Let \(u:[0,1]\to\mathbb R\) be a \(C^2\) function satisfying \[ u''(x)=e^x u(x). \]
- Show that if \(0<x_0<1\), then \(u\) cannot have a positive local maximum at \(x_0\), and cannot have a negative local minimum there.
- If \(u(0)=u(1)=0\), prove that \(u(x)\equiv0\) on \([0,1]\).
Solution
1. Part 1: \(u\) cannot have a positive local maximum at an interiorpoint.
Suppose \(0<x_0<1\) and \(u\) has a positive local maximum at \(x_0\). Since \(u\) is \(C^2\), the second-derivative test gives \[ u''(x_0)\le0. \] On the other hand, the differential equation gives \[ u''(x_0) = e^{x_0}u(x_0) > 0, \] because \(e^{x_0}>0\) and \(u(x_0)>0\). This is a contradiction.
2. Part 1: \(u\) cannot have a negative local minimum at an interiorpoint.
Suppose \(0<x_0<1\) and \(u\) has a negative local minimum at \(x_0\). The second-derivative test gives \[ u''(x_0)\ge0. \] But \[ u''(x_0) = e^{x_0}u(x_0) < 0, \] since \(u(x_0)<0\). This is again a contradiction.
3. Under the additional hypothesis \(u(0)=u(1)=0\), the function\(u\) cannot take a positive value.
If \(u(x)>0\) for some \(x\in(0,1)\), continuity on the compact interval \([0,1]\) gives a point \(x_0\) where \(u\) attains its maximum. That maximum is positive. Since the endpoint values are both \(0\), the maximizer satisfies \(0<x_0<1\). It is therefore a positive local maximum, contradicting step 1.
4. Under the same boundary conditions, the function \(u\) cannottake a negative value.
If \(u(x)<0\) for some \(x\in(0,1)\), continuity gives a point \(x_0\) where \(u\) attains its minimum on \([0,1]\). That minimum is negative, so the zero endpoint values force \(0<x_0<1\). This is a negative local minimum, contradicting step 2.
5. Part 2 gives\[ \boxed{u(x)\equiv0\text{ on }[0,1]}. \]
By steps 3 and 4, \(u\) takes neither positive nor negative values. Hence \(u(x)=0\) for every \(x\in[0,1]\).
6. Q.E.D.
Steps 1 and 2 prove part 1, and step 5 proves part 2.