# Finding Dispersion relation

1. Sep 4, 2011

### autobot.d

$\imath\frac{\partial u}{\partial t} + \frac{\partial^2 u}{\partial x^2}=0$

$\left(x,t\right) = \int^{\infty}_{-\infty}A\left(k\right)e^{\imath\left(kx-wt\right)}dk$

$u\left(x,0\right)=\delta\left(x\right)$

This is what I am working with. I am supposed to find the dispersion relation. So far I have gotten

$A\left(k\right) = \frac{1}{2\pi}\int^{\infty}_{-\infty}\delta\left(x\right)e^{-\imath\left(kx\right)}dx = \frac{1}{2\pi}$

plugging this in to u(x,t) do I work with

$u\left(x,t\right) =\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{\imath\left(kx-wt\right)}dk$

This is where I am stuck. I know w(k) is the dispersion relation. If I put in the pde do I just deal with

$\imath \left(-\imath w\right) + \frac{d^{2}u}{dt^{2}} = w +\frac{d^{2}u}{dt^{2}}=0$

not sure how to pull out the dispersion equation or if I am even going the right route. Any clues on how to proceed would be most appreciated. Solving this equation does not seem to get me to where I want to be. Thanks!

Last edited: Sep 4, 2011
2. Sep 4, 2011

### autobot.d

Am I making this too hard on myself. I just saw a similar problem and according to this book I should just get

$w-k^2=0\Rightarrow w=k^2$

I figured this out I believe.

Last edited: Sep 4, 2011
3. Sep 5, 2011

### hunt_mat

You have:
$$i\frac{\partial u}{\partial t}+\frac{\partial^{2}u}{\partial x^{2}}dk$$
Write:
$$u=\int_{-\infty}^{\infty}A(k)e^{i(kx-\omega (k)t)}$$
So we just differentiate!
$$\frac{\partial u}{\partial t}=\int_{-\infty}^{\infty}(-i\omega (k))A(k)e^{i(kx-\omega (k)t)}dk,\quad\frac{\partial^{2}u}{\partial x^{2}}=\int_{-\infty}^{\infty}A(k)(-k^{2})e^{i(kx-\omega (k)t)}dk$$
Insert the above into your equation to obtain the dispersion relation.

4. Sep 7, 2011

### autobot.d

Is that dk a typo? Not sure what it is there for.

My professor said to just use

$\phi\left(x,t\right)=A\left(k\right)e^\left(i\left(kx-wt\right)\right)$

to find the dispersion relation, therefore leaving me with

$w=k^2$

and that the integral is just the "sum" of all the solutions.

Thanks for the help!

5. Sep 7, 2011

### hunt_mat

You have to integrate over all possible wavenumbers k, so when do the substitution as I suggested you can get everything to one side as:
$$\int_{-\infty}^{\infty}(\omega (k)-k^{2})A(k)e^{i(kx-\omega (k)t)}dk=0$$
The only way for this integral to be zero of if the integrand is zero and hence you have your dispersion relation.

6. Sep 7, 2011

### autobot.d

Ahhhh...makes sense now. Thank you for the clarification.