Use a reverse method of characteristics to get the trajectories equation

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 $u = u(x_1,x_2)$: $$-1 = \partial_1 u \cdot \left( b_1 - \frac{b_2 x_1}{\sqrt{x_1^2 + x_2^2}}\right) + \partial_2 u \cdot \left( - \frac{b_2 x_2}{\sqrt{x_1^2 + x_2^2}}\right)$$
1. Willie Wong (https://math.stackexchange.com/users/1543/willie-wong), The yacht race problem from L.C.Evans’s PDE book 2nd edition chapter 10, URL (version: 2011-08-24): https://math.stackexchange.com/q/59494

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