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The integro-differential equation takes the form

$$c′(t)+iac(t)+\int_{0}^{\infty} f(t−\tau)c(\tau)d\tau=0$$

in which

$$f(t−\tau)=\frac{\pi}{4}∫_{0}^{\infty} J(\omega)e^{−i\omega(t−\tau)} d\omega,$$

$$J(ω)=2\pi \alpha \omega_{c}^{1−b}\omega^{b}\Theta(\omega_{c} −\omega),$$

and $a,\alpha,\omega_{c},b>0$ are all real constants. $\Theta(x)$ is the usual step function. To solve this equation, it seems the Laplace transform does not work.

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  • $\begingroup$ Why do you say that Laplace transform techniques don't work? After making use of the half-line convolution property of the Laplace transform, your first equation becomes $\sigma\mathcal{L}c(\sigma)+ia\mathcal{L}c(\sigma)+\mathcal{L}f(\sigma)\mathcal{L}c(\sigma)-c(0) = 0$, or in other words: $\mathcal{L}c(\sigma) = \frac{c(0)}{\sigma+ia+\mathcal{L}f(\sigma)}$. So all you need to do is evaluate the Laplace transform of $f$. $\endgroup$ – Cameron Williams Jun 19 '13 at 1:58
  • $\begingroup$ I'm curious about $J(\omega)$. You have s in your t-domain? $\endgroup$ – Zen Jun 19 '13 at 1:59
  • $\begingroup$ @Zen I think $s$ is just an arbitrary constant, not the same as the $s$ for the Laplace transform domain. $\endgroup$ – Cameron Williams Jun 19 '13 at 2:02
  • $\begingroup$ Ah, I read it "a... are real constants, s>0". Then it would make sense why the OP couldn't get anywhere trying to take the LT of something already in s! I must be slightly dyslexic! Still, poor choice of notation. $\endgroup$ – Zen Jun 19 '13 at 2:06
  • $\begingroup$ @Zen, it is really unfortunate notation to say the least. $\endgroup$ – Cameron Williams Jun 19 '13 at 2:11
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It seems to me that Laplace transform leads to a dead-end indeed. So I am posting an community-wiki answer, anyone is welcome to edit it.

Let's replace your constant $s$ by $b$ here. Taking Laplace transform of: $$c'(t)+iac(t)+\int_{0}^{\infty} f(t−\tau)c(\tau)d\tau=0$$ gives: $$ \newcommand{\L}[1]{\mathcal{L}\{#1\}}s\L{c}-c(0) + ia \L{c} + \L{f}\cdot \L{c} = 0.\tag{1} $$ Now onto computing $\L{f}$, first rewritting the $f(t)$ as: $$ f(t)=\frac{\alpha\pi^2}{2}\int_{0}^{\infty} \omega_{c}^{1−b}\omega^{b}\Theta(\omega_{c} −\omega)e^{−i\omega t} d\omega. $$ The step function is: $$ \Theta(\omega_{c} −\omega)=\left\{\begin{aligned} 1\quad \text{when } \omega_{c} −\omega\ge 0, \\ 0\quad \text{when } \omega_{c} −\omega <0. \end{aligned}\right. $$ Hence, $f$ is: $$ f(t)=\frac{\alpha\pi^2}{2}\int_{0}^{\omega_c} \omega_{c}^{1−b}\omega^{b}e^{−i\omega t}\, d\omega. $$ The Laplace transform is with respect to $t$, and we can interchange it with the integral with respect to $\omega$ for everything is smooth here: $$ \L{f}(s) = \frac{\alpha\pi^2}{2}\omega_{c}^{1−b}\int_{0}^{\omega_c} \omega^{b} \L{e^{−i\omega t} }(s)\,d\omega = \frac{\alpha\pi^2}{2}\omega_{c}^{1−b}\int_{0}^{\omega_c} \omega^{b} \frac{1}{s-i\omega}\,d\omega. $$ For general $b>0$ (the $s$ in the question), the integral $$ \int_{0}^{\omega_c} \omega^{b} \frac{1}{s-i\omega}\,d\omega $$ involves special function. (WolframAlpha gives hypergeometric function)

If $b=1,2,\ldots$, we can have a closed form of above integral. Let's try letting $b=1$: $$ \L{f}(s) = \frac{\alpha\pi^2}{2} \left(\omega_c + is\ln\Big|\frac{s}{s-i\omega_c} \Big|\right). $$ Plugging back to (1): $$ \L{c} = \frac{c(0)}{s + ia + \L{f}}, $$ and it doesn't look like we can get a reasonable inverse Laplace transform out of the expression, I am suggesting do certain qualitative analysis of this equation and then numerically solve it if you would like to know what the solution is like for different parameters.

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