[proofplan] We parametrize the hypersurface near $y_0$, lift the initial data to $U \times I$, and solve the characteristic ODE system for the position and height variables. The noncharacteristic hypothesis says exactly that the projected characteristic map has invertible derivative at the initial point, so the inverse function theorem supplies local coordinates by characteristic time and initial surface parameter. Defining $u$ by pulling the height component back through these characteristic coordinates gives the required $C^1$ solution. Local uniqueness follows because any other solution with the same initial trace produces lifted curves solving the same ODE system, hence agrees with the constructed characteristic family by uniqueness for ODEs. [/proofplan]

[step:Parametrize the initial hypersurface and lift the data to phase space]Choose an [open set](/page/Open%20Set) $\Theta \subset \mathbb{R}^{n-1}$, a parameter $\theta_0 \in \Theta$, and a $C^1$ embedded parametrization \begin{align*} Y: \Theta \to \Gamma \end{align*} of a relatively open neighbourhood of $y_0$ in $\Gamma$, with $Y(\theta_0)=y_0$ and with $DY_{\theta_0}: \mathbb{R}^{n-1} \to \mathbb{R}^n$ injective. After replacing $\Theta$ by a smaller open neighbourhood of $\theta_0$, we may assume $Y(\Theta) \subset U$. Define the lifted initial map \begin{align*} H: \Theta \to U \times I, \quad \theta \mapsto (Y(\theta), g(Y(\theta))). \end{align*} Since $Y$ and $g$ are $C^1$, the map $H$ is $C^1$.[/step]

[guided]The hypersurface data must be converted into initial data for an ordinary differential equation. Since $\Gamma$ is a $C^1$ embedded hypersurface and $y_0 \in \Gamma$, there is a local $C^1$ parametrization \begin{align*} Y: \Theta \to \Gamma \end{align*} from an open set $\Theta \subset \mathbb{R}^{n-1}$, with $Y(\theta_0)=y_0$ for some $\theta_0 \in \Theta$. The derivative \begin{align*} DY_{\theta_0}: \mathbb{R}^{n-1} \to \mathbb{R}^n \end{align*} is injective because $Y$ parametrizes an embedded hypersurface. Define the tangent space to the hypersurface at $y_0$ by \begin{align*} T_{y_0}\Gamma := \operatorname{Range}(DY_{\theta_0}) \subset \mathbb{R}^n. \end{align*} This definition is independent of the chosen embedded parametrization, and in the present parametrization it says exactly that the image of $DY_{\theta_0}$ is $T_{y_0}\Gamma$. We also shrink $\Theta$ if necessary so that $Y(\Theta) \subset U$. This is legitimate because $U$ is open and $Y(\theta_0)=y_0 \in U$. The initial graph in phase space $U \times I$ is then encoded by the map \begin{align*} H: \Theta \to U \times I, \quad \theta \mapsto (Y(\theta), g(Y(\theta))). \end{align*} The map $H$ is $C^1$ because it is built from the $C^1$ map $Y$ and the $C^1$ trace function $g: \Gamma \to I$. This lifted map records both the starting point $Y(\theta)$ and the starting value $g(Y(\theta))$ for the characteristic issuing from that point.[/guided]

