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Math Help - Heat equation? how to solve?

  1. #1
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    Heat equation? how to solve?

    Hello
    I have one problem with solve heat equation with integral Fourier,
    This is problem
    du/dt-d(2)u/dx^2= 2t when 0<x< \pi , 1 when x>0
    u(x,0)=2x when o<x<1 , -1 when x>1
    u(0,t)=t-1 when t \geqslant 0

    I write u(x,t)=F(x)G(t) and try to find answer!
    but I can not find a(w) and b(w) and also G(t)
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  2. #2
    Member Ruun's Avatar
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    You should starting by taking the Fourier transform of u(x,t) with respecto to x, this is \mathcal{F}(u)(\xi,t)=U(\xi,t) and apply the properties of the Fourier transform of the derivative.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Hamed View Post
    Hello
    I have one problem with solve heat equation with integral Fourier,
    This is problem
    du/dt-d(2)u/dx^2= 2t when 0<x< \pi , 1 when x>0
    u(x,0)=2x when o<x<1 , -1 when x>1
    u(0,t)=t-1 when t \geqslant 0

    I write u(x,t)=F(x)G(t) and try to find answer!
    but I can not find a(w) and b(w) and also G(t)
    I cannot understand what it is you are trying to solve. It is very hard to read.

    Do you mean

    \frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2 }=g(x,t)

    Where

    g(x,t)=\begin{cases} 2t, \text{ if } 0 \le x \le \pi \\ 1, \text{ if } x > \pi \end{cases}


    u(x,0)=\begin{cases} 2x, \text{ if } 0 \le x \le 1 \\ -1, \text{ if } x > 1\end{cases}

    u(0,t)=t-1

    So I am assuming that the function is not defined for x < 0. If that is the cases you should use the Fourier sine transform since you do not know the value of the normal derivative at x=0. Also note that the transform of the right hand side will be a distribution as the function is not in L^1
    Last edited by TheEmptySet; June 3rd 2011 at 06:23 AM. Reason: grammer
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  4. #4
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    Thanks,
    But I want to use integral of Fourier to solve this problem, I can solve it when
    du/dt-d(2)u/dx^2=0
    but now I don't know what should I do!!!
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  5. #5
    Member Ruun's Avatar
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    The procedure is similar. You can divide the problem in two parts, one for each domain of x.

    When the problem is homogeneous yo take the Fourier transform of the equation u(x,t), wich means just (using TheEmptySet notation for the right hand side)

    \mathcal{F}[ Heat Equation ](\xi,t)=\mathcal{F}\left[\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2 }\right](\xi,t)=\mathcal{F}[g](\xi,t)

    and eventually you arrive at a First Order Ordinary Differential Equation for U(\xi,t) treating \xi as constant. In the Homogeneus problem \mathcal{F}[g](\xi,t)=0. In your problem this will not be true.

    You solve both problems, one for each domain of x and finally paste both solutions by continuity.

    The point of using Foutier Transform is that your Heat Problem became an EDO, way easier to solve.
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