Exactness of 2nd-Order Differential Equations via Differential Forms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:17:54Z http://mathoverflow.net/feeds/question/22910 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22910/exactness-of-2nd-order-differential-equations-via-differential-forms Exactness of 2nd-Order Differential Equations via Differential Forms Cam McLeman 2010-04-28T23:37:11Z 2011-08-13T22:15:57Z <p>This (probably very elementary) question came up the last time I taught differential equations, and I've been toying with it for a while with no success:</p> <p>A 1st-order differential equation $M(x,y)dx+N(x,y)dy=0$ is exact if $$M(x,y)dx+N(x,y)dy=f_x(x,y)dx+f_y(x,y)dy$$ for some differentiable function $f(x,y)$ defined on the domain of $\omega$. In this case, we easily arrive at an implicitly-defined solution to the differential equation. Importantly, there is a nice test for exactness stemming from Clairaut's theorem -- for everywhere smooth $M$ and $N$ (for simplicity/laziness...obvious generalizations abound), the differential equation is exact iff $N_y=M_x$. Of course, this procedure is easily re-interpreted as saying that by the triviality of $H^1(\mathbb{R}^2)$, a one-form is closed if and only if it is exact.</p> <p>Now let's move one degree higher. Boyce and Di Prima define a 2nd-order differential equation $P(x)y''+Q(x)y'+R(x)y=0$ to be exact if there exists a differentiable function $f(x,y)$ such that the differential equation can be written</p> <p>$$P(x)y''+Q(x)y'+R(x)y=[P(x)y']'+[f(x)y]'=0.$$</p> <p>The analogous expression to Clairaut's theorem seems to be that (again, for sufficiently smooth inputs) an equation of that form is exact iff $P''(x)-Q'(x)+R(x)=0.$ Of importance is that such forms can be integrated once to leave us with a 1st-order differential equation. So we've successfully lowered the degree of our problem.</p> <p>This feels to me very much like an analogous $H^2$ calculation. We have a condition on some coefficients that very much looks like an alternating sum coming from a $d$ map on forms, and lets us conclude that the equation "comes from" a one-degree-smaller differential equation. </p> <p>But! (and here's the question) I can't seem to fit any 2-forms into this picture that would explain this analogy. Presumably there's some big story here linking the two notions of exactness about which I'd love to be enlightened.</p> <p>Side remark: I once received a partial response that there might be a link with Cartan tableau, which I've been unsuccessful in pursuing, if that helps spark an idea.</p> http://mathoverflow.net/questions/22910/exactness-of-2nd-order-differential-equations-via-differential-forms/71724#71724 Answer by Robert Bryant for Exactness of 2nd-Order Differential Equations via Differential Forms Robert Bryant 2011-07-31T14:13:04Z 2011-08-13T22:15:57Z <p>What you are looking for nowadays goes by the name of the <em>Rumin complex</em> and is defined on any contact manifold. Moreover, there is a vast generalization of this that sometimes goes by the name of 'the variational bicomplex' and sometimes by the name 'characteristic cohomology'. Here is a brief description that is suited for the question you asked:</p> <p>On $M = \mathbb{R}^3$ with coordinates $x,y_0,y_1$, consider the differential ideal $\mathcal{I}\subset\Omega^\ast(M)$ generated by the $1$-form $\omega = dy_0 - y_1\ dx$, i.e., $\mathcal{I}$ is the set of linear combinations of all multiples of $\omega$ and $d\omega$. Note that $\mathcal{I}$ is a homogeneous ideal and equals $\Omega^\ast(M)$ in degrees 2 and 3. Because $\mathcal{I}$ is closed under exterior derivative, it is a sub-complex of $\bigl(\Omega^\ast(M),d\bigr)$. Thus, there is a graded quotient complex, call it $\bigl(\mathcal{Q},\bar d\bigr)$, that vanishes in degrees above $1$. Note that $\mathcal{Q}^0 = \Omega^0(M)= C^\infty(M)$, since $\mathcal{I}$ vanishes in degree $0$.</p> <p>Now, say that an element $\phi \in \mathcal{Q}^1$ is <em>exact</em> if $\phi = \bar d f$ for some $f\in \mathcal{Q}^0 = \Omega^0(M)= C^\infty(M)$. Unfortunately, unlike $\bigl(\Omega^\ast(M),d\bigr)$, the complex $\bigl(\mathcal{Q},\bar d\bigr)$ is not locally exact in positive degree. In fact, $\bar d \phi =0$ for all $\phi\in\mathcal{Q}^1$, even though $\bar d: \mathcal{Q}^0\to \mathcal{Q}^1$ is not onto.</p> <p>Let me pause just a moment to explain how this fits into your question. Your equation $P(x)y'' + Q(x)y' +R(x)y = 0$ is encoded as the $1$-form $\phi = P(x) dy_1 + (Q(x)y_1 + R(x) y_0) dx$ (which represents the same class as the $1$-form $P(x) dy_1 + Q(x)dy_0 + R(x) y_0 dx$ in $\mathcal{Q}^1$), and you are asking when there is a function $f(x,y_0,y_1)$ such that $\phi = \bar d f$. (You should verify that $f = P(x) y_1 + (Q(x)-P'(x))y_0$ works when your equation is satisfied and that, otherwise, nothing does.)</p> <p>Now, how can we test for exactness in this sense? This is where the Rumin complex (aka the variational bicomplex, etc.) comes in. It turns out that there is a way to embed the operator $\bar d:\mathcal{Q}^0\to\mathcal{Q}^1$ into a complex that provides a fine resolution of the constant sheaf, the same way that the exterior derivative does for the full complex of exterior differential forms. </p> <p>What you do is this: Let $\mathcal{E}^2\subset\Omega^2(M)$ be the set of multiples of $\omega$ and let $\mathcal{E}^3=\Omega^3(M)$. </p> <p>We now want to define a complex $$0\longrightarrow \mathcal{Q}^0 \buildrel{\bar d}\over\longrightarrow \mathcal{Q}^1 \buildrel{D}\over\longrightarrow \mathcal{E}^2 \buildrel{d}\over\longrightarrow \mathcal{E}^3 \longrightarrow 0.$$ The map from $\mathcal{E}^2$ to $\mathcal{E}^3$ is the usual exterior derivative, so the only thing left to define is the map $D:\mathcal{Q}^1\to \mathcal{E}^2$. To do this, we first define a (first-order) operator $\delta: \mathcal{Q}^1\to\Omega^1(M)$, by requiring that $\delta(\phi)$ be a 1-form representing $\phi$ in the quotient complex and that $d\bigl(\delta\phi\bigr)$ lie in $\mathcal{E}^2$, i.e., that it be a multiple of $\omega$. (I'll let you write down the formula for $\delta$ in local coordinates.) Now, define $D\phi$ to be $d\bigl(\delta\phi\bigr)$. (Sounds almost trivial doesn't it?) The operator $D$ is easily verified to be second order and linear.</p> <p>Now, it is not hard to verity that this complex is locally exact in positive degrees. (It also gives a fine resolution of the constant sheaf, so its cohomology on $M$ is canonically isomorphic to the deRham cohomology of $M$.) In particular, the local condition that $\phi\in\mathcal{Q}^1$ be exact is that $D\phi=0$. </p> <p>You should verify (after you have defined $D$) that this reproduces your condition precisely in the linear case you asked about.</p>