[step:Solve the characteristic system with $C^1$ dependence on the initial parameter]Define the characteristic vector field \begin{align*} F: U \times I \to \mathbb{R}^{n+1}, \quad (x,z) \mapsto (a(x,z), b(x,z)). \end{align*} After replacing $I$ by a relatively open interval in $\mathbb{R}$ containing $g(y_0)$ if necessary, the phase domain $U \times I$ is open in $\mathbb{R}^{n+1}$. Since $a$ and $b$ are $C^1$, the map $F$ is $C^1$, hence continuous and locally Lipschitz on compact subsets of $U \times I$. Choose a closed ball $K \subset U \times I$ centred at $H(\theta_0)$ and contained in $U \times I$. On $K$, the functions $F$ and its derivative are bounded, so the boundedness, Lipschitz, and $C^1$ vector-field hypotheses of the [Picard Lindelof theorem with $C^1$ dependence on initial data](/theorems/69) hold for initial points in a smaller neighbourhood of $H(\theta_0)$. The lifted initial point $H(\theta_0)=(y_0,g(y_0))$ lies in this open phase domain. Applying the [Picard Lindelof theorem with $C^1$ dependence on initial data](/theorems/69) to the $C^1$ vector field $F$ and the $C^1$ initial map $H$, there exist $\varepsilon>0$ and an open neighbourhood $\Theta_1 \subset \Theta$ of $\theta_0$ such that the initial value problem \begin{align*} \frac{\partial}{\partial s}(X(s,\theta), Z(s,\theta)) = F(X(s,\theta), Z(s,\theta)) \end{align*} with initial condition \begin{align*} (X(0,\theta), Z(0,\theta)) = (Y(\theta), g(Y(\theta))) \end{align*} has a unique $C^1$ solution \begin{align*} (X,Z): (-\varepsilon,\varepsilon) \times \Theta_1 \to U \times I. \end{align*} Equivalently, \begin{align*} \partial_s X(s,\theta) = a(X(s,\theta), Z(s,\theta)) \end{align*} and \begin{align*} \partial_s Z(s,\theta) = b(X(s,\theta), Z(s,\theta)). \end{align*}[/step]

[guided]We now turn the geometric initial data into an ODE problem in phase space. The unknown phase variable is a pair $(x,z) \in U \times I$, where $x$ is the spatial position and $z$ is the value transported along the characteristic. Define \begin{align*} F: U \times I \to \mathbb{R}^{n+1}, \quad (x,z) \mapsto (a(x,z), b(x,z)). \end{align*} The characteristic equation is then the autonomous ODE whose velocity at $(x,z)$ is $F(x,z)$. We must verify the hypotheses of the ODE theorem before applying it. After shrinking $I$ to a relatively open interval in $\mathbb{R}$ containing $g(y_0)$, the set $U \times I$ is open in $\mathbb{R}^{n+1}$. Since $a$ and $b$ are $C^1$, the vector field $F$ is $C^1$. In finite-dimensional Euclidean space, a $C^1$ vector field is locally Lipschitz: on any compact ball $K \subset U \times I$, the derivative of $F$ is bounded, and the mean value estimate gives a Lipschitz constant on $K$. The same compactness also bounds $F$ on $K$. Thus the boundedness, Lipschitz, and $C^1$ vector-field hypotheses in the [Picard Lindelof theorem with $C^1$ dependence on initial data](/theorems/69) are satisfied after choosing a compact ball around $H(\theta_0)$ and shrinking the initial parameter neighbourhood so that $H(\theta)$ remains in that ball. The initial point for the characteristic labelled by $\theta$ is \begin{align*} H(\theta)=(Y(\theta),g(Y(\theta))). \end{align*} Because $H$ is $C^1$, the $C^1$ dependence conclusion in the linked ODE theorem gives solutions depending $C^1$ on both the time variable and the initial parameter. Therefore there exist $\varepsilon>0$ and an open neighbourhood $\Theta_1 \subset \Theta$ of $\theta_0$ such that the initial value problem \begin{align*} \frac{\partial}{\partial s}(X(s,\theta), Z(s,\theta)) = F(X(s,\theta), Z(s,\theta)) \end{align*} with \begin{align*} (X(0,\theta), Z(0,\theta)) = (Y(\theta), g(Y(\theta))) \end{align*} has a unique $C^1$ solution \begin{align*} (X,Z): (-\varepsilon,\varepsilon) \times \Theta_1 \to U \times I. \end{align*} Writing out the two components of this vector equation gives \begin{align*} \partial_s X(s,\theta) = a(X(s,\theta), Z(s,\theta)) \end{align*} and \begin{align*} \partial_s Z(s,\theta) = b(X(s,\theta), Z(s,\theta)). \end{align*} These are exactly the characteristic equations associated to the quasilinear PDE.[/guided]

