Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Differential equations of the form $M\,dx+N\,dy=0$ such that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ are said to be exact, because the left hand side of our equation is the exact differential of some function f, and the DE has a solution of the form $f(x,y)=c$, where c is a constant. Standard textbooks give us a procedure for solving such equations that essentially amounts to using the identity $f=\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy$. Specifically, we integrate M (respectively N, &c.) with respect to x to get $f=\int\frac{\partial f}{\partial x}\: dx+C(y)$, and then differentiate this with respect to y and do algebra to find C'(y), which we can then integrate to find C(y), which we can substitute back into our expression for f.

It occurred to me that this technique should generalize---that is, when instead of $M\,dx+N\,dy=0$ such that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, we have $\sum_{i}M_{i}\,dx_{i}=0$ where the $M_{i}$ satisfy the criteria for being an exact differential, then we can iteratively apply the same procedure to solve the equation. And so I came up with the following---

Theorem (?). A differential equation of the form $\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}=0$ for some function f has an implicit solution of the form $f=\sum_{i=1}^{n}a_{i}=c$ where $a_{0}=0$, $a_{i}=\int\left(\frac{\partial f}{\partial x_{i}}-\frac{\partial}{\partial x_{i}}\left(\sum_{j=1}^{i-1}a_{j}\right)\right)\: dx_{i}$ for $i\in\mathbb{N}_{+}$, and c is an arbitrary constant.

Proof (?). By induction on the number of variables n. The theorem is true for n=1 because $f=\int\,(\frac{\partial f}{\partial x}-0)\,dx$ by the fundamental theorem of calculus.

To complete the induction, we need to show that if $f(x_{1},...,x_{n})=\sum_{i=1}^{n}a_{i}=c$ is a solution to $\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}=0$, then $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n+1}a_{i}=c$ is a solution to $\sum_{i=1}^{n+1}\frac{\partial f}{\partial x_{i}}=0$. Think of $f(x_{1},...,x_{n})$ as being the "special case" of $f(x_{1},...,x_{n+1})$ where $x_{n+1}$ is being "treated like a constant." (My thinking here is not nearly as precise as it should be and I'm overloading the function name f, but hopefully you understand the intuition to which I am appealing.) So we can say that $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n}a_{i}+C(x_{n+1})$.

Then we proceed analogously as in the two-variable case. Differentiating by $x_{n+1}$ and applying algebra yields $C'(x_{n+1})=\frac{\partial f}{\partial x_{n+1}}-\frac{\partial}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)$, and then integrating with respect to $x_{n+1}$ yields $C(x_{n+1})=\int\left(\frac{\partial f}{\partial x_{n+1}}-\frac{\partial f}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)\right)\: dx_{n+1}$.

Then substituting into our earlier expression for $f(x_{1},...,x_{n+1})$ we get $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n}a_{i}+\int\left(\frac{\partial f}{\partial x_{n+1}}-\frac{\partial f}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)\right)\: dx_{n+1}$, which by virtue of the definition of $a_{i}$ is equivalent to $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n+1}a_{i}$. But by the principle of induction, this is quod erat demonstrandum.

Example. In the n=3 case (and naming our variables x, y, and z), we get $f(x,y,z)=\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy$ $+\int(\frac{\partial f}{\partial z}-\frac{\partial}{\partial z}(\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy))\, dz$, which is seen to be an identity by performing the integrations. What's going on is that $a_{1}=\int\frac{\partial f}{\partial x}\: dx=f$, and all following terms are zero by design, e.g., $a_{2}=f-f$.

So my question is (and I offer my sincerest apologies if this is not an appropriate question, or if my poor exposition has rendered my intentions virtually unreadable) does this seem basically correct, or am I doing something wrong? And how do I fix my sloppy reasoning in the inductive step ("special case" ... "treated like a constant")? I would be much obliged for any input.

share|improve this question

1 Answer 1

up vote 6 down vote accepted

The idea is essentially correct, but what you wrote needs some editing. You have independently rediscovered some beautiful mathematics! Bravo!

Your Theorem (as stated) has a much easier proof, but in your proof, you address an important point not mentioned in the theorem statement. There are two key ideas you really need to separate. Once you do that, I think you will see how to clean up your arguments.

In the following, I assume that you are familiar with differential forms and the exterior derivative, so that expressions like $$ df = \sum \frac{\partial f}{\partial x_i} d x_i$$ seem reasonable to you, but also that you know something about what $d( Adx + Bdy + Cdz)$ is. If not, let me know in a comment, and I will gladly rewrite this in $\mathbb R^3$ using the curl operator ($\nabla \times$).

First, about the proof of the theorem you stated. You need to think about what you mean by the differential equation $$A dx + Bdy = 0.$$ Obviously, you can "divide through" by $dx$ and then get an ordinary differential equation written in the normal way. Another way of seeing it is to say that you are looking for a parametrized curve $( x(t), y(t) )$ such that the tangent vector $(x'(t), y'(t))$ satisfies $$ 0 = A d x + B dy = A x'(t) dt + B y'(t) dt = (Ax'(t) + By'(t)) dt.$$ In other words, we want a curve $(x(t), y(t))$ so that its tangent vector is orthogonal to the vector $(A(x,y), B(x,y))$. If you happen to know that $(A, B) = \nabla f$ for some function $f$ (i.e. $df = Adx + Bdy$) then you have implicitly defined solutions given by $f = c$ for any constant $c$, since $\nabla f$ is orthogonal to level sets of $f$!

In higher dimensions, this picture generalizes perfectly. The solution to an equation given by $$ Adx + B dy + Cdz = 0$$ is going to be a surface in $\mathbb R^3$. If the differential form $Adx + Bdy + Cdz = df$ then by the same reasoning as above, level sets of $f$ will give you surfaces that satisfy this equation.

Now for the important point you address in your proof, but which you do not mention in the theorem statement (but you do mention in the text above the theorem statement). This is something called the Poincaré Lemma. A special case of this says that in $\mathbb R^n$, if you have a differential 1-form $\alpha$ and $d\alpha = 0$ then $\alpha = df$ for some function $f$. You have all the key ideas of the proof of this "lemma" lurking in your proof. (It's called a lemma, but it is a key result in differential geometry and differential topology.)

(The Poincaré Lemma does not hold if you work in a space with "holes". E.g. consider the 1-form given by $$\alpha := -\frac{y}{x^2+y^2} dx + \frac{x}{x^2+y^