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Solution of Helmholtz equation

  1. Aug 3, 2009 #1
    hi guys..
    I'm trying desperately to solve the following Helmholtz equation:

    *(all parameters are known)

    [tex]
    \frac{\partial^2 E_z}{\partial x^2}+ \frac{\partial^2 E_z}{\partial y^2} +j\omega\sigma E_z=0
    [/tex]
    (Ez is a scalar of course)
    in the boundaries -inf<x<inf, -inf<y<0

    with the following boundary conditions:
    Ez(y->-inf)=0; [tex] Ez(y=0)=\frac{jwI}{4*pi}*log(\frac{(d+x)^2+a^2}{(d-x)^2+a^2}) [/tex]
    Ez(x->inf or -inf)=0

    the proposed solution is of the form Ez=sin(Kx*x)*exp(Ky*y)

    the main problem here is to find a sine/Fourier series expansion of the boundary condition.

    anyone have an idea??
     
    Last edited: Aug 3, 2009
  2. jcsd
  3. Aug 3, 2009 #2

    gabbagabbahey

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    Hi yyuy1, welcome to PF!:smile:

    This forum supports [itex]\LaTeX[/itex], which makes it easy to write equations clearer. A short introduction to using [itex]\LaTeX[/itex] on PF can be found in this thread.


    Are those supposed to be second order partial derivatives by any chance (the Helmholtz equation is second order after all)? Specifically, is your equation supposed to be

    [tex]\frac{\partial E_z}{\partial x}+ \frac{\partial E_z}{\partial y} +j\omega\sigma E_z=0[/tex]

    or

    [tex]\frac{\partial^2 E_z}{\partial x^2}+ \frac{\partial^2 E_z}{\partial y^2} +j\omega\sigma E_z=0[/tex]

    or something else entirely?

    Why is your proposed solution of this form? Does this proposed solution go to zero as [itex]x[/itex] approaches [itex]\pm \infty[/itex]? If not, how can you expect to fit it to your boundary conditions? Does it even satisfy your original differential equation?
     
    Last edited: Aug 3, 2009
  4. Aug 3, 2009 #3
    thanks for the quick response gabbagabbahey
    unfortunately I currently have no Latex capabilities. hopefully I will manage to learn it in the future.

    The solution will approach 0 in [tex]y\rightarrow\pm\infty[/tex] due to behavior of the boundary condition on Ez(y=0).
    my problem is in finding an appropriate expansion..

    Ohh, and yes, it's supposed to be a second partial derivative..
     
  5. Aug 3, 2009 #4

    gabbagabbahey

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    I'm not sure what you mean here. For starters, the solution is only supposed to be defined on the interval [itex]y\in (-\infty,0][/itex], so [itex]y[/itex] cannot approach [itex]+\infty[/itex].

    What you want is for your solution to approach zero for [itex]x\rightarrow\pm\infty[/itex] and [itex]y\rightarrow -\infty[/itex]. So far, you have not stated any restrictions on [itex]k_x[/itex] and [itex]k_y[/itex], so you cannot say whether your proposed solution accomplishes this or not...
     
  6. Aug 3, 2009 #5
    okay, I think I understand your point (what I meant was [itex]x\rightarrow\pm\infty[/itex])

    how do you suggest I approach this problem?
    I am having difficulty with the fact that the boundary condition is not periodic.
    (hence it is impossible to find a Fourier series expansion for it)
    thanks again
     
  7. Aug 3, 2009 #6

    gabbagabbahey

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    Okay, I think I see what you meant now. But, just because [itex]E_z(y=0)=\frac{j\omega I}{4\pi}\ln\left(\frac{(d+x)^2+a^2}{(d-x)^2+a^2}\right) [/itex] clearly approaches zero for [itex]x\to \pm \infty[/itex] doesn't mean your proposed solution does....what is [itex]\sin(\pm k_x* \infty)[/itex]?

    Well, how did you approach the problem? Specifically, how did you come up with your proposed form of the solution in the first place?
     
  8. Aug 3, 2009 #7
    well, the proposed solution was sort of a guess,
    but take in mind that the sine part was supposed to be a sum of a sine series and therefore I would hope to make sure it does converge properly.
     