[step:Use the noncharacteristic condition to invert the projected characteristic map]Define the projected characteristic map \begin{align*} \Phi: (-\varepsilon,\varepsilon) \times \Theta_1 \to U, \quad (s,\theta) \mapsto X(s,\theta). \end{align*} The set $(-\varepsilon,\varepsilon) \times \Theta_1$ is an open subset of $\mathbb{R} \times \mathbb{R}^{n-1}=\mathbb{R}^n$, and $\Phi$ is $C^1$ because $X$ is $C^1$. At $(0,\theta_0)$ its derivative has columns \begin{align*} \partial_s \Phi(0,\theta_0) = a(y_0,g(y_0)) \end{align*} and, for each $i \in \{1,\dots,n-1\}$, \begin{align*} \partial_{\theta_i}\Phi(0,\theta_0) = \partial_{\theta_i}Y(\theta_0). \end{align*} The vectors $\partial_{\theta_1}Y(\theta_0),\dots,\partial_{\theta_{n-1}}Y(\theta_0)$ span $T_{y_0}\Gamma$. Since $a(y_0,g(y_0)) \cdot \nu(y_0) \neq 0$ and $\nu(y_0)$ is normal to $T_{y_0}\Gamma$, the vector $a(y_0,g(y_0))$ is not tangent to $\Gamma$ at $y_0$. Hence the $n$ displayed columns form a basis of $\mathbb{R}^n$. By the [Inverse Function Theorem](/theorems/51), applied to the $C^1$ map $\Phi$ on the open subset $(-\varepsilon,\varepsilon) \times \Theta_1 \subset \mathbb{R}^n$ and using the invertibility of $D\Phi_{(0,\theta_0)}$, there are an open neighbourhood \begin{align*} Q \subset (-\varepsilon,\varepsilon) \times \Theta_1 \end{align*} of $(0,\theta_0)$ and an open neighbourhood $V_0 \subset U$ of $y_0$ such that \begin{align*} \Phi|_Q: Q \to V_0 \end{align*} is a $C^1$ diffeomorphism. Let \begin{align*} \Psi_0: V_0 \to Q \end{align*} denote its inverse.[/step]

[guided]The inverse function theorem is the point at which the noncharacteristic hypothesis is used. The theorem requires a $C^1$ map between open subsets of Euclidean spaces with an invertible derivative at the point under consideration. Here the domain $(-\varepsilon,\varepsilon) \times \Theta_1$ is open in $\mathbb{R} \times \mathbb{R}^{n-1}=\mathbb{R}^n$, and the map \begin{align*} \Phi: (-\varepsilon,\varepsilon) \times \Theta_1 \to U \end{align*} is $C^1$ because the characteristic position map $X$ is $C^1$. We compute the derivative at $(0,\theta_0)$ by differentiating the two kinds of variables separately. Differentiating in the characteristic time direction gives \begin{align*} \partial_s\Phi(0,\theta_0)=\partial_sX(0,\theta_0)=a(y_0,g(y_0)). \end{align*} Differentiating in the surface parameters gives, for each $i \in \{1,\dots,n-1\}$, \begin{align*} \partial_{\theta_i}\Phi(0,\theta_0)=\partial_{\theta_i}Y(\theta_0). \end{align*} The vectors $\partial_{\theta_1}Y(\theta_0),\dots,\partial_{\theta_{n-1}}Y(\theta_0)$ span the tangent space $T_{y_0}\Gamma$. The normal vector $\nu(y_0)$ is orthogonal to this tangent space, and the hypothesis \begin{align*} a(y_0,g(y_0))\cdot \nu(y_0)\ne 0 \end{align*} says that $a(y_0,g(y_0))$ has a nonzero