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Math Help - Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)

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    Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)

    Solution for PDE

    u/t=k ∂/∂x (∂u/∂x)

    I.C.
    → u(x,0)=12 cos(9πx⁄L)-7 sin(4πx⁄L)
    B.C.→
    u(-L,t)=u(L,t) & ∂u/∂x (-L,t)=∂u/∂x(L,t)




    Last edited by tykim; August 6th 2012 at 06:21 AM. Reason: type error
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    Re: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)

    Quote Originally Posted by tykim View Post
    Solution for PDE

    u/t=k ∂/∂x (∂u/∂x)

    I.C.
    → u(x,0)=12 sin(9πx⁄L)-7 sin(4πx⁄L)
    B.C.→
    u(-L,t)=u(L,t) & ∂u/∂x (-L,t)=∂u/∂x(L,t)




    Just so we're clear, is this your DE? \displaystyle \begin{align*} \frac{\partial u}{\partial t} = k\,\frac{\partial ^2 u}{\partial x^2} \end{align*}

    Also in your boundary conditions, are you writing the function \displaystyle \begin{align*} u \end{align*} and its derivatives as \displaystyle \begin{align*}  u(t, x) \end{align*}?
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    Re: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)

    Dear Prove IT

    Correct.

    Function u and its derivatives as u(x, t)
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    Re: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)

    I expect you can solve this with separation of variables.

    Assume you can write \displaystyle \begin{align*} u = T(t)\,X(x) \end{align*}, then we have \displaystyle \begin{align*} \frac{\partial u}{\partial t} &= \frac{dT}{dt}\,X \end{align*} and \displaystyle \begin{align*} \frac{\partial^2 u}{\partial x^2} = T\,\frac{d^2X}{dx^2} \end{align*}.

    Substituting into the DE gives

    \displaystyle \begin{align*} \frac{\partial u}{\partial t} &= k\,\frac{\partial ^2 u}{\partial x^2} \\ \frac{dT}{dt}\,X &= k\,T\,\frac{d^2X}{dx^2} \\ \frac{1}{k\,T}\,\frac{dT}{dt} &= \frac{1}{X}\,\frac{d^2X}{dx^2}\end{align*}

    Now notice that the LHS is only a function of t, and the RHS is only a function of x, so for them to be equal, they must be equal to the same constant value \displaystyle \begin{align*} -\lambda \end{align*}.

    \displaystyle \begin{align*} \frac{1}{k\,T}\,\frac{dT}{dt} &= -\lambda \\ \int{\frac{1}{k\,T}\,\frac{dT}{dt}\,dt} &= \int{-\lambda\,dt} \\ \frac{1}{k}\int{\frac{1}{T}\,dT} &= -\lambda \,t + C_1 \\ \frac{1}{k}\ln{|T|} + C_2 &= -\lambda\,t + C_1 \\ \frac{1}{k}\ln{|T|} &= -\lambda \, t + C_1 - C_2 \\ \ln{|T|} &= -k\,\lambda\,t + k\left(C_1 - C_2\right) \\ |T| &= e^{-k\,\lambda\,t + k\left(C_1 - C_2\right)} \\ |T| &= e^{k\left(C_1 - C_2\right)}e^{-k\,\lambda\,t} \\ T &= A\,e^{-k\,\lambda \, t} \textrm{ where } A = \pm e^{k\left(C_1 - C_2\right)} \end{align*}

    We also have

    \displaystyle \begin{align*} \frac{1}{X}\,\frac{d^2X}{dx^2} &= -\lambda \\ \frac{d^2X}{dx^2} &= -\lambda\,X \\ \frac{d^2X}{dx^2} + \lambda\,X &= 0 \\ \\ m^2 + \lambda &= 0 \\ m^2 &= -\lambda \\ m &= \pm i\sqrt{\lambda} \\ \\ X &= B\sin{\left(\sqrt{\lambda}\,x\right)} + C\cos{\left(\sqrt{\lambda}\,x\right)} \end{align*}

    Now see what you can do with the Boundary and Initial conditions to evaluate \displaystyle \begin{align*} \lambda \end{align*}.

    This might help you further...
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    Re: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)

    Dear Prove IT,

    Thanks for your kind guidance.

    There may be type error, and I corrected as below. Sorry for that.

    I.C.→ u(x,0)=12 cos(9πx⁄L)-7 sin(4πx⁄L)

    Could you let me know how to get root-square of (lambda)


    Thanks
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