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Math Help - help PDE's!

  1. #1
    degoman
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    help PDE's!

    I'm not so sure how to type my problems out legibly on here so here's a document with the problems i need help with:
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  2. #2
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    1) \mbox{div}(\mbox{grad} u) = \nabla^2 u where u=u(x,y,z)

    By definition,
    \mbox{grad} u = \frac{\partial u}{\partial x}\bold{i} + \frac{\partial u}{\partial y}\bold{j} + \frac{\partial u}{\partial z}\bold{k}

    By definition,
    \mbox{div} \left( \frac{\partial u}{\partial x}\bold{i} + \frac{\partial u}{\partial y}\bold{j} + \frac{\partial u}{\partial z}\bold{k} \right) = \frac{\partial^2 u}{\partial x^2}+\frac{\partial u^2}{\partial y^2}+\frac{\partial u^2}{\partial z^2}
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  3. #3
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    u(x,t) = e^{at}\sin bx

    Then,
    u_t = ae^{at}\sin bx

    u_x = be^{at}\cos bx
    u_{xx} = -b^2 e^{at}\sin bx

    Thus,
    u_t - u_xx = ae^{at}\sin bx +b^2 e^{at}\sin bx = (a+b^2)e^{at}\sin bx = 0

    Make each factor equal to zero:
    1) \ \ \ a+b^2 = 0 \implies a=-b^2
    2) \ \ \ e^{at} = 0 \mbox{ cannot be since }e^{at}>0
    3) \ \ \ \sin bx = 0 \implies b = 2\pi n \mbox{ for }n\in \mathbb{Z}
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  4. #4
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    4)I think you made a mistake. That is not the solution to:
    \nabla^2 u = \frac{1}{k^2}u_t

    5)The boring what is to substitute that into the PDE and verify that it solves it, i.e. adds up to zero. A more exciting way is to use the Cauchy-Riemann equations. The prinipal logarithm \ln z = \ln |z| + i\arg(z) \mbox{ with }-\pi<\arg z \leq \pi is holomorphic on \mathbb{C} - (-\infty,0]. That means \mbox{Re} \ln z \mbox{ and }\mbox{Im} \ln z are harmonic functions.
    We can write,
    \ln |z| + i \arg(z) = \ln \sqrt{x^2+y^2} + i\left( \tan^{-1} \frac{y}{x} +k \right)
    That means the real part is \ln \sqrt{x^2+y^2} solves Laplace's equation.
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  5. #5
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    3)I am really not sure about this one, but I think I know what I am doing. First, in your problem you have the Dirichlet problem in 3 variables, is that a mistake? I will do it in two variables.

    \frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2} = \lambda u

    Seperate variables and write u(x,y) = X(x)Y(y).

    Then by substituting this into the PDE we have:
    X''Y + XY'' = \lambda XY

    Divide through by XY to get:

    \frac{X''}{X} + \frac{Y''}{Y} = \lambda \implies \frac{X''}{X} = \lambda - \frac{Y''}{Y}

    And hence by the seperation of variables techinque we have a system of ODE:

    1) \ \ \ \frac{X''}{X} = - k \implies X'' + kX = 0
    2) \ \ \ \lambda - \frac{Y''}{Y} = - k \implies Y'' -(k+\lambda) Y = 0

    Look at the first equation together with the boundary value problem u(x,y) =0 \mbox{ on }\partial \Omega implies that X(0) = 0 \mbox{ and }X(L)=0 for L>0 means non-trivial solutions can only exist iff k>0, that is the solutions are sines and cosines.

    Look at the second equation by similar reasoning -(k+\lambda) > 0 \implies k + \lambda < 0. So if k>0 (by the first equation) it must mean that \lambda < 0 for otherwise k+\lambda >0 and that will lead to trivial solutions to the two point boundary value problem.
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