## Chapter 4Burger’s PDE

Solve $$u_{t}+u\ u_{x}=Du_{xx} \tag{1}$$

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 $$u\left ( x,t\right ) =-2D\frac{\phi _{x}}{\phi } \tag{2}$$ 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

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

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

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

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 $$\phi \left ( x,t\right )$$

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

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

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

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

(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