PDE - Change of variables

**angels_symphony** · January 18th 2008, 09:51 PM

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

**Peritus** · January 19th 2008, 12:26 AM

$ \begin{array}{l} u_t - \left( {u_x } \right)^2 = u_{xx} \\ \\ now\;we\;use\;the\;suggested\;substitution \\ \end{array} $

$ \begin{array}{l} v = e^u \to u = \ln v \\ \\ u_t = \frac{\partial }{{\partial t}}\ln v = \frac{{v_t }}{v} \\ \end{array} $

$ \begin{array}{l} u_x = \frac{\partial }{{\partial x}}\ln v = \frac{{v_x }}{v} \\ \\ 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 \\ \end{array} $

now substituting these results into the pde we get:

$ \begin{array}{l} \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 \\ \\ \Leftrightarrow v_t = v_{xx} \\ \end{array} $

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

**earboth** · January 19th 2008, 02:00 AM

Originally Posted by Peritus

$ \begin{array}{l} u_t - \left( {u_x } \right)^2 = u_{xx} \\ \\ now\;we\;use\;the\;suggested\;substitution \\ \end{array} $

$ \begin{array}{l} v = e^u \to u = \ln v \\ \\ u_t = \frac{\partial }{{\partial t}}\ln v = \frac{{v_t }}{v} \\ \end{array} $

$ \begin{array}{l} u_x = \frac{\partial }{{\partial x}}\ln v = \frac{{v_x }}{v} \\ \\ 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 \\ \end{array} $

now substituting these results into the pde we get:

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

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

Hello,

probably you meant:

$\begin{array}{l} \frac{{v_t }}{v} - \left( {\frac{{v_x }}{v}} \right)^2 = \frac{{v_{xx} }}{v} - \left( {\frac{{v_x }}{v}} \right)^2 \\ \\ \iff~ v_t = v_{xx} \\ \end{array} $

**Peritus** · January 19th 2008, 02:28 AM

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)

**earboth** · January 19th 2008, 04:43 AM

Originally Posted by Peritus

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

**angels_symphony** · January 19th 2008, 09:52 AM

Hey thanks! Umm just one questions...why does u_t = u_xx?

**Peritus** · January 19th 2008, 10:26 AM

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} $

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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