[proofplan] We write the characteristic equations as one autonomous ordinary differential equation on $U \times I$ with a $C^k$ vector field. Since the initial point of the lifted characteristic depends $C^k$ on $(\lambda,\theta)$, smooth dependence for ordinary differential equations gives a common $C^k$ characteristic flow on a small parameter box. The noncharacteristic condition implies that the spatial characteristic map $(s,\theta)\mapsto x_{\lambda_0}(s,\theta)$ has invertible derivative at $(0,\theta_0)$, and continuity preserves this invertibility for nearby $\lambda$. A parameter-dependent inverse function theorem then supplies a $C^k$ inverse $(\lambda,x)\mapsto (s_\lambda(x),\theta_\lambda(x))$, and composing it with the lifted characteristic height $z_\lambda$ gives the desired $C^k$ family of solutions. [/proofplan]

[step:Rewrite the characteristic equations as a parameter-dependent initial value problem]Define the lifted vector field \begin{align*} F: U \times I \to \mathbb{R}^{n+1} \end{align*} by $F(x,z)=(a(x,z),b(x,z))$. Since $a$ and $b$ are $C^k$, the map $F$ is $C^k$. Define the initial-data map \begin{align*} Q: \Lambda \times \Theta \to U \times I \end{align*} by $Q(\lambda,\theta)=(Y(\theta),h(\lambda,\theta))$. Because $Y$ and $h$ are $C^k$, the map $Q$ is $C^k$. For each fixed $(\lambda,\theta)$, let $J_{\lambda,\theta}\subset\mathbb{R}$ denote the maximal open interval containing $0$ on which the lifted characteristic curve is defined. The lifted characteristic curve is the solution \begin{align*} \gamma_{\lambda,\theta}: J_{\lambda,\theta} \to U \times I \end{align*} of the ordinary differential equation \begin{align*} \frac{d}{ds}\gamma_{\lambda,\theta}(s)=F(\gamma_{\lambda,\theta}(s)). \end{align*} It satisfies the initial condition \begin{align*} \gamma_{\lambda,\theta}(0)=Q(\lambda,\theta). \end{align*} Writing \begin{align*} \gamma_{\lambda,\theta}(s)=(x_\lambda(s,\theta),z_\lambda(s,\theta)) \end{align*} recovers exactly the characteristic system in the theorem statement. By the $C^k$ smooth-dependence refinement of the [Picard-Lindelof theorem](/theorems/69), applied to the $C^k$ vector field $F$ and the $C^k$ initial-data map $Q$, there exist open neighbourhoods $\Lambda_1 \subset \Lambda$ of $\lambda_0$, $\Theta_1 \subset \Theta$ of $\theta_0$, and an interval $J=(-\varepsilon,\varepsilon)$ with $\varepsilon>0$ such that the solution map $\Gamma: \Lambda_1 \times J \times \Theta_1 \to U \times I$, defined by $\Gamma(\lambda,s,\theta)=\gamma_{\lambda,\theta}(s)$, is $C^k$. The hypotheses are satisfied because $F$ is $C^k$, the initial point $Q(\lambda,\theta)$ depends $C^k$ on $(\lambda,\theta)$, and $Q(\lambda_0,\theta_0)\in U\times I$. Let $X: \Lambda_1 \times J \times \Theta_1 \to U$ be defined by $X(\lambda,s,\theta)=x_\lambda(s,\theta)$, and let $Z: \Lambda_1 \times J \times \Theta_1 \to I$ be defined by $Z(\lambda,s,\theta)=z_\lambda(s,\theta)$. These are the component maps of $\Gamma$. Then $X$ and $Z$ are $C^k$.[/step]

