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evinda

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Hello! (Wave)

I want to find the dispersion relation for the solutions in the form $u(x,t)=e^{i(kx-\omega t)}, k, \omega>0$ of the following partial differential equations:

Which of the above equations are dispersion equations?

For the first differential equation, I have tried the following so far:

We suppose that $u(x,t)=e^{i(kx-\omega t)}, k, \omega>0$ is a solution of $u_t+au_x=du_{xx}$.

We have: $u_t(x,t)=- \omega i e^{i(kx-\omega t)} \\ u_x=ik e^{i(kx-\omega t)} \\ u_{xx}=-k^2 e^{i(kx- \omega t)}$

Thus, it has to hold: $- \omega i e^{i(kx-\omega t)}+ a ik e^{i(kx - \omega t)}=-d k^2 e^{i(kx-\omega t)}$

or equivalently $(aik- \omega i +d k^2) e^{i(kx-\omega t)}=0 \ \ \forall x, t \in \mathbb{R}$.

So it has to hold: $aik- \omega i +d k^2=0$.

$u(x,t)= e^{i(kx-\omega t)}$ is a solution of $u_t+au_x=du_{xx}$ iff $aik- \omega i +d k^2=0$.If we would look for a solution of the form $A \cos(kx- \omega t)$, we would continue by writing the solution in the form of a traveling wave.In our case, do we use the fact that $e^{i(kx-\omega t)}=\cos(kx- \omega t)+i \sin(kx-\omega t)$?

If so, then would we say the following?

$aik- \omega i +d k^2=0 \Rightarrow \frac{\omega}{k}=a-dk$ and so $u(x,t)= \cos \left( k \left( x-(a-dk)t \right)\right)+i sin \left( k \left( x-(a-dk)t \right)\right)$.

Thus, solutions of the differential equation that correspond to different wavenumbers "travel" with different velocities and thus $u_t+au_x=du_{xx}$ is a dispersion equation.

Or do we have to continue in an other way?

I want to find the dispersion relation for the solutions in the form $u(x,t)=e^{i(kx-\omega t)}, k, \omega>0$ of the following partial differential equations:

- $u_t+au_x=du_{xx}$
- $i u_t+u_{xx}=0$
- $u_{tt}=au_{xx}$, where $a,d>0$.

Which of the above equations are dispersion equations?

For the first differential equation, I have tried the following so far:

We suppose that $u(x,t)=e^{i(kx-\omega t)}, k, \omega>0$ is a solution of $u_t+au_x=du_{xx}$.

We have: $u_t(x,t)=- \omega i e^{i(kx-\omega t)} \\ u_x=ik e^{i(kx-\omega t)} \\ u_{xx}=-k^2 e^{i(kx- \omega t)}$

Thus, it has to hold: $- \omega i e^{i(kx-\omega t)}+ a ik e^{i(kx - \omega t)}=-d k^2 e^{i(kx-\omega t)}$

or equivalently $(aik- \omega i +d k^2) e^{i(kx-\omega t)}=0 \ \ \forall x, t \in \mathbb{R}$.

So it has to hold: $aik- \omega i +d k^2=0$.

$u(x,t)= e^{i(kx-\omega t)}$ is a solution of $u_t+au_x=du_{xx}$ iff $aik- \omega i +d k^2=0$.If we would look for a solution of the form $A \cos(kx- \omega t)$, we would continue by writing the solution in the form of a traveling wave.In our case, do we use the fact that $e^{i(kx-\omega t)}=\cos(kx- \omega t)+i \sin(kx-\omega t)$?

If so, then would we say the following?

$aik- \omega i +d k^2=0 \Rightarrow \frac{\omega}{k}=a-dk$ and so $u(x,t)= \cos \left( k \left( x-(a-dk)t \right)\right)+i sin \left( k \left( x-(a-dk)t \right)\right)$.

Thus, solutions of the differential equation that correspond to different wavenumbers "travel" with different velocities and thus $u_t+au_x=du_{xx}$ is a dispersion equation.

Or do we have to continue in an other way?

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