# Solve Dispersion Equations: Wave Solutions

• MHB
• evinda
In summary, the conversation discusses finding the dispersion relation for solutions of various partial differential equations in the form $u(x,t)=e^{i(kx-\omega t)}, k, \omega>0$. It is determined that $u_t+au_x=du_{xx}$ is a dispersion equation since the velocity $v=\frac{\omega}{k}$ depends on the wave number $k$. It is also concluded that $i u_t+u_{xx}=0$ is not a dispersion equation because the velocity does not depend on $k$. Finally, it is shown that $u_{tt}=au_{xx}$ is not a dispersion equation because it does not have solutions in the form $u(x,t)=e^{i(k evinda Gold Member MHB Hello! (Wave) I want to find the dispersion relation for the solutions in the form$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}$is a dispersion equation. Or do we have to continue in an other way? Last edited: evinda said:$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$). There is an imaginary part, but that will translate in a phase shift. (Mmm) I like Serena said: Btw, you've lost an$i$in$\frac{\omega}{k}=a-dk$. (Worried) Oh yes, right... (Tmi) I like Serena said: Since it follows that$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$or not, right? We don't have to write the solution in the form of a traveling wave, or have I understood it wrong? (Thinking) Also for the second and third differential equation is it as follows? • We suppose that$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$is not a dispersion equation? evinda said: So this means that in order to deduce if we have a dispersion equation, it suffices to check if$\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)}$. In physics the latter is the more common representation with the understanding that what we actually observe is the real part of it. (Emo) evinda said: So the velocity is$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) I like Serena said: 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) I like Serena said: Doesn't that mean that it is a dispersion equation? (Wondering) Oh yes, you are right! (Blush) I like Serena said: 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. evinda said: 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)

I like Serena said:
Seems good to me. (Nod)

Nice... Thank you! (Party)

## What is a dispersion equation?

A dispersion equation is a mathematical equation that describes the relationship between the frequency and wavelength of a wave. It is used to solve for the dispersion relation, which determines the behavior of waves in a specific medium.

## What are wave solutions?

Wave solutions are the solutions to a dispersion equation, which describe the behavior of waves in a particular medium. They show how the amplitude, frequency, and wavelength of a wave are related to each other.

## How do you solve dispersion equations?

Dispersion equations can be solved using various mathematical techniques, such as Fourier transforms, Laplace transforms, and Green's functions. The specific method used depends on the type of wave and the properties of the medium.

## What factors affect the dispersion of waves?

The dispersion of waves can be affected by various factors, such as the properties of the medium (e.g. density, elasticity), the frequency of the wave, and the geometry of the system. Additionally, boundary conditions and external forces can also influence wave dispersion.

## How are dispersion equations used in science?

Dispersion equations are used in various fields of science, such as physics, engineering, and mathematics, to understand and predict the behavior of waves in different systems. They are particularly useful for studying the propagation of electromagnetic waves, sound waves, and seismic waves.

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