[guided]The characteristic system is easiest to control if we regard the spatial component $x$ and the unknown value $z$ as a single point in $U \times I$. Define \begin{align*} F: U \times I \to \mathbb{R}^{n+1} \end{align*} by $F(x,z)=(a(x,z),b(x,z))$. This is a $C^k$ map because both component maps $a:U\times I\to\mathbb{R}^n$ and $b:U\times I\to\mathbb{R}$ are $C^k$. The initial condition also depends on parameters. We package it as the map \begin{align*} Q: \Lambda \times \Theta \to U \times I \end{align*} defined by $Q(\lambda,\theta)=(Y(\theta),h(\lambda,\theta))$. The map $Q$ is $C^k$ because $Y$ and $h$ are $C^k$. For fixed $(\lambda,\theta)$, let $J_{\lambda,\theta}\subset\mathbb{R}$ denote the maximal open interval containing $0$ on which the lifted characteristic curve is defined, and define \begin{align*} \gamma_{\lambda,\theta}: J_{\lambda,\theta} \to U \times I \end{align*} to be the solution of \begin{align*} \frac{d}{ds}\gamma_{\lambda,\theta}(s)=F(\gamma_{\lambda,\theta}(s)). \end{align*} It satisfies the initial condition \begin{align*} \gamma_{\lambda,\theta}(0)=Q(\lambda,\theta). \end{align*} If we write $\gamma_{\lambda,\theta}(s)=(x_\lambda(s,\theta),z_\lambda(s,\theta))$, then the single lifted equation is precisely the pair of characteristic equations \begin{align*} \frac{\partial x_\lambda}{\partial s}(s,\theta)=a(x_\lambda(s,\theta),z_\lambda(s,\theta)) \end{align*} and \begin{align*} \frac{\partial z_\lambda}{\partial s}(s,\theta)=b(x_\lambda(s,\theta),z_\lambda(s,\theta)). \end{align*} The initial condition for the spatial component is \begin{align*} x_\lambda(0,\theta)=Y(\theta). \end{align*} The initial condition for the height component is \begin{align*} z_\lambda(0,\theta)=h(\lambda,\theta). \end{align*} We now use the $C^k$ smooth-dependence refinement of the [Picard-Lindelof theorem](/theorems/69). Its hypotheses are satisfied: the vector field $F$ is $C^k$, the initial point $Q(\lambda,\theta)$ depends $C^k$ on $(\lambda,\theta)$, and the initial point $Q(\lambda_0,\theta_0)=(Y(\theta_0),h(\lambda_0,\theta_0))$ lies in the [open set](/page/Open%20Set) $U\times I$. Therefore, after shrinking to neighbourhoods $\Lambda_1$ of $\lambda_0$ and $\Theta_1$ of $\theta_0$, and after choosing an interval $J=(-\varepsilon,\varepsilon)$, the solution exists for all $(\lambda,s,\theta)\in \Lambda_1\times J\times \Theta_1$ and defines a $C^k$ map $\Gamma: \Lambda_1 \times J \times \Theta_1 \to U \times I$ by $\Gamma(\lambda,s,\theta)=\gamma_{\lambda,\theta}(s)$. Taking coordinate projections gives $C^k$ maps $X: \Lambda_1 \times J \times \Theta_1 \to U$ and $Z: \Lambda_1 \times J \times \Theta_1 \to I$ defined by $X(\lambda,s,\theta)=x_\lambda(s,\theta)$ and $Z(\lambda,s,\theta)=z_\lambda(s,\theta)$. This common parameter box is the point of invoking smooth dependence: it gives one domain on which all nearby characteristics are defined and vary $C^k$ in every variable.[/guided]

[step:Use the noncharacteristic condition to invert the spatial characteristic map] For each $\lambda \in \Lambda_1$, define $X_\lambda: J \times \Theta_1 \to U$ by $X_\lambda(s,\theta)=X(\lambda,s,\theta)=x_\lambda(s,\theta)$. At $(s,\theta)=(0,\theta_0)$ and $\lambda=\lambda_0$, the derivative with respect to $(s,\theta)$ has columns \begin{align*} \partial_s X_{\lambda_0}(0,\theta_0) = a(Y(\theta_0),h(\lambda_0,\theta_0)) \end{align*} and \begin{align*} \partial_{\theta_j}X_{\lambda_0}(0,\theta_0) = \partial_{\theta_j}Y(\theta_0), \qquad 1 \leq j \leq