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I have to solve the following integral equation \begin{align*} \int_{-\infty}^\infty e^{-y^2} \log \left( \int_{-\infty}^\infty e^{-(y-x-t)^2} f(t) dt\right) dy=-cx^2 \end{align*} where $c$ is some constant. We have to solve for $f(t)$. Also, assume, if needed, that $\int_{-\infty}^\infty f(t) dt < \infty$

The only thin I noticed that integral inside log is a convolution and so is the outside integral.

I don't know much about integral equations so any reference would be great too. Thanks

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  • $\begingroup$ What are the bounds on the integrals? $\endgroup$
    – JimmyK4542
    Dec 24 '14 at 1:57
  • $\begingroup$ Thanks. I edited my question. $\endgroup$
    – Boby
    Dec 24 '14 at 2:01
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Clearly, this problem is meant to be solved using Fourier Transform methods.

I'm using the definition that the Fourier transform of $f(t)$ is $F(\omega) = \displaystyle\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt$.

Accordingly, the inverse Fourier transform of $F(\omega)$ is $f(t) = \dfrac{1}{2\pi}\displaystyle\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega$.

Define $g(z) = \log \left( \displaystyle\int_{-\infty}^\infty e^{-(-z-t)^2} f(t) dt\right)$. Then, we have $\displaystyle\int_{-\infty}^\infty e^{-y^2}g(x-y)\,dy=-cx^2$.

The left hand side is simply the convolution of $e^{-x^2}$ and $g(x)$.

The Fourier transform of $e^{-x^2}$ is $\sqrt{\pi}e^{-\omega^2/4}$ and the Fourier transform of $-cx^2$ is $2\pi c\delta''(\omega)$.

By the convolution property, we get $\sqrt{\pi}e^{-\omega^2/4}G(\omega) = 2\pi c\delta''(\omega)$, i.e. $G(\omega) = 2c\sqrt{\pi}e^{\omega^2/4}\delta''(\omega)$.

From Wikipedia, the Dirac Delta function satisfies $x\delta'(x) = -\delta(x)$. Differentiation yields $x\delta''(x)+\delta'(x) = -\delta'(x)$, i.e. $x\delta''(x) = -2\delta'(x)$. Therefore, $x^2\delta''(x) = -2x\delta'(x) = 2\delta(x)$. Then, since $x^n\delta(x) = 0$ for integers $n \ge 1$, we have $x^n\delta''(x) = 0$ for integers $n \ge 3$.

Using Taylor expansion on $e^{\omega^2/4}$, we get $G(\omega) = 2c\sqrt{\pi}e^{\omega^2/4}\delta''(\omega) = c\sqrt{\pi}(2\delta''(\omega)+\delta(\omega))$.

Taking the inverse Fourier transform yields $g(z) = -\dfrac{c}{\sqrt{\pi}}\left(z^2-\dfrac{1}{2}\right)$.

Now, we've reduced the problem to solving $\displaystyle\int_{-\infty}^\infty e^{-(-z-t)^2} f(t)\,dt = e^{-\tfrac{c}{\sqrt{\pi}}\left(z^2-\tfrac{1}{2}\right)}$.

Using the substitution $z \to -z$, we get $\displaystyle\int_{-\infty}^\infty e^{-(z-t)^2} f(t)\,dt = e^{\tfrac{c}{2\sqrt{\pi}}}e^{-\tfrac{c}{\sqrt{\pi}}z^2}$.

The left hand side is simply the convolution of $e^{-z^2}$ and $f(z)$.

The Fourier transform of $e^{-z^2}$ is $\sqrt{\pi}e^{-\tfrac{\omega^2}{4}}$ and the Fourier transform of $e^{\tfrac{c}{2\sqrt{\pi}}}e^{-\tfrac{c}{\sqrt{\pi}}z^2}$ is $e^{\tfrac{c}{2\sqrt{\pi}}}\sqrt{\tfrac{\pi\sqrt{\pi}}{c}}e^{-\tfrac{\sqrt{\pi}}{c}\tfrac{\omega^2}{4}}$

By the convolution property, we get $\sqrt{\pi}e^{-\tfrac{\omega^2}{4}}F(\omega) = e^{\tfrac{c}{2\sqrt{\pi}}}\sqrt{\tfrac{\pi\sqrt{\pi}}{c}}e^{-\tfrac{\sqrt{\pi}}{c}\tfrac{\omega^2}{4}}$.

Now, solve for $F(\omega)$, and take the inverse Fourier transform to get the answer.

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  • $\begingroup$ Thank you very much. Very nice answer. Could you point me to some books that solve integral equations? $\endgroup$
    – Boby
    Dec 24 '14 at 4:41
  • $\begingroup$ Unfortunately, I don't know of any books which go into depth on solving integral equations. Perhaps someone else reading this knows of a good book on the topic. $\endgroup$
    – JimmyK4542
    Dec 24 '14 at 4:54
  • $\begingroup$ Well. Thanks a lot anyway. $\endgroup$
    – Boby
    Dec 24 '14 at 4:56
  • $\begingroup$ Question. Would the the path of Laplace transform be equivalent? $\endgroup$
    – Boby
    Dec 24 '14 at 5:01
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    $\begingroup$ @Boby F.G. Tricomi's "Integral Equations" is a slim but easy-to-follow introductory text. $\endgroup$ Dec 24 '14 at 5:25

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