{.}\)
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.
(b)
Applying Theorem 1.2 to \(w\text{,}\) conclude that either \(u\) is a constant multiple of \(\cos x\) or else
\begin{equation*}
u(x) \lt \sqrt 2 \max\big(u(\tfrac \pi4),u(-\tfrac \pi4)\big) \cos x
\end{equation*}
for all \(x \in (-\frac \pi4, \frac \pi 4)\text{.}\)
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.
