Results 1 to 4 of 4

Math Help - heat equation

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    58

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,325
    Thanks
    9
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2009
    Posts
    58
    Thanks !!

    Are the 3 conditions the same thing that you would do for the wave equation?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,325
    Thanks
    9
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2D heat equation
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: April 17th 2010, 10:41 AM
  2. Heat problem with a heat equation
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: November 25th 2009, 09:40 AM
  3. the heat equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 10th 2009, 04:41 AM
  4. heat equation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: April 18th 2009, 03:50 AM
  5. Heat equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 6th 2008, 04:47 AM

Search Tags


/mathhelpforum @mathhelpforum