[proofplan]
Subtract the two solutions to reduce the assertion to a vanishing statement for their difference along the given characteristic. The chain rule then converts the partial differential equation along that curve into a scalar linear ordinary differential equation. An integrating factor shows that the transported difference remains zero after it is zero at the initial point.
[/proofplan]
[step:Subtract the two transport equations]
Define the function $z: U \to \mathbb{R}$ by $z(x) = u(x) - v(x)$. Since $u, v \in C^1(U)$, we have $z \in C^1(U)$. For each $x \in U$, linearity of the gradient and of multiplication by $c(x)$ gives
\begin{align*}
b(x) \cdot \nabla z(x) + c(x)z(x)
= b(x) \cdot \nabla u(x) + c(x)u(x) - b(x) \cdot \nabla v(x) - c(x)v(x).
\end{align*}
Since both $u$ and $v$ solve the same homogeneous transport equation, the right side is zero:
\begin{align*}
b(x) \cdot \nabla z(x) + c(x)z(x)=0.
\end{align*}
The hypothesis $u(X(s_0)) = v(X(s_0))$ is exactly $z(X(s_0)) = 0$.
[/step]
[step:Convert the transport equation into a scalar equation along the characteristic]
Define the scalar function $y: I \to \mathbb{R}$ by $y(s) = z(X(s))$. Because $z \in C^1(U)$ and $X \in C^1(I; U)$, the chain rule gives $y \in C^1(I)$ and, for every $s \in I$,
\begin{align*}
y'(s)=\nabla z(X(s))\cdot X'(s).
\end{align*}
Since $X'(s) = b(X(s))$, this becomes
\begin{align*}
y'(s)=\nabla z(X(s))\cdot b(X(s)).
\end{align*}
The Euclidean dot product is symmetric, and the transport equation for $z$ at the point $X(s) \in U$ gives
\begin{align*}
y'(s)=b(X(s))\cdot \nabla z(X(s))=-c(X(s))z(X(s)).
\end{align*}
By the definition of $y$, this is
\begin{align*}
y'(s)=-c(X(s))y(s).
\end{align*}
[guided]
We now restrict the partial differential equation to the single curve $X$. Define $y: I \to \mathbb{R}$ by $y(s) = z(X(s))$. This function records the value of the difference $z = u - v$ along the characteristic. The regularity assumptions are exactly what is needed for the chain rule: $z \in C^1(U)$ and $X \in C^1(I; U)$, so $y \in C^1(I)$ and $y'(s) = \nabla z(X(s)) \cdot X'(s)$ for every $s \in I$.
The defining property of a characteristic is that its velocity equals the vector field: $X'(s) = b(X(s))$. Substituting this into the chain-rule identity gives $y'(s) = \nabla z(X(s)) \cdot b(X(s))$.
Since the Euclidean dot product is symmetric, this equals $b(X(s)) \cdot \nabla z(X(s))$. The function $z$ satisfies the homogeneous transport equation $b(x) \cdot \nabla z(x) + c(x)z(x) = 0$ for every $x \in U$. Evaluating this identity at $x = X(s)$ gives $b(X(s)) \cdot \nabla z(X(s)) = -c(X(s))z(X(s))$.
Finally, by the definition of $y$, we have $z(X(s)) = y(s)$. Therefore $y'(s) = -c(X(s))y(s)$. Thus the transport equation along $X$ has become a scalar linear equation for $y$.
[/guided]
[/step]
[step:Use an integrating factor to force the scalar solution to vanish]
Define the [continuous function](/page/Continuous%20Function) $a: I \to \mathbb{R}$ by $a(s) = c(X(s))$. The preceding step gives $y'(s) + a(s)y(s) = 0$ for every $s \in I$, and the initial condition is $y(s_0) = 0$.
Fix $s \in I$. Let $J \subset I$ be the closed interval with endpoints $s_0$ and $s$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Define $A: J \to \mathbb{R}$ by
\begin{align*}
A(t)=\int_{s_0}^{t} a(r)\, d\mathcal{L}^1(r),
\end{align*}
where the integral is oriented in the usual one-dimensional sense. Since $a$ is continuous on $J$, the function $A$ is $C^1$ on $J$ and satisfies $A'(t) = a(t)$ for $t$ in the interior of $J$.
Define $M: J \to \mathbb{R}$ by $M(t) = e^{A(t)}y(t)$. For every interior point $t$ of $J$, the product rule gives
\begin{align*}
M'(t)
= e^{A(t)}A'(t)y(t) + e^{A(t)}y'(t).
\end{align*}
Using $A'(t)=a(t)$ and $y'(t)+a(t)y(t)=0$, this becomes
\begin{align*}
M'(t)=e^{A(t)}(a(t)y(t)+y'(t))=0.
\end{align*}
If $s=s_0$, then $y(s)=0$ immediately. Otherwise, $J$ has nonempty interior, and the [mean value theorem](/theorems/186) applied on each subinterval of $J$ shows that $M'(t)=0$ on the interior forces $M$ to be constant on all of $J$. Since $A(s_0) = 0$ and $y(s_0) = 0$, we have $M(s_0) = e^{A(s_0)}y(s_0) = 0$. Therefore $M(s) = 0$. Since $e^{A(s)} > 0$, it follows that $y(s) = 0$.
[/step]
[step:Translate vanishing of the difference back to equality of the original solutions]
The point $s \in I$ was arbitrary, so $y(s) = 0$ for every $s \in I$. By the definitions of $y$ and $z$, $0 = y(s) = z(X(s)) = u(X(s)) - v(X(s))$. Thus $u(X(s)) = v(X(s))$ for every $s \in I$, which is the desired local uniqueness along the characteristic.
[/step]
