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evinda
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MHB
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Hello! (Wave)
Definition
A partial differential equation is called dispersion equation if it allows solutions in the form of a wave function and furthermore solutions in the form of a wave function with different wavenumbers have different velocities.
The relation between the cyclic frequency and the wavenumber at a wave function that is a solution of a pde is called dispersion relation.
I want to check if the transport equation $u_t+au_x=0, a \in \mathbb{R}$ is a dispersion equation.
I have tried the following:
We suppose that $u(x,t)=A \cos(kx-\omega t)$ is a solution of $u_t+a u_x=0, a \in \mathbb{R}$.
We have: $u_t= A \omega \sin(kx -\omega t) \\ u_x=-A k \sin(kx- \omega t)$
Thus it has to hold : $A \omega \sin(kx-\omega t)-a A k \sin(kx- \omega t)=0$ or equivalently $A(\omega-ak) sin(kx-\omega t)=0$.
So it has to hold: $\omega-ak=0$.
$u(x,t)=A \cos(kx- \omega t)$ is a solution of the transport equation iff $\omega-ak=0$.
We fix $\omega, k$ for which the above algebraic equation is satisfied.
We write the solution in the form of a traveling wave.
We have $u(x,t)= A \cos(kx-\omega t)=A \cos \left( k \left( x- \frac{\omega}{k}t \right)\right)$
Since $\omega-ak=0$ we have $\frac{\omega}{k}=a$.
So $u(x,t)=A \cos\left( k(x-at)\right)$ is a traveling wave with velocity $a$.
So we see that if $k_1, k_2>0$ wavenumbers with $k_1 \neq k_2$ then we have solutions in the form of a wave function
$$u_1(x,t)=A \cos(k_1(x-at)) \\ u_2(x,t)=A \cos(k_2(x-at))$$
that "travel" with the same velocity $a$.
Therefore, we conclude that the transport equation $u_t+au_x=0, a \in \mathbb{R}$ is not a dispersion equation.Am I right or have I done something wrong? (Thinking)
Definition
A partial differential equation is called dispersion equation if it allows solutions in the form of a wave function and furthermore solutions in the form of a wave function with different wavenumbers have different velocities.
The relation between the cyclic frequency and the wavenumber at a wave function that is a solution of a pde is called dispersion relation.
I want to check if the transport equation $u_t+au_x=0, a \in \mathbb{R}$ is a dispersion equation.
I have tried the following:
We suppose that $u(x,t)=A \cos(kx-\omega t)$ is a solution of $u_t+a u_x=0, a \in \mathbb{R}$.
We have: $u_t= A \omega \sin(kx -\omega t) \\ u_x=-A k \sin(kx- \omega t)$
Thus it has to hold : $A \omega \sin(kx-\omega t)-a A k \sin(kx- \omega t)=0$ or equivalently $A(\omega-ak) sin(kx-\omega t)=0$.
So it has to hold: $\omega-ak=0$.
$u(x,t)=A \cos(kx- \omega t)$ is a solution of the transport equation iff $\omega-ak=0$.
We fix $\omega, k$ for which the above algebraic equation is satisfied.
We write the solution in the form of a traveling wave.
We have $u(x,t)= A \cos(kx-\omega t)=A \cos \left( k \left( x- \frac{\omega}{k}t \right)\right)$
Since $\omega-ak=0$ we have $\frac{\omega}{k}=a$.
So $u(x,t)=A \cos\left( k(x-at)\right)$ is a traveling wave with velocity $a$.
So we see that if $k_1, k_2>0$ wavenumbers with $k_1 \neq k_2$ then we have solutions in the form of a wave function
$$u_1(x,t)=A \cos(k_1(x-at)) \\ u_2(x,t)=A \cos(k_2(x-at))$$
that "travel" with the same velocity $a$.
Therefore, we conclude that the transport equation $u_t+au_x=0, a \in \mathbb{R}$ is not a dispersion equation.Am I right or have I done something wrong? (Thinking)
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