The connection between differential forms and ODE

Solution 1:

Solving an exact differential equation can be interpreted as finding the integral curves of a one-dimensional distribution defined by an exact form. Let me describe the relation:

Let $\omega$ be a differential one-form defined on an open subset $\Omega \subseteq \mathbb{R}^2$ and assume that $\omega$ doesn't vanish at any point of $\Omega$. Then $\ker(\omega)$ defines a one-dimensional distribution on $\Omega$. For each $p \in \Omega$, the subspace $\ker(\omega_p)$ is a one-dimensional subspace of $T_p(\mathbb{R}^2) \cong \mathbb{R}^2$ so we can think of $\omega$ as defining a field of lines on $\Omega$. A curve $\alpha \colon I \rightarrow \Omega$ is called an integral curve of $\omega$ if $\omega_{\alpha(t)}(\dot{\alpha}(t)) = 0$ for all $t \in I$.

If we write $\omega$ explicitly as $\omega = g(x,y)dx + h(x,y)dy$ then $\alpha(t) = (x(t),y(t))$ is an integral curve of $\omega$ if

$$ g(x(t),y(t)) \dot{x}(t) + h(x(t),y(t)) \dot{y}(t) \equiv 0. $$

If $\omega$ is exact, then $\omega = df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$ for some $f \colon \Omega \rightarrow \mathbb{R}$. The function $f$ is called a potential for $\omega$ and can be used to find integral curves of $\omega$ as follows. Let $(x_0, y_0) \in \Omega$ and set $C = f(x_0,y_0)$. Since $df = \omega$ has constant rank one, the level set $f^{-1}(C)$ is a one dimensional submanifold of $\Omega$. If $\alpha(t)$ is any parametrization of (a portion of) $f^{-1}(C)$ near $(x_0,y_0)$ then $f(\alpha(t)) = C$ and so, by differentiating, we get

$$ df_{\alpha(t)}(\dot{\alpha}(t)) = \omega_{\alpha(t)}(\dot{\alpha}(t)) \equiv 0. $$

This means that to find integral curves of $\omega$, we can find the level sets of $f$ and they will, implicitly, give us integral curves.

Now, let us assume that we are given a first order differential equation of the form $h(x,y(x))y'(x) = g(x,y(x))$. By performing formal manipulation, we have

$$ h(x,y) \frac{dy}{dx} = g(x,y) \implies g(x,y)dx - h(x,y) dy = 0. $$

This can be interpreted rigorously as saying that if we define a one form $\omega$ by $$\omega = g(x,y)dx - h(x,y) dy$$ then the graph $(x,y(x))$ of a solution of the first order differential equation will be an integral curve of $\omega$. Conversely, any integral curve of $\omega$ which can be expressed as $(x,y(x))$ will be a solution of the first order differential equation. Thus, instead of solving the original equation, we can instead find the integral curves of $\omega$. If $\omega$ is exact, we can find a potential $f$ for $\omega$ (determined uniquely up to a constant on connected domains) and then the level sets of $f$ will given an implicit description for the solutions of the original equation.

If $\omega$ is not exact, it might still be closed and so locally exact. Then, by finding local potentials we can find local solutions of our equation. If $\omega$ is not closed, we can try and find a non-zero function $\mu$ such that $\mu \omega$ is closed. Since the distribution defined by $\omega$ and $\mu \omega$ is the same, integral curves of $\mu \omega$ will also be integral curves of $\omega$ and they will allow us to find solutions. Such $\mu$ is called an integrating factor and can always be found locally. Note that almost all the methods one usually learns of solving first order differential equations are particular cases of the discussion above (this works for linear equations, linear non homogeneous equations, separable equations, exact equations, etc).