Characteristic Direction Field of a First-Order Partial Differential Equation

Sep 26—2025

This post is a continuation of the previous post. We left off having defined a symplectic structure on the contact planes of dimension 2n in $J^1$ .

The symplectic structure is given by the 2-form $d\alpha|_{\alpha = 0}$ ; we have shown that it is defined up to a scalar multiple on each contact plane.

Claim: the skew-orthogonal complement of a hyperplane is a 1-dimensional subspace, a line.

Step 1: Show that there exists a basis $\xi_1, ..., \xi_n, \zeta_1, ..., \zeta_n$ for the symplectic space such that:

$d\alpha(\xi_i, \xi_j) = 0 = d\alpha(\zeta_i, \zeta_j)$

and $d\alpha(\xi_i, \zeta_j) = \delta_{i,j}$

Let $v_1, ..., v_{n}, v_{2n}$ be a basis for the symplectic space.

We know from the previous post that $d\alpha$ is not degenerate. Hence, if we let $\xi_1 = v_1$ , there exists $v_k$ such that $d\alpha(\xi_1, v_k) \neq 0$ . Notice that $d\alpha(\xi_1, \xi_1) = 0$ . WLOG, let $v_k = v_2$ .

Then, let $\zeta_1 = \frac{v_2}{d\alpha(\xi_1, v_2)}$ .

Key step: just like in the Gram-Schmidt process where we apply projection onto the orthogonal complement of a subspace to obtain a vector orthogonal to all our previous vectors, here we apply symplectic projection.

Set $\xi_2 := v_3 - d\alpha(v_3, \zeta_1)\xi_1 + d\alpha(v_3, \xi_1)\zeta_1$ .

Then, $d\alpha(\xi_2, \xi_2) = 0$ (we can check this simply by plugging it into the algebraic expression. It also intuitively makes sense because integrating $\alpha$ forward and backward along the vector self-cancels.

$d\alpha(\xi_2, \zeta_1) = d\alpha[v_3 - d\alpha(v_3, \zeta_1)\xi_1 + d\alpha(v_3, \xi_1)\zeta_1, \zeta_2]$

$= d\alpha[v_3, \zeta_1) - d\alpha(v_3, \zeta_1) d\alpha(\xi_1, \zeta_1) + d\alpha(v_3, \xi_1) d\alpha(\zeta_1, \zeta_1]$

$= d\alpha(v_3, \zeta_2)(1 - d\alpha(\xi_1, \zeta_2))$

$= 0$

and

$d\alpha(\xi_3, \xi_1) = d\alpha[v_3 - d\alpha(v_3, \zeta_2)\xi_1 + d\alpha(v_3, \xi_1)\zeta_2, \xi_1]$

$= d\alpha(v_3, \xi_1) - d\alpha(v_3, \zeta_1) d\alpha(\xi_1, \xi_1) + d\alpha(v_3, \xi_1) d\alpha(\zeta_1, \xi_1)$

$= d\alpha(v_3, \xi_1)(1 + d\alpha(\zeta_2, \xi_1))$

$= 0$

Thus, $\xi_2$ is in the orthogonal complement of $\text{span}\{\xi_1, \zeta_1 \}$ . We can use the same method as before to find $\zeta_2$ . Repeating the process, we will be able to find the n pairs of $\xi_k$ ‘s and $\zeta_k$ ‘s.

Now that we have found an orthonormal basis, the claim follows trivially.

Notice: since $d\alpha(\xi, \xi) = 0$ , $\xi$ lies inside its own skew-orthogonal complement.

This observation tells us that we can obtain a unique direction from each of the (2n-1)-dimension subspace obtained from the intersection of the contact plane and the tangent plane by taking its skew-orthogonal complement. Hence, we obtain a direction field in $J^1$ .

The next step: given a first-order PDE $F(x, y, p) = 0$ , what are the expressions for this direction field?

Miscellaneous

Characteristic Direction Field of a First-Order Partial Differential Equation

Share this:

Leave a Reply Cancel reply