Scalar conservation law

Consider the scalar equation

$$\frac{\partial u}{\partial t} + x^{1/3} \frac{\partial u}{\partial x} = c(t), x\in[0,1], t > 0$$ (1)

where $c(t)$ is a source term. This equation can be solved exactly using the method of characteristics, which tells us that

$$\frac{\textrm{d}u}{\textrm{d}t} = c(t) \textrm{ along } \frac{\textrm{d}x}{\textrm{d}t} = x^{1/3}$$ (2)

The characteristics are given by integrating the second equation above

$$x^{2/3} = \frac{2}{3} t + K$$ (3)

where $K$ is a constant. These are plotted in figure below. The red line is the characteristic with $K=0$.

Figure 3: Characteristics in the x-t plane

Now let $(x,t)$ be a point on some characteristic and we want to find the foot of the characteristic $(x_o,t_o)$ ie., the point where the characteristic hits the x-axis or the t-axis when drawn backward in time. From the figure we see that if

$$x^{2/3} - \frac{2}{3} t \ge 0$$

then the characteristic hits the x-axis and

$$x_o(x,t) = \left[ x^{2/3} - \frac{2}{3} t \right]^{3/2}, t_o(x,t)=0$$ (4)

and if

$$x^{2/3} - \frac{2}{3} t \le 0$$

then the characteristic hits the t-axis and

$$x_o(x,t) = 0, t_o(x,t) = t - \frac{3}{2} x^{2/3}$$ (5)

The solution can now be easily written. Integrating backwards we have

$$u(x,t) = u(x_o(x,t), t_o(x,t)) + \int^t_{t_o(x,t)} c(s) \textrm{d}s$$ (6)

The characteristics tell us the correct boundary conditions; the initial condition $u(x,0)$ and the boundary condition $u(0,t)$ must be specified.