Can exterior calculus be used to solve differential equations?

Question

I know one can formulate partial differential equations in terms of exterior derivatives, etc but I have been wondering for a while now how one might be able to use that formalism to extrapolate information about the solutions.

Answer 1

I'll write a very quick motivation for de Rham cohomology.

Say we have some manifold $M$ (or open set $M \subset \mathbb R^n$ if you will) and that we want to solve the differential equation $$ du = v $$ on $M$, where $v$ is a $k$-form (so $k = 1$ has us searching for a function $u$ that solves the equation, higher $k$ have us searching for $(k-1)$-forms).

If this equation has a solution, then we necessarily have $$ dv = d^2 u = 0. $$ This gives a necessary condition for our problem to have a solution, but not a sufficient one, since there may be "obstructions" to finding a solution of $du = v$ even if $dv = 0$.

To get some sense of these obstructions, we form a quotient vector space $$ H^k(M,\mathbb R) := \{ \text{$k$-forms v} \mid dv = 0 \} / \{du \mid \text{$u$ is a $(k-1)$-form}\}. $$ This is the $k$-th de Rham cohomology group of the manifold $M$. If our problem $du = v$ is well-posed, that is if $d v = 0$ so it has a chance of having a solution, then $v$ defines an element $[v]$ in $H^k(M,\mathbb R)$. Our problem also has a solution if and only if $[v] = 0$ in this space. A remarkable and nontrivial fact is that these cohomology groups are finite-dimensional if the space $M$ is compact.

This may not seem very useful, since it is in general quite difficult to show that some specific $v$ defines a zero cohomology class on $M$. What makes this formalism useful is that in many cases one can use geometric properties of the manifold $M$ to prove that some or all of its nontrivial cohomology groups (those with $k > 0$) are zero, and thus that the differential equation that interests us has a solution.

These geometric properties come in two large categories; topological and differential-geometric. The first includes facts like that if $M$ is contractible, then all its cohomology groups are zero. The second is known as Hodge theory and rests on the Hodge isomorphism theorem, which says that if you pick a Riemannian metric on $M$, then the space of harmonic $k$-forms is isomorphic to $H^k(M,\mathbb R)$ (if $M$ is compact). Both can give valuable information about the cohomology groups and the geometry of $M$, and the second has extremely important connections to algebraic geometry.

In short, passing to cohomology replaces the problem of solving one particular differential equation with the problem of solving all differential equations on a manifold, which has connections to the geometry of the underlying space. Knowledge of the geometry can then perhaps help us to solve our original problem.

Since you asked for an explicit example of these calculations, we can consider the $1$-dimensional torus $S^1 = \mathbb R / \mathbb Z$. Actually I'll cheat and not do any explicit calculations at all, but leave them to you to figure out. The torus admits a flat Riemannian metric $g$, induced by the standard Euclidean metric on $\mathbb R$. Hodge theory will tell you that $$ H^1(S^1, \mathbb R) \cong \mathbb R, $$ and that and explicit isomorphism between the two is $\mathbb R \to H^1(S^1,\mathbb R)$, $\lambda \mapsto \lambda dx$, where $dx$ is the constant $1$-form on $\mathbb R$ (see Hodge isomorphism theorem).

Now let $dV$ be the volume form of the Riemannian metric $g$, which we'll suppose to be of volume 1. I claim that we cannot solve the differential equation $$ df = dV $$ on $S^1$. If we could, then Stokes' theorem would give us $$ \int_{S^1} dV = \int_{S^1} df = \int_{\partial S^1} f = 0 $$ because $S^1$ has no boundary. But by hypothesis $$ \int_{S^1} dV = \mathrm{Vol}(S^1,g) = 1, $$ which gives a contradiction.