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Thread: Linear PDEq in two variables

  1. #1
    Forum Admin topsquark's Avatar
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    Linear PDEq in two variables

    Here's the equation:
    \frac{\partial v}{\partial t} - \frac{\partial v}{\partial z} = 0

    Boundary conditions: Not much. v has to be bounded for all values of t, z.

    There's a simple way to approach this...I just can't think of it. And yes, I could look up solution by characteristic curves but I'm convinced there is an easier way.

    As it happens, I like the solution v = constant, but I need to see if there are any others.

    Thanks for the help!

    -Dan
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    MHF Contributor

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    Re: Linear PDEq in two variables

    This is the same as \frac{\partial v}{\partial t}= \frac{\partial v}{\partial z}.

    One method: which is really based on the "characteristics" which are z+ t= constant and z- t= constant. Let x= z+ t and y= z- t. Then, adding the two equations, x+ y= 2z so z= \frac{x+ y}{2}. Subtracting the two equations, x- y= 2t so t= \frac{x- y}{2}. \frac{\partial v}{\partial t}= \frac{\partial v}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial v}{\partial y}\frac{\partial y}{\partial t}= \frac{1}{2}\frac{\partial v}{\partial x}- \frac{1}{2}\frac{\partial v}{\partial y}.
    \frac{\partial v}{\partial z}= \frac{\partial v}{\partial x}\frac{\partial x}{\partial z}+ \frac{\partial v}{\partial y}\frac{\partial y}{\partial z}= \frac{1}{2}\frac{\partial v}{\partial x}+ \frac{1}{2}\frac{\partial v}{\partial y}.

    So the differential equation becomes \frac{\partial v}{\partial y}= 0. The integral, with respect to y is a costant. But, since taking the partial derivative with respect to y treats x as a constant, that "constant" might be any function of x. That is, v(x, y)= F(x) where F can be any differentiable function of x. Since x= t+ z, We have v(t, z)= F(t+ z). That is, for F any differentiable function of a single variable, v(x, t)= F(t+ z) satisfies the differential equation.

    What that function, F, is depends upon those boundary conditions of which you say "Not much"!
    Last edited by HallsofIvy; Jun 19th 2017 at 09:59 AM.
    Thanks from topsquark and romsek
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    Forum Admin topsquark's Avatar
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    Re: Linear PDEq in two variables

    Thanks again, HoI. I don't know why I didn't think of it earlier, but a useful solution based on your form for the answer is a 2D Gaussian:
    v(t, z) = A e^ { -i(t + z)^2 / \sigma ^2 }.

    -Dan
    Last edited by topsquark; Jun 26th 2017 at 02:36 PM.
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