Originally Posted by

**Peritus** $\displaystyle

\begin{array}{l}

u_t - \left( {u_x } \right)^2 = u_{xx} \\

\\

now\;we\;use\;the\;suggested\;substitution \\

\end{array}

$

$\displaystyle

\begin{array}{l}

v = e^u \to u = \ln v \\

\\

u_t = \frac{\partial }{{\partial t}}\ln v = \frac{{v_t }}{v} \\

\end{array}

$

$\displaystyle

\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:

$\displaystyle \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).