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I need to solve a system of Volterra integro-diff equation of form

$$ y(t) = x(t) - \int_{0}^{t} k(t-\tau) y'(\tau) \;\mathrm{d}\tau $$

where kernel is of form $$ k(t-\tau) = P(t)Q(\tau) $$

Is it possible to find solution without Laplace transforms? Thank you.

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We can substitute your kernel definition to yield

$$ y(t) = x(t) - \int_0^{t} P(t)Q(\tau)y'(\tau) d\tau $$

Note the independence of $t$ and $\tau$ allows us to write:

$$ y(t) = x(t) - P(t)\int_0^{t} Q(\tau)y'(\tau) d\tau $$

We can then differentiate to find:

$$ y'(t) = x'(t) - P'(t)\left( \int_0^{t} Q(\tau)y'(\tau) d\tau\right) - P(t)Q(t)y'(t) $$

Note that

$$ y(t) = x'(t) - P(t)\int_0^{t} Q(\tau)y'(\tau) d\tau $$ Informs us that

$$ \frac{y(t)-x'(t)}{P(t)} = -\int_0^{t} Q(\tau)y'(\tau) d\tau$$

Thus it follows that

$$ y'(t) = x'(t) - P'(t)\frac{y(t)-x'(t)}{P(t)} - P(t)Q(t)y'(t) $$

Which can be rearranged to

$$ \left(1 + P(t)Q(t) \right)y'(t) + \frac{P'(t)}{P(t)}y(t)+\left(\frac{P'(t)}{P(t)}-1 \right)x'(t) = 0 $$

Which is a standard First Order Linear ODE in terms of y. The solution to which is given via the technique of integrating factors. (Message on comment if you want me to show you the closed form as well. For now I will omit)

Or if you really want the full equation. We divide through by $\left(1 + P(t)Q(t) \right)$ to yield

$$ y'(t) + \frac{\frac{P'(t)}{P(t)}}{\left(1 + P(t)Q(t) \right)}y(t) = \frac{\left(1-\frac{P'(t)}{P(t)} \right)}{\left(1 + P(t)Q(t) \right)} x'(t) $$

From here it follows as per the techniques described in:

https://en.wikipedia.org/wiki/Linear_differential_equation#First-order_equation_with_variable_coefficients

That the general solution is

$$ y = \frac{\int_{t_0}^{t} \frac{\left(1-\frac{P'(t)}{P(t)} \right)}{\left(1 + P(t)Q(t) \right)} x'(t) e^{\int \frac{\frac{P'(t)}{P(t)}}{\left(1 + P(t)Q(t) \right)}} }{e^{\int \frac{\frac{P'(t)}{P(t)}}{\left(1 + P(t)Q(t) \right)}}}$$

Time to hit it with some serious algebraic simplification. The integrals without specified bounds can be set to any lower bound (the results will cancel). The $t_0$ lower bound is made explicit where it is necessary to be specified.

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  • $\begingroup$ Thanks very much - an x'(t) got dropped but looks good. $\endgroup$ – Joshua Gordis Jun 15 '15 at 22:16
  • $\begingroup$ oops! will correct $\endgroup$ – frogeyedpeas Jun 15 '15 at 22:44
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    $\begingroup$ So pleasant to read such well-written work! Endorsed! $\endgroup$ – Robert Lewis Jun 15 '15 at 23:34
  • $\begingroup$ @Robert Lewis, why thank you! $\endgroup$ – frogeyedpeas Jun 16 '15 at 13:42
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Solution can also be obtained (with a lot more work) by integrating by parts, differentiating and then integrating by parts again, recognizing that y(0) = x(0) = 0 (at least in my case).

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