normal component. Hence $a(y_0,g(y_0))$ is not tangent to $\Gamma$ at $y_0$. Therefore the time direction together with the $n-1$ surface directions forms a basis of $\mathbb{R}^n$, so $D\Phi_{(0,\theta_0)}$ is invertible. The [Inverse Function Theorem](/theorems/51) now gives open neighbourhoods $Q$ of $(0,\theta_0)$ and $V_0$ of $y_0$ such that \begin{align*} \Phi|_Q: Q \to V_0 \end{align*} is a $C^1$ diffeomorphism. Its inverse \begin{align*} \Psi_0: V_0 \to Q \end{align*} assigns to each nearby point its unique characteristic time and initial surface parameter.[/guided]

[step:Define the solution by characteristic coordinates and verify the initial condition]Because $Q$ is open and contains $(0,\theta_0)$, there are $\delta>0$ and an open neighbourhood $\Theta_2 \subset \Theta_1$ of $\theta_0$ such that \begin{align*} (-\delta,\delta) \times \Theta_2 \subset Q. \end{align*} Shrink the hypersurface parametrization domain to $\Theta_2$. Since $Y: \Theta_2 \to \Gamma$ is a parametrization of a relatively open neighbourhood of $y_0$ in $\Gamma$, and since $V_0$ is open in $U$, there is an open neighbourhood $V \subset V_0$ of $y_0$ such that \begin{align*} V \cap \Gamma \subset Y(\Theta_2). \end{align*} Let \begin{align*} \Psi := \Psi_0|_V: V \to Q. \end{align*} Write \begin{align*} \Psi(x) = (\sigma(x), \vartheta(x)) \end{align*} where $\sigma: V \to \mathbb{R}$ and $\vartheta: V \to \Theta_1$ are $C^1$ maps. Define \begin{align*} u: V \to I, \quad x \mapsto Z(\sigma(x),\vartheta(x)). \end{align*} The function $u$ is $C^1$ because $Z$, $\sigma$, and $\vartheta$ are $C^1$. If $y \in V \cap \Gamma$, then $y \in Y(\Theta_2)$, so there is $\theta \in \Theta_2$ with $y=Y(\theta)$. Since $(0,\theta) \in Q$ and $\Phi(0,\theta)=Y(\theta)=y$, the inverse relation for $\Psi=\Psi_0|_V$ gives $\Psi(y)=(0,\theta)$. Therefore \begin{align*} u(y)=Z(0,\theta)=g(Y(\theta))=g(y). \end{align*} Thus $u=g$ on the whole initial hypersurface patch $V \cap \Gamma$.[/step]

[guided]The inverse map from the preceding step gives characteristic coordinates for points near $y_0$. Because $Q$ is open and contains $(0,\theta_0)$, we can choose $\delta>0$ and an open neighbourhood $\Theta_2 \subset \Theta_1$ of $\theta_0$ such that \begin{align*} (-\delta,\delta) \times \Theta_2 \subset Q. \end{align*} We shrink the initial hypersurface patch to $Y(\Theta_2)$. Since $Y(\Theta_2)$ is a relatively open neighbourhood of $y_0$ in $\Gamma$ and $V_0$ is open in $U$, we may choose an open neighbourhood $V \subset V_0$ of $y_0$ satisfying \begin{align*} V \cap \Gamma \subset Y(\Theta_2). \end{align*} This ensures that every initial point of the final hypersurface patch is represented by one of the parameters for which the characteristic chart is valid. Let \begin{align*} \Psi := \Psi_0|_V: V \to Q. \end{align*} Since $\Psi_0$ is a $C^1$ inverse map, $\Psi$ is $C^1$. Write \begin{align*} \Psi(x) = (\sigma(x), \vartheta(x)) \end{align*} where $\sigma: V \to \mathbb{R}$ and $\vartheta: V \to \Theta_1$ are the time and initial-parameter components. Define the candidate solution by transporting the characteristic height back to physical space: \begin{align*} u: V \to I, \quad x \mapsto Z(\sigma(x),\vartheta(x)). \end{align*} The function $u$ is $C^1$ because it is a composition of the $C^1$ maps $Z$, $\sigma$, and $\vartheta$. Now verify the initial condition. Let $y \in V \cap \Gamma$. By the choice of $V$, there is $\theta \in \Theta_2$ such that $y=Y(\theta)$. Since $(0,\theta) \in Q$ and \begin{align*} \Phi(0,\theta)=X(0,\theta)=Y(\theta)=y, \end{align*} the inverse relation for $\Psi$ gives \begin{align*} \Psi(y)=(0,\theta). \end{align*} Therefore \begin{align*} u(y)=Z(0,\theta)=g(Y(\theta))=g(y). \end{align*} Thus $u=g$ on $V \cap \Gamma$, which is exactly the prescribed initial trace on the final local hypersurface patch.[/guided]

[step:Differentiate along characteristics to obtain the quasilinear equation]Fix $x \in V$, and write $\Psi(x)=(s,\theta)$. Since $x=\Phi(s,\theta)=X(s,\theta)$, the defining identity for $u$ gives \begin{align*} u(X(s,\theta)) = Z(s,\theta). \end{align*} Differentiate this identity with respect to $s$. The chain rule gives \begin{align*} \nabla u(X(s,\theta)) \cdot \partial_s X(s,\theta) = \partial_s Z(s,\theta). \end{align*} Using the characteristic equations, \begin{align*} \nabla u(X(s,\theta)) \cdot a(X(s,\theta), Z(s,\theta)) = b(X(s,\theta), Z(s,\theta)). \end{align*} Since $Z(s,\theta)=u(X(s,\theta))$ and $X(s,\theta)=x$, this becomes \begin{align*} a(x,u(x)) \cdot \nabla u(x) = b(x,u(x)). \end{align*} Because $x \in V$ was arbitrary, $u$ solves the stated equation on $V$.[/step]

[guided]The reason for defining $u$ through the inverse map $\Psi$ is that every point $x \in V$ has a unique representation as a point reached by one characteristic. Write this representation as \begin{align*} \Psi(x)=(s,\theta). \end{align*} Then \begin{align*} x=X(s,\theta) \end{align*} and, by definition of $u$, \begin{align*} u(X(s,\theta))=Z(s,\theta). \end{align*} Now differentiate this identity with respect to the characteristic time $s$. The maps involved are $C^1$, so the chain rule applies and gives \begin{align*} \nabla u(X(s,\theta)) \cdot \partial_s X(s,\theta)=\partial_s Z(s,\theta). \end{align*} The characteristic system was chosen precisely so that its right-hand side matches the coefficients in the PDE: \begin{align*} \partial_s X(s,\theta)=a(X(s,\theta),Z(s,\theta)) \end{align*} and \begin{align*} \partial_s Z(s,\theta)=b(X(s,\theta),Z(s,\theta)). \end{align*} Substituting these two identities into the differentiated equation yields \begin{align*} \nabla u(X(s,\theta)) \cdot a(X(s,\theta),Z(s,\theta))=b(X(s,\theta),Z(s,\theta)). \end{align*} Finally, because $u(X(s,\theta))=Z(s,\theta)$, the same identity can be written entirely in terms of the point $x=X(s,\theta)$: \begin{align*} a(x,u(x)) \cdot \nabla u(x)=b(x,u(x)). \end{align*} Thus the function reconstructed from the characteristic height satisfies the quasilinear first-order equation at every point reached by the local characteristic coordinates.[/guided]

