# Numerical solution to an integral equation

I know this question mayn't be new, but here is my problem: I have to solve this integral equation (numerically, of course, except from very few special cases):

$$\phi_{\nu}(t) = 1 - q\int_0^t \frac{d\psi_{\nu}}{d t'} \phi_{\nu}(t - t')\ dt'$$

Where of course I know what $\psi$ is. I can choose $q = 1$.

I tried to take a look at some past answers, but I did not find what I was looking for. The problem may be that this is a convolution integral.

• I recently looked into a similar matrix equation. The proposed solution is that if your model allows it to change the lower boundary of the integral to be $t-c$ with $c$ a constant, then you can discretize the integral using some formula (e.g. Trapezoidal rule) and solve the obtained delay equation. – Anton Antonov Jun 10 '17 at 13:40
• Perhaps, this question belongs in Mathematics instead of Mathematica.SE, – bbgodfrey Jul 12 '17 at 4:38
• @bbgodfrey Since I wanted to have a mathematica code to solve that, I don't see how it should belong to Math SE, people over there won't provide me for a code in that sense... – xyzt Jul 12 '17 at 12:44
• Mathematica certainly can reproduce the calculation by @CraigTracy, but no further progress can be made without an expression for g[s]. Given that expression, it may be possible to use InverseLaplaceTransform to obtain the desired answer. Am alternative approach is to discretize your integral equation to form a matrix equation, which can be inverted to obtain a numerical answer. – bbgodfrey Jul 12 '17 at 13:01

This problem can be solved using the Laplace transform of a function $f$: enter code here$$\hat{f}(s):=\int_0^\infty e^{-s t} f(t)\, dt.$$ Your equation can be written as enter code here$$f(t)=1-q\int_0^t f(t-t') g(t')\, dt'.$$ Taking the Laplace transform of both sides and using the convolution property of the Laplace transform one gets enter code here$$\hat{f}(s) = 1/s - q\, \hat{f}(s)\,\hat{g}(s).$$ Solving this for $\hat{f}$ gives enter code here$$\hat{f}(s)=(1/s) \left(1+q\,\hat{g}(s)\right)^{-1}.$$ Now use the inverse transform to find $f(t)$.