= Solution
Write $F_p,F_q$ evaluated at the fixed pair $(p(s),q(s))$. Differentiating in $\tau$ gives $x_\tau=F_p$, $y_\tau=F_q$, and $u_\tau=pF_p+qF_q$, proving $u_\tau=px_\tau+qy_\tau$.
Differentiate the other parameter. The difference of the required sides is
$$
u_s-px_s-qy_s=(u_0'-px_0'-qy_0')+\tau(p'F_p+q'F_q).
$$
The first bracket is zero by the prescribed initial compatibility condition, and the second is zero by differentiating $F(p(s),q(s))=0$. Thus $u_s=px_s+qy_s$ as well. These calculations use the differentiability of the supplied parametric functions.
If the coordinate <Jacobian determinant> $J=x_sy_\tau-y_sx_\tau$ is nonzero, the <inverse function theorem> gives local coordinates $(x,y)$. The <chain rule> then makes $(u_x-p,u_y-q)$ solve a homogeneous two-by-two <linear system> with determinant $J$. Therefore
$$
\boxed{u_x=p,\qquad u_y=q.}
$$
This is the <method of characteristics> for a first-order equation depending only on the gradient.
Back to article page