n-1. \end{align*} The vectors $\partial_{\theta_1}Y(\theta_0),\dots,\partial_{\theta_{n-1}}Y(\theta_0)$ span the tangent space of the parametrised hypersurface at $y_0$, and $\nu(Y(\theta_0))$ is normal to that tangent space. Since \begin{align*} a(Y(\theta_0),h(\lambda_0,\theta_0)) \cdot \nu(Y(\theta_0)) \neq 0, \end{align*} the vector $a(Y(\theta_0),h(\lambda_0,\theta_0))$ is not tangent to the hypersurface. Hence the $n$ vectors \begin{align*} a(Y(\theta_0),h(\lambda_0,\theta_0)), \quad \partial_{\theta_1}Y(\theta_0), \quad \dots, \quad \partial_{\theta_{n-1}}Y(\theta_0) \end{align*} form a basis of $\mathbb{R}^n$. Therefore \begin{align*} \det D_{(s,\theta)}X_{\lambda_0}(0,\theta_0) \neq 0. \end{align*} Because $X$ is $C^k$ and $k\geq 1$, the map \begin{align*} (\lambda,s,\theta) \mapsto \det D_{(s,\theta)}X(\lambda,s,\theta) \end{align*} is continuous. After shrinking $\Lambda_1$, $J$, and $\Theta_1$, we may assume \begin{align*} \det D_{(s,\theta)}X(\lambda,s,\theta)\neq 0 \end{align*} for every $(\lambda,s,\theta)\in \Lambda_1\times J\times \Theta_1$. [/step]

[step:Apply the inverse function theorem with parameters]Define the augmented map $\mathcal{X}: \Lambda_1 \times J \times \Theta_1 \to \Lambda_1\times U$ by $\mathcal{X}(\lambda,s,\theta)=(\lambda,X(\lambda,s,\theta))$. Since $X$ is $C^k$, the map $\mathcal{X}$ is $C^k$. Its derivative at $(\lambda_0,0,\theta_0)$ is block triangular: the derivative in the $\lambda$ variables is the identity on the parameter component, and the derivative in the $(s,\theta)$ variables on the spatial component is $D_{(s,\theta)}X_{\lambda_0}(0,\theta_0)$. Hence $D\mathcal{X}_{(\lambda_0,0,\theta_0)}$ is invertible because $D_{(s,\theta)}X_{\lambda_0}(0,\theta_0)$ is invertible. By the [Inverse Function Theorem](/theorems/51), after shrinking there are open neighbourhoods $\Lambda_2\subset\Lambda_1$ of $\lambda_0$, $J_2\subset J$ of $0$, $\Theta_2\subset\Theta_1$ of $\theta_0$, and an open neighbourhood $W\subset\Lambda_1\times U$ of $(\lambda_0,y_0)$ such that $\mathcal{X}:\Lambda_2\times J_2\times\Theta_2\to W$ is a $C^k$ diffeomorphism onto its image. Since $W$ is open in $\Lambda_1\times U$ and contains $(\lambda_0,y_0)$, we may shrink $\Lambda_2$ and choose an open neighbourhood $V\subset U$ of $y_0$ so that $\Lambda_2\times V\subset W$. On this product neighbourhood, the inverse has the form $(\lambda,x)\mapsto (\lambda,S(\lambda,x))$ because the first component of $\mathcal{X}$ is $\lambda$. Thus $S: \Lambda_2\times V\to J_2\times\Theta_2$ is $C^k$, and $V\subset X_\lambda(J_2\times\Theta_2)$ for every $\lambda\in\Lambda_2$. Thus there exist open neighbourhoods $J_2\subset J$ of $0$, $\Theta_2\subset\Theta_1$ of $\theta_0$, $\Lambda_2\subset\Lambda_1$ of $\lambda_0$, an open neighbourhood $V\subset U$ of $y_0$, and a $C^k$ map $S: \Lambda_2 \times V \to J_2 \times \Theta_2$ defined by $S(\lambda,x)=(s_\lambda(x),\theta_\lambda(x))$, such that, for every $\lambda\in\Lambda_2$, the restricted map \begin{align*} X_\lambda|_{J_2\times\Theta_2}: J_2 \times \Theta_2 \to X_\lambda(J_2\times\Theta_2) \end{align*} is a $C^k$ diffeomorphism onto its open image, $V\subset X_\lambda(J_2\times\Theta_2)$, and \begin{align*} (X_\lambda|_{J_2\times\Theta_2})^{-1}(x)=(s_\lambda(x),\theta_\lambda(x)) \end{align*} for every $(\lambda,x)\in\Lambda_2\times V$. Equivalently, \begin{align*} X(\lambda,s_\lambda(x),\theta_\lambda(x))=x \end{align*} for every $(\lambda,x)\in\Lambda_2\times V$.[/step]

