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Math Help - PDE - Change of variables

  1. #1
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    PDE - Change of variables

    Hi, I'm having problems understanding this question and I'm not entierly sure what it even means so some help would be really appreciated!

    Question:
    Using the change of dependent variable: v=e^u, find a second order linear partial differential equation for v(x,t) [This is that I don't understand..what does it mean by "for v(x,t)?] from the following nonlinear partial differential equation for u(x,t):

    u_t - (u_x)^2 = u_xx
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\begin{array}{l}<br />
 u_t  - \left( {u_x } \right)^2  = u_{xx}  \\ <br />
  \\ <br />
 now\;we\;use\;the\;suggested\;substitution \\ <br />
 \end{array}<br />


     <br /> <br />
\begin{array}{l}<br />
 v = e^u  \to u = \ln v \\ <br />
  \\ <br />
 u_t  = \frac{\partial }{{\partial t}}\ln v = \frac{{v_t }}{v} \\ <br />
 \end{array}<br /> <br />

    <br /> <br />
\begin{array}{l}<br />
 u_x  = \frac{\partial }{{\partial x}}\ln v = \frac{{v_x }}{v} \\ <br />
  \\ <br />
 u_{xx}  = \frac{\partial }{{\partial x}}u_x  = \frac{\partial }{{\partial x}}\frac{{v_x }}{v} = \frac{{v_{xx} }}{v} - v_x \frac{{v_x }}{{v^2 }} = \frac{{v_{xx} }}{v} - \left( {\frac{{v_x }}{v}} \right)^2  \\ <br />
 \end{array}<br /> <br />

    now substituting these results into the pde we get:


    <br /> <br />
\begin{array}{l}<br />
 \frac{{{\rm v}_{\rm t} }}{{\rm v}} - \left( {\frac{{v_x }}{v}} \right)^2  = \frac{{v_{xx} }}{v} - \left( {\frac{{v_x }}{v}} \right)^2  \\ <br />
  \\ <br />
  \Leftrightarrow v_t  = v_{xx}  \\ <br />
 \end{array}<br /> <br /> <br />

    as promised we've got ourselves a second order linear partial differential equation for v(x,t).
    Last edited by Peritus; January 19th 2008 at 02:27 AM.
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by Peritus View Post
    <br />
\begin{array}{l}<br />
 u_t  - \left( {u_x } \right)^2  = u_{xx}  \\ <br />
  \\ <br />
 now\;we\;use\;the\;suggested\;substitution \\ <br />
 \end{array}<br />


     <br /> <br />
\begin{array}{l}<br />
 v = e^u  \to u = \ln v \\ <br />
  \\ <br />
 u_t  = \frac{\partial }{{\partial t}}\ln v = \frac{{v_t }}{v} \\ <br />
 \end{array}<br /> <br />

    <br /> <br />
\begin{array}{l}<br />
 u_x  = \frac{\partial }{{\partial x}}\ln v = \frac{{v_x }}{v} \\ <br />
  \\ <br />
 u_{xx}  = \frac{\partial }{{\partial x}}u_x  = \frac{\partial }{{\partial x}}\frac{{v_x }}{v} = \frac{{v_{xx} }}{v} - v_x \frac{{v_x }}{{v^2 }} = \frac{{v_{xx} }}{v} - \left( {\frac{{v_x }}{v}} \right)^2  \\ <br />
 \end{array}<br /> <br />

    now substituting these results into the pde we get:


    \begin{array}{l}<br /> <br />
\begin{array}{l}<br />
 \frac{{v_t }}{v} - \left( {\frac{{v_x }}{v}} \right)^2  = \frac{{v_{xx} }}{v} - \left( {\frac{{v_x }}{v}} \right)^2  \\ <br />
  \\ <br />
  <  =  > v_t  = v_{xx}  \\ <br />
 \end{array}<br /> <br />

    as promised we've got ourselves a second order linear partial differential equation for v(x,t).

    Hello,

    probably you meant:

     \begin{array}{l}<br />
 \frac{{v_t }}{v} - \left( {\frac{{v_x }}{v}} \right)^2  = \frac{{v_{xx} }}{v} - \left( {\frac{{v_x }}{v}} \right)^2  \\ <br />
  \\ <br />
  \iff~ v_t  = v_{xx}  \\ <br />
 \end{array}<br /> <br />
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  4. #4
    Senior Member Peritus's Avatar
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    Thanks earthbot, I'm still having problems with Latex, could you tell me what program do you use to post mathematical expressions ? (I use TeXaide)
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  5. #5
    Super Member
    earboth's Avatar
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    Quote Originally Posted by Peritus View Post
    Thanks earthbot, I'm still having problems with Latex, could you tell me what program do you use to post mathematical expressions ? (I use TeXaide)
    Hello,

    have a look here: http://www.mathhelpforum.com/math-he...-tutorial.html
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  6. #6
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    Hey thanks! Umm just one questions...why does u_t = u_xx?
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  7. #7
    Senior Member Peritus's Avatar
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    Hey thanks! Umm just one questions...why does u_t = u_xx?
    nobody said it does, as you can see in my post:

    v_t  = v_{xx} <br />

    and it follows from the suggested substitution, again as you can see from my post. Once you'll find v you'll be able to find u, through the use of : u = \ln v
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