Chapter 4
Burger’s PDE

Solve \begin{equation} u_{t}+u\ u_{x}=Du_{xx} \tag{1} \end{equation}

BC\begin{align*} u\left ( 0,t\right ) & =0\qquad t>0\\ u\left ( L,t\right ) & =0\qquad t>0 \end{align*}

Initial conditions\[ u\left ( x,0\right ) =f\left ( x\right ) \qquad 0<x<L \]

Where \(D\) is the diffusion constant.

Solution

Using Cole-Hopf, let \begin{equation} u\left ( x,t\right ) =-2D\frac{\phi _{x}}{\phi } \tag{2} \end{equation} where \(\phi \equiv \phi \left ( x,t\right ) \). Rewriting equation (1) as

\begin{align} u_{t} & =Du_{xx}-u\ u_{x}\nonumber \\ & =\left ( Du_{x}-\frac{u^{2}}{2}\right ) _{x} \tag{3} \end{align}

Substituting (2) into (3) gives

\begin{align} \left ( -2D\frac{\phi _{x}}{\phi }\right ) _{t} & =\left [ D\left ( -2D\frac{\phi _{x}}{\phi }\right ) _{x}-\frac{1}{2}\left ( -2D\frac{\phi _{x}}{\phi }\right ) ^{2}\right ] _{x}\nonumber \\ -2D\left ( \frac{\phi _{x}}{\phi }\right ) _{t} & =\left ( -2D^{2}\left ( \frac{\phi _{x}}{\phi }\right ) _{x}-2D^{2}\left ( \frac{\phi _{x}}{\phi }\right ) ^{2}\right ) _{x}\nonumber \\ -2D\left ( \frac{\phi _{x}}{\phi }\right ) _{t} & =-2D^{2}\left ( \left ( \frac{\phi _{x}}{\phi }\right ) _{x}+\left ( \frac{\phi _{x}}{\phi }\right ) ^{2}\right ) _{x} \tag{4} \end{align}

But \[ \left ( \frac{\phi _{x}}{\phi }\right ) _{t}=-\frac{1}{\phi ^{2}}\phi _{t}\phi _{x}+\frac{1}{\phi }\phi _{xt}\]

And

\[ \left ( \frac{\phi _{t}}{\phi }\right ) _{x}=-\frac{1}{\phi ^{2}}\phi _{x}\phi _{t}+\frac{1}{\phi }\phi _{tx}\]

Therefore \(\left ( \frac{\phi _{x}}{\phi }\right ) _{t}=\left ( \frac{\phi _{t}}{\phi }\right ) _{x}\). Using this in LHS of (4) gives

\begin{equation} -2D\left ( \frac{\phi _{t}}{\phi }\right ) _{x}=-2D^{2}\left ( \left ( \frac{\phi _{x}}{\phi }\right ) _{x}+\left ( \frac{\phi _{x}}{\phi }\right ) ^{2}\right ) _{x} \tag{5} \end{equation}

And \begin{align*} \left ( \frac{\phi _{x}}{\phi }\right ) _{x}+\left ( \frac{\phi _{x}}{\phi }\right ) ^{2} & =-\frac{1}{\phi ^{2}}\phi _{x}^{2}+\frac{\phi _{xx}}{\phi }+\frac{\phi _{x}^{2}}{\phi ^{2}}\\ & =\frac{\phi _{xx}}{\phi } \end{align*}

Using the above in the RHS of (5) gives

\begin{equation} -2D\left ( \frac{\phi _{t}}{\phi }\right ) _{x}=-2D^{2}\left ( \frac{\phi _{xx}}{\phi }\right ) _{x} \tag{6} \end{equation}

Integrating both side w.r.t. \(x\) gives

\[ -2D\frac{\phi _{t}}{\phi }=-2D^{2}\frac{\phi _{xx}}{\phi }+G\left ( t\right ) \]

Where \(G\left ( t\right ) \) is the constant of integration since \(\phi \equiv \phi \left ( x,t\right ) \). The above simplifies to the heat PDE in \(\phi \left ( x,t\right ) \)

\begin{equation} \phi _{t}=D\phi _{xx}+G\left ( t\right ) \phi \tag{6A} \end{equation}

Let \begin{equation} \psi =\phi e^{-\int G\left ( t\right ) dt} \tag{6B} \end{equation}

Then \begin{align*} \psi _{t} & =\phi _{t}e^{-\int G\left ( t\right ) dt}-\phi G\left ( t\right ) e^{-\int G\left ( t\right ) dt}\\ & =\left ( \phi _{t}-\phi G\left ( t\right ) \right ) e^{-\int G\left ( t\right ) dt} \end{align*}

But from (6A), we see that \(\phi _{t}-\phi G\left ( t\right ) =D\phi _{xx}\). Therefore the above becomes

\[ \psi _{t}=D\phi _{xx}e^{-\int G\left ( t\right ) dt}\]

