Differentiating a convolution integral

Question

I'm trying to turn the integro-differential equation

$\phi'(t) + \phi(t) = \int_0^t \sin{(t - \xi)} \, \phi(\xi) \, \mathrm{d} {\xi}$

into the differential equation

$\phi'''(t) + \phi''(t) + \phi'(t) = 0$

through differentiation.

I noticed that $\frac{d^2}{dt^2} \left( \sin{t} \ast \phi(t) \right) = \left( \frac{d^2}{dt^2} \sin{t} \right) \ast \phi(t) = -\sin{t} \ast \phi(t)$, which gives us the equation:

$\phi'''(t) + \phi''(t) = \int_0^t (-\sin{(t - \xi)}) \, \phi(\xi) \, \mathrm{d} {\xi} = - \int_0^t \sin{(t-\xi)} \, \phi(\xi) \, \mathrm{d} {\xi} = -(\phi'(t) + \phi(t))$

Then we can rewrite this as

$\phi'''(t) + \phi''(t) + \phi'(t) + \phi(t) = 0$, which is almost what I want except for the $\phi(t)$ term.

Is this the right way of going about it or should I be trying something else?

Your differentiation is not right. Note that the integral bounds also depend on t. — Peder, Feb 03 '14 at 23:54
Note that, if you differentiate the $ \int_0^t \sin{(t - \xi)} , \phi(\xi) , \mathrm{d} {\xi} $ twice you should get $ -\int_0^t \sin{(t - \xi)} , \phi(\xi) , \mathrm{d} {\xi} + \phi(t) $ which is your mistake. — Mhenni Benghorbal, Feb 04 '14 at 00:23

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

Here is an approach. If you use this technique, then it is easy to derive the result. We use the Leibniz's rule, for differentiating under the integral sign, to differentiate the equation

$$ \phi'(t) + \phi(t) = \int_0^t \sin{(t - \xi)} \, \phi(\xi) \, \mathrm{d} {\xi} \longrightarrow (1)$$

twice which gives

$$ \phi'''(t) + \phi''(t) = -\int_0^t \sin{(t - \xi)} \, \phi(\xi) \, \mathrm{d} {\xi} + \phi(t) \longrightarrow (2).$$

Adding the two equations yields

$$ \phi'''(t) + \phi''(t)+\phi'(t)=0. $$

Differentiating a convolution integral

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