Solution to 2D- Burgers' equation with a source

Question

Does anyone know of any solution to the 2D Burgers' equation

$$ u_t + (u^2/2)_x + (u^2/2)_y = \beta u $$ $$ u(t=0,x,y) = h_0(x,y)$$ For some constant $\beta $. The closest I've gotten is by following the work in

https://arxiv.org/pdf/1503.09079.pdf

but I am having trouble generalizing to 2D. Any ideas?

Answer 1

We write the quasilinear PDE in the form $u_t + u u_x + u u_y = \beta u$, and we obtain the characteristic equations \begin{equation} \frac{dt}{1} = \frac{dx}{u} = \frac{dy}{u} = \frac{du}{\beta u}. \end{equation} We solve these as

1. $\frac{dt}{1} = \frac{du}{\beta u}$: $u = u_0 e^{\beta t}$,
2. $\frac{dt}{1} = \frac{dx}{u}$: $dx \stackrel{1.}{=} u_0 e^{\beta t} dt$, $x = x_0 + u_0 \frac{e^{\beta t}-1}{\beta} \stackrel{1.}{=} x_0 + u \frac{1-e^{-\beta t}}{\beta}$,
3. $\frac{dt}{1} = \frac{dy}{u}$: $dy \stackrel{1.}{=} u_0 e^{\beta t} dt$, $y = y_0 + u_0 \frac{e^{\beta t}-1}{\beta} \stackrel{1.}{=} y_0 + u \frac{1-e^{-\beta t}}{\beta}$,

where the initial values $(x_0,y_0,u_0)$ are all given at $t=0$. From the initial condition we now obtain \begin{equation} u_0 = u(0,x_0,y_0) = h_0(x_0,y_0) \stackrel{2., 3.}{=} h_0\left( x - \frac{1-e^{-\beta t}}{\beta} u, y - \frac{1-e^{-\beta t}}{\beta} u \right), \end{equation} so that \begin{equation} u \stackrel{1.}{=} u_0 e^{\beta t} = e^{\beta t} h_0\left( x - u \frac{1-e^{-\beta t}}{\beta}, y - u \frac{1-e^{-\beta t}}{\beta} \right). \end{equation} This is correct for $\beta \neq 0$, if $h_0$ is differentiable.

We obtain the correct expression for $\beta = 0$ by letting $\beta \rightarrow 0$: \begin{equation} u = h_0\left( x - u t, y - u t\right), \end{equation} which is the familiar expression for the Burgers' equation without source term.