1
$\begingroup$

We have differential equation $3s(t)-2s''(t)=r(t)\,$ and $s(t)$ is convolution $s=g*r\,$ where $g(t)=ae^{-b\left | t \right |}\,$ $\\a,b\in\mathbb R+$ Solve constans a and b.

I tried to solve function $s(t)$ with fourier transform and I got $s(t)=\int_{R}e^{i2\pi tf}\frac{\hat{r}(f)}{3+8\pi ^{2}f^{2}}df$ Now I have no idea how to calculate this integral. Is there some simplier way?

$\endgroup$

migrated from mathematica.stackexchange.com Sep 26 '14 at 17:31

This question came from our site for users of Wolfram Mathematica.

3
$\begingroup$

Taking both equations to Fourier space gives you $$3 \hat s(k) + 8 \pi ^2 k^2 \hat s(k) = \hat r(k) \text{ and} \\ \hat s(k)=\hat g(k) \cdot \hat r (k).$$ Now substituting the second into the first gives you $$(3 + 8\pi^2 k^2) \hat g(k) \cdot \hat r(k)= \hat r(k),$$ so clearly you want the constants to be such that $\hat g(k) = (3 + 8 \pi^2 k^2)^{-1} $.

The fact you can choose constants to do this follows from that the fourier transform of $e^{-\alpha |x|}$ for $\mathrm{Re}(\alpha) > 0$ is $2\alpha/(\alpha^2 + 4 \pi^2 k^2)$. To see this, note that $$\begin{align} \int_{-\infty}^\infty e^{-\alpha |x|} e^{-2\pi k x} \, dx &= \int_0^\infty e^{-(\alpha + 2\pi ik) x}\, dx +\int_{-\infty}^0e^{-(-\alpha + 2\pi ik)x}\, dx \\ &= \Big[\frac{-1}{\alpha + 2 \pi ik}e^{-(\alpha + 2\pi ik)x} \Big]^{\infty}_0 + \Big[\frac{-1}{-\alpha + 2 \pi ik}e^{-(-\alpha + 2\pi ik)x} \Big]^{0}_{-\infty}\\ &= \frac{1}{\alpha + 2 \pi ik}-\frac{1}{2\pi ik - \alpha} \\ &= \frac{2 \alpha}{\alpha^2 + 4 \pi^2 k^2}. \end{align}$$The way to go from here would be just pull out a $2$ from the denominator to make the coefficient of $k^2$ match, pick the right $\alpha$ to make the denominator right, then multiply by a constant to get the numerator right.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy