Solving the Sturm-Liouville System with Different Boundary Conditions

[the_dialogue]

Homework Statement

If I have the following Sturm-Liouville system:

$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}+\lambda_{k,m}^2)\phi_{k,m}(r,\phi)=0, (a<r<b,0<=\phi<=2\pi$$

$$\phi_{k,m}(r,\phi)=0, (r=a,0<=\phi<=2\pi$$

$$\phi_{k,m}(r,\phi)=0, (r=b, 0<=\phi<=2\pi$$

I'm told the solution to this is the following:
$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}<br /> <br /> \phi_{k,m}(r,\phi)=C_{k,m}(r){\colv{cos(m\phi}{sin(m\phi)}, (k=1,2...; m=0,1,2...),$$

where

$$C_{k,m}(r)=J_m(\lambda_{k,m}r)Y_m(\lambda_{k,m}a)-J_m(\lambda_{k,m}a)Y_m(\lambda_{k,m}r)$$

and $$(\lambda_{k,m}r)$$ is found by setting $$C_{k,m}(b)=0$$.

Homework Equations

Now I'm trying to solve the same system at different boundary conditions:

$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}+\lambda_{k,m}^2)\phi_{k,m}(r,\phi)=0, (r<b,0<=\phi<=2\pi$$

$$\phi_{k,m}(r,\phi)=0, (r=b, 0<=\phi<=2\pi)$$

With this system, where the constraints at $$a$$ have been removed (i.e. a=0), how should I approach solving for $$\phi_{k,m}(r,\phi)$$?

 

[fzero]

Homework Statement

If I have the following Sturm-Liouville system:

$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}+\lambda_{k,m}^2)\phi_{k,m}(r,\phi)=0, (a<r<b,0<=\phi<=2\pi$$

$$\phi_{k,m}(r,\phi)=0, (r=a,0<=\phi<=2\pi$$

$$\phi_{k,m}(r,\phi)=0, (r=b, 0<=\phi<=2\pi$$

I'm told the solution to this is the following:
$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}<br /> <br /> \phi_{k,m}(r,\phi)=C_{k,m}(r){\colv{cos(m\phi}{sin(m\phi)}, (k=1,2...; m=0,1,2...),$$

where

$$C_{k,m}(r)=J_m(\lambda_{k,m}r)Y_m(\lambda_{k,m}a)-J_m(\lambda_{k,m}a)Y_m(\lambda_{k,m}r)$$

and $$(\lambda_{k,m}r)$$ is found by setting $$C_{k,m}(b)=0$$.

Homework Equations

Now I'm trying to solve the same system at different boundary conditions:

$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}+\lambda_{k,m}^2)\phi_{k,m}(r,\phi)=0, (r<b,0<=\phi<=2\pi$$

$$\phi_{k,m}(r,\phi)=0, (r=b, 0<=\phi<=2\pi)$$

With this system, where the constraints at $$a$$ have been removed (i.e. a=0), how should I approach solving for $$\phi_{k,m}(r,\phi)$$?

You will still need a second boundary condition to specify the solution. It could be that you want the normal derivative at r = a to vanish, but without knowing the physical situation it's tough to guess.

You can use the r=b boundary condition to solve


$$C_{k,m}(r)=J_m(\lambda_{k,m}r)Y_m(\lambda_{k,m}b)-J_m(\lambda_{k,m}b)Y_m(\lambda_{k,m}r),$$

but you need one more boundary condition to solve for the $$(\lambda_{k,m}r)$$ .

 

[the_dialogue]

Thanks for the response.

If the edge @ r=b is clamped, then the slope should of course be zero. In 1-dimension, this would be e.g. dr/dx=0 -- would this be equivalent to dr/dn=0?

You will still need a second boundary condition to specify the solution. It could be that you want the normal derivative at r = a to vanish, but without knowing the physical situation it's tough to guess.

You can use the r=b boundary condition to solve


$$C_{k,m}(r)=J_m(\lambda_{k,m}r)Y_m(\lambda_{k,m}b)-J_m(\lambda_{k,m}b)Y_m(\lambda_{k,m}r),$$

but you need one more boundary condition to solve for the $$(\lambda_{k,m}r)$$ .

 

[fzero]

Thanks for the response.

If the edge @ r=b is clamped, then the slope should of course be zero. In 1-dimension, this would be e.g. dr/dx=0 -- would this be equivalent to dr/dn=0?

Clamping an edge fixes the value of the function, not it's slope. You can usually argue that the slope must vanish at an unclamped end, because the energy of the motion must be reflected from such a point.

 

[the_dialogue]

Yes. But from the geometry of my application, I know the slope must be zero at the edges. If this is so, how can the above Sturm system be solved?

Thank you for your help!

Clamping an edge fixes the value of the function, not it's slope. You can usually argue that the slope must vanish at an unclamped end, because the energy of the motion must be reflected from such a point.

 

[fzero]

Yes. But from the geometry of my application, I know the slope must be zero at the edges. If this is so, how can the above Sturm system be solved?

Thank you for your help!

If it's actually valid to also require the derivative at r=b to vanish, then that would give a condition to allow you to solve for the eigenvalues.

 

[the_dialogue]

Hmm I see. Unfortunately that has been my question from the beginning. Could you help me with this particular problem?

Also, is "derivative 0 at b" equivalent to "dr/dn=0"? If not, how does one describe this condition?

If it's actually valid to also require the derivative at r=b to vanish, then that would give a condition to allow you to solve for the eigenvalues.

 

[fzero]

Hmm I see. Unfortunately that has been my question from the beginning. Could you help me with this particular problem?

