solve an integral equation

Question

Given a continuous function $f$ and an equation: $$ \int_0^1 e^{xf(t)} dt -1 = 0 , $$ check if a positive solution $x$ exists and find it if it exists.

Answer 1

If $f(t)=0$, any $x$ is a solution. Otherwise, let $$g(x) = \int_0^1 e^{xf(t)} dt -1$$, then $$g''(x)=\int_0^1 f^2(t)e^{xf(t)} dt > 0$$, therefore $g$ is convex. On the other hand, $g(0)=0$. Therefore, $g(x)=0$ has a positive root iff $g'(0)<0$ and $g(x)>0$ for some $x$.

The first condition is $$ \int_0^1 f(t)dt < 0 $$

The second condition never obeys if $f(t)\le 0$ everywhere on $[0,1]$. If there is a point $t \in [0,1]$ for which $f(t)>0$, there is an interval $[a,b]$ around $t$ at which $f>f(t)/2$. It is easy to prove that for sufficiently large $x$, $\int_a^b e^{xf(t)} dt$ can be made arbitrary large, so the second condition obeys.

So the complete condition is either $f(t)=0$ at the whole $[0,1]$ (1), or $ \int_0^1 f(t)dt < 0 $ (2), and $f(t)>0$ for some $t\in [0,1]$ (3).

In case of (1), any $x$ is a root. In case of (2) and (3), $x$ can be found as a zero of any convex function.