Consider the PDE $$u_t+uu_x=u_{xx}$$
I want to show that the substitution $u(x,t)=-C(\ln v(x,t))_x$ transforms the PDE into the heat equation and show that $C=1/2.$
Let $\ln(v(x,t))=p$.
Then $u_t=-Cp_{xt}$, $u_x=-Cp_{xx}$, $u_{xx}=-Cp_{xxx}$. I can write $uu_{x}=C^2\frac{\partial}{\partial x}(p_x^2)$ which then gives me $$C\frac{\partial}{\partial x}\left(p_{xx}+\frac{C}{2}p_x^2-p_t\right)=0$$
What do I do now though?
Best Answer
Let's summarize the derivation in Whitham's Linear and Nonlinear Waves, $\S 4.1$: The Cole-Hopf Transformation
Burgers' equation
The Burgers equation is the simplest example of a partial differential equation demonstrating both diffusive and nonlinear propagation effects. $$ u_{\color{red}{t}} + u u_{\color{blue}{x}} = \nu u_{\color{blue}{xx}} \tag{1} $$ Coloring distinguishes $\color{red}{time}$ derivatives from $\color{blue}{space}$ derivatives.
Cole-Hopf transform
The nonlinear transform of Cole and Hopf is, $$ u = -2\nu \frac{\varphi_{\color{blue}{x}}}{\varphi}, $$ similar to the Thomas transformation of exchange equations. The transformation is resolved into two steps.
Step 1
Set $$u = \psi_{\color{blue}{x}},$$ and equation (1) becomes $$ \psi_{\color{red}{t}} + \frac{1}{2}\left(\psi_{\color{blue}{x}}\right)^{2} = \nu \psi_{\color{blue}{xx}} \tag{2} $$
Step 2
Set $$\psi = -2\nu \ln \varphi,$$ and equation (2) reduces to the heat equation $$ \boxed{ \varphi_{\color{red}{t}} = \nu \varphi_{\color{blue}{xx}} } \tag{3} $$ The nonlinear term has vanished because of the nonlinear transformation.
Transforming heat equation solutions to Burgers equation solution
The initial value problem which starts with a spatial waveform $$ u = F(x), \qquad t =0 $$ transforms into $$ \varphi = \Phi(x) = e^{-\frac{1}{2\nu}\int_{0}^{x} F(\varphi)d\eta}, \qquad t = 0 $$ The heat equation solution for this IVP is $$ \varphi = \frac{1}{\sqrt{4\pi \nu t}} \int_{-\infty}^{\infty} \Phi(\eta)e^{-\frac{(x-\eta)^{2}}{4\nu t}} d\eta $$
Going back to the Cole-Hopf transform to recover the solution function $u$: $$ u(x,t) = \frac {\int_{-\infty}^{\infty} \frac{x-\eta}{t} e^{-\frac{G}{2\nu}} d\eta} {\int_{-\infty}^{\infty} e^{-\frac{G}{2\nu}} d\eta} $$ with the function $$ G\left(\eta; x, t \right) =\frac{\left( x - \eta \right)^{2}}{2t} + \int_{\eta}^{0} F\left(\xi \right) d\xi $$