This is an exercise in Evans, Partial Differential Equations (2nd edition), Chapter 10 Conservation laws.

## Problem

There is one yacht starting at $(x_1,0)$ when $t=0$, which is sailing toward positive direction of $x$-axis with a constant velocity $b_1$, another yacht is starting at $(0,x_2)$ when $t=0$, and is always sailing toward the first yacht with a constant speed $b_2 > b_1$.
Question: Find the equation $u(x_1,x_2)$ satisfies, where $u(x_1,x_2)$ is the time that the second yacht catches up with the first yacht.

## Solution

The solution is courtesy of Willie Wong on Math.SE ^{1}.

We derive the equation for $u$ by starting with the equations of the trajectories of the two yachts, and "reversing" the method of characteristics.
Let $(X_1,0)$ be the coordinates of the first yacht, and $(X_2,Y_2)$ be the coordinates of the second yacht. You have
$$
\begin{aligned}
\frac{d}{dt}X_1 &= b_1 \\
\frac{d}{dt}(X_2, Y_2) &= \frac{b_2}{\sqrt{(X_1 - X_2)^2 + (Y_2)^2}} (X_1-X_2, -Y_2)
\end{aligned}
$$
or
$$ \frac{d}{dt}(X_1 - X_2, Y_2) = \left( b_1 - \frac{b_2(X_1-X_2)}{\sqrt{(X_1 - X_2)^2 + (Y_2)^2}}, \frac{-b_2 Y_2}{\sqrt{(X_1 - X_2)^2 + (Y_2)^2}}\right) $$
Observe that starting with $X_1 = x_1, X_2 = 0, Y_2 = x_2$, the trajectory will overlap with a shifted trajectory of the initial conditions $x'_1 = X_1(t)- X_2(t)$ and $x'_2 = Y_2(t)$. So we have that
$$ u(X_1(t) - X_2(t), Y_2(t)) = s + u(X_1(s+t) - X_2(s+t), Y_2(s+t)) $$
We now take the total derivative relative to $s$, and we end up with
$$ 0 = 1 + du \circ \frac{d}{dt}(X_1 - X_2, Y_2) $$
or the equation for $u = u(x_1,x_2)$:
$$ -1 = \partial_1 u \cdot \left( b_1 - \frac{b_2 x_1}{\sqrt{x_1^2 + x_2^2}}\right) + \partial_2 u \cdot \left( - \frac{b_2 x_2}{\sqrt{x_1^2 + x_2^2}}\right) $$

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