Solving the Helmholtz Equation with Neumann Boundary Condition in a Unit Circle

[Larsson]

solve the Helmholtzequation laplace(u) + lamda*u = 0 in the unit circle with neumann condition for r = 1

I don't get very far on this problem. What I start with is rewriting the problem with how laplace looks in this case.

laplace(u) = (1/r) d/dr(r * du/dr) + (1/r^2) * d^2u/dtheta^2)

so the equation will be

R'/R + rR''/R + theta''/(r*theta) + lambda = 0

but that's as far as I get

 

[J77]

Could you write it in tex...

The next step will be to separate variables.

 

[Larsson]

no, I'm sorry, I cant. Dont know how tex works.

I know that the next thing is to separate the variables, but I don't get it right, that is as far as I get.

 

[J77]

What's your capital R?

First write: $$u(r,\theta)=u_r(r)u_\theta(\theta)$$

(multiplying the equation by $$r^2$$ to get it in a separable form)

 

[Larsson]

Yeah, well that's what I did. But I wrote it as u(r,Theta) = R(r)*Theta(theta). And that's how I got R'/R + rR''/R + theta''/(r*theta) + lambda = 0

I just applied the laplace on it and divided everything with R*Theta

 

[J77]

In the form you have, you need an r in front of the $$\lambda$$ - then multiply up by the second r (the one beneath the $$\theta''$$)...

 

[Larsson]

rR'/R + r^2R''/R + theta''/theta + r^2*lambda = 0

but I still don't see how to continue. I would really appreciate if you would show some steps on how to solve this, since I'm kind of lost. In the other assignmensts I've solved it've only been like X''/X = Y'/Y. But here it looks much more complicated (atleast for my untrained eye)

probably the next step would be something like rR'/R + r^2R''/R + r^2*lambda = -theta''/theta = k

-theta''/theta = k is "easy" to solve, bt rR'/R + r^2R''/R + r^2*lambda = k is a bit worse

 

[HallsofIvy]

rR'/R + r^2R''/R + theta''/theta + r^2*lambda = 0

but I still don't see how to continue. I would really appreciate if you would show some steps on how to solve this, since I'm kind of lost. In the other assignmensts I've solved it've only been like X''/X = Y'/Y. But here it looks much more complicated (atleast for my untrained eye)

Yeah, in fact it looks a lot like Bessel's equation! Of course, you know the general solutions to Bessel's equation: the two independent solutions are J_i(r) and I_i(r), Bessel's functions of the first and second kind, of order i.

probably the next step would be something like rR'/R + r^2R''/R + r^2*lambda = -theta''/theta = k

-theta''/theta = k is "easy" to solve, bt rR'/R + r^2R''/R + r^2*lambda = k is a bit worse

Yes, -theta"/theta= k is easy to solve- theta"= -k theta gives either exponential (if k is negative) or sine and cosine (if k is positive). Since the solutions had better be periodic, with period a multiple of $2\pi$ you have an idea of what k (and therefore i in J_i and I_i) had better be.

 

[J77]

rR'/R + r^2R''/R + theta''/theta + r^2*lambda = 0

but I still don't see how to continue. I would really appreciate if you would show some steps on how to solve this, since I'm kind of lost. In the other assignmensts I've solved it've only been like X''/X = Y'/Y. But here it looks much more complicated (atleast for my untrained eye)

probably the next step would be something like rR'/R + r^2R''/R + r^2*lambda = -theta''/theta = k

-theta''/theta = k is "easy" to solve, bt rR'/R + r^2R''/R + r^2*lambda = k is a bit worse

Was giving a presentation yesterday so couldn't follow this thread but adding to Halls of Ivy...

I'll first write your stuff in latex so we get a better view:

$$r^2 \frac{1}{R}\frac{\partial^2 R}{\partial r^2} + r \frac{1}{R}\frac{\partial R}{\partial r} + \frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2} + r^2\lambda = 0$$

Separation of variables gives:

$$\frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2}=const.=-m^2$$

which gives:

$$r^2 \frac{\partial^2 R}{\partial r^2} + r \frac{\partial R}{\partial r} + (r^2\lambda-m^2)R = 0$$

This last term is Bessels equation: The boundary conditions (geometry of problem) give you which functions are valid - you should consider both normal and modified kind, both 1st and 2nd of.

Again, $$\Theta$$ is periodic and therefore satisfies $$Ae^{\pm im\theta}$$

The above isn't the full answer but should help...

 

[Larsson]

Was giving a presentation yesterday so couldn't follow this thread but adding to Halls of Ivy...

I'll first write your stuff in latex so we get a better view:

$$r^2 \frac{1}{R}\frac{\partial^2 R}{\partial r^2} + r \frac{1}{R}\frac{\partial R}{\partial r} + \frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2} + r^2\lambda = 0$$

Separation of variables gives:

$$\frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2}=const.=-m^2$$

which gives:

$$r^2 \frac{\partial^2 R}{\partial r^2} + r \frac{\partial R}{\partial r} + (r^2\lambda-m^2)R = 0$$

This last term is Bessels equation: The boundary conditions (geometry of problem) give you which functions are valid - you should consider both normal and modified kind, both 1st and 2nd of.

