1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding a solution to Laplace's equation

  1. Dec 2, 2015 #1

    H Smith 94

    User Avatar
    Gold Member

    So here I have Laplace's equation with non-homogeneous, mixed boundary conditions in both [itex]x[/itex] and [itex]y[/itex].

    1. The problem statement, all variables and given/known data

    Solve Laplace's equation \begin{equation}\label{eq:Laplace}\nabla^2\phi(x,y)=0\end{equation} for the following boundary conditions:
    1. [itex]\phi(0, y)=2[/itex];
    2. [itex]\phi(1, y)=0[/itex];
    3. [itex]\phi(x, 0)=0[/itex];
    4. [itex]\frac{\partial}{\partial y}\phi(x,1)=1[/itex].
    Here, [itex]\phi[/itex] is the potential.

    2. Relevant equations

    In two dimensions:[tex]\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}.[/tex]

    3. The attempt at a solution

    As far as I can tell, a solution exists in \begin{equation}\label{eq:phi1}\phi(x,y)=\phi_0(x) + \Delta\phi(x,y),\end{equation} where [itex]\phi_0(x)[/itex] satisfies the [itex]x[/itex]-boundaries, in that [tex]\phi_0(x)=A_0+B_0 x[/tex] and [itex]\Delta\phi(x,y)[/itex] satisfies the [itex]y[/itex]-boundaries, with a separable solution of the form [tex]\Delta\phi(x,y)=\sum_{n}\chi_n(x)\gamma_n(y).[/tex]

    Applying boundary condition 1: [tex]\phi_0(0,y) = 2 \implies \boxed{A_0 = 2},[/tex] so now [itex]\phi_0= 2+B_0 x [/itex].

    Applying boundary condition 2: [tex]\phi_0(1,y) = 2+ B_0 = 0 \implies \boxed{B_0 = -2}.[/tex] Hence, have that \begin{equation}\phi_0(x,y) = 2-2x.\end{equation}

    Now, equation [itex]\ref{eq:phi1}[/itex] can be substituted into Laplace's equation ([itex]\ref{eq:Laplace}[/itex]) to yield [tex] \frac{\partial^2\Delta\phi(x,y)}{\partial x^2} + \frac{\partial^2\Delta\phi(x,y)}{\partial y^2} = 0.[/tex] This has the separable solution in [itex]\Delta\phi(x,y) = \chi(x)\gamma(y)[/itex]: [tex]\frac{\mathrm{d}^2\chi}{\mathrm{d}x^2} = \kappa \chi(x) \\ \frac{\mathrm{d}^2\gamma}{\mathrm{d}y^2} = -\kappa \gamma(y)[/tex] so that [tex]\begin{align}\chi(x) = E \cos{\kappa x} + F \sin{\kappa x} \\ \gamma(y) = G e^{\kappa y} + H e^{-\kappa y}, \end{align}[/tex] for which I found using the homogeneous boundary conditions produced from [itex]\phi_0(x)[/itex] that [itex]F = -E[/itex] and, in turn, [itex]E = 0[/itex] -- so [tex]\chi(x) = 0\ \forall x \implies \boxed{\Delta\phi(x,y) = 0}[/tex] and so \begin{equation}\boxed{\boxed{\phi(x,y)=2 - 2x}},\end{equation} which just cannot be correct!

    I have tried solving the solution directly using separation of variables with all types of attempted forms of solutions and---after pages and pages of working---have come up with nothing. I feel like there's a very simple but essential point I'm missing here.

    Thanks in advance!
     
    Last edited: Dec 2, 2015
  2. jcsd
  3. Dec 2, 2015 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You are fine until the point you claim that the homogeneous boundary conditions give ##E = - F = 0##. This is not correct.
     
  4. Dec 2, 2015 #3

    H Smith 94

    User Avatar
    Gold Member

    Hi!

    For this step, I used the boundary conditions ##\Delta\phi(0,y) = \Delta\phi(1,y) = 0##, originating from 1 and 2. Was this incorrect?
     
  5. Dec 2, 2015 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, that is correct. Your solution of the resulting equations is not.
     
  6. Dec 2, 2015 #5

    H Smith 94

    User Avatar
    Gold Member

    Right you are!

    I had that ##e^{\xi} - e^{-\xi} = 2 \cosh{\xi}##—which would imply the only way ##\chi(1)=0## is if ##E = 0## since ##\cosh{x}\not=0## for any ##x##—when instead, ##e^{\xi} - e^{-\xi} = 2 \sinh{\xi} \implies e^{\kappa x} - e^{-\kappa x} = 2 \sinh{\kappa x}##, so [tex]\chi(x) = 2E \sinh{\kappa x}[/tex] which means [tex] \chi(1) = 2E\sinh{\kappa} = 0[/tex] which is only true for either ##E = 0## (trivial) or for ##\kappa = 0##.

    hyper011.gif
    Fig 1.
    Blue line is ##\sinh{x}##; red line is ##\cosh{x}##.​

    Does this look like a good place to continue from? The implications of ##\kappa = 0## on ##\gamma(y)## also worries me.
     
  7. Dec 2, 2015 #6

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, you are not using the actual form you wrote down for the solution in the x-direction. Also remember that you must check all values of ##\kappa##.

    There is also an error in your solution, the argument of the functions is not ##\kappa x##, this would give ##\kappa^2## as a factor when differentiating twoce instead of ##\kappa##, which is in your differential equations.
     
  8. Dec 2, 2015 #7

    H Smith 94

    User Avatar
    Gold Member

    Oh yes! I hadn't even noticed that! Thank you for pointing this out.

    This was a typo on my part -- sorry! I also should have the opposite signs on the ##\kappa## for each ##x## and ##y## solution in order to adequately satisfy the solutions ##\chi## and ##\gamma## I gave.

    Thank you for all your help -- you've been of great assistance!
     
  9. Dec 2, 2015 #8

    H Smith 94

    User Avatar
    Gold Member

    Solution:
    For anyone reading this who is interested in the solution:

    Applying the new homogeneous boundary conditions in ##x##:
    1. ##\Delta\phi(0,y) = 0 \implies \boxed{F = 0};##
    2. ##\Delta\phi(1,y) = 0 \implies \boxed{\kappa = n \pi}## for ##n\in\mathbb{N}##.
    means that \begin{equation} \chi(x) = E \sin{n\pi x}.\end{equation} We can hence determine ##G## and ##H## using the original boundary conditions 3 and 4. 3 finds that ##\boxed{A = -B}##, so \begin{equation}\gamma(y) = A\sinh{n\pi y}.\end{equation} Hence, using the separable solution and summing over all possible (infinite) solutions, we find that \begin{equation}\label{eq:phisol} \boxed{\boxed{ \Delta\phi(x,y) = \lim_{M\rightarrow\infty} \sum_{n=1}^{M} c_n\,\sin{n\pi x}\, \sinh{n\pi y} }},\end{equation} where ##c_n \equiv A_nE_n## and ##M## is simply included for a computational application.

    Now, we use 4 to calculate the ##c_n## coefficient. We differentiate \ref{eq:phisol}, apply Fourier's trick (as David J. Griffiths calls it) and hence find that \begin{equation} \label{eq:cnsol} \boxed{ \boxed{ c_n = \frac{ 2 (1 - \cos{ n \pi }) }{ ( n \pi )^2 \cosh{ n \pi } } } }.\end{equation}

    Using this solutions, we find that for ##M = 20##, the potential looks like the attached graph.
     

    Attached Files:

    Last edited: Dec 2, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted