Before I can state the actual problem I have to list some preceding definitions from § 4.10 in Bishop's book:
Given is the first-order PDE $F(x^1, \dots, x^n, p_1, \dots, p_n, z) = 0$ with $p_i = \frac{\partial z}{\partial x^i}, \, i = 1, \dots, n$ on $\mathbb R^{d}, d=2n+1$, and $F:\mathbb R^d \to \mathbb R$ is $\mathcal C^{\infty}$. The solution is an $n$-dimensional submanifold of $\mathbb R^d$ parametrised as $\{(x^1, \dots, x^n, z = f(x^1, \dots, x^n), p_1, \dots, p_n)\}$.
Associated to this is the two-dimensional codistribution $\Delta = \mathrm{span}\{\omega^0 = \mathrm d F, \omega^1 = \mathrm d z - p_i \mathrm d x^i\}$ (span taken pointwise), of which said submanifold is an integral submanifold, and the characteristic vector-field $X = f (G^i \partial_{x^i} - (F_i + p_i F_z) \partial_{p_i} + p_i G^i \partial_z)$, where $F_i = \partial_{x^i} F, \; G^i = \partial_{p_i} F, \; F_z = \partial_z F$, and any smooth functions $f\neq 0$ can be chosen. By construction $\omega^0(X) = 0 = \omega^1(X)$. And furthermore $X$ was constructed precisely in such a way that $i(X) \mathrm d \omega^1 = f (\omega^0 - F_z \omega^1) $, that is $X$ is the uniquely defined vector field for which $i(X) \mathrm d \omega^1 \in \Delta$. For sake of simplicity I will assume $f \equiv 1$ from now on.
The actual problem:
If $\Delta$ and $X$ are as above, show that there are local bases $\theta^0, \theta^1$ of $\Delta$ such that $i(X) \mathrm d \theta^0 = 0$ and $i(X) \mathrm d \theta^1 = 0$.
I believe that I could use $\theta^0 = \omega^0$ as it is closed by construction ($\mathrm d \omega^0 = \mathrm d^2 F = 0$ $\Rightarrow i(X) \mathrm d \omega^0 = i(X) 0 = 0$). But I have no clue how I can choose $\theta^1$. Somehow I should be able to make use of the fact that $i(X) \mathrm d \omega^1 \in \Delta$.
Any hints?
Update and solution attempt:
Pick a coordinate chart $\mu: U \to \mathbb R^d, \, n \mapsto (x^1 n, \dots, x^d n)$, centered at $m \in U$ (i.e. $\mu m = 0$), with associated coordinate basis $\{\partial_i\}$ which satisfies $X = \partial_1$ (this is always possible per Theorem 3.5.1 in the book and the fact that $X$ is nowhere zero) . Now choose on $U$ the ansatz $\theta^1 = \alpha \omega^0 + \beta \omega^1$ with $\alpha, \beta \in \mathcal C^{\infty}(U)$.
Evaluating the constraint on $\theta^1$:
$$\mathrm d \theta^1 = \mathrm d \alpha \wedge \omega^0 + \mathrm d \beta \wedge \omega^1 + \beta \mathrm d \omega^1, \quad$$ where $\mathrm d \omega^0 = 0$ was used.
$$ \begin{align} 0 &= i(\partial_1)\mathrm d \theta^1\\ &= (i(\partial_1)\mathrm d \alpha )\wedge \omega^0 - \mathrm d \alpha \wedge (i(\partial_1)\omega^0) + (i(\partial_1)\mathrm d \beta) \wedge \omega^1 - \mathrm d \beta \wedge (i(\partial_1)\omega^1) + \beta \, i(\partial_1)\mathrm d \omega^1\\ &= (i(\partial_1)\mathrm d \alpha )\wedge \omega^0 + (i(\partial_1)\mathrm d \beta) \wedge \omega^1 + \beta \, i(\partial_1)\mathrm d \omega^1\\ &= (\partial_1 \alpha) \omega^0 + (\partial_1 \beta) \omega^1 + \beta \, (\omega^0 -F_z \omega^1)\\ &= (\partial_1 \alpha + \beta)\omega^0 + (\partial_1 \beta - F_z \beta)\omega^1, \end{align}$$
where the defining properties of $X$ were used. Since $\omega^0, \omega^1$ are assumed to be linearly independent it follows that $\partial_1 \alpha + \beta = 0$ and $\partial_1 \beta - F_z \beta = 0$ have to hold.
Thus for fixed coordinate values $(x^2, \dots, x^n)$ (i.e. along the flow of $X$), $\alpha$ and $\beta$ have to satisfy the ODE $$\begin{align} \alpha' &= - \beta\\ \beta' &= \beta F_z \end{align} $$ where $f' = \partial_1 f$. Hence solutions can be constructed as follows: Let $\pi: \mathbb{R}^n \to \mathbb R^{n-1}: (x^1, \dots, x^n) \mapsto (x^2, \dots, x^n)$. Pick any smooth non-zero $\tilde{\alpha}, \tilde{\beta}: \mathbb R^{n-1} \to \mathbb R$ s.t. the domains of $\tilde{\alpha} \circ \pi \circ \mu$ and $\tilde{\beta} \circ \pi \circ \mu$ contain $U$. Then the solutions are given with respect to the coordinate chart $\mu$ as $$\begin{align} \beta \circ \mu^{-1}(x^1, \dots, x^n) &= \tilde{\beta}(x^2, \dots, x^n) e^{\int_0^{x^1}F_z \circ \mu^{-1}(s, x^2, \dots, x^n) \mathrm d s}\\ \alpha \circ \mu^{-1}(x^1, \dots, x^n) &= \tilde{\alpha}(x^2, \dots, x^n) - \int_0^{x^1}\beta \circ \mu^{-1}(s, x^2, \dots, x^n) \mathrm d s. \end{align}$$
Existence and uniqueness theorems of ODE's guarantee that on some open $V \subset U$ $\alpha, \beta$ are well-defined. On that $V$ $\{\theta^0, \theta^1\}$ is a local basis for $\Delta$ that satisfies the stipulated restrictions.
I would highly appreciate to receive comments on the validity of my solution, and whether it could be improved in any way. Thanks in advance.