Cfrgtkky

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

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

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

The idea is to prove that if there exists $x_0in (a,b)$ such that $f(x_0)ne0$ (let's assume that $f(x_0)>0$), we can prove that there exists $x_1in (a,b)$ 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 $left[x_1,bright]$.

So my questions are:

1. Is there any other solution of this problem?
   
   


2. Can we solve this differential equation problem?

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 $g$.

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.

If there is some $x_0$ with $f(x)>0$ then there is also some $x_1$ with $f(x_1)=max_{xin[a,b]}f(x)$. 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.