Solution of convolution Integral Equation

Question

I have an integral equation as follows:

$f(t)=\int_0^t \sqrt{\tau } g(t-\tau ) \, d\tau$

For a prescribed continuous function $f(t)$, How can I determine an analytical solution for $g(t)$? It should be noted that the derivative of $\sqrt{\tau }$ is discontinuous at $t=0$.

score 1 · Answer 1 · answered Oct 22 '19 at 05:59

1

If you take $t\geq 0$, then we have by Laplace transform properties that

$$F(s) = \frac{\sqrt{\pi}}{2}s^{-\frac{3}{2}}G(s)$$

Then we can find $g$ on $[0,\infty)$ by claiming

$$g(t) = \mathcal{L}^{-1}\left\{\frac{2}{\sqrt{\pi}}s^{\frac{3}{2}}F(s)\right\}$$

answered Oct 22 '19 at 05:59

Ninad Munshi

34,407

Thanks, the solution can be written as $\int_0^t f(t-\tau ) \left(\mathcal{L}_s^{-1}\left\frac{2 s^{3/2}}{\sqrt{\pi }}\right\right) , d\tau $. However, $\mathcal{L}_s^{-1}\left\frac{2 s^{3/2}}{\sqrt{\pi }}\right$ can not be calculated. – Mohammad Tahaye Abadi Oct 22 '19 at 06:29
It doesn't have to. You can try to compute the inverse transform directly for certain kinds of $f$. Kind of like how convolving with derivatives of delta is exactly the derivative of the function. The convolution property is only true if both reverse transforms exist. That doesn't mean the reverse transform of the product doesn't exist, necessarily. – Ninad Munshi Oct 22 '19 at 09:52

Solution of convolution Integral Equation

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