[guided]We now turn local invertibility for each fixed $\lambda$ into a jointly $C^k$ inverse depending on $\lambda$. The relevant map is $X: \Lambda_1 \times J \times \Theta_1 \to U$, defined by $X(\lambda,s,\theta)=x_\lambda(s,\theta)$. For every point in the shrunk box, the derivative of $X$ with respect to the variables being inverted, namely $(s,\theta)$, is an invertible [linear map](/page/Linear%20Map) \begin{align*} D_{(s,\theta)}X(\lambda,s,\theta): \mathbb{R}\times\mathbb{R}^{n-1}\to \mathbb{R}^n. \end{align*} Rather than citing a separate parameter-dependent theorem, we obtain the uniform statement from the ordinary [Inverse Function Theorem](/theorems/51). Define the augmented map $\mathcal{X}:\Lambda_1\times J\times\Theta_1\to\Lambda_1\times U$ by $\mathcal{X}(\lambda,s,\theta)=(\lambda,X(\lambda,s,\theta))$. This map is $C^k$ because $X$ is $C^k$. Its derivative at $(\lambda_0,0,\theta_0)$ is block triangular: the parameter block is the identity on $\mathbb{R}^m$, and the spatial block in the variables $(s,\theta)$ is $D_{(s,\theta)}X_{\lambda_0}(0,\theta_0)$. Since that spatial block is invertible, the full derivative of $\mathcal{X}$ is invertible. The inverse function theorem therefore gives, after shrinking, a $C^k$ inverse to $\mathcal{X}$ on a neighbourhood of $(\lambda_0,y_0)$ in $\Lambda_1\times U$. Now use the fact that the target neighbourhood is open in $\Lambda_1\times U$ and contains $(\lambda_0,y_0)$. After shrinking $\Lambda_2$ and choosing an open neighbourhood $V\subset U$ of $y_0$, we may ensure $\Lambda_2\times V$ lies inside that target neighbourhood. Because the first component of $\mathcal{X}(\lambda,s,\theta)$ is exactly $\lambda$, its inverse on this product has the form $(\lambda,x)\mapsto (\lambda,S(\lambda,x))$. This gives a single $C^k$ inverse map $S:\Lambda_2\times V\to J_2\times\Theta_2$ defined by $S(\lambda,x)=(s_\lambda(x),\theta_\lambda(x))$. For each fixed $\lambda$, this says that the restriction $X_\lambda|_{J_2\times\Theta_2}$ is a $C^k$ diffeomorphism onto its open image, that $V$ lies inside that image, and that \begin{align*} (X_\lambda|_{J_2\times\Theta_2})^{-1}(x)=(s_\lambda(x),\theta_\lambda(x)) \end{align*} for every $x\in V$. Thus \begin{align*} X(\lambda,s_\lambda(x),\theta_\lambda(x))=x \end{align*} for every $(\lambda,x)\in\Lambda_2\times V$. This is the step that lets us solve for the characteristic label $\theta$ and characteristic time $s$ as $C^k$ functions of the physical point $x$ and the external parameter $\lambda$.[/guided]

[step:Define the solution by pulling back the lifted characteristic height]Define $u: \Lambda_2 \times V \to I$ by $u(\lambda,x)=Z(\lambda,s_\lambda(x),\theta_\lambda(x))$. For each $\lambda\in\Lambda_2$, define $u_\lambda: V \to I$ by $u_\lambda(x)=u(\lambda,x)$. Since $Z$ is $C^k$ and $(\lambda,x)\mapsto (s_\lambda(x),\theta_\lambda(x))$ is $C^k$, the composition defining $u$ is $C^k$. It remains to verify the initial condition. Shrink $\Theta_2$ further, if necessary, so that $Y(\Theta_2)\subset V$. If $\theta\in\Theta_2$, then \begin{align*} X(\lambda,0,\theta)=x_\lambda(0,\theta)=Y(\theta). \end{align*} Since $S(\lambda,\cdot)$ is the inverse of the restricted map $X_\lambda|_{J_2\times\Theta_2}$ on $V$, this gives \begin{align*} s_\lambda(Y(\theta))=0, \qquad \theta_\lambda(Y(\theta))=\theta. \end{align*} Therefore $u_\lambda(Y(\theta))=Z(\lambda,0,\theta)=z_\lambda(0,\theta)=h(\lambda,\theta)$. Thus the constructed family has the prescribed initial data. Finally, we verify the equation. The inverse identity gives \begin{align*} u_\lambda(X(\lambda,s,\theta))=Z(\lambda,s,\theta) \end{align*} for $(\lambda,s,\theta)\in\Lambda_2\times J_2\times\Theta_2$. Since $k\geq 1$, differentiating this identity with respect to $s$ and using the chain rule gives \begin{align*} \nabla u_\lambda(X(\lambda,s,\theta))\cdot \partial_s X(\lambda,s,\theta)=\partial_s Z(\lambda,s,\theta). \end{align*} The characteristic equations identify $\partial_s X(\lambda,s,\theta)$ with $a(X(\lambda,s,\theta),Z(\lambda,s,\theta))$ and $\partial_s Z(\lambda,s,\theta)$ with $b(X(\lambda,s,\theta),Z(\lambda,s,\theta))$. For $x\in V$, set $(s,\theta)=(s_\lambda(x),\theta_\lambda(x))$. Since $X(\lambda,s_\lambda(x),\theta_\lambda(x))=x$ and $Z(\lambda,s_\lambda(x),\theta_\lambda(x))=u_\lambda(x)$, substitution yields \begin{align*} a(x,u_\lambda(x))\cdot \nabla u_\lambda(x)=b(x,u_\lambda(x)). \end{align*} Thus $u_\lambda$ is a classical solution with the prescribed initial data.[/step]

[guided]The inverse branch is defined only on $\Lambda_2\times V$, so the solution must be defined on that same parameter neighbourhood. Define $u: \Lambda_2 \times V \to I$ by \begin{align*} u(\lambda,x)=Z(\lambda,s_\lambda(x),\theta_\lambda(x)). \end{align*} For each $\lambda\in\Lambda_2$, define $u_\lambda:V\to I$ by $u_\lambda(x)=u(\lambda,x)$. The map $Z$ is $C^k$ from the ODE smooth-dependence step, and the inverse branch $S(\lambda,x)=(s_\lambda(x),\theta_\lambda(x))$ is $C^k$ from the parameter-dependent inverse theorem. Therefore their composition is $C^k$, so $(\lambda,x)\mapsto u_\lambda(x)$ is $C^k$ on $\Lambda_2\times V$. We now check the initial condition. Shrink $\Theta_2$ further, if necessary, so that $Y(\Theta_2)\subset V$. For $\theta\in\Theta_2$, the characteristic initial condition gives $X(\lambda,0,\theta)=x_\lambda(0,\theta)=Y(\theta)$. Since $S(\lambda,\cdot)$ is the chosen inverse branch of $X_\lambda$ on $V$, applying this inverse to $Y(\theta)$ gives $s_\lambda(Y(\theta))=0$ and $\theta_\lambda(Y(\theta))=\theta$. Hence $u_\lambda(Y(\theta))=Z(\lambda,0,\theta)=z_\lambda(0,\theta)=h(\lambda,\theta)$. Finally we verify that $u_\lambda$ satisfies the first-order PDE. The defining identity is \begin{align*} u_\lambda(X(\lambda,s,\theta))=Z(\lambda,s,\theta), \end{align*} because $S$ is the inverse branch of $X_\lambda$ and $u_\lambda$ was defined by composing $Z$ with that inverse branch. Since $k\geq 1$, all functions in this identity are differentiable. Differentiating with respect to the characteristic time $s$ gives \begin{align*} \nabla u_\lambda(X(\lambda,s,\theta))\cdot \partial_s X(\lambda,s,\theta)=\partial_s Z(\lambda,s,\theta). \end{align*} The characteristic system identifies the two $s$-derivatives: \begin{align*} \partial_s