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Math Help - Help with Heat Equation with nonhomogeneous boundary conditions

  1. #1
    Super Member 11rdc11's Avatar
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    Help with Heat Equation with nonhomogeneous boundary conditions

    So I'm not sure how to start on this problem.

    u_{t} = u_{xx} +1
     u(0,t) = 1 ; u_{x}(1,t) = 0
    u(x,0)=1

    t>0; 0<x<1

    Where im having my difficulty is finding the steady state solution.

    v''(x) = -\frac{1}{\beta}P(x) with P(x) =1

    Where do I go from here?
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  2. #2
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    If you want steay state (i.e. t \to \infty) then solve

    u_{xx} + 1 = 0 subject to your BC's.
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  3. #3
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    The steady state solution depends only on x s(x).
    Inserting we get

    <br />
s_{xx}=-1<br />

    <br />
s(0)=1 \; and \; s_x(1)=0.<br />

    Then we find the solution

    <br />
u(x,t)=s(x)+v(x,t)<br />

    defining initial and boundary conditions to v(x,t).
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  4. #4
    Super Member 11rdc11's Avatar
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    r^2=-1

    r = \pm i

    s(x) = c_{1}\cos{x} +c_{2}\sin{x}

    s'(x) = -c_{1}\sin{x} +c_{2}\cos{x}

    s(0) = 1 ; s'(1)=0

    so

    1 = c_{1}\cos{0} +c_{2}\sin{0}

    so

    c_{1} = 1

    0 = -\sin{1} +c_{2}\cos{1}

    \sin{1} = c_{2}\cos{1}

    \tan{1} = c_{2}

    s(x) = \cos(x) +\tan(1)\sin{x}
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  5. #5
    Super Member 11rdc11's Avatar
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    What would be my next step?

    v_{t} = \beta v_{xx}

    v(0,t) = 0; v_{x}(1,t) = 0

    v(x,0) = 1- s(x)

    Separation of variables?

    v(x,t) = X(x)T(t)

    X''(x) - KX(x) = 0

    T'(t) - \beta KT(t) = 0

    Case 1 -- K=0

    X(x)= C_{1}x+C_{2}

    using boundary condtions I got

    0 = C_{1} = C_{2}

    so there exist no nontrivial solution

    Case 2 --- K > 0

    X(x) = C_{1}e^{\sqrt{K}x} +C_{2}e^{-\sqrt{K}x}

    using boundary conditions also produces nontrivial solutions.

    0 = C_{1} = C_{2}

    Case 3 ---- K < 0

    X(x) = C_{1}\cos{\sqrt{-K}x} + C_{2}\sin{\sqrt{-K}x}

    and

    X'(x) = -C_{1}\sqrt{-K}\sin{\sqrt{-K}x} + C_{2}\sqrt{-K}\cos{\sqrt{-K}x}

    and this is where I run into problems again

    0 = C_1

    which just leaves me with

     0 = C_{2}\sqrt{-K}\cos{\sqrt{-K}}

    so my question is if

    C_2 = 0

    produces a nontirvial answer

    \cos{\sqrt{-K}} = \cos{\frac{(2n+1)\pi}{2}} for n = 0,1,2,3,...

    K = - \frac{(2n+1)^2 {\pi}^2}{4}

    my question is what does

    \sqrt{-K} = 0

    have?

    Can I just consider that a nontrivial solution with C_2 = 0?

    And am I approaching the problem correctly so far? Thanks in advance
    Last edited by 11rdc11; November 13th 2010 at 10:53 PM.
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  6. #6
    Super Member 11rdc11's Avatar
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    My last step to finishing if everything is correct is to

    v(x,t) = \sum_{n = 0}^{\infty} c_n e^{(- \frac{(2n+1)^2 {\pi}^2}{4})t}\sin{\frac{(2n+1)\pi x}{2}}

    1 - (\cos(x) +\tan(1)\sin{x}) = c_n \sin{\frac{(2n+1)\pi x}{2}}

    which then gives me

    u(x,t) = s(x) + v(x,t)
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  7. #7
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    Quote Originally Posted by 11rdc11 View Post
    r^2=-1

    r = \pm i

    s(x) = c_{1}\cos{x} +c_{2}\sin{x}

    s'(x) = -c_{1}\sin{x} +c_{2}\cos{x}

    s(0) = 1 ; s'(1)=0

    so

    1 = c_{1}\cos{0} +c_{2}\sin{0}

    so

    c_{1} = 1

    0 = -\sin{1} +c_{2}\cos{1}

    \sin{1} = c_{2}\cos{1}

    \tan{1} = c_{2}

    s(x) = \cos(x) +\tan(1)\sin{x}
    Look at the ODE again. What you solved was s_{xx} + s = 0.
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  8. #8
    Super Member 11rdc11's Avatar
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    Thanks Danny,

    So the correct

    s(x) = 1 + x - \frac{x^2}{2}

    so

    1-(1 + x - \frac{x^2}{2}) =  c_n \sin{\frac{(2n+1)\pi x}{2}}

    Now im not sure how to finish up. Don't know how to attempt a fourier series with the term \sin{\frac{(2n+1)\pi x}{2}}

    Do I just do the same as when I'm dealing with with \sin{\frac{n\pi x}{L}}?

    c_n = \frac{2}{L} \int_{0}^{L}f(x)\sin{\frac{n\pi x}{L}}

    but instead

    c_n = \frac{2}{L} \int_{0}^{L}f(x)\sin{\frac{(2n+1)\pi x}{2L}}

    Once again, thanks in advance
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  9. #9
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    Your problem is

    u_t = u_{xx} + 1

    u(0,t) = 1, u_x(1,t) = 0, u(x,0) = 1.

    The first boundary condition is a problem so we'll change this problem into a new (do-able) problem.

    Let u = v + ax + b and choose a and b such that the new problem is

    v_t = v_{xx} + 1

    v(0,t) = 0, v_x(1,t) = 0, v(x,0) = ?

    Without the source term ( Q = 1), separation of a variables would lead to what you have

    \displaystyle v(x,t) = \sum_{n = 0}^{\infty} c_n e^{(- \frac{(2n+1)^2 {\pi}^2}{4})t} \; \sin{\frac{(2n+1)\pi x}{2}}

    Hence we look for a solution of the form

    \displaystyle v(x,t) = \sum_{n = 0}^{\infty} T_n(t) \sin {\frac{(2n+1)\pi x}{2}}

    noting that we'll need a Fourier series of the form

    \sum_{n = 0}^{\infty} q_n \sin{\frac{(2n+1)\pi x}{2}}

    for the source term Q = 1

    PS. Yes on

    c_n = \frac{2}{L} \int_{0}^{L}f(x)\sin{\frac{(2n+1)\pi x}{2L}}
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