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2nd order ODE boundary value constant input-- stuck

  1. Aug 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Uxx - SU = A ; 0<x<1
    Boundary conditions :
    Ux(0) = 0
    U(1) = 0
    3. The attempt at a solution
    I tried to set a new variable W = u + A, I can get rid of the A in the main equation and U(1) becomes = 1.
    If I set U= C*esqrt(S)x into the equation, its a trivial solution because of the boundary conditions C=0
    If I plug U= -C*cos(sqrt(s)x) in, x=nπ if sqrt(s)=1,2,3.. but x <1, so I dont think this is viable...

    How do I go about solving this analytically? Can someone point me in the right direction? Do I need to use Laplace transform or something?
    Books answer = (A/S) * (cosh(x√s)-1)/cosh(√s)
     
  2. jcsd
  3. Aug 16, 2016 #2

    Ray Vickson

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    Homework Helper

    You have made several errors.

    The DE reads as ##D_x^2 U -S(U + A/S)=0##, so ##v = U +A/S## satisfies the homogeneous equation ##D_x^2 v = Sv##. What do the boundary conditions on ##U## become for ##v##? How can you determine the constants ##a,b## so that
    $$ v = a \cosh(x \sqrt{S}) + b \sinh(x \sqrt{S})$$
    satisfies the boundary conditions?
     
  4. Aug 16, 2016 #3

    Twigg

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    Gold Member

    Your first approach (using Ray's notation), with ## v = ce^{\sqrt{s}x} ## was two or so steps away from being equivalent to Ray's. From the characteristic equation ##k^{2} - s = 0## you have two solutions, ##k = +\sqrt{s}## and ##k = -\sqrt{s}##. That means your general solution is a linear combination of both, ##v = c_{1} e^{\sqrt{s}x} + c_{2} e^{-\sqrt{s}x}##. Notice what happens when you plug that solution into the boundary condition ##v'(0) = 0##, and keep in mind that ##cosh(x) = \frac{1}{2} (e^{x} + e^{-x})##. It doesn't matter if you use the combination of exponentials I suggested or the combinations of cosh and sinh that Ray suggested. They give the same result, once you convert between the two with the identity I mentioned.
     
  5. Aug 19, 2016 #4
    Thanks, got it.
     
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