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Math Help - heat equation

  1. #1
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    heat equation

    Hi! Can someone please help me with the heat equation, my working out is below. The question is: Solve
    <br />
\frac {du}{dt} = \frac {d^2u}{dx^2}
    with the boundary conditions:
    <br />
0<=x<=L, T>=0, u(x,0) =f(x), u_x(0,t)=u_x(L,t)=0.<br />

    My working out is:
    <br />
X'' + sX = 0

    So first case is:  s>0
    when  s= 0 The general Solution of ODE for  x
    X(x) = A e^{ xs^{\frac {1}{2}}} +  Be^{ xs^{\frac {-1}{2}}}

    Then i differentiate it and got
    <br />
X'(x) = A s^{\frac {1}{2}}e^{ xs^{\frac {1}{2}}} -  Bs^{\frac {1}{2}}e^{ xs^{\frac {-1}{2}}}

    From the boundary conditions:
    X'(0) = A s^{\frac {1}{2}} -  Bs^{\frac {1}{2}}
    X'(L) = A s^{\frac {1}{2}}e^{ Ls^{\frac {1}{2}}} -  Bs^{\frac {1}{2}}e^{ Ls^{\frac {-1}{2}}}

    And this is where i get lost
    I tried solving simultaneously but I don't know how to. Can someone please help me? Thank you
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  2. #2
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    Quote Originally Posted by gconfused View Post
    Hi! Can someone please help me with the heat equation, my working out is below. The question is: Solve
    <br />
\frac {du}{dt} = \frac {d^2u}{dx^2}
    with the boundary conditions:
    <br />
0<=x<=L, T>=0, u(x,0) =f(x), u_x(0,t)=u_x(L,t)=0.<br />

    My working out is:
    <br />
X'' + sX = 0

    So first case is:  s>0
    when  s= 0 The general Solution of ODE for  x
    X(x) = A e^{ xs^{\frac {1}{2}}} + Be^{ xs^{\frac {-1}{2}}}

    Then i differentiate it and got
    <br />
X'(x) = A s^{\frac {1}{2}}e^{ xs^{\frac {1}{2}}} - Bs^{\frac {1}{2}}e^{ xs^{\frac {-1}{2}}}

    From the boundary conditions:
    X'(0) = A s^{\frac {1}{2}} - Bs^{\frac {1}{2}}
    X'(L) = A s^{\frac {1}{2}}e^{ Ls^{\frac {1}{2}}} - Bs^{\frac {1}{2}}e^{ Ls^{\frac {-1}{2}}}

    And this is where i get lost
    I tried solving simultaneously but I don't know how to. Can someone please help me? Thank you
    It's actually better to consider s = - \omega^2, 0 \;\text{and}\; \omega^2

    The first case is  X'' - \omega^2 X = 0 whose solution is X = A e^{\omega x} + Be^{- \omega x} as you said. Sub in the BC gives

    X'(0) = A\omega - B\omega = 0
     <br />
X'(L) = A \omega e^{\omega L} - B \omega e^{-\omega L}<br />

    The only solution to this pair is A = B = 0 which leads to the trivial solution X = 0 so u = 0. So on to the next case s = 0 - you're up.
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  3. #3
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    Thanks !!

    Are the 3 conditions the same thing that you would do for the wave equation?
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  4. #4
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    Quote Originally Posted by gconfused View Post
    Thanks !!

    Are the 3 conditions the same thing that you would do for the wave equation?
    Depends on the boundary conditions but typically yes, there are.
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