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Inhomogeneous Dirichlet problem in a rectangle

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the problem
    \begin{cases}
    u_{xx} + u_{yy} = y, \; \; 0 < x < 2, \; \; 0 < y < 1\\
    u(x,0)=0, \; \; u(x,1)=0\\
    u(0,y)=y-y^3, \; \; y(2,y)=0.
    \end{cases}
    2. Relevant equations
    N/A

    3. The attempt at a solution
    I start by trying to find a steady-state solution ##u_0(y)## (or I guess a solution independent of ##x## since we don't have a time variable). We have
    ##u_0''=y \Longrightarrow y = \frac{y^3}{6}+ay+b##. The boundary values ##u(x,0)=0, \; \; u(x,1)=0## give us that ##u_0 = \frac{y^3-y}{6}##.

    Setting ##u = v+u_0## we end up with the new homogeneous problem
    \begin{cases}v_{xx}+v_{yy}=0\\
    v(x,0) = 0, \; \; v(x,1)=0\\
    v(0,y) = \frac{7}{6}(y-y^3), \; \; v(2,y)=\frac{y-y^3}{6}\end{cases}
    Separation of variables give us
    ##X(x) = c_1\cosh \lambda x + c_2 \sinh \lambda x## and
    ##Y(y) = c_3\cos \lambda y + c_3\sin \lambda y##.
    The boundary conditions ##v(x,0) = 0, \; \; v(x,1)=0## give us that ##c_3=0## and ##\lambda = n\pi##. We then have solutions of the form
    ##v(x,y) = \sum_1^\infty \left( a_n\cosh n\pi x + b_n \sinh n\pi x \right) \sin n\pi y##.
    Imposing the boundary condition ##v(0,y) = \frac{7}{6}(y-y^3)## we have that
    ##\frac{7}{6}(y-y^3) = \sum_1^\infty a_n \sin n\pi y##. Calculating the Fourier coefficients we have that
    ##a_n = \frac{7}{3}\int_0^1 (y-y^3)\sin n\pi y dy = \frac{14 (-1)^{n-1}}{\pi^3 n^3}##.
    So we have
    ##v(x,y) = \sum_1^\infty \left( \frac{14 (-1)^{n-1}}{\pi^3 n^3}\cosh n\pi x + b_n \sinh n\pi x \right) \sin n\pi y##. Imposing the other boundary condition ##v(2,y)=\frac{y-y^3}{6}## we get
    ##\frac{y-y^3}{6} =\sum_1^\infty \left( \frac{14 (-1)^{n-1}}{\pi^3 n^3}\cosh n\pi 2 + b_n \sinh n\pi 2 \right) \sin n\pi y## at which point I'm stuck not knowing how to proceed.

    The answer should be
    ##u(x,y) = \frac{1}{6}(y^3-y)+\frac{2}{\pi^3}\sum_1^\infty \frac{(-1)^{n-1}}{n^3\sinh 2n \pi} (\sinh n\pi x + 7 \sinh n\pi(2-x) )\sin n\pi y##.
    I guess they choose the eigenfunction basis to the Sturm-Liouville problem as
    ##\sinh n\pi x## and ##\sinh n\pi (2-x)## instead which work just as well. Is there a way I can get the answer in this form or do I have to redo every calculation with these eigenfunctions?
     
  2. jcsd
  3. Oct 18, 2015 #2

    pasmith

    User Avatar
    Homework Helper

    Well, you know that [tex]
    \frac{14(-1)^{n-1}}{n^3 \pi^3}\cosh 2n\pi + b_n \sinh 2n\pi[/tex] must be equal to the [itex]\sin n\pi y[/itex] coefficient in the fourier series for [itex](y - y^3)/6[/itex] - which you know, because you've just calculated the fourier series of [itex]7(y - y^3)/6[/itex]. Then, if you wish, use the hyperbolic sine addition rule [tex]\sinh (a + b) = \sinh a \cosh b + \cosh a \sinh b[/tex] to express your linear combination of hyperbolic sines and cosines as a linear combination of [itex]\sinh n\pi x[/itex] and [itex]\sinh n\pi(2-x)[/itex].

    But the natural choice of eigenfunctions here is indeed [itex]\sinh n\pi x \sin n\pi y[/itex] and [itex]\sinh n\pi (2 - x) \sin n\pi y[/itex]: The boundary condition on [itex]x = 2[/itex] is just [itex]\frac17[/itex] times that on [itex]x = 0[/itex]. Thus if [itex]X_{n,1}[/itex] and [itex]X_{n,2}[/itex] are linear combinations of [itex]e^{n\pi x}[/itex] and [itex]e^{-n\pi x}[/itex] subject to [itex]X_{n,1}(0) = 1[/itex], [itex]X_{n,1}(2) = 0[/itex] and [itex]X_{n,2}(0) = 0[/itex], [itex]X_{n,2}(1) = 1[/itex] then [tex]
    v(x,y) = \sum_{n=1}^\infty a_n (X_{n,1}(x) + \tfrac17 X_{n,2}(x)) \sin n\pi y[/tex] where [itex]a_n[/itex] is obtained from the fourier expansion on [itex]x = 0[/itex]. And on investigation we find [tex]
    X_{n,1}(x) = \frac{\sinh n\pi (2 - x)}{\sinh 2n\pi}, \\
    X_{n,2}(x) = \frac{\sinh n\pi x}{\sinh 2n\pi}. [/tex]
     
  4. Oct 18, 2015 #3
    The alternative eigenfunctions really made it a lot easier. How would I go about finding them in the first place? or are they just something you get from experience where someone just happened upon them? Our book does indeed suggest them for a similar example problem but doesn't explain how they got them in the first place.

    As for transforming the the ##\cosh ## and ##\sinh## I think I got it right
    ##\frac{2(-1)^{n-1}}{\pi^3 n^3} = \frac{14 (-1)^{n-1}}{n^3\pi^3}\cosh 2n\pi + b_n \sinh 2n\pi##
    Rearranging we have
    ##b_n= \frac{2(-1)^{n-1}(1-7\cosh 2n\pi)}{\pi^3 n^3 \sinh 2n\pi}##. Our expansion is then
    ##v(x,y) = \frac{2}{\pi^3}\sum_1^\infty \frac{(-1)^{n-1}\sin n\pi y}{n^3 \sinh 2n\pi} \left(7 \sinh 2n\pi \cosh n\pi x + (1-7\cosh 2n\pi)\sinh n\pi x \right)##
    So I want to show that ##(\sinh 2n\pi \cosh n\pi x-\cosh 2n\pi \sinh n\pi x) = \sinh n\pi (2-x)## which follow immediately from your addition rule.
     
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