[proofplan] The Hilbert form can be written in Hamiltonian coordinates as \begin{align*} \omega=-H(x,y,P)\,dx+P\cdot dy. \end{align*} If it is closed, exactness of closed $1$-forms on simply connected domains gives a potential $S$ with $dS=\omega$. Comparing the $dy$ and $dx$ components gives $\nabla_yS=P$ and the Hamilton-Jacobi equation. Conversely, any such $S$ satisfies $dS=\omega$, so $d\omega=d^2S=0$. [/proofplan]

[step:Rewrite the Hilbert form in Hamiltonian coordinates] By definition, \begin{align*} \omega=L(x,y,q(x,y))\,dx+P(x,y)\cdot(dy-q(x,y)\,dx). \end{align*} Expanding the right-hand side gives \begin{align*} \omega=\bigl(L(x,y,q(x,y))-P(x,y)\cdot q(x,y)\bigr)\,dx+P(x,y)\cdot dy. \end{align*} The Hamiltonian identity along the field is \begin{align*} H(x,y,P(x,y))=P(x,y)\cdot q(x,y)-L(x,y,q(x,y)). \end{align*} Substituting this identity gives \begin{align*} \omega=-H(x,y,P(x,y))\,dx+P(x,y)\cdot dy. \end{align*} [/step]

[step:Integrate a closed Hilbert form to a Hamilton-Jacobi potential]Assume $d\omega=0$. The coefficient $P$ is $C^1$ by hypothesis, and the coefficient $-H(x,y,P(x,y))$ is equal to $L(x,y,q(x,y))-P(x,y)\cdot q(x,y)$, which is $C^1$ because $L$ is $C^2$ and $q$ and $P$ are $C^1$. Thus the coefficients of $\omega$ are $C^1$. Since $D$ is simply connected, [Closed 1-Forms on Simply Connected Domains are Exact](/theorems/3618) gives a $C^2$ function $S:D\to\mathbb{R}$ such that \begin{align*} dS=\omega. \end{align*} Writing \begin{align*} dS=\partial_xS\,dx+\nabla_yS\cdot dy \end{align*} and comparing with \begin{align*} \omega=-H(x,y,P(x,y))\,dx+P(x,y)\cdot dy \end{align*} gives \begin{align*} \nabla_yS(x,y)=P(x,y) \end{align*} and \begin{align*} \partial_xS(x,y)=-H(x,y,P(x,y)). \end{align*} Substituting $P=\nabla_yS$ into the second identity gives \begin{align*} \partial_xS(x,y)+H(x,y,\nabla_yS(x,y))=0. \end{align*}[/step]

[guided]First rewrite the Hilbert form directly. By definition, \begin{align*} \omega=L(x,y,q(x,y))\,dx+P(x,y)\cdot(dy-q(x,y)\,dx). \end{align*} Expanding gives \begin{align*} \omega=\bigl(L(x,y,q(x,y))-P(x,y)\cdot q(x,y)\bigr)\,dx+P(x,y)\cdot dy. \end{align*} Since \begin{align*} H(x,y,P(x,y))=P(x,y)\cdot q(x,y)-L(x,y,q(x,y)), \end{align*} this becomes \begin{align*} \omega=-H(x,y,P(x,y))\,dx+P(x,y)\cdot dy. \end{align*} The coefficient $P$ is $C^1$. The coefficient $-H(x,y,P(x,y))$ is the same as $L(x,y,q(x,y))-P(x,y)\cdot q(x,y)$, so it is $C^1$ because $L$ is $C^2$ and $q$ and $P$ are $C^1$. Thus $\omega$ is a $C^1$ one-form. The closedness condition $d\omega=0$ is exactly the integrability condition for finding a potential. Since $D$ is simply connected, the exactness theorem for closed $1$-forms applies and gives $S$ with \begin{align*} dS=\omega. \end{align*} Now the whole argument is coefficient comparison. The differential of $S$ is \begin{align*} dS=\partial_xS\,dx+\nabla_yS\cdot dy. \end{align*} The Hilbert form in Hamiltonian coordinates is \begin{align*} \omega=-H(x,y,P(x,y))\,dx+P(x,y)\cdot dy. \end{align*} Equality of these two $1$-forms forces the $dy$ coefficients to agree, so \begin{align*} \nabla_yS=P. \end{align*} It also forces the $dx$ coefficients to agree, so \begin{align*} \partial_xS=-H(x,y,P). \end{align*} Replacing $P$ by $\nabla_yS$ gives the Hamilton-Jacobi equation \begin{align*} \partial_xS+H(x,y,\nabla_yS)=0. \end{align*}[/guided]

[step:Differentiate an existing Hamilton-Jacobi potential back to closedness] Conversely, suppose $S\in C^2(D)$ satisfies \begin{align*} \nabla_yS(x,y)=P(x,y) \end{align*} and \begin{align*} \partial_xS(x,y)+H(x,y,\nabla_yS(x,y))=0. \end{align*} Then \begin{align*} \partial_xS(x,y)=-H(x,y,P(x,y)). \end{align*} Therefore \begin{align*} dS=-H(x,y,P(x,y))\,dx+P(x,y)\cdot dy. \end{align*} By the Hamiltonian-coordinate expression for the Hilbert form, the right-hand side is $\omega$. Hence $\omega=dS$. Since exterior differentiation satisfies $d^2S=0$ for every $C^2$ function $S$, it follows that \begin{align*} d\omega=d(dS)=0. \end{align*} [/step]