Solving Inviscid Burgers' using Similarity

Question

I want to solve the inviscid Burgers' equation:

$$\begin{equation} \frac{\partial u}{\partial t} + u\frac{\partial u }{\partial x} = 0 \end{equation}$$ I want to reduce the PDE to an ODE by saying that$\ u $ is a function of:

$$\ u = u(x, t , u_o)$$

I expressed $\ u $ as a dimensionless quantity in terms of length (x), time(t) and the initial velocity (u0):

$$\ [u] = [x^a t^b u_o^c]$$

Yielding the similarity variable $\eta $:

$$\eta = \frac{x}{tu_0}$$

I then defined the following function:

$$u = u_0(\frac{x}{tu_0})^a = u_0F(\eta)$$

Subsequently carrying out the change of variables:

$$\begin{equation} \frac{\partial u}{\partial t} = F'(\eta)\frac{\partial \eta}{\partial t} = F'(\eta)(\frac{-x}{t^2}) \end{equation}$$

$$\begin{equation} \frac{\partial u}{\partial x} = F'(\eta)\frac{\partial \eta}{\partial x} = F'(\eta)(\frac{1}{t}) \end{equation}$$

Plugging it back into the differential equation I end up with a strange ODE:

$$\ F(\eta)F'(\eta) = \eta F'(\eta) $$

I think somewhere I made an error. If I need to show more steps please let me know. Any help would be appreciated.

Answer 1

Let us derive self-similar solutions by making the similarity Ansatz $$ x = |t|^{\alpha} \xi \qquad\text{and}\qquad u = |t|^{\beta} U(\xi)\, . $$ Under this assumption, we therefore have $\xi = x |t|^{-\alpha}$, and \begin{aligned} u_t(x,t) &= \beta t^{\beta-1} U(\xi) - \alpha t^{\beta-1} \xi U'(\xi) \\ u_x(x,t) &= t^{\beta-\alpha} U'(\xi) \end{aligned} for positive times, so that the PDE rewrites as $$ \beta U(\xi) - \alpha \xi U'(\xi) + t^{\beta - \alpha + 1} U(\xi) U'(\xi) = 0 \, . $$ The previous equation reduces to an ODE if $\beta = \alpha-1$, which is assumed from now on. Analytical solutions can be obtained for various scaling parameters $\alpha$. Due to the problem's invariance by uniform dilatation of space and time $(x,t) \mapsto (k x,k t)$, a natural scaling is $\alpha = 1$. This scaling corresponds to $\xi = x/t$, and leads to the differential equation $(U(\xi)-\xi)\, U'(\xi) = 0$ in OP. Non-trivial solutions are $U(\xi) = \xi$, i.e. $u(x,t) = x/t$. A self-similar solution of this form is called a rarefaction wave, a.k.a simple wave.