[step:Prove local uniqueness by comparing lifted characteristic curves]Let $\widetilde V \subset U$ be an open neighbourhood of $y_0$, and let $\widetilde u \in C^1(\widetilde V;I)$ solve the same equation with $\widetilde u=g$ on $\widetilde V \cap \Gamma$. Choose $\eta>0$ and an open neighbourhood $\Theta_3 \subset \Theta_1$ of $\theta_0$ such that the product set \begin{align*} Q_1:=(-\eta,\eta)\times \Theta_3 \end{align*} has compact closure contained in $Q$, satisfies $X(Q_1) \subset V \cap \widetilde V$, and satisfies $Y(\Theta_3) \subset \widetilde V \cap \Gamma$. This is possible by continuity of $X$, continuity of $Y$, the fact that $(0,\theta_0) \in Q$, and openness of $V \cap \widetilde V$ and $\widetilde V \cap \Gamma$ relative to $\Gamma$. For this product neighbourhood, set \begin{align*} \Theta_{Q_1}:=\Theta_3. \end{align*} Then $\Theta_{Q_1}$ is an open neighbourhood of $\theta_0$ in $\Theta_1$. For each fixed $\theta \in \Theta_{Q_1}$, define the curve \begin{align*} \widetilde X_\theta: J_\theta \to \widetilde V \end{align*} as the solution of \begin{align*} \frac{d}{ds}\widetilde X_\theta(s)=a(\widetilde X_\theta(s),\widetilde u(\widetilde X_\theta(s))) \end{align*} with initial condition \begin{align*} \widetilde X_\theta(0)=Y(\theta), \end{align*} on a sufficiently small interval $J_\theta \subset \mathbb{R}$ containing $0$. Define \begin{align*} \widetilde Z_\theta: J_\theta \to I, \quad s \mapsto \widetilde u(\widetilde X_\theta(s)). \end{align*} By the chain rule and the PDE for $\widetilde u$, \begin{align*} \frac{d}{ds}\widetilde Z_\theta(s)=\nabla \widetilde u(\widetilde X_\theta(s)) \cdot a(\widetilde X_\theta(s),\widetilde u(\widetilde X_\theta(s)))=b(\widetilde X_\theta(s),\widetilde Z_\theta(s)). \end{align*} Thus the lifted curve \begin{align*} s \mapsto (\widetilde X_\theta(s),\widetilde Z_\theta(s)) \end{align*} solves the same ODE system as \begin{align*} s \mapsto (X(s,\theta),Z(s,\theta)). \end{align*} At $s=0$ both curves equal \begin{align*} (Y(\theta),g(Y(\theta))). \end{align*} By uniqueness in the [Picard Lindelof theorem](/theorems/69) for the characteristic ODE system, after restricting to a smaller common interval in $s$ if necessary, \begin{align*} (\widetilde X_\theta(s),\widetilde Z_\theta(s))=(X(s,\theta),Z(s,\theta)). \end{align*} For each $\theta$ near $\theta_0$, the two lifted curves agree on an open interval around $0$. To make the interval uniform, use the same compact set for all initial data. Define \begin{align*} K_1:=\{(X(s,\theta),Z(s,\theta)):(s,\theta)\in \overline{Q_1}\}\subset U\times I. \end{align*} The set $K_1$ is compact because $\overline{Q_1}$ is compact and $(X,Z)$ is continuous. Since $F$ is $C^1$, there is an open neighbourhood $N_1\subset U\times I$ of $K_1$ on which $F$ is bounded and Lipschitz. Applying the local-flow and uniqueness part of the [Picard Lindelof theorem](/theorems/69) on $N_1$, and shrinking to a smaller product neighbourhood $Q'\subset Q_1$ of $(0,\theta_0)$ if necessary, all lifted solutions starting from $(Y(\theta),g(Y(\theta)))$ with $(0,\theta)\in Q'$ are uniquely defined on the common $s$-interval occurring in $Q'$. Hence, for every $(s,\theta) \in Q'$, \begin{align*} \widetilde u(X(s,\theta))=Z(s,\theta)=u(X(s,\theta)). \end{align*} Setting \begin{align*} W:=\Phi(Q') \end{align*} gives an open neighbourhood $W \subset V \cap \widetilde V$ of $y_0$ in the same characteristic image, and $u=\widetilde u$ on $W$. This proves the asserted local uniqueness.[/step]

