# Fourier transform with mixed derivatives/ 2nd order ODE

## Homework Statement

Hi, So I'm suppose to solve the following problem:

$$\left.\frac{d^{2}u}{dt^{2}}-4\frac{d^{3}u}{dt dx^{2}}+3\frac{d^{4}u}{dx^{4}}=0$$

$$\left.u(x,0) = f(x)$$
$$\left.\frac{du}{dt}(x,0) = g(x)$$

## The Attempt at a Solution

First I use fourier transform on the given expression so that I get the following:

Fourier transform of $$\left.\frac{d^{2}u}{dt^{2}}(x,t) = \frac{d^{2}\widehat{u}}{dt^{2}}(\omega ,t)$$

Fourier transform of $$\left.\frac{du}{dt}(x,t) = \frac{d\widehat{u}}{dt}(\omega ,t)$$

Fourier transform of $$\left.\frac{d^{2}u}{dx^{2}}(x,t) = \left(i\omega\right)^{2}\widehat{u}(\omega ,t) = -\left(\omega\right)^{2}\widehat{u}(\omega ,t)$$

Fourier transform of $$\left.\frac{d^{4}u}{dx^{2}}(x,t) = \left(i\omega\right)^{4}\widehat{u}(\omega ,t) = \left(\omega\right)^{4}\widehat{u}(\omega ,t)$$

Which means me overall expression after transform is:
$$\left.\frac{d^{2}\widehat{u}}{dt^{2}}(\omega ,t)+4\left(\omega\right)^{2}\frac{d\widehat{u}}{dt}(\omega ,t)+3\left(\omega\right)^{4}\widehat{u}(\omega ,t)=0$$

Now assuming I did that correctly, the next step I think I should proceed with is to solve for $$\left.\widehat{u}(\omega ,t)$$. I don't remember how to solve this type of ODE, I was reading a couple of sites and it says I should use a characteristic equation which would I assume then be, $$\left.\lambda^{2}+4\omega^{2} \lambda +3\omega$$ where $$\lambda$$ is just an arbitrary symbol to denote a quadratic equation. I looked for the roots and used it along with the general expression of the 2nd order ODE to get
$$\left.\widehat{u}(\omega ,t)=c_{1}+c_{2}e^{-4\omega^{2}t}$$
But it seems to be incorrect since I took the derivative and plugged it back into my fourier transform expression and did not get a 0 for my answer so...Any guidance would be much appreciated!!! Thanks!

tiny-tim
Homework Helper
HI jianxu!

(have an omega: ω )
Which means me overall expression after transform is:
$$\left.\frac{d^{2}\widehat{u}}{dt^{2}}(\omega ,t)+4\left(\omega\right)^{2}\frac{d\widehat{u}}{dt}(\omega ,t)+3\left(\omega\right)^{4}\widehat{u}(\omega ,t)=0$$

… I should use a characteristic equation which would I assume then be, $$\left.\lambda^{2}+4\omega^{2} \lambda +3\omega^4$$ where $$\lambda$$ is just an arbitrary symbol to denote a quadratic equation. I looked for the roots and used it along with the general expression of the 2nd order ODE to get
$$\left.\widehat{u}(\omega ,t)=c_{1}+c_{2}e^{-4\omega^{2}t}$$

No, the roots are ω2 = -1 and -3, so the general solution is c1e2t + c2e-3ω2t

Hi TinyTim, thanks for the reply! I just realized where I made my mistake! Thanks very much for the help!

Hello!

I've been working on this problem and was wondering if someone could check if I've done the rest of this problem correctly!

So after finding the roots, I apply the initial conditions where:
$$\left.\widehat{u}\left(\omega,0\right) = \widehat{f}\left(\omega\right)$$

since t = 0, I have:
$$\left.\widehat{u}\left(\omega,0\right) = \widehat{f}\left(\omega\right) = C_{1}+C_{2}$$

For $$\left.\frac{d\widehat{u}}{dt}$$:
$$\left.\frac{d\widehat{u}}{dt}= -\omega C_{1}e^{-\omega^{2}t}-3\omega C_{2}e^{-3\omega^{2}t}$$
applying initial conditions:
$$\left.\frac{d\widehat{u}}{dt}\left(\omega,0\right)= -\omega C_{1}-3\omega C_{2} = \widehat{g}\left(\omega\right)$$

Now I solved for the constants using elimination and got:
$$\left.C_{1}= \frac{3\widehat{f}(\omega)}{2} + \frac{\widehat{g}(\omega)}{2\omega^{2}}$$
and:
$$\left.C_{2}= -\frac{\widehat{f}(\omega)}{2} - \frac{\widehat{g}(\omega)}{2\omega^{2}}$$

Therefore our $$\left.\widehat{u}(\omega ,t) = \left(\frac{3\widehat{f}(\omega)}{2} + \frac{\widehat{g}(\omega)}{2\omega^{2}}\right)e^{-\omega^{2}t}+ \left(-\frac{\widehat{f}(\omega)}{2} - \frac{\widehat{g}(\omega)}{2\omega^{2}}\right)e^{-3\omega^{2}t}$$

Now I apply inverse fourier transformation so that:
$$\left.u(x,t) = \widehat{f^{-1}}(\omega)$$

Which means:
$$\left.u(x,t) = \frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}(\widehat{u}(\omega ,t))e^{i\omega x}d\omega$$

that would be the solution seeing that we don't know what $$\left.\widehat{f}(\omega )$$ or $$\left.\widehat{g}(\omega )$$ are.

Thanks!!