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Separation of variables and potential

  1. Sep 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A potential satisfies ##\nabla^2 Φ = 0## in the 2d slab ## -\inf < x < \inf ##, ##-b < y < b ##, with boundary conditions ## Φ(x, +b) = +V_s(x)## on the top and ##Φ(x, b) = -V_s(x)## on the bottom, where


    ##V_s (x)= -V_0 for -a<x<0##
    ##V_s (x)=+V_0 for 0<x<a##
    (a) what is ##\Phi(x,y)## in the interior?
    2. Relevant equations
    ##\nabla^2 \phi=0##
    ##\nabla^2 = \frac{\partial ^2}{\partial x^2} + \frac{\partial ^2}{\partial y^2} ##


    3. The attempt at a solution
    ##\Phi = X(x)Y(y)##

    then using separation of variables we arrive at the following:
    ##\frac{1}{X}\frac{\partial^2 X}{\partial x^2 } + \frac{1}{Y}\frac{\partial^2 Y}{\partial y^2 }=0##
    ##\frac{1}{Y}\frac{\partial^2 Y}{\partial y^2 }=-\alpha^2## which gives us ##Y(y)= Asin \alpha y +Bcos \alpha y ##
    ##\frac{1}{X}\frac{\partial^2 X}{\partial x^2 }=\alpha ^2## and solution of its is ##X(x)=Ce^{\alpha x}+De^{-\alpha x}##
    applying the periodic boundary conditions to the Y solution will give us :
    ##Y(b)=V_s =A sin \alpha b +B cos \alpha b ##
    ##Y(-b)=-V_s =-A sin \alpha b +B cos \alpha b ##
    so summing these up will give B=0

    my question is have I gotten it right till here and how I go about finding the ##\alpha## and the boundary conditions for ##X(x)##? I appreciate any ideas or hints.
     
    Last edited: Sep 19, 2015
  2. jcsd
  3. Sep 19, 2015 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Hi sayebms,

    ##Y(-b)=-V_s =-A sin \alpha b +B cos \alpha b ## does not match the boundary condition you gave in the problem statement.

    Also, if the product of 2 numbers is zero, does that guarantee the the first number must be zero? If not, how do you justify claiming ##2B cos \alpha b = 0## means ##B=0##?
     
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