[guided]We compare the constructed solution with another solution by lifting both to curves in phase space. The possible difficulty is that uniqueness for ordinary differential equations is initially local in time and could depend on the starting parameter $\theta$. To make the comparison uniform, first shrink to an explicit product neighbourhood in the characteristic coordinates. Choose $\eta>0$ and an open neighbourhood $\Theta_3\subset \Theta_1$ of $\theta_0$ such that \begin{align*} Q_1:=(-\eta,\eta)\times \Theta_3 \end{align*} has compact closure contained in $Q$, satisfies $X(Q_1) \subset V \cap \widetilde V$, and has initial points $Y(\Theta_3)$ contained in $\widetilde V \cap \Gamma$. This choice is allowed because $X$ and $Y$ are continuous, $(0,\theta_0)\in Q$, and the target sets are open in their respective ambient spaces. For this product neighbourhood, set \begin{align*} \Theta_{Q_1}:=\Theta_3. \end{align*} Then $\Theta_{Q_1}$ is an open neighbourhood of $\theta_0$ in $\Theta_1$. Fix $\theta \in \Theta_{Q_1}$. Let \begin{align*} \widetilde X_\theta: J_\theta \to \widetilde V \end{align*} be the solution of \begin{align*} \frac{d}{ds}\widetilde X_\theta(s)=a(\widetilde X_\theta(s),\widetilde u(\widetilde X_\theta(s))) \end{align*} with initial condition $\widetilde X_\theta(0)=Y(\theta)$. Define \begin{align*} \widetilde Z_\theta: J_\theta \to I, \quad s \mapsto \widetilde u(\widetilde X_\theta(s)). \end{align*} Because $\widetilde u$ is $C^1$ and solves the same PDE, the chain rule gives \begin{align*} \frac{d}{ds}\widetilde Z_\theta(s)=\nabla \widetilde u(\widetilde X_\theta(s))\cdot a(\widetilde X_\theta(s),\widetilde u(\widetilde X_\theta(s)))=b(\widetilde X_\theta(s),\widetilde Z_\theta(s)). \end{align*} Thus the lifted curve $s \mapsto (\widetilde X_\theta(s),\widetilde Z_\theta(s))$ solves the same characteristic ODE system as $s \mapsto (X(s,\theta),Z(s,\theta))$. At $s=0$ both curves have the same initial value because $\widetilde u=g$ on $\widetilde V \cap \Gamma$: \begin{align*} (\widetilde X_\theta(0),\widetilde Z_\theta(0))=(Y(\theta),g(Y(\theta)))=(X(0,\theta),Z(0,\theta)). \end{align*} Uniqueness in the [Picard Lindelof theorem](/theorems/69) for the characteristic ODE system therefore gives equality of the lifted curves for small $s$. The final shrinking makes this small interval independent of $\theta$ near $\theta_0$. Define the compact phase-space trace of the constructed characteristics by \begin{align*} K_1:=\{(X(s,\theta),Z(s,\theta)):(s,\theta)\in \overline{Q_1}\}\subset U\times I. \end{align*} This set is compact because $\overline{Q_1}$ is compact and $(X,Z)$ is continuous. Since $F$ is $C^1$, we may choose an open neighbourhood $N_1\subset U\times I$ of $K_1$ on which $F$ is bounded and Lipschitz. The local-flow and uniqueness part of the [Picard Lindelof theorem](/theorems/69), applied on $N_1$, gives a common existence and uniqueness neighbourhood for all initial data $(Y(\theta),g(Y(\theta)))$ with $\theta$ sufficiently close to $\theta_0$. After replacing $Q_1$ by a smaller open product neighbourhood $Q'$ of $(0,\theta_0)$, both lifted curves stay in $N_1$ and uniqueness applies on the whole $s$-interval represented in $Q'$. Consequently, for every $(s,\theta) \in Q'$, \begin{align*} \widetilde u(X(s,\theta))=Z(s,\theta)=u(X(s,\theta)). \end{align*} The image $W=\Phi(Q')$ is open because $\Phi|_Q$ is a diffeomorphism, contains $y_0$, and lies in the same characteristic image. Hence $u=\widetilde u$ on $W$, which is the asserted local uniqueness.[/guided]