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Verification of solution to Heat Equation

  1. Oct 8, 2011 #1
    1. verify that u(t,x,y)=e-λtsin(αt)cos(βt) (for arbitrary α, β and with λ=α22) satisfies the 2-D Heat Equation.



    2. ut=Δu



    3. I began with:
    Δu=uxx+uyy.
    note the equation does not contain variable "x"
    so uxx=0 i.e. Δu=uyy

    uy=e-λtsin(αt){-βsin(βt)}
    uyy=e-λtsin(αt){-β2cos(βt)}

    next I found ut
    ut=cos(βy) {e-λtαcos(αt)+sin(αt)-λe-λt}

    I have tried to reduce both equations but don't see how they are equal. I also have tried using the λ=α22 to re-write the eq. Any suggestions?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 8, 2011 #2

    LCKurtz

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    Your equation is transcribed incorrectly. Your original equation should be:

    u(t,x,y)=e-λtsin(αx)cos(βy)
     
    Last edited: Oct 8, 2011
  4. Oct 8, 2011 #3
    The original equation was posted incorrectly.

    The actual question states:
    u(t,x,y)=e-λtsin(αt)cos(βy)

    My apologies
     
  5. Oct 8, 2011 #4
    The heat equation in 2D is the Fourier equation. I think the answer they want you to check out is:

    u(x,y,t) = e**(-lambda*t) * sin(alpha*x) * cos(beta*y)

    Take the partial derivatives and plug them back into the original equation and you'll see that the equation is satisfied if lambda = alpha**2 + beta**2. The original equation is the Laplacian of u equals the partial of u with respect to time.
     
  6. Oct 8, 2011 #5
    I checked the problem statement

    The question actually states:
    u(t,x,y)=e-λtsin(αt)cos(βy)

    Could the instructor possibly have made a typo??

    When I set ut=uyy (from simplifying Δu) I was able to substitute: λ=22

    My last line reads:
    αcos(αt)+β2sin(αt)=(α22)sin(αt)

    Any thoughts?
     
  7. Oct 8, 2011 #6

    LCKurtz

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    Of course he could have. That t in the sine term should be x. Then it all works.
     
  8. Oct 8, 2011 #7
    the last substitution should read λ=α22
     
  9. Oct 8, 2011 #8
    Thx for your help.
     
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