Mixing
Helmholtz equation with moving boundary in the plane
$begingroup$
Let us assume we have a unit disk $Dsubsetmathbb{R}^2$ s.t. $vec{0}in D$. To obtain the eigenfrequencies and eigenmodes (or eigenvalues and -functions if you like) we must solve $Delta psi + lambda^2 psi = 0$ with appropriate boundary conditions.
This can be done analytically using the ansatz $psi(r,phi)=R(r)Phi(phi)$ to obtain Bessel's equation. After a long but simple derivation one obtains the roots of Bessel's functions of the first kind as eigenvalues.
Here's my problem then:
Assuming we start to rotate the disk in the plane around the origin will result in a moving boundary.
Can we still determine the eigenvalues of the unit disc analytically?
I know that one can obtain the eigenvalues employing coordinate transformations which will modify the equation, transforming it into a coordinate frame where the boundary is at rest and hence Dirichlet conditions may be used. But I'm interested specifically in an analytic solution where the differential equation is not modified.
Does such a solution exist?
ordinary-differential-equations differential-geometry pde eigenvalues-eigenvectors eigenfunctions
edited Jan 11 at 12:52
Dylan
12.7k31026
asked Jan 9 at 16:30
NoxNox
534313
$endgroup$
add a comment |
$begingroup$
Let us assume we have a unit disk $Dsubsetmathbb{R}^2$ s.t. $vec{0}in D$. To obtain the eigenfrequencies and eigenmodes (or eigenvalues and -functions if you like) we must solve $Delta psi + lambda^2 psi = 0$ with appropriate boundary conditions.
This can be done analytically using the ansatz $psi(r,phi)=R(r)Phi(phi)$ to obtain Bessel's equation. After a long but simple derivation one obtains the roots of Bessel's functions of the first kind as eigenvalues.
Here's my problem then:
Assuming we start to rotate the disk in the plane around the origin will result in a moving boundary.
Can we still determine the eigenvalues of the unit disc analytically?
I know that one can obtain the eigenvalues employing coordinate transformations which will modify the equation, transforming it into a coordinate frame where the boundary is at rest and hence Dirichlet conditions may be used. But I'm interested specifically in an analytic solution where the differential equation is not modified.
Does such a solution exist?
ordinary-differential-equations differential-geometry pde eigenvalues-eigenvectors eigenfunctions
edited Jan 11 at 12:52
Dylan
12.7k31026
asked Jan 9 at 16:30
NoxNox
534313
$endgroup$
$begingroup$
If the equation itself doesn't depend on time, then the eigenvalues remain the same. The final solution just has time as a parameter
$endgroup$
– Dylan
Jan 11 at 4:43
$begingroup$
I can argue, that the eigenvalues should remain the same from a physical point of view (changing how I look at the system, not the system itself). I need either a math. proof of identity of EV or a way to derive an explicit expression of the EV for moving boundary.
$endgroup$
– Nox
Jan 16 at 9:35
$begingroup$
Apply separation of variables: $psi = R(r)Phi(phi)T(t)$. If you plug this in the original equation, the time parts would all cancel out, and you get same equations for $R(r)$ and $Phi(phi)$ as before. The only thing that changes is the coefficients in the series solution would be dependent on $t$
$endgroup$
– Dylan
Jan 16 at 11:16
add a comment |
$begingroup$
Let us assume we have a unit disk $Dsubsetmathbb{R}^2$ s.t. $vec{0}in D$. To obtain the eigenfrequencies and eigenmodes (or eigenvalues and -functions if you like) we must solve $Delta psi + lambda^2 psi = 0$ with appropriate boundary conditions.
This can be done analytically using the ansatz $psi(r,phi)=R(r)Phi(phi)$ to obtain Bessel's equation. After a long but simple derivation one obtains the roots of Bessel's functions of the first kind as eigenvalues.
Here's my problem then:
Assuming we start to rotate the disk in the plane around the origin will result in a moving boundary.
Can we still determine the eigenvalues of the unit disc analytically?
I know that one can obtain the eigenvalues employing coordinate transformations which will modify the equation, transforming it into a coordinate frame where the boundary is at rest and hence Dirichlet conditions may be used. But I'm interested specifically in an analytic solution where the differential equation is not modified.
Does such a solution exist?
ordinary-differential-equations differential-geometry pde eigenvalues-eigenvectors eigenfunctions
edited Jan 11 at 12:52
Dylan
12.7k31026
asked Jan 9 at 16:30
NoxNox
534313
$endgroup$
Let us assume we have a unit disk $Dsubsetmathbb{R}^2$ s.t. $vec{0}in D$. To obtain the eigenfrequencies and eigenmodes (or eigenvalues and -functions if you like) we must solve $Delta psi + lambda^2 psi = 0$ with appropriate boundary conditions.
This can be done analytically using the ansatz $psi(r,phi)=R(r)Phi(phi)$ to obtain Bessel's equation. After a long but simple derivation one obtains the roots of Bessel's functions of the first kind as eigenvalues.
Here's my problem then:
Assuming we start to rotate the disk in the plane around the origin will result in a moving boundary.
Can we still determine the eigenvalues of the unit disc analytically?
