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Math Help - Pde

  1. #1
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    Pde

    Hi, just working on some problems but I'm getting stuck.

    1)In trying to express the Laplacian in parabolic cylindrical coordinates, I can't get u = a function of x and y or v = a function of x and y...so I'm stuck on how to keep on going. (Here the parabolic cylindrical coordinates are given as (u,v,z) and rectangular coordinates as (x,y,z)

    2)Find a transformation of the form: u(x,t) = v(x,t)e^(qx-wt) where a and w have to be determined, that would bring or transform the following equation for u(x,t): u_t-au_xx-bu_x-cu = 0 where a b and c are non zero constants to a diffusion partial differential equation.

    Thanks!
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  2. #2
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    Quote Originally Posted by angels_symphony View Post
    Hi, just working on some problems but I'm getting stuck.

    1)In trying to express the Laplacian in parabolic cylindrical coordinates, I can't get u = a function of x and y or v = a function of x and y...so I'm stuck on how to keep on going. (Here the parabolic cylindrical coordinates are given as (u,v,z) and rectangular coordinates as (x,y,z)

    Mr F says: Try reading this and see how things go .....

    2)Find a transformation of the form: u(x,t) = v(x,t)e^(qx-wt) where q and w have to be determined, that would bring or transform the following equation for u(x,t): u_t-au_xx-bu_x-cu = 0 where a b and c are non zero constants to a diffusion partial differential equation. Mr F edit in red.

    Thanks!
    2) Well, your not asking for much here!

    First note that the diffusion p.d.e. is \frac{\partial \phi}{\partial t} = k \frac{\partial^2 \phi}{\partial x^2} where k is a constant (the thermal diffusivity).


    Now note that if \displaystyle u(x, t) = v(x, t) e^{qx - wt} then (using the product rule):


    1. \displaystyle \frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} e^{qx - wt} - w v e^{qx - wt} = \left( \frac{\partial v}{\partial t} - wv \right) e^{qx - wt}.


    2. \displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial x} e^{qx - wt} + q v e^{qx - wt} = \left( \frac{\partial v}{\partial x} + q v \right) e^{qx - wt}.


    3. \displaystyle \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left( \frac{\partial u}{\partial x} \right) = \left( \frac{\partial^2 v}{\partial x^2} + q \frac{\partial v}{\partial x} \right) e^{qx - wt} + q \left( \frac{\partial v}{\partial x} + q v \right) e^{qx - wt}


    \displaystyle = \left( \frac{\partial^2 v}{\partial x^2} + 2q \frac{\partial v}{\partial x} + q^2 v\right) e^{qx - wt}.


    Substitute all that lot into \displaystyle \frac{\partial u}{\partial t} - a \frac{\partial^2 u}{\partial x^2} - b \frac{\partial u}{\partial x} - cu = 0 and divide through by the common (non-zero) factor of \displaystyle e^{qx - wt}:


    \displaystyle \left( \frac{\partial v}{\partial t} - wv \right) - a \left( \frac{\partial^2 v}{\partial x^2} + 2q \frac{\partial v}{\partial x} + q^2 v\right) - b \left( \frac{\partial v}{\partial x} + q v \right) - cv = 0


    \displaystyle \Rightarrow \frac{\partial v}{\partial t} = a \frac{\partial^2 v}{\partial x^2} + \frac{\partial v}{\partial x} (2aq + b) + v(c + aq^2 + bq + w).


    Comparing this with the diffusion equation, you clearly require the coefficients of \displaystyle \frac{\partial v}{\partial x} and v to equal zero:


    2aq + b = 0 \Rightarrow q = -\frac{b}{2a} .... (1)


    c + aq^2 + bq + w = 0 .... (2)


    Substitute (1) into (2) and solve for w:

    \displaystyle w = -c - aq^2 - bq = -c + \frac{b^2}{4a} = \frac{b^2 - 4ac}{4a}.


    There are probably numerous typos in the above - work through it carefully, double check everything .....
    Last edited by mr fantastic; February 6th 2008 at 03:44 AM. Reason: 'numerous' can now be replaced 'some'.
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  3. #3
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    Quote Originally Posted by angels_symphony View Post
    1)In trying to express the Laplacian in parabolic cylindrical coordinates, I can't get u = a function of x and y or v = a function of x and y...so I'm stuck on how to keep on going. (Here the parabolic cylindrical coordinates are given as (u,v,z) and rectangular coordinates as (x,y,z)
    x(r,\theta)=r\cos\theta \mbox{ and }y(r,\theta) = r\sin \theta

    Then, (confirm this by chain rule),
    \frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = \frac{\partial ^ 2 u}{\partial r^2} + \frac{1}{r} \frac{\partial  u}{ \partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}
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