But from (6B), we see that \(\phi _{xx}e^{-\int G\left ( t\right ) dt}=\psi _{xx}\), therefore the above becomes

\[ \psi _{t}=D\psi _{xx}\]

Which is the heat PDE. The original BC and initial conditions are now transformed to \(\psi \) to solve the above. Since \(u=-2D\frac{\phi _{x}}{\phi }\), then solving this first for \(\phi \)

\begin{align*} \frac{\phi _{x}}{\phi } & =-\frac{1}{2D}u\\ \frac{\partial \phi }{\partial x}\frac{1}{\phi } & =-\frac{1}{2D}u\\ \frac{d\phi }{\phi } & =-\frac{1}{2D}udx \end{align*}

Integrating gives

\begin{align*} \ln \phi & =-\frac{1}{2D}\int _{0}^{x}u\ ds+C_{0}\\ \phi \left ( x,t\right ) & =Ce^{-\frac{1}{2D}\int _{0}^{x}u\ ds} \end{align*}

Since \(u=-2D\frac{\phi _{x}}{\phi }\), then the constant \(C\) cancels out. Then we it can be set to any value as it does not affect the solution. Let \(C=1\) and the above becomes

\begin{equation} \phi \left ( x,t\right ) =e^{-\frac{1}{2D}\int _{0}^{x}u\ ds} \tag{7} \end{equation}

(7) is now used to transform the initial conditions. When \(u\left ( x,0\right ) =f\left ( x\right ) \) the above becomes

\begin{align*} \phi \left ( x,0\right ) & =e^{-\frac{1}{2D}\int _{0}^{x}u\left ( s,0\right ) \ ds}\\ & =e^{-\frac{1}{2D}\int _{0}^{x}f\left ( s\right ) \ ds} \end{align*}

Since from (6B), \(\psi \left ( x,t\right ) =\phi e^{-\int G\left ( t\right ) dt}=\phi e^{-\int _{0}^{t}G\left ( s\right ) ds}\) therefore

\begin{align*} \psi \left ( x,0\right ) & =\phi \left ( x,0\right ) e^{0}\\ & =\phi \left ( x,0\right ) \\ & =e^{-\frac{1}{2D}\int _{0}^{x}f\left ( s\right ) \ ds} \end{align*}

To transform the boundary conditions, \(u=-2D\frac{\phi _{x}}{\phi }\) is used. When \(u\left ( 0,t\right ) =0\) then \(0=-2D\frac{\phi _{x}\left ( 0,t\right ) }{\phi \left ( 0,t\right ) }\) or

\[ \phi _{x}\left ( 0,t\right ) =0 \]

But \(\psi =\phi e^{-\int _{0}^{t}G\left ( s\right ) ds}\), then \(\psi _{x}=\phi _{x}e^{-\int _{0}^{t}G\left ( s\right ) ds}\) and therefore

\begin{align*} \psi _{x}\left ( 0,t\right ) & =\phi _{x}\left ( 0,t\right ) e^{-\int _{0}^{t}G\left ( s\right ) ds}\\ & =0 \end{align*}

Similarly, when \(u\left ( L,t\right ) =0\) then \(0=-2D\frac{\phi _{x}\left ( L,t\right ) }{\phi \left ( L,t\right ) }\) or

\[ \phi _{x}\left ( L,t\right ) =0 \]

Which gives

\[ \psi _{x}\left ( L,t\right ) =0 \]

Hence the heat PDE to solve is

\[ \psi _{t}=D\psi _{xx}\]

BC\begin{align*} \psi _{x}\left ( 0,t\right ) & =0\qquad t>0\\ \psi _{x}\left ( L,t\right ) & =0\qquad t>0 \end{align*}

Initial conditions\[ \psi \left ( x,0\right ) =e^{-\frac{1}{2D}\int _{0}^{x}f\left ( s\right ) \ ds}\qquad 0<x<L \]

The above heat PDE is now solved for \(\psi \left ( x,t\right ) \). This solution is transformed back to \(u\left ( x,t\right ) \). First using \(\psi =\phi e^{-\int G\left ( t\right ) dt}\) to find \(\phi \left ( x,t\right ) \), then using \(u\left ( x,t\right ) =-2D\frac{\phi _{x}}{\phi }\), to find \(u\left ( x,t\right ) \).

So in summary, there are two transformations needed. Going from \(u\left ( x,t\right ) \rightarrow \phi \left ( x,t\right ) \) uses Cole-Hopf. Going from \(\phi \left ( x,t\right ) \rightarrow \psi \left ( x,t\right ) \) uses \(\psi \left ( x,t\right ) =\phi \left ( x,t\right ) e^{-\int G\left ( t\right ) dt}.\) It is \(\psi \left ( x,t\right ) \) which is solved as the heat PDE \(\psi _{t}=D\psi _{xx}\) and not \(\phi \left ( x,t\right ) \), which is just an intermediate transformation.

 4.1 Example 1