I know that one can obtain the eigenvalues employing coordinate transformations which will modify the equation, transforming it into a coordinate frame where the boundary is at rest and hence Dirichlet conditions may be used. But I'm interested specifically in an analytic solution where the differential equation is not modified.
Does such a solution exist?
ordinary-differential-equations differential-geometry pde eigenvalues-eigenvectors eigenfunctions
ordinary-differential-equations differential-geometry pde eigenvalues-eigenvectors eigenfunctions
edited Jan 11 at 12:52
Dylan
12.7k31026
asked Jan 9 at 16:30
NoxNox
534313
edited Jan 11 at 12:52
Dylan
12.7k31026
asked Jan 9 at 16:30
NoxNox
534313
edited Jan 11 at 12:52
Dylan
12.7k31026
edited Jan 11 at 12:52
Dylan
12.7k31026
edited Jan 11 at 12:52
Dylan
12.7k31026
12.7k31026
asked Jan 9 at 16:30
NoxNox
534313
asked Jan 9 at 16:30
NoxNox
534313
asked Jan 9 at 16:30
NoxNox
534313
534313
$begingroup$
If the equation itself doesn't depend on time, then the eigenvalues remain the same. The final solution just has time as a parameter
$endgroup$
– Dylan
Jan 11 at 4:43
$begingroup$
I can argue, that the eigenvalues should remain the same from a physical point of view (changing how I look at the system, not the system itself). I need either a math. proof of identity of EV or a way to derive an explicit expression of the EV for moving boundary.
$endgroup$
– Nox
Jan 16 at 9:35
$begingroup$
Apply separation of variables: $psi = R(r)Phi(phi)T(t)$. If you plug this in the original equation, the time parts would all cancel out, and you get same equations for $R(r)$ and $Phi(phi)$ as before. The only thing that changes is the coefficients in the series solution would be dependent on $t$
$endgroup$
– Dylan
Jan 16 at 11:16
add a comment |
$begingroup$
If the equation itself doesn't depend on time, then the eigenvalues remain the same. The final solution just has time as a parameter
$endgroup$
– Dylan
Jan 11 at 4:43
$begingroup$
I can argue, that the eigenvalues should remain the same from a physical point of view (changing how I look at the system, not the system itself). I need either a math. proof of identity of EV or a way to derive an explicit expression of the EV for moving boundary.
$endgroup$
– Nox
Jan 16 at 9:35
$begingroup$
Apply separation of variables: $psi = R(r)Phi(phi)T(t)$. If you plug this in the original equation, the time parts would all cancel out, and you get same equations for $R(r)$ and $Phi(phi)$ as before. The only thing that changes is the coefficients in the series solution would be dependent on $t$
$endgroup$
– Dylan
Jan 16 at 11:16
$begingroup$
If the equation itself doesn't depend on time, then the eigenvalues remain the same. The final solution just has time as a parameter
$endgroup$
– Dylan
Jan 11 at 4:43
$begingroup$
If the equation itself doesn't depend on time, then the eigenvalues remain the same. The final solution just has time as a parameter
$endgroup$
– Dylan
Jan 11 at 4:43
$begingroup$
I can argue, that the eigenvalues should remain the same from a physical point of view (changing how I look at the system, not the system itself). I need either a math. proof of identity of EV or a way to derive an explicit expression of the EV for moving boundary.
$endgroup$
– Nox
Jan 16 at 9:35
$begingroup$
I can argue, that the eigenvalues should remain the same from a physical point of view (changing how I look at the system, not the system itself). I need either a math. proof of identity of EV or a way to derive an explicit expression of the EV for moving boundary.
$endgroup$
– Nox
Jan 16 at 9:35
$begingroup$
Apply separation of variables: $psi = R(r)Phi(phi)T(t)$. If you plug this in the original equation, the time parts would all cancel out, and you get same equations for $R(r)$ and $Phi(phi)$ as before. The only thing that changes is the coefficients in the series solution would be dependent on $t$
$endgroup$
– Dylan
Jan 16 at 11:16
$begingroup$
Apply separation of variables: $psi = R(r)Phi(phi)T(t)$. If you plug this in the original equation, the time parts would all cancel out, and you get same equations for $R(r)$ and $Phi(phi)$ as before. The only thing that changes is the coefficients in the series solution would be dependent on $t$
$endgroup$
– Dylan
Jan 16 at 11:16
add a comment |
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$begingroup$
If the equation itself doesn't depend on time, then the eigenvalues remain the same. The final solution just has time as a parameter
$endgroup$
– Dylan
Jan 11 at 4:43
$begingroup$
I can argue, that the eigenvalues should remain the same from a physical point of view (changing how I look at the system, not the system itself). I need either a math. proof of identity of EV or a way to derive an explicit expression of the EV for moving boundary.
$endgroup$
– Nox
Jan 16 at 9:35
$begingroup$
Apply separation of variables: $psi = R(r)Phi(phi)T(t)$. If you plug this in the original equation, the time parts would all cancel out, and you get same equations for $R(r)$ and $Phi(phi)$ as before. The only thing that changes is the coefficients in the series solution would be dependent on $t$
$endgroup$
– Dylan
Jan 16 at 11:16