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Thread: 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.

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

    $\displaystyle t>0; 0<x<1$

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

    $\displaystyle v''(x) = -\frac{1}{\beta}P(x)$ 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. $\displaystyle t \to \infty$) then solve

    $\displaystyle u_{xx} + 1 = 0$ 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

    $\displaystyle
    s_{xx}=-1
    $

    $\displaystyle
    s(0)=1 \; and \; s_x(1)=0.
    $

    Then we find the solution

    $\displaystyle
    u(x,t)=s(x)+v(x,t)
    $

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

    $\displaystyle r = \pm i$

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

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

    $\displaystyle s(0) = 1 ; s'(1)=0$

    so

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

    so

    $\displaystyle c_{1} = 1$

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

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

    $\displaystyle \tan{1} = c_{2}$

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

    $\displaystyle v_{t} = \beta v_{xx}$

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

    $\displaystyle v(x,0) = 1- s(x)$

    Separation of variables?

    $\displaystyle v(x,t) = X(x)T(t)$

    $\displaystyle X''(x) - KX(x) = 0$

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

    Case 1 -- K=0

    $\displaystyle X(x)= C_{1}x+C_{2}$

    using boundary condtions I got

    $\displaystyle 0 = C_{1} = C_{2}$

    so there exist no nontrivial solution

    Case 2 --- K > 0

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

    using boundary conditions also produces nontrivial solutions.

    $\displaystyle 0 = C_{1} = C_{2}$

    Case 3 ---- K < 0

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

    and

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

    and this is where I run into problems again

    $\displaystyle 0 = C_1$

    which just leaves me with

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

    so my question is if

    $\displaystyle C_2 = 0$

    produces a nontirvial answer

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

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

    my question is what does

    $\displaystyle \sqrt{-K} = 0$

    have?

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

    And am I approaching the problem correctly so far? Thanks in advance
    Last edited by 11rdc11; Nov 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

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

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

    which then gives me

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

    $\displaystyle r = \pm i$

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

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

    $\displaystyle s(0) = 1 ; s'(1)=0$

    so

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

    so

    $\displaystyle c_{1} = 1$

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

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

    $\displaystyle \tan{1} = c_{2}$

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

    So the correct

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

    so

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

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

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

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

    but instead

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

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

    $\displaystyle u_t = u_{xx} + 1$

    $\displaystyle u(0,t) = 1, u_x(1,t) = 0, u(x,0) = 1$.

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

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

    $\displaystyle v_t = v_{xx} + 1$

    $\displaystyle v(0,t) = 0, v_x(1,t) = 0, v(x,0) = ?$

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

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

    Hence we look for a solution of the form

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

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

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

    for the source term $\displaystyle Q = 1$

    PS. Yes on

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