Integral equation of the form $\int_{-\infty}^{\infty} e^{-a t^4} g(x,t) dt = e^{-b x^4}$

Question

How to solve an integral equation of the following form \begin{align} \int_{-\infty}^{\infty} e^{-a t^4} g(x,t) dt = e^{-b x^4} \end{align} where $a$ and $b$ are some positive constants.

I am not very familiar with this subject any suggestions you might have would be great.

Answer 1

An easy solution can be obtained by making $g(x,t)$ degenerate: $$ g(x,t)={1\over2}\exp(-bx^4)\exp(at^4-|t|) $$ which is susceptible to the generalization $$ g(x,t)={1\over C}\exp(-bx^4)\exp(at^4)f(t) $$ where $f$ is integrable over $\mathbb R$ and $$ \int_{-\infty}^{\infty}f(t)dt=C\neq0 $$

Answer 2

Hah! This is actually a specific example of something in my research! (My work attacks a more general set of integral equations, in some sense.) Let's go for something nontrivial (unlike previous answers/comments).

If you consider what I like to call a diagonal kernel, i.e. $g(x,t) = f(xt)$ for some $f$ and assume $g$ is real analytic, then this is very easily solvable. Since we want to get $\exp(-bx^4)$ after doing the integral, it stands to reason that $g$ should only consist of powers of the form $t^{4l}$.

Write $g(x,t) = \sum\limits_{l=0}^{\infty} c_l x^{4l} t^{4l}$, then we have

$$ e^{-bx^4} = \int_{-\infty}^{\infty} \sum_{l=0}^{\infty} c_l x^{4l} t^{4l} e^{-at^4}\,dt.$$

Interchanging sum and limit (which can be justified after we compute what $c_l$ has to be by appealing to Fubini-Tonelli) this becomes

$$ e^{-bx^4} = \sum_{l=0}^{\infty}2 c_l x^{4l} \int_0^{\infty} t^{4l}e^{-at^4}\,dt.$$

Letting $u = at^4$, $du = 4at^3\,dt$ so we have

$$ e^{-bx^4} = \sum_{l=0}^{\infty}2 c_l x^{4l} \int_0^{\infty} \frac{t^{4l}}{4at^3} e^{-u}\,du = \sum_{l=0}^{\infty} \frac{1}{2a} c_l x^{4l} \int_0^{\infty} \left(\frac{u}{a}\right)^{l-\frac{3}{4}} e^{-u}\,du.$$

The latter integral can be recognized as $\Gamma\left(l+\frac{1}{4}\right)$, yielding

$$ e^{-bx^4} = \frac{1}{2a^{\frac{1}{4}}}\sum_{l=0}^{\infty} a^{-l}c_l x^{4l} \Gamma\left(l+\frac{1}{4}\right). $$

Equating coefficients in the two power series gives that

$$ c_l = \frac{2(-1)^l a^{l+\frac{1}{4}}b^l}{l!\Gamma\left(l+\frac{1}{4}\right)}.$$

Plugging this back into $g$ we have

$$ g(x,t) = \sum_{l=0}^{\infty} \frac{2(-1)^la^{l+\frac{1}{4}}b^l}{l!\Gamma\left(l+\frac{1}{4}\right)} x^{4l}t^{4l}.$$

This looks pretty terrible, however it is not! This actually has a closed form solution in terms of a Bessel function:

$$ g(x,t) = 2 a^{\frac{5}{8}} b^{\frac{3}{8}} |xt|^{\frac{3}{2}} J_{-\frac{3}{4}}\left(2\sqrt{ab} x^2t^2\right). $$

What is nice is that if $a=b$, then you can use this to define a unitary integral transform (defined on a dense subspace of $L^2$ of course).