  9. Aug 3, 2009 #8

    gabbagabbahey

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    Okay, it turns out that your "guess" is more or less correct, but guessing is typically no way to solve physics/math problems!:smiles:

    Have you learned the method of 'separation of variables' (SoV) yet?
     
  10. Aug 3, 2009 #9
    yes, I have. Actually, It's not really a total guess, it is pretty intuitive ...
    anyhow, I assumed that the problem was separable Ez=X(x)Y(y), and since the boundary condition on X is odd it should be expandable to a sine column. The Y part is probably a decaying exponent due to it's boundary conditions.
     
  11. Aug 3, 2009 #10

    gabbagabbahey

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    Okay, but be specific. Which of the 4 boundary conditions led you to this conclusion? How did you rule out rising or decaying exponentials and constant and linear terms?

    Again, which of the 4 boundary conditions led you to this conclusion?

    Are there any obvious restrictions on the values of [itex]k_x[/itex] and [itex]k_y[/itex]? Are they dependent or independent of eachother and why? Are they real valued or complex valued? How can you tell?
     
  12. Aug 3, 2009 #11
    the logarithmic boundary condition is odd, which led me to believe that the solution X(x) is expandable to a sine column.
    the boundary condition [itex]E_z(y\rightarrow-\infty)=0 [/itex] together with the logarithmic boundary condition on X mean that Y should be a positive constant in y=0 and decay to 0 as [itex]y\rightarrow-\infty [/itex].
    the real question is, how do I turn the logarithm into a sine column..
     
  13. Aug 3, 2009 #12

    gabbagabbahey

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    Good.
    Okay, but [itex](\alpha+\beta y +\gamma \sin(\kappa y)+\delta \cos(\kappa y))e^{||k_y||y}[/itex] satisfies that just as well as [itex]\alpha e^{||k_y||y}[/itex] does right?...this brings me back to my previous question:


    Try dealing with these questions first before you move on to applying the y=0 boundary condition and finding the series coefficients....after all, until you are sure that the individual terms of your series are of the correct form, you stand little chance of finding the coefficients of the series.
     
  14. Aug 3, 2009 #13
    well, I do know that [itex] {K_x}^2-{K_y}^2=j\omega\sigma[/itex]
     
  15. Aug 3, 2009 #14

    gabbagabbahey

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    good... now, what do the two boundary conditions as [itex]x\to\pm\infty[/itex] tell you about [itex]k_x[/itex]....specifically, is it real or complex?
     
  16. Aug 3, 2009 #15
    well, then Kx must be complex. the real part of [tex]Kx^2[/tex] will cancel out with the
    [tex]Ky^2[/tex] and the imaginary part will be [tex]j\omega\sigma[/tex].
    is this correct?
     
  17. Aug 3, 2009 #16

    gabbagabbahey

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    Well...

    [tex]\sin(K_x x)=\frac{e^{jK_x x}-e^{-jK_x x}}{2j}[/tex]

    so if [itex]K_x[/itex] has a non-zero imaginary part, X(x) will contain rising and falling exponentials....what happens to a rising exponential as [itex]x\to\infty[/itex]? What happens to a falling exponential as [itex]x\to-\infty[/itex]? Do the boundary conditions allow this?
     
  18. Aug 3, 2009 #17
    aha. does this mean that Kx must be real or that sin(Kx*x) is not the right solution?
     
  19. Aug 3, 2009 #18

    gabbagabbahey

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    It means that if sin(Kx*x) is the right solution, then Kx must be real.

    You've already reasoned that the solution must be of the form sin(Kx*x) from looking at the y=0 boundary condition, so therefor Kx must be real.

    Hence Ky is_____
     
  20. Aug 3, 2009 #19
    ky is complex, so that the real part of [tex]ky^2 [/tex] cancels out with [tex]kx^2[/tex] and then the imaginary part is [tex]j\omega\sigma[/tex]

    *(by the way, thanks allot for all this help)
     
  21. Aug 3, 2009 #20
    well, ky is obviously positive in order for the function to decay as it goes to y -> -inf
     
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