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Math Help - IBVP with a nonzero BCs

  1. #1
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    IBVP with a nonzero BCs

    I must solve this initial boundary value problem by transforming it into homogeneous BCs.

    \[\\(PDE)\; u_{t}=\alpha ^{2}u_{xx} \: \: \: \: \: \: \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \;   0<x<1\\<br />
\\<br />

    (BCs)\; \left\{\begin{matrix}<br />
u(0,t)=1) & 0<t< \infty \\<br />
u_{x}(1,t)+hu(1,t)=1 & 0<t< \infty<br />
\end{matrix}\right.\\

    (IC)\;\: \:  u(x,0)=sin(\pi x)+x\; \: \: \: \: \: \: \: \: \: \: 0\leq x\leq 1

    ---> My work

    I set \[s(x,t)=a(t)x+b(t)(1-x)+U(x,t)\]<br />
where S(x,t) is the steady state and U(x,t) is the transient

    So i substitute s(x,t) into the BCs
    \[s(0,t)=1<br />
    s_{x}(1,t)+hs(1,t)=1

    These two equations can give us b(t)=1 and a(t)=2/1-h ?
    and therefore u(x,t)=2x/(1-h) + (1-x) + U(x,t) = 1 - (1+h/1-h)x + U(x,t)



    MY PROBLEM is, how did they get b(t)=1 and a(t) = 2/1-h.
    I know i have have to make s(0,t) = to something like a(t) + b(t) = 1 and s_x(1,t) = ?something? so that i can plug it into the \[<br />
s_{x}(1,t)+hs(1,t)=1\]<br />
but I am not sure what s and s_x should be.
    Can someone please tell me what they are so I can finally get a(t) and b(t).
    Then I can finally finish solving this problem.

    Thanks in advance!
    Last edited by wilday86; September 25th 2010 at 07:24 PM.
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  2. #2
    MHF Contributor
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    The problem is the BC's i.e. =1 part. If

    u(x,t) = a(t)x + b(t)(1-x) + U(x,t)

    then u(0,x) = b(t) + U(0,t) = 1 so choosing b(t) = 1 gives U(0,t) = 0

    Further u_x(1,t) + hu(1,t) = a(t)-b(t) +U_x(1,t) + h(a(t) + U(1,t)) = 1
    and choosing  a(t) - b(t) + h a(t) = 1 gives U_x(1,t) + hU(1,t) = 0.

    With b(t) = 1 solving a(t) - b(t) + h a(t) = 1 for a(t) gives what you want.
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