MHB Solve Dispersion Equations: Wave Solutions

[evinda]

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:

• $u_t+au_x=du_{xx}$
• $i u_t+u_{xx}=0$
• $u_{tt}=au_{xx}$, 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?

 

[I like Serena]

$u(x,t)= e^{i(kx-\omega t)}$ 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?

Hey! (Blush)

I think you are already there. (Nod)
The solution $Ae^{i(kx-\omega t)}$ already represents a harmonic solution.
Btw, you've lost an $i$ in $\frac{\omega}{k}=a-dk$. (Worried)
Since it follows that $v = \frac{\omega}{k}=a-dik$, we see that velocity depends on wave number (since $d \ne 0$). There is an imaginary part, but that will translate in a phase shift. (Mmm)

 

[evinda]

Btw, you've lost an $i$ in $\frac{\omega}{k}=a-dk$. (Worried)

Oh yes, right... (Tmi)

Since it follows that $v = \frac{\omega}{k}=a-dik$, we see that velocity depends on wave number (since $d \ne 0$). There is an imaginary part, but that will translate in a phase shift. (Mmm)

So this means that in order to deduce if we have a dispersion equation, it suffices to check if $\frac{\omega}{k}$ depends on $k$ or not, right?
We don't have to write the solution in the form of a traveling wave, or have I understood it wrong? (Thinking)

 

[evinda]

Also for the second and third differential equation is it as follows?

• We suppose that $u(x,t)=e^{i(kx-\omega t)}, k, \omega>0$ is a solution of $i u_t+u_{xx}=0$.
  
  Thus, it has to hold: $ - \omega i^2 e^{i(kx - \omega t)}-k^2 e^{i(kx- \omega t)}=0$
  
  or equivalently $(\omega-k^2) e^{i(kx- \omega t)}\ \ \forall x, t \in \mathbb{R}$.
  
  So it has to hold: $\omega-k^2=0$.
  
  $u(x,t)= e^{i(kx-\omega t)}$ is a solution of $i u_t+u_{xx}=0$ iff $\omega=k^2$.
  
  So the velocity is $v=\frac{\omega}{k}=k$ and thus it depends on $k$ and so we deduce that $i u_t+u_{xx}=0$ is not a dispersion equation.
• We suppose that $u(x,t)=e^{i(kx-\omega t)}, k, \omega>0$ is a solution of $u_{tt}=au_{xx}, a>0$.
  
  Then $u_{tt}(x,t)=\omega^2 e^{i(kx-\omega t)}$.
  
  Thus, it has to hold: $\omega^2 e^{i(kx-\omega t)}=a (-k^2 e^{i(kx-\omega t)})$
  
  or equivalently $e^{i(kx-\omega t)} (\omega^2+k^2 a)=0$
  
  So it has to hold: $\omega^2+k^2 a=0 \Rightarrow \omega^2=-k^2 a<0$, contradiction.
  
  Is it right? If so does this mean that $u_{tt}=au_{xx}, a>0$ does not have solutions of the form $u(x,t)=e^{i(kx-\omega t)}, k, \omega>0$? If so, then do we deduce that $u_{tt}=au_{xx}, a>0$ is not a dispersion equation?

 

[I like Serena]

So this means that in order to deduce if we have a dispersion equation, it suffices to check if $\frac{\omega}{k}$ depends on $k$ or not, right?

Yes. (Smile)

We don't have to write the solution in the form of a traveling wave, or have I understood it wrong? (Thinking)

We do. It's just that a traveling wave can be represented both as $A\cos(kx-\omega t)$ or as $Ae^{i(kx-\omega t)}$. In physics the latter is the more common representation with the understanding that what we actually observe is the real part of it. (Emo)

 

[I like Serena]

So the velocity is $v=\frac{\omega}{k}=k$ and thus it depends on $k$ and so we deduce that $i u_t+u_{xx}=0$ is not a dispersion equation.

Doesn't that mean that it is a dispersion equation? (Wondering)

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

Then $u_{tt}(x,t)=\omega^2 e^{i(kx-\omega t)}$.

Isn't it like this:
$$u_{tt}(x,t) = (-i\omega)^2e^{i(kx-\omega t)} = -\omega^2 e^{i(kx-\omega t)}$$
(Wasntme)

 

[evinda]

Yes. (Smile)

We do. It's just that a traveling wave can be represented both as $A\cos(kx-\omega t)$ or as $Ae^{i(kx-\omega t)}$. In physics the latter is the more common representation with the understanding that what we actually observe is the real part of it. (Emo)

Ah, I see... (Smile)

Doesn't that mean that it is a dispersion equation? (Wondering)

Oh yes, you are right! (Blush)

Isn't it like this:
$$u_{tt}(x,t) = (-i\omega)^2e^{i(kx-\omega t)} = -\omega^2 e^{i(kx-\omega t)}$$
(Wasntme)

Oh yes, right... So is it as follows? (Thinking)

We suppose that $u(x,t)=e^{i(kx- \omega t)}, \omega, k>0$ is a solution of the differential equation $u_{tt}=au_{xx}, a>0$.
Then $u_{tt}(x,t)=-\omega^2 e^{i(kx-\omega t)}$ and $u_{xx}(x,t)=-k^2 e^{i(kx-\omega t)}$.

Thus, it has to hold: $\omega^2 e^{i(kx- \omega t)}=a k^2 e^{i(kx- \omega t)}$ or equivalently $(\omega^2-a k^2) e^{i(kx- \omega t)}=0 \ \ \forall x,t \in \mathbb{R}$

So it has to hold: $\omega^2=a k^2$.

$u(x,t)=e^{i(kx- \omega t)}$ is a solution of $u_{tt}=au_{xx}, a>0$ iff $\omega^2=ak^2 \Rightarrow \frac{\omega}{k}= \sqrt{a}$.

Thus $u(x,t)= e^{i(k(x-\sqrt{a}t))}$.

We see that the velocity that is equal to $\frac{k}{ \omega}=\sqrt{a}$ doesn't depend on the wave number.

Thus $u_{tt}=au_{xx}, a>0$ is not a dispersion equation.

 

[I like Serena]

We see that the velocity that is equal to $\frac{k}{ \omega}=\sqrt{a}$ doesn't depend on the wave number.

Thus $u_{tt}=au_{xx}, a>0$ is not a dispersion equation.

Seems good to me. (Nod)

 

[evinda]

Seems good to me. (Nod)

Nice... Thank you! (Party)