Again, $$\Theta$$ is periodic and therefore satisfies $$Ae^{\pm im\theta}$$

The above isn't the full answer but should help...

Ok guys, thanks for all the help. Here's what I got. How do you guys get the tex code, do ypu write it by hand or do you have any help software?

I didnt do it exactly as you wrote J77, but it's the same way of thinking.

$$\frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2}=const.=-k$$

which I solved to T = A*exp(-i*sqrt(k)*t) + B*exp(-i*sqrt(k)*t)
T(t) = T(t+2*pi)
A*exp(-i*sqrt(k)*t) + B*exp(-i*sqrt(k)*t) =
A*exp(-i*sqrt(k)*(t+2pi)) + B*exp(-i*sqrt(k)*(t+2pi))

exp(i*sqrt(k)*t) = exp(i*sqrt(k)*(t+2*pi))
sqrt(k)*t = sqrt(k)*t + 2pi*sqrt(k) + 2pi*n
sqrt(k) = -n


The R part get the solution

R(r) = a*J_-n(sqrt(lambda)*r) + b*Y_-n(sqrt(lambda)*r)

and the T part (former theta part, looks so greasy to write theta without tex) get the solution

T(t) = A*exp(-i*n*t) + B*exp(i*n*t)

and u(r,t) = R(r)*T(t)

now the problem is how to fit the neumann condition u(1,t) = 0 to this.
And I also want to know how to do for dirichlet condition for r = 1

 

[JohanL]

If u is bounded at r=0 then b=0 and with boundary conditions u(1,t)=0 you get

$$u(1,t)=a*J_m(k)*T(t)=0$$

$$k=\lambda^2$$

from which

$$J_m(k)=0$$

and

$$k=\gamma_{ms}$$

the s:th root of the m:th besselfunction

its similar for neumann conditions but u get it for the derivative of J. But there are some useful relations between J and J'. For homogenous dirichlet conditions you get the solution

$$u(r,t)=\sum_{m=0}^\infty C_m*J_m(\gamma_{ms}r)*T(t)$$

And you need a second boundary condition to determine the constants C_m.
More about how to find C_m you can read in the litterature for bessel series.

If u isn't bounded at r=0 or isn't defined at r=0
you get a system of linear equations for the coefficients a and b, and a "determinant-equation" gives the solution.

 

[Larsson]

ok, so in my case I would get sqrt(lambda) = my_ms for dirichlet where my_ms is the zeroes for the bessel function. In this case it will be
sqrt(lambda)= 2.405 ; 3.382; 5.136; 5.520 ...
and for neumann
sqrt(lambda) = 0; 1.841: 3.054; 3.832...
can anyone check if these values are right so I can be sure that I got this right

this was wihot having to care about the z-axis. But let's try with a z-axis this time, then I get

R''/R + R'/rR + T''/r^2T + Z''/Z + lambda = 0 how do I separete this when I have 3 variabels? doesn't that get complicated?

Edit:

I think I got it

R''r^2/R + R'r/R + Z''r^2/Z + r^2*lambda = -T''/T = k

and -T''/T = k is the same as in the first problem, and therefor I get that k = n^2 where n is an integer

now I can rewrite the remaining part as

R''/R + R'/Rr + lambda - (n/r)^2 = -Z''/Z = C ( another constant)

and -Z''/Z = C ought to give me that c is an integer just as T''/T = k told me that. So C = j^2 where j is an integer

now we have R''/R + R'/Rr + lambda - j^2 - (n/r)^2 = 0

R'' + R'/r + R(lambda - j^2 - (n/r)^2) = 0

which have the solution R = a*J_n(sqrt(lambda-j^2)*r) + bY_n(sqrt(lambda-j^2)*r)

Edit2:

Just figured out that -Z''/Z = C doesn't say that C = j^2, the only reason that that worked for T is that I forced 2pi periodicity.

In this case I have a rod with the radius R_0 and length L and it's closed at Z = 0 and opened at Z = L. so this would give me that Z = cos(pi/L (1/2 + j)*z) and C = (pi/2L + pi*j/L)^2 where j is an integer. Is this correct?

 

[JohanL]

ok, so in my case I would get sqrt(lambda) = my_ms for dirichlet where my_ms is the zeroes for the bessel function.?

yupp, s:th zero for the m:th besselfunction

R''/R + R'/rR + T''/r^2T + Z''/Z + lambda = 0 how do I separete this when I have 3 variabels? doesn't that get complicated?

well, depending on the separation constants for Z and T you could get either the ordinary bessel functions as solutions for R or the modified
bessel functions I_n and K_n. These functions you can say is the ordinary bessel functions but with complex argument. But they arent oscillating functions like J_n and N_n(or Y_n) they behave more like exponential functions.

Edit:

I think I got it

R''r^2/R + R'r/R + Z''r^2/Z + r^2*lambda = -T''/T = k

and -T''/T = k is the same as in the first problem, and therefor I get that k = n^2 where n is an integer

usually you use, k^2, as the separation constant. And in this case it has to be negative for T too. Thats because T is periodic. With a negative separation constant you get cos/sin as the solutions, and with a positive you get exp or sinh/cosh, which are not periodic. For Z you should get sinh.