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Homework Help: Solving the Helmholtzequation

  1. May 22, 2006 #1
    solve the Helmholtzequation laplace(u) + lamda*u = 0 in the unit circle with neumann condition for r = 1

    I dont 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 thats as far as I get
     
  2. jcsd
  3. May 22, 2006 #2

    J77

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    Could you write it in tex...

    The next step will be to separate variables.
     
  4. May 22, 2006 #3
    no, I'm sorry, I cant. Dont know how tex works.

    I know that the next thing is to separate the variables, but I dont get it right, that is as far as I get.
     
  5. May 22, 2006 #4

    J77

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    What's your capital R?

    First write: [tex]u(r,\theta)=u_r(r)u_\theta(\theta)[/tex]

    (multiplying the equation by [tex]r^2[/tex] to get it in a separable form)
     
  6. May 22, 2006 #5
    Yeah, well thats what I did. But I wrote it as u(r,Theta) = R(r)*Theta(theta). And thats 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
     
  7. May 22, 2006 #6

    J77

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    In the form you have, you need an r in front of the [tex]\lambda[/tex] - then multiply up by the second r (the one beneath the [tex]\theta''[/tex])...
     
  8. May 22, 2006 #7
    rR'/R + r^2R''/R + theta''/theta + r^2*lambda = 0

    but I still dont see how to continue. I would really appriciate 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
     
    Last edited: May 22, 2006
  9. May 22, 2006 #8

    HallsofIvy

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    Science Advisor

    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 Ji(r) and Ii(r), Bessel's functions of the first and second kind, of order i.

    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 [itex]2\pi[/itex] you have an idea of what k (and therefore i in Ji and Ii) had better be.
     
  10. May 24, 2006 #9

    J77

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    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:

    [tex]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[/tex]

    Separation of variables gives:

    [tex]\frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2}=const.=-m^2[/tex]

    which gives:

    [tex]r^2 \frac{\partial^2 R}{\partial r^2} + r \frac{\partial R}{\partial r} + (r^2\lambda-m^2)R = 0[/tex]

    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, [tex]\Theta[/tex] is periodic and therefore satisfies [tex]Ae^{\pm im\theta}[/tex]

    The above isn't the full answer but should help...
     
  11. May 24, 2006 #10
    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.

    [tex]\frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2}=const.=-k[/tex]

    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
     
  12. May 26, 2006 #11
    If u is bounded at r=0 then b=0 and with boundary conditions u(1,t)=0 you get

    [tex]u(1,t)=a*J_m(k)*T(t)=0[/tex]

    [tex]k=\lambda^2[/tex]

    from which

    [tex]J_m(k)=0[/tex]

    and

    [tex]k=\gamma_{ms}[/tex]

    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

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

    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 isnt bounded at r=0 or isnt defined at r=0
    you get a system of linear equations for the coefficients a and b, and a "determinant-equation" gives the solution.
     
    Last edited: May 26, 2006
  13. May 28, 2006 #12
    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 lets 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? doesnt 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 doesnt 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?
     
    Last edited: May 29, 2006
  14. May 31, 2006 #13
    yupp, s:th zero for the m:th besselfunction

    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.

    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.
     
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