4

I need to show, that the following PDE does not have a solution:

\begin{align} u_x + u_t &= 0 \\ u(x,t) &= x , \forall x,t: x^2 + t^2 = 1 \end{align}

My attempt:

It's first-order linear PDE with constant coefficients, so I thought about using the basic method of characteristics:

\begin{align} u(x,t) &= f(bx - ay)\\ u(x,t) &= f(x - t) = x \end{align}

But, it implies, that $t = 0$. And there is the problem, I guess, but I'm not sure about this part.

So, how can I prove, that this equation does not have a solution?

Eenoku
  • 894

1 Answers1

6

Assume a solution exists. The characteristic curves of the PDE $u_x + u_t = 0$ are the straight lines $x = C - t$. A straight line through the origin intersects the unit circle at the two points, $(x_0,t_0)$ and $(-x_0,-t_0)$. Since this line is a characteristic curve we have $u(x_0,t_0) = u(-x_0,-t_0)$. However the initial condition stipulate that $u(x,t) = x$ on the unit circle so

$$u(x_0,t_0) = x_0~~~~\text{and}~~~~u(-x_0,-t_0) = -x_0$$

which gives us a contradiction.

Winther
  • 24,478