2
$\begingroup$

In a variety of contexts, I have noticed hints of a strong connection between exact differential equations and machinery from multivariable calculus.

From another question, I have gathered that the geometry of finding the general solution to an exact differential equation $M(x,\ y)\ dx + N(x,\ y)\ dy = 0$ includes finding the potential function of $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$, that is, the surface in $\mathbb{R^3}$ whose gradient is the vector field $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$. What is missing from this description? For example, we can write the LHS of a not-necessarily-exact $M(x,\ y)\ dx + N(x,\ y)\ dy = 0$ as $\begin{bmatrix}M(x,\ y) \\ N(x,\ y) \end{bmatrix} \cdot \begin{bmatrix}dx \\ dy\end{bmatrix}$, which expresses the differential of work done by $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$, and thus shows promise for extending the above geometric picture. In fact, since work is a path function, I suspect the Wikipedia quote

In mathematics, an integrating factor is a function that is chosen to facilitate the solving of a given equation involving differentials. It is commonly used to solve ordinary differential equations, but is also used within multivariable calculus when multiplying through by an integrating factor allows an inexact differential to be made into an exact differential (which can then be integrated to give a scalar field). This is especially useful in thermodynamics where temperature becomes the integrating factor that makes entropy an exact differential.

is misleading, in that the ordinary differential equations and multivariable calculus uses are actually identical, which if correct, might allow geometry from other path functions (such as the referenced thermodynamics example) to be leveraged.

A big part of what I'm trying to remedy is that it's not clear to me why finding the potential function of $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$ should solve the ODE. I understand the algebraic rationale that the solution technique undoes the multivariable chain rule (divide through by $dx$ then integrate both sides), but is the geometric picture just an afterthought, or does it offer its own independent rationale for the solution technique?

$\endgroup$
2
$\begingroup$

Interesting question, althought I am not exactly sure what kind of answer you want. If this answer is not fully satisfactory, please leave a comment.

Why finding the potential function solves the differential equation. Suppose we have a potential function $\phi$ for $(M(x,y),N(x,y))$. Then all the curves $\phi(x,y)=c$ with $c$ a constant form a solution of the ODE. Indeed, if we take the total derivative of $\phi(x,y)=c$ we get $$ \phi_x(x,y) dx + \phi_y(x,y) dy = 0, \\ M(x,y) dx + N(x,y) dy = 0. $$ This is a rather formal explanation.

The physics picture. The differential $Mdx + N dy$ can indeed be regarded as the infinitesimal amount of work done by a field $\vec F=(M(x,y),N(x,y))$. This picture can help you understand intuitively why $F(x,y)=c$ solves the ODE $Mdx + Ndy =0$.

Note that a potential in physics is a scalar function $\phi$ such that $-\nabla \phi =\vec{F}=(M,N)$; one adds a minus sign. If such a potential exists, we say that the vector field is exact or, more commonly in physics, conservative. Well-known examples of conservative vector fields are the gravitational force field and electric field in electrostatics.

Conservative vector fields are characterized by the fact that the line integral of them only depend on the end points. Physically this means that the work done by the fields does not depend on the path, only on the beginning and end position. So if $(M(x,y),N(x,y))=\vec{F} = -\nabla \phi$, then $$ \int_C M(x,y)dx + N(x,y)dy = \int_C \vec{F} = -(\phi(p_2) - \phi(p_1)) $$ where $C$ is a curve going from $p_1$ to $p_2$. This line integral can be interpreted as the work done by the field $\vec F$ along the path $C$. This line integral (i.e. the total amount of work) is zero iff the potential of the endpoints are the same.

Back to the ODE. Now let us look back at the ODE $M(x,y)dx + N(x,y)dy = 0$. Physically this equation means that the infinitesimal work should be zero. The solution curves $\phi(x,y)=c$ are the equipotential curves. These curves are exactly the paths along which the work is zero for every infinitesimal step.

$\endgroup$
  • $\begingroup$ This answer led to the ideas I was searching for. My thoughts are still kind of scattered, so feel free to reorganize if you incorporate them into your answer. Finding the potential function of the given vector field yields a surface in $R^3$, which is the geometry of the general solution. A particular solution is a curve in $R^2$ obtained by slicing $R^3$ parallel to the $xy$-plane. Then projecting the vector field up or down onto that slice yields an instance of $R^2$ with exactly the vector field and curve needed for the dot product interpretation. $\endgroup$ – user10478 Mar 20 '19 at 18:29
  • 1
    $\begingroup$ The key insight was that this vector field, being the gradient of the surface, points in the direction of steepest ascent, its negation in the direction of steepest descent, and perpendicular vectors in the directions of zero ascent. So for any vector in the field, dotting it with a new vector $d\vec r = \begin{bmatrix}dx \\ dy\end{bmatrix}$ serves dually as a measure of work and a measure of steepness of ascent (the steeper the ascent in the direction which $d\vec r$ points, the greater the dot product). Mnemonically, it takes work to climb the surface. $\endgroup$ – user10478 Mar 20 '19 at 18:30
  • 1
    $\begingroup$ Since the ODE states that $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix} \cdot d\vec r = 0$, what we need is for $d\vec r$ to always be perpendicular to the field, which is exactly the direction tangent to the curve. So if we restrict $d\vec r$ to always be tangent to the curve, thus perpendicular to the vector field, this ensures that the ODE remains satisfied, and simultaneously that we don't ascent or descent the surface, popping out of our slice. $\endgroup$ – user10478 Mar 20 '19 at 18:30
  • $\begingroup$ I guess that by a surface in $R^3$ you just mean the graph of the potential function $\phi$. I would say that the insight in your comment is pretty complete. My answer was rather in the context of physics, but your story is very geometrical, as you asked in your OP. I like it. $\endgroup$ – Ernie060 Mar 20 '19 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.