I am considering the following PDE \begin{equation} u_t + \alpha \, u u_x - \beta \, u_{xx} = 0 \end{equation} with $\alpha, \beta > 0$. When $\alpha = 1$, this is the viscous Burgers' equation. Given a solution for a case with $\alpha = 1$, I would like to know how this is changed by assuming $\alpha \neq 1$.
I will certainly benefit from the writing of the new solution but also from the explanation of the methodology to obtain that. So in the future I can adapt the reasoning to new initial and boundary conditions.
The domain is $x \in [-\infty,\infty]$ and $t \in [0,\infty[$.
At the boundary: $$ u(-\infty,t) = 1 \\ u(\infty,t) = 0 $$ The initial conditions describe a step change at the origin: $$ u(x,0) = 1 \textrm{ for } x \leq 0 \\ u(x,0) = 0 \textrm{ for } x \geq 0 $$
The solution for the viscous Burgers equation is $$ u(x,t) = \frac { % numerator \int_{-\infty}^{\infty} \frac {x - \xi} {t} e^{-G} \, d\xi }{ % denominator \int_{-\infty}^{\infty} e^{-G} \, d\xi } % end fraction $$ with $$ G(\xi; x,t) = \frac{1}{2\beta} \left( \int_0^\xi u(\xi',0) \, d\xi' + \frac {(x-\xi)^2} {2t} \right) $$
Thanks in advance
Hint:
Approach $1$:
Let $\begin{cases}x_1=x\\t_1=\alpha t\end{cases}$ ,
Then $u_x=u_{x_1}(x_1)_x+u_{t_1}(t_1)_x=u_{x_1}$
$u_{xx}=(u_{x_1})_{x_1}(x_1)_x+(u_{x_1})_{t_1}(t_1)_x=u_{x_1x_1}$
$u_t=u_{x_1}(x_1)_t+u_{t_1}(t_1)_t=\alpha u_{t_1}$
$\therefore\alpha u_{t_1}+\alpha uu_{x_1}-\beta u_{x_1x_1}=0$
$u_{t_1}+uu_{x_1}-\dfrac{\beta}{\alpha}u_{x_1x_1}=0$
Which converts to the viscous Burgers' equation.
Approach $2$:
Let $\begin{cases}x_2=\dfrac{x}{\alpha}\\t_2=t\end{cases}$ ,
Then $u_x=u_{x_2}(x_2)_x+u_{t_2}(t_2)_x=\dfrac{u_{x_2}}{\alpha}$
$u_{xx}=(u_{x_2})_{x_2}(x_2)_x+(u_{x_2})_{t_2}(t_2)_x=\dfrac{u_{x_2x_2}}{\alpha^2}$
$u_t=u_{x_2}(x_2)_t+u_{t_2}(t_2)_t=u_{t_2}$
$\therefore u_{t_2}+uu_{x_2}-\dfrac{\beta}{\alpha^2}u_{x_2x_2}=0$
Which converts to the viscous Burgers' equation.