# The yacht race problem from L.C.Evans's PDE book 2nd edition chapter 10

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.

My thought was writing out two integral equations first, which show the relations of the total traveled distances in horizontal and vertical direction of those two yachts. \begin{aligned} \int^{u(x_1,x_2)}_0 b_2^{(x_1)}(t;x_1,x_2)\,dt &= x_1+ b_1 u(x_1,x_2) \\ \int^{u(x_1,x_2)}_0 b_2^{(x_2)}(t;x_1,x_2)\,dt &= x_2 \end{aligned} where $b_2^{(x_i)}(t)$ is the velocity $b_2$ in the direction $x_i$, but trying to differentiate with respect to $x_1$ and $x_2$ give me some weird looking results.

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You probably mean that the second yacht has constant speed, and not constant velocity. :) – Willie Wong Aug 24 '11 at 15:40
@Willie Wong: Yes, thanks for the correction – Shuhao Cao Aug 24 '11 at 18:24

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{align} \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{align}

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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