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Thread: Substitution in partial differential equation

  1. #1
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    Substitution in partial differential equation

    The setting is the following: we have \frac{\partial^2\phi}{\partial^2 y} - \frac{\partial^2\phi}{\partial^2 x} = 0 and if we reduce this to a first order pde by substituting u = \frac{\partial\phi}{\partial x} \text{, } v = \frac{\partial\phi}{\partial y}\text{ and } \frac{\partial}{\partial x} = \xi \text{, } \frac{\partial}{\partial y} = \zeta, we get the following equations: \xi \dot{v} + \zeta \dot{u} = 0 \text{ and } -\xi \dot{u} + \zeta \dot{v} = 0. My question is why is there a \dot{u} and a \dot{v} after the substitution? Should it not be just u and v without a dot? Thank you.
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    Re: Substitution in partial differential equation

    What does \frac{\partial}{\partial x} = \xi mean? The left hand side is an operator. Similarly for \frac{\partial}{\partial y} = \zeta.
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    Re: Substitution in partial differential equation

    We defined \xi \text{ to be } \frac{\partial}{\partial{x}} . So we set xi :=d/dx. Similarly for d/dy, and plugged it into the first differential equation. Thank you.
    Last edited by laguna; Oct 3rd 2016 at 07:03 AM.
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    Re: Substitution in partial differential equation

    Why substitute the operator for another symbol?
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    Re: Substitution in partial differential equation

    Well thats not really the question But you can just replace the symbols by the operator if you don't like it. Can you help me understand why u,v is replaced by u^dot, v^dot? What am I missing?
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    Re: Substitution in partial differential equation

    Quote Originally Posted by laguna View Post
    The setting is the following: we have \frac{\partial^2\phi}{\partial^2 y} - \frac{\partial^2\phi}{\partial^2 x} = 0 and if we reduce this to a first order pde by substituting u = \frac{\partial\phi}{\partial x} \text{, } v = \frac{\partial\phi}{\partial y}\text{ and } \frac{\partial}{\partial x} = \xi \text{, } \frac{\partial}{\partial y} = \zeta, we get the following equations: \xi \dot{v} + \zeta \dot{u} = 0 \text{ and } -\xi \dot{u} + \zeta \dot{v} = 0. My question is why is there a \dot{u} and a \dot{v} after the substitution? Should it not be just u and v without a dot? Thank you.
    I have no idea why those substitutions are necessary, but this question is just screaming for separation of variables.

    Let $\displaystyle \begin{align*} \phi = X\left( x \right)Y\left( y \right) \end{align*}$ which gives

    $\displaystyle \begin{align*} \frac{\partial \phi}{\partial x } &= X'\left( x \right) Y\left( y \right) \\ \frac{\partial ^2 \phi}{\partial x^2} &= X''\left( x \right) Y\left( y \right) \\ \\ \frac{\partial \phi}{\partial y} &= X\left( x \right) Y'\left( y \right) \\ \frac{\partial ^2 \phi}{\partial y^2 } &= X\left( x \right) Y''\left( y \right) \end{align*}$

    so substituting into the DE gives

    $\displaystyle \begin{align*} X\left( x \right) Y'' \left( y \right) - X''\left( x \right) Y \left( y \right) &= 0 \\ X\left( x \right) Y''\left( y \right) &= X''\left( x \right) Y\left( y \right) \\ \frac{ Y''\left( y \right) }{Y\left( y \right) } &= \frac{X''\left( x \right) } {X\left( x \right) } \end{align*}$

    The only way for the LHS and RHS to be equal is if they both equal the same constant value (k) so by doing that you get two second order constant coefficient DEs you can solve for x and y.
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