I have a differential equation of the form

$$ \frac{\mathrm{d}g}{\mathrm{d}x} = f(x) + \int_0^x g(y)f(x-y)\mathrm{d}y + \alpha g(x). $$

$f$ is a monotonous decreasing function, satisfying $\int_0^\infty f(x)\mathrm{d}x = 1$.

To solve it, I thought of getting the second derivative to handle the integral but the convolution does not let me get rid of it.

Laplace transform might work here, but I am not sure the integral term on the right side qualifies as convolution such that I can use Laplace transform of $f$.

Before trying to find the Laplace transform of $f$, which would be pretty messy, I wanted to ask if this is the best way to approach such a differential equation?

  • 2
    $\begingroup$ Not sure if this is gonna work, but I would differentiate both side and see what you get; to differentiate the integral, you can use Leibniz's rule: en.wikipedia.org/wiki/Leibniz_integral_rule $\endgroup$
    – Chee Han
    Commented Aug 1, 2023 at 22:27
  • $\begingroup$ @CheeHan I thought of that but differentiating the convolution-like part does not eliminate the integral. It gives me $\int_0^x g(y)f^\prime(x-y)\mathrm{d}y+f(0)g(x)$, if I did it correctly. $\endgroup$
    – ck1987pd
    Commented Aug 1, 2023 at 22:37
  • $\begingroup$ How messy is $f(x)$? $\endgroup$
    – Cesareo
    Commented Aug 1, 2023 at 22:44
  • $\begingroup$ @Cesareo I haven't exactly calculated that. I have two versions though. One version that is for large $x$ involves $\vartheta_4$, and one version for small $x$ involves $_3F_2$. It is messy, at least for me :) Do you think it will help if you know $f(x)$? $\endgroup$
    – ck1987pd
    Commented Aug 1, 2023 at 22:49
  • $\begingroup$ Do you just want to prove that a solution exists or do you need a (more or less) explicit solution formula? $\endgroup$
    – junjios
    Commented Aug 1, 2023 at 22:56

1 Answer 1


The equation is linear in $g(x)$, and its special form does allow for a solution using the Laplace transform, using the convolution theorem and denoting

$$ (F,G)(s):=\int_0^\infty (f,g)(t)e^{-st}dt$$

we obtain by directly Laplace transforming the equation:

$$sG(s)-g(0)=F(s)+F(s)G(s)+\alpha G(s)$$

which can then be solved for $G$


I don't think that there is a better way to analytically constructing solutions for integrodifferential equations like this, since most of them don't even have an analytical solution. Approximating with piecewise constant $f$ should be easy since the resulting equation has simple solutions in terms of exponentials, all one needs to do is to determine the constants of integration by demanding continuity of the solution at each point where $f$ has a jump discontinuity. Numerically inverting the Laplace transform is also possible for arbitrary, reasonably well behaved $f$.

  • $\begingroup$ Thanks a lot. This is what I reached as well with the Laplace transform. I'll wait a few days before accepting this answer. $\endgroup$
    – ck1987pd
    Commented Aug 1, 2023 at 23:39

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