# Fourier transform with mixed derivatives/ 2nd order ODE

1. Jun 18, 2009

### jianxu

1. The problem statement, all variables and given/known data
Hi, So I'm suppose to solve the following problem:

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

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

2. Relevant equations

3. 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!

2. Jun 18, 2009

### tiny-tim

HI jianxu!

(have an omega: ω )
No, the roots are ω2 = -1 and -3, so the general solution is c1e2t + c2e-3ω2t

3. Jun 18, 2009

### jianxu

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

4. Jun 19, 2009

### jianxu

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!!