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Define $f \in C^{2}\left[a,b\right]$ satisfying $f''(x)=e^xf(x)$. Show that $f''(x)=e^xf(x)$ with $f(a)=f(b)=0$ makes $f\equiv 0$ $\forall x\in [a,b]$.

Actually, I figure out a solution as follows: (just taking about the idea)

We can prove a general conclusion:if $f \in C^{2}\left[a,b\right]$ satisfying $f''(x)=g(x)f(x)$ where $g(x) \in C^{0}\left[a,b\right]$ satisfying $g(x)>0$, and $f(a)=f(b)=0$, we have $f\equiv 0$ $\forall x \in [a,b]$.

The idea is to prove that if there exists $x_0\in (a,b)$ such that $f(x_0)\ne0$ (let's assume that $f(x_0)>0$), we can prove that there exists $x_1\in (a,b)$ such that $f(x_1)>0$, $f'(x_1)>0$ and $f''(x_1)>0$. And through this conclusion we can easily get that $f(x)$ will be strictly monotonically increasing in the interval $\left[x_1,b\right]$.

So my questions are:

  1. Is there any other solution of this problem?
  2. Can we solve this differential equation problem?
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Assume that $f$ is not identically zero on the interval, without loss of generality $f(x_0) > 0$ for some $x_0 \in (a, b)$. Then $f$ attains its maximum $M>0$ at some point $x_1 \in (a, b)$. At the maximum we necessarily have $$ f'(x_1) = 0 \, , \quad f''(x_1) \le 0 \, , $$ which is a contradiction to the assumption that $$ f''(x_1) = e^{x_1} f(x_1) = e^{x_1} M > 0 \, . $$

The same solution works with the more general assumption that $f''(x)=g(x)f(x)$ with a strictly positive function $g$.

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We can actually find a closed-form solution, sort of. Let $y = 2 e^{x/2}$. Then we have $$ \frac{df}{dx} = \frac{dy}{dx} \frac{df}{dy} = e^{x/2} \frac{df}{dy} = \frac{y}{2} \frac{df}{dy} $$ and $$ \frac{d^2f}{dx^2} = \frac{dy}{dx} \frac{d}{dy} \left( \frac{df}{dx} \right) = \frac{y}{2} \frac{d}{dy} \left( \frac{y}{2} \frac{df}{dy} \right) = \frac{y^2}{4} \frac{d^2f}{dy^2} + \frac{y}{4} \frac{df}{dy}. $$ The differential equation then becomes $$ \frac{y^2}{4} \frac{d^2f}{dy^2} + \frac{y}{4} \frac{df}{dy} = \frac{y^2}{4} f, $$ or $$ y^2 f'' + y f' - y^2 f = 0 $$ which is a modified Bessel ODE with $n = 0$. The general solution is therefore $$ f(y) = A I_0(y) + B K_0(y), $$ or $$ f(x) = A I_0(2 e^{x/2}) + B K_0(2 e^{x/2}), $$ where $I_0$ and $K_0$ are the modified Bessel functions.

To show that the function $f(x)$ cannot vanish at two points, we note that $I_0(2e^{x/2})$ is strictly increasing while $K_0(2e^{x/2})$ is strictly decreasing, and that both functions are strictly positive everywhere. If the function $f(x)$ vanishes at one point, it must therefore be the case that either $A$ and $B$ both vanish, or that they are of opposite sign but non-zero. But if they are of opposite sign and non-zero, this means that the combined function $f(x)$ is either strictly increasing or strictly decreasing, and thus cannot vanish anywhere else. Thus, if the $f(x)$ vanishes at two points, it is zero everywhere.

Disclaimer: I wouldn't have come up with the above "closed-form" solution on my own; I ran the differential equation through Mathematica, saw the result, and reverse-engineered the solution.

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If there is some $x_0$ with $f(x)>0$ then there is also some $x_1$ with $f(x_1)=\max_{x\in[a,b]}f(x)$. Because $f$ is differentiable, $f'(x_1)=0$. But then also $f''(x_1)=g(x_1)f(x_1)>0$ so that locally $f(x_1+s)=f(x_1)+\frac12f''(x_1)s^2+o(s^2)$, which provides values larger than $f(x_1)$, for instance using $f(x_1+s)\ge f(x_1)+\frac14f''(x_1)s^2$ for $|s|<\delta$ for some small $δ>0$, in contradiction to the construction. Thus the first assumption is wrong, there is no such $x_0$.

Then repeat the same argument for $-f(x)$ to find that only $f=0$ remains.

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