I have obtained an integral equation that models a particular physical phenomenon. I wish to convert it into a differential equation so that I can solve it. Here's the equation defined for $z\geq z_0$: \begin{equation} g(z-z_0)=\int_{z_0}^{z}dt~f(t-z_0)g(z-t) \end{equation} After reading post-1 and Fundamental theorem of calculus, I thought this was the answer: \begin{equation} \frac{dg(z-z_0)}{dz}=f(z-z_0)g(0) \end{equation} but obviously this is incorrect, since upon integration again, $g(0)$ will be treated as a constant and so we do not get back the original integral equation. I have also read post-2, post-3, post-4, but still can't figure out the correct way. Any help is appreciated. Thanks in advance.

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    $\begingroup$ I am afraid what you want is not possible since the integral will remain after derivation. Observation: If $z_0 = 0$ and you apply the Laplace transform to your equation, you get $G(s) = F(s)G(s)$, that is $F(s) = 1$ or $G(s) = 0$. $\endgroup$ – Friedrich Philipp Mar 16 '17 at 6:39

$$g(z-z_0)=\int_{z_0}^{z}f(t-z_0)g(z-t)dt $$

Let $z-t = y$ $$g(z-z_0)=\int_{0}^{z-z_0} f(z-z_0 -y)g(y)dy $$

This implies that for $z-z_0 = t \geq 0$

$$g(t) = (g*f)(t)$$

Then apply Laplace to both sides

$$G(s) = G(s) F(s)$$

I think you can go from here.


As Friedrich Philipp commented.

If you apply the fundamental theorem of calculus $$\begin{equation} g(z-z_0)=\int_{z_0}^{z}~f(t-z_0)\,g(z-t)\,dt \end{equation}$$ $$\begin{equation} \frac{dg(z-z_0)}{dz}= f(z-z_0)\,g(0)+\int_a^z f(t-z_0)\, g'(z-t) \, dt \end{equation}$$


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