1-Graph of a Solution to a 1st-Order PDE

Oct 2—2025

Claim: Let $\Gamma^n$ be the 1-graph of a solution of $F(x, y, p) = 0$ . Suppose a characteristic passes through a point of this graph. Then, the characteristic lies entirely in $\Gamma^{n}$ .

Proof:

Let $V^{2n}$ be the zero set of $F(x, y, p)$ and $K^{2n}_z$ the contact plane at $z$ . Let $z \in \Gamma^n$ be a regular point (i.e. the contact plane of $z$ is not tangent to $\Gamma^n$ ) and $\xi$ be the vector of the characteristic direction at point z.

We can make the following 3 observations:

1) $T_z \Gamma^n \subset T_z V^{2n}$ . This is because $\Gamma^n$ is a sub-manifold of $V^{2n}$ . (Interesting: this gives us a new way of geometrically finding the solutions (not that this is any easier). The sub-manifolds of $V^{2n}$ that are tangent to the field of contact planes at every point (is this enough?) and their projections onto the x-plane (parallel in the other direction) are invertible maps give us the 1-graphs of the solutions.

Thinking about: can we find an example of a maximal integral sub-manifold that is not a solution to a first-order PDE?

2) $T_z \Gamma^n \subset K^{2n}_z$

3) $T_z \Gamma^n$ is an isotropic subspace of $K^{2n}_z$ endowed with the symplectic structure, i.e. for any $u, v \in T_z \Gamma^n$ , $d\alpha(u, v) = 0$ .

Proof of 3):

Let’s call the solution that our 1-graph describes $\phi$ .

Key: $\Gamma^n$ is the image of the map $\Psi: x \mapsto (x, \phi(x), \phi_x(x))$ .

It follows that the tangent space at $z = (x, \phi(x), \phi_x(x))$ is spanned by the column vectors of the Jacobian of $\Psi$ , i.e.

$\{ X_j = \partial_{x_j} + \phi_{x_j}\partial{y} + \sum_{k=1}^n \phi_{x_k, x_j} \partial{p_k}, j = 1, ..., n \}$

It suffices to show that for any $X_i, X_j$ , $d\alpha(X_i, X_j) = 0$ .

Recall: $d\alpha((\dot{x}, \dot{y}, \dot{p}), (x', y', p') = \dot{x}p' - \dot{p}x'$ .

$d\alpha(X_i, X_j) = -\sum_{k=1}^n (dp_k(X_i)dx_k(X_j) - dp_k(X_j)dx_k(X_i)$ (*)

Notice that $dx_k(X_j) = \delta_{k,j}$ and $dp_k(X_j) = \phi_{x_k, x_j}$ .

And so,

(*) $= - \sum_{k=1}^n(\phi_{x_k, x_j}\delta_{k,i} - \phi_{x_k, x_i}\delta_{k,j})$

$= -\phi_{x_i, x_j} + \phi_{x_j, x_i}$

$= 0$ (by the symmetries of the partial derivatives.

This proves 3).

Now, suppose for a contradiction $\xi$ is not in $T_z \Gamma^n$ . By definition of $\xi$ being characteristic, $\xi$ is the skew-orthogonal complement of $T_z V^{2n} \cap K^{2n}_z$ . Thus, by 2), we also know that $\xi$ is skew-orthogonal complement to $T_z \Gamma^{n}$ (since $T_z \Gamma^{n}$ lies in $T_z V^{2n} \cap K^{2n}_z$ ).

Hence, $\text{span} \{ \xi, T_z \Gamma^n \}$ is an (n+1)-dimensional isotropic subspace of a 2n-dimensional symplectic space.

Is this possible? No!

The dimension of the isotropic subspace CANNOT exceed n. This is because an isotropic subspace is contained in its skew-orthogonal complement (obvious). If the dimension of the isotropic subspace is m, then its skew-orthogonal complement must also be m. As discussed in a previous post, the skew-orthogonal complement is also given by $2n-m$ (just like the orthogonal complement in Euclidean space). And so, we have $2n-m \geq m$ ; hence, $m \leq n$ .