Well in the problem from your 1st post with boundary conditions at r=a,b the r=a condition was used to specify the eigenvalues as solutions to an equation. It's sufficiently complicated that there might be no closed form solution, but you can argue that the solutions exist and then find numerical approximations.

Also, is "derivative 0 at b" equivalent to "dr/dn=0"? If not, how does one describe this condition?

The normal derivative is

$$\nabla \phi \cdot \hat{n}.$$

If $$\hat{n}=\hat{r}$$, this is just $$\partial_r \phi$$. Just find the normal derivative from your geometry.

 

[the_dialogue]

I apologize -- I was in error in a few of my previous posts. 'fzero' was of course correct in saying that the slope cannot be zero at the boundaries.

I have a membrane that is clamped at the outside edge and is otherwise a full/regular membrane.

What is the corresponding Sturm-Liouville system and it's solution? 1 condition is that the value of displacement at the boundary = 0. What is the 2nd?

 

[fzero]

When I wrote the solution in post #2 for some reason I thought there would still be a 2nd boundary condition. If the only boundary condition is at r=b, then the Bessel function of the 2nd kind cannot appear, since it has a singularity at the origin. It should be very easy now for you to write down the solution and interpret the boundary condition.

 

[the_dialogue]

When I wrote the solution in post #2 for some reason I thought there would still be a 2nd boundary condition. If the only boundary condition is at r=b, then the Bessel function of the 2nd kind cannot appear, since it has a singularity at the origin. It should be very easy now for you to write down the solution and interpret the boundary condition.

Now if I have (the Bessel equation):
$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+(\lambda^2 -n^2\frac{1}{r^2})u_{k,m}(r,\phi)=0, (0<r<b,0<=\phi<=2\pi)$$

$$\phi_{k,m}(r,\phi)=0, (r=b, 0<=\phi<=2\pi)$$

The solution would be:
$$u_{k,m}(r,\phi)=AJ_n(\lambda r)+BY_n(\lambda r)$$

where to fulfill the condition that the infinite displacement is impossible, I can say that
$$B=0$$ so
$$u_{k,m}(r,\phi)=AJ_n(\lambda r)$$

But what shall I do if I have the following equation, instead of the above Bessel form:
$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}+\lambda_{k,m}^2)u_{k,m}(r,\phi)=0, (0<r<b,0<=\phi<=2\pi)$$
$$\phi_{k,m}(r,\phi)=0, (r=b, 0<=\phi<=2\pi)$$

I'm told the solution is something like:

$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}<br /> \u_{k,m}(r,\phi)=C_{k,m}(r){\colv{cos(m\phi}{sin (m\phi)}, (k=1,2...; m=0,1,2...),$$
$$C_{k,m}(r)=?$$

But I don't know where this comes from, nor what $$C_{k,m}(r)$$ should be.

Thanks for your help!

 

[fzero]

The equation

$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}+\lambda_{k,m}^2)u_{k,m}(r,\phi)=0, (0<=\phi<=2\pi)$$

is separable, with the angular part solved by the trig functions as in the first post. For a given harmonic, n, the radial function is precisely the Bessel equation

$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+(\lambda^2 -n^2\frac{1}{r^2})u_{k,m}(r,\phi)=0, (0<=\phi<=2\pi)$$

 

[the_dialogue]

The equation

$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}+\lambda_{k,m}^2)u_{k,m}(r,\phi)=0, (0<=\phi<=2\pi)$$

is separable, with the angular part solved by the trig functions as in the first post. For a given harmonic, n, the radial function is precisely the Bessel equation

$$(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+(\lambda^2 -n^2\frac{1}{r^2})u_{k,m}(r,\phi)=0, (0<=\phi<=2\pi)$$

Oh I see. So then would this be correct?

$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}<br /> <br /> \u_{k,m}(r,\phi)=J_{m}(\lambda_{k,m}r){\colv{cos(m\phi}{sin (m\phi)}, (k=1,2...; m=0,1,2...)$$

Would $$\lambda_{k,m}$$ be determined by calculating $$J_{m}(\lambda_{k,m}b)=0$$?

 

[fzero]

Oh I see. So then would this be correct?

$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}<br /> <br /> u_{k,m}(r,\phi)=J_{m}(\lambda_{k,m}r){\colv{cos(m\phi}{sin (m\phi)}, (k=1,2...; m=0,1,2...)$$

Would $$\lambda_{k,m}$$ be determined by calculating $$J_{m}(\lambda_{k,m}b)=0$$?

I believe that's right. The $$\lambda_{k,m}$$ are the zeros of the Bessel functions.

 

[the_dialogue]

I believe that's right. The $$\lambda_{k,m}$$ are the zeros of the Bessel functions.

In:
$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}<br /> <br /> u_{k,m}(r,\phi)=J_{m}(\lambda_{k,m}r){\colv{cos(m\phi}{sin (m\phi)}, (k=1,2...; m=0,1,2...)$$

But what does it mean for u to be = to a column vector? Shouldn't the phi solution just be $$cos(m\phi)+sin(m\phi)$$?

 

[fzero]

In:
$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}<br /> <br /> u_{k,m}(r,\phi)=J_{m}(\lambda_{k,m}r){\colv{cos(m\phi}{sin (m\phi)}, (k=1,2...; m=0,1,2...)$$

But what does it mean for u to be = to a column vector? Shouldn't the phi solution just be $$cos(m\phi)+sin(m\phi)$$?

The expression you are writing down is one for the eigenvectors of the system and is perfectly fine. If you want to write down the most general solution, then you would write an expression down which was a linear combination of all eigenvectors.