First order sketches

mark

Let's suppose that $u$ satisfies the first order IVP

$$u_t + 2u_x = 0, \: u(x,0) = u_0(x),$$

where $u_0$ is a function whose graph looks something like this:

Sketch the graphs of $u(x,t)$ for $t=1,2,3$.

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Suppose we change the PDE to $u_t + 2u_x = -u$. How will this change the graphs?

AbS

The solution to the PDE is

$$u = u_0(x - 2t),$$

so at each time interval, the graph above will shift to the right by two units. Here are sketches of the graph at each of the three times:

If we changed the PDE to $u_t + 2u_x = -u$, the solution becomes

$$u = u_0(x - 2t)e^{-t},$$

so $u$ will decay over time, approaching zero at large values of $t$.