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Solve the inhomogeneous Volterra integral equation and hence find its resolvent kernel : $$Y(t)=F(t)+\int_0^t (t^2-x^2)Y(x)dx$$

We know that if the kernel $K(t,x)$ is of the form $K(t-x)$, then it can be solved easily by using Laplace transform. But the given kernel is of the form $K(t^2-x^2)$. So we use the transformations $t^2=u$ and $x^2=v$. After transformation, the equation takes up the form $$Y(\sqrt{u})=F(\sqrt{u})+\int_0^u (u-v)Y(\sqrt{v})d(\sqrt{v}) \\ \implies Y(\sqrt{u})=F(\sqrt{u})+\int_0^u (u-v)\frac{Y(\sqrt{v})}{2\sqrt{v}}dv$$ Now if we use $Y(\sqrt{u})=y(u)$ and $F(\sqrt{u})=f(u)$ then we get $$y(u)=f(u)+\int_0^u \frac{u-v}{2\sqrt{v}}y(v)dv$$ and that's where I'm stuck. Because instead of the "nice" kernel $u-v$, we now have a somewhat "ugly" kernel $\displaystyle{\frac{u-v}{2\sqrt{v}}}$ which is not Laplace transformable. Can someone take a look what's happening here? Is the trasnformation I started with is wrong? If so, what's the general transformation for this type of kernel? Thanks in advance.

am_11235...
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1 Answers1

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Taking the derivative w.r.t. $t$ on both sides, we get

$$Y'(t) = F'(t) + (t^2-t^2)Y(t) + \int_0^t 2tY(x)dx = F'(t) + 2t\int_0^t Y(x) dx$$

Then taking Laplace Transforms on both sides,

$$sy(s) - Y(0) = sf(s) - F(0) + 2\frac{y(s)-sy'(s)}{s^2}$$

$$ \implies y'(s) + \left(\frac{1}{2}s^2-\frac{1}{s}\right)y(s) = \frac{1}{2}s^2f(s) + \frac{1}{2}s\left(Y(0)-F(0)\right) = \frac{1}{2}s^2f(s)$$

This can then be solved with the integrating factor method in the frequency domain.

Ninad Munshi
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  • This is OK. But this answer is incomplete, as you can't find the resolvent kernel in this process. Moreover, this is not the approach I'm asking. I want to use the Laplace transform at the start (I mean, after the transformation), not after differentiating the equation. In general, if you were given a kernel, say, $K(t,x)=\frac{1}{t^2-x^2}$, then would your method work? – am_11235... Sep 22 '19 at 07:52
  • Yes, I'm expecting that general answer for $K(t^2-x^2)$. – am_11235... Sep 22 '19 at 08:19