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Math Help - Method of Transformations for Partial Diff. Eqns

  1. #1
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    Method of Transformations for Partial Diff. Eqns

    Hey guys,

    I have a problem I am trying to solve by method of transformation. This is the problem:

    PDE --- U(t) = Uxx

    B.C's --- U(0,t) = 0 0<t<infinity
    U(1,t) = 1

    I.C. --- U(x,0) = x^2

    Ok, so I know that I can't use method of tranformation yet, b/c in order to use this method, the B.C's have to be = 0. I have to first do the transformation to make the B.C's = 0. All goes well with the transformation until I get to the I.C. transformation. This is what I get:

    I.C. U(x,0) = x^2 - x b/c U(x,t) = f(x) - k1 (1 - x/L) - k2 (x/L)

    Thing is I don't know if this is right or not. I don't know if U(x,t) will always equal the above equation or will it always change. If so, how do I figure out the new I.C. in my example above. I am really confused on this part.

    Thanks for the help!
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  2. #2
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    Quote Originally Posted by spearfish View Post
    Hey guys,

    I have a problem I am trying to solve by method of transformation. This is the problem:

    PDE --- U(t) = Uxx

    B.C's --- U(0,t) = 0 0<t<infinity
    U(1,t) = 1

    I.C. --- U(x,0) = x^2

    Ok, so I know that I can't use method of tranformation yet, b/c in order to use this method, the B.C's have to be = 0. I have to first do the transformation to make the B.C's = 0. All goes well with the transformation until I get to the I.C. transformation. This is what I get:

    I.C. U(x,0) = x^2 - x b/c U(x,t) = f(x) - k1 (1 - x/L) - k2 (x/L)

    Thing is I don't know if this is right or not. I don't know if U(x,t) will always equal the above equation or will it always change. If so, how do I figure out the new I.C. in my example above. I am really confused on this part.

    Thanks for the help!
    If u = v + x, then the PDE becomes

    v_t = v_{xx}

    The BC's are v(0,t) = 0, v(x,t) = 0 and the IC v(x,0) = x^2 - x. Once you solve this PDE then use your tranformation to find u.
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  3. #3
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    got it! Thanks Danny!
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