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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.

Adrian Keister
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zxman
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  • 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 $f$. – Cameron L. Williams Jun 19 '13 at 01:58
  • I'm curious about $J(\omega)$. You have s in your t-domain? – Zen Jun 19 '13 at 01:59
  • @Zen I think $s$ is just an arbitrary constant, not the same as the $s$ for the Laplace transform domain. – Cameron L. Williams Jun 19 '13 at 02:02
  • 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. – Zen Jun 19 '13 at 02:06
  • @Zen, it is really unfortunate notation to say the least. – Cameron L. Williams Jun 19 '13 at 02:11
  • @Cameron Williams Thank you very much for your answer. The adoption of Laplace transform is obvious. but, how can I get $\mathcal{L}f(\sigma)$? The integration for $\omega$ is not easy. – zxman Jun 19 '13 at 09:05
  • @zxman you again have a convolution in the definition of $f$ so apply the convolution theorem for the Laplace transform. – Cameron L. Williams Jun 19 '13 at 19:20
  • @CameronWilliams $f$ is not a convolution. – Shuhao Cao Jun 19 '13 at 19:22
  • @ShuhaoCao it is given by a convolution due to the definition of $J$ including the Heaviside function. – Cameron L. Williams Jun 19 '13 at 19:23
  • @CameronWilliams Oh yes, I should have looked more carefully. – Shuhao Cao Jun 19 '13 at 19:24
  • @ShuhaoCao no worries. I actually stared at this for a little bit last night before I figured out how to do the Laplace transform of it. The hidden convolution threw me off as well. – Cameron L. Williams Jun 19 '13 at 19:27
  • @CameronWilliams hmm....but the Laplace transform is done in $t$ not in $\omega$. – Shuhao Cao Jun 19 '13 at 19:30
  • This equation is taken from the reference Physical Review A 87, 052139 (2013). I have asked the author how can he figured out this question. He wrote to me that they solved the equation using Fortran program and the Laplace transform is not easy for this problem. I am very grateful if you can present a Fortran program for this problem. – zxman Jun 20 '13 at 02:08

1 Answers1

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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.

Shuhao Cao
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