Analyze : $u_t-u^2u_x +cu =0, u(x,0)=g(x)$

Question

Analyze : $u_t-u^2u_x +cu =0 $, $ u(x,0)=g(x)$. From This we have following

$$\begin{align} \frac{dt}{ds} &=1 \\ \frac{du}{ds} &=c \\ \frac{dx}{ds} &=-u^2 \end{align}$$

then how to analyze?

Answer 1

Note: The original solution to this problem was wrong. I have edited the solution to correct this.

Firstly, one of your equations is incorrect; it should be

$$\frac{du}{ds} = -cu$$

Eliminating $ds$ yields the parameterisation invariant form of the ODEs. We have

$$\frac{dt}{1} = \frac{dx}{-u^{2}} = \frac{du}{-cu}$$

Solving the first and third ratios yields

$$u e^{ct} = C_{1}$$

Solving the second and third ratios yields

$$\frac{1}{2} u^{2} - c x = C_{2}$$

Noting that $C_{2} = f(C_{1})$, we have

$$\frac{1}{2} u^{2} - c x = f(u e^{ct}) \tag1$$

Using the initial condition, we find

\begin{align} \frac{1}{2} g^{2}(x) - c x &= \frac{1}{2} u^{2}(x, 0) - c x \\ &= f(u(x, 0)) \\ &= f(g(x)) \end{align}

Let $X = g(x) \implies x = g^{-1}(X)$ provided $g$ is invertible and hence

\begin{align} f(X) &= \frac{1}{2} X^{2} - c g^{-1}(X) \\ \implies f(u e^{ct}) &= \frac{1}{2} (u e^{ct})^{2} - c g^{-1}(u e^{ct}) \end{align}

Substituting into $(1)$ and rearranging yields

$$u = e^{-ct} g \left( \frac{1}{2 c} u^{2} \left( e^{2ct} - 1 \right) + x \right)$$

which we can check satisfies the PDE and initial condition.