X(\lambda,s,\theta)=a(X(\lambda,s,\theta),Z(\lambda,s,\theta)) \end{align*} and \begin{align*} \partial_s Z(\lambda,s,\theta)=b(X(\lambda,s,\theta),Z(\lambda,s,\theta)). \end{align*} For an arbitrary $x\in V$, put $(s,\theta)=(s_\lambda(x),\theta_\lambda(x))$. The inverse identity gives $X(\lambda,s_\lambda(x),\theta_\lambda(x))=x$, and the definition of $u_\lambda$ gives $Z(\lambda,s_\lambda(x),\theta_\lambda(x))=u_\lambda(x)$. Substituting these two identities into the differentiated characteristic identity yields \begin{align*} a(x,u_\lambda(x))\cdot \nabla u_\lambda(x)=b(x,u_\lambda(x)). \end{align*} Thus $u_\lambda$ is a classical solution, it has the prescribed initial data, and the solution family depends $C^k$ on $(\lambda,x)$.[/guided]

[step:Record the common neighbourhood and conclude smooth dependence]The preceding construction used only shrinking of $\Lambda_1$, $\Theta_1$, $J$, and $V$ to $\Lambda_2$, $\Theta_2$, $J_2$, and $V$. Since the lifted solution map $\Gamma: \Lambda_2 \times J_2 \times \Theta_2 \to U \times I$ takes values in $U\times I$, all lifted characteristic curves in the final box remain in $U\times I$. The determinant condition \begin{align*} \det D_{(s,\theta)}X(\lambda,s,\theta)\neq 0 \end{align*} holds throughout the same box by construction, so each spatial characteristic map has invertible Jacobian on that box. We use only the inverse branch on the common neighbourhood $V\subset X_\lambda(J_2\times\Theta_2)$; we do not require the whole fixed box $J_2\times\Theta_2$ to map onto $V$. The map $u: \Lambda_2 \times V \to I$, defined by $u(\lambda,x)=u_\lambda(x)$, is $C^k$, and each $u_\lambda$ satisfies \begin{align*} u_\lambda(Y(\theta))=h(\lambda,\theta) \end{align*} for $\theta\in\Theta_2$. Renaming the shrunk neighbourhoods as $\Lambda$, $\Theta$, and $V$ gives the theorem.[/step]

[guided]The final point is bookkeeping: all earlier shrinkings must be compatible with the statement of the theorem. We have shrunk the original parameter neighbourhoods and characteristic-time interval to $\Lambda_2$, $\Theta_2$, and $J_2$, and we have chosen one physical neighbourhood $V\subset U$ of $y_0$. The lifted solution map $\Gamma:\Lambda_2\times J_2\times\Theta_2\to U\times I$ still takes values in $U\times I$, so every lifted characteristic used in the construction remains inside the domain where $a$ and $b$ are defined. The same final box also preserves the nondegeneracy needed for inversion. By construction, \begin{align*} \det D_{(s,\theta)}X(\lambda,s,\theta)\neq 0 \end{align*} for every $(\lambda,s,\theta)\in\Lambda_2\times J_2\times\Theta_2$. Thus the spatial characteristic map has invertible Jacobian throughout the parameter box. We only use the inverse branch on the common neighbourhood $V\subset X_\lambda(J_2\times\Theta_2)$, so the proof does not require the fixed box $J_2\times\Theta_2$ to map exactly onto $V$. The map $u:\Lambda_2\times V\to I$, defined by $u(\lambda,x)=u_\lambda(x)$, is $C^k$, and the initial-data verification gives \begin{align*} u_\lambda(Y(\theta))=h(\lambda,\theta) \end{align*} for every $\theta\in\Theta_2$. Renaming $\Lambda_2$, $\Theta_2$, and $V$ as the final neighbourhoods $\Lambda$, $\Theta$, and $V$ gives exactly the smooth-dependence statement.[/guided]