Solving $P(x,y)dx + Q(x,y)dy =0$: interpretation in terms of forms

Question

I asked a similar question here which I will formulate more sharply:

When we write a differential equation as $P(x,y)dx + Q(x,y)dy = 0$, what is the interpretation in terms of differential forms?

(I suppose the language of differential forms is the proper one to understand it.) Suppose we can separate into $\alpha(x)dx + \beta(y)dy = 0$. We integrate to find a relation between $x$ and $y$. What is the interpretation of this action in terms of differential forms? At first I thought we were flowing along the vector field $\alpha(x)dx + \beta(y)dy$, and the relation between $x$ and $y$ describes the flow lines. But then I realized $\alpha(x)dx + \beta(y)dy$ is a covector field, not a vector field, so this interpretation is not correct.

When we integrate the right-hand side and get a constant, what is the justification of that in terms of forms?

Answer 1

Very briefly for now, this falls within a course on Exterior Differential Systems; the given differential forms generate Ideal and the solution is given by its integral manifold. A good book about it is "Exterior Differential Systems", R.L. Bryant S.S. Chern R.B. Gardner H.L. Goldschmidt P.A. Griffiths.

Answer 2

So in my other answer, we established that this corresponds to a particular line integral: if $\vec V(\vec r) = P(\vec r) \hat x + Q(\vec r) \hat y$, then the integral takes the form

$$\oint_{\partial M} \vec V \cdot d\vec \ell = 0$$

over some closed path $\partial C$. That the integral is over come closed path is important; this effectively removes the constant of integration you were worried about.

Of course, the generalized Stokes theorem gives us an integral identity, transforming this to a surface integral:

$$\oint_{\partial M} \vec V \cdot d\vec \ell = \int_M \nabla \times \vec V \, dA$$

If the space the vector field $\vec V$ lives in is two-dimensional (as suggested by the coordinate dependence), then there is a 1-form-field $v = P e^x + Q e^y$ (where $e^x, e^y$) are the basis covectors to get

$$\oint_{\partial M} v = \int_M dv$$

where $d$ is the exterior derivative.

Answer 3

Let

$$ \Omega^{1}(\Bbb{R}^{2}) \ni \omega = P(x,y)\,dx + Q(x,y)\,dy $$

be a 1-form on (an open subset of) $\Bbb{R}^{2}$, and think about the differential equation

$$ P(x,y) \, dx + Q(x, y) \, dy = 0. \tag{1} $$

This equations asks us to find the relationship between $x$ and $y$ that makes LHS vanish.

Suppose we have find such a nice relation between $x$ and $y$. Then it will induce a curve, say $\gamma$, in $\Bbb{R}^{2}$. Thus we may rephrase the problem as follows: Find a curve $\gamma$ such that $\omega = 0$ along $\gamma$.

So it remains to elaborate this idea. The most important part is, what does the phrase '$\omega$ along $\gamma$' mean? This is realized by the concept of pullback. I won't explain here how it is defined (or you can google it!).

Returning to the original problem, let us expand out the pullback $\gamma^{*}\omega$. We have

\begin{align*} \gamma^{*}\omega &= (P \circ \gamma) \, (\gamma^{*}dx) + (Q\circ \gamma) \, (\gamma^{*}dy) = \bigg\{ (P \circ \gamma) \frac{d\gamma^{1}}{dt} + (Q \circ \gamma) \frac{d\gamma^{2}}{dt} \bigg\} \, dt. \end{align*}

So the equation $\gamma^{*}\omega = 0$ is actually equivalent to

$$ (P \circ \gamma) \frac{d\gamma^{1}}{dt} + (Q \circ \gamma) \frac{d\gamma^{2}}{dt} = 0, $$

or if we implicitly write $(x, y) = (\gamma^{1}(t), \gamma^{2}(t))$, then it further simplifies to a more traditional form:

$$ P(x, y) \frac{dx}{dt} + Q(x, y) \frac{dy}{dt} = 0. \tag{2} $$

So you actually see the direct relationship between (1) and (2).

Answer 4

The interpretation is the following: given a differential 1-form $\omega=Pdx+Qdy$ in the plane, you are asked to find its integral curves, i.e. 1-dim submanifolds of $\mathbb R^2$ whose tangent line at each point is annihilated by the 1-form. For example, the integral curves of $xdx+ydy$ are the concentric circles around the origin. At a point where $\alpha\neq 0$ either $P$ or $Q$ do not vanish, hence the same holds in a neighborhood of the point. If say $Q$ does not vanish, then the integral curves are graphs of solutions to the ODE $dy/dx=-P/Q$.

Answer 5

Such a differential equation is called an Exact Differential Equation. The equation is solved by finding a multiplicative "integrating" factor $R$ such that the following form is exact: $$ R(Qdx+Pdy) = QRdx+PRy=0. $$ In other words, $$ QRdx + PRdy = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y} dx = 0. $$ for some function $f$. That's what it has to do with forms. Equivalently, $$ \frac{\partial f}{\partial x}=QR,\;\;\; \frac{\partial f}{\partial y}=PR. $$ In such a case, the requirement $Qdx+Pdy=0$ means $df=0$ or, equivalently, $f$ is constant (at least locally). Then, by the implicit function theorem $f(x,y)=C$ can be solved for $x=x(y)$ or $y=y(x)$ in a neighborhood of $(x_{0},y_{0})$ for which $f(x_{0},y_{0})=0$ if $\frac{\partial f}{\partial x}(x_{0},y_{0})\ne 0$ or if $\frac{\partial f}{\partial y}(x_{0},y_{0})\ne 0$. If such $R$ exists, then it factors out of the equation when considering the ratios of derivatives of $f$. If you can solve for $y=y(x)$, then $f(x,y(x))=C$ and $y(x_{0})=y_{0}$, which gives $$ \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}=0, $$ or $$ \frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = -\frac{QR}{PR}=-\frac{Q}{P}. $$ Similarly, if $x=x(y)$ satisfies $f(x(y),y)=C$ and $x(y_{0})=x_{0}$, then $$ \frac{dx}{dy} = -\frac{P}{Q}. $$ That is how one arrives at (and justifies) the final form of equation $$ Q+P\frac{dy}{dx}=0\;\;\mbox{ or } \;\; Q\frac{dx}{dy}+P=0. $$ The process generalizes to more variables through the notion of exactness. One is then able to solve for one variable in terms of the others by setting $f=0$.

Here is a nice introduction to relate geometry, exactness, and this ODE:
http://tutorial.math.lamar.edu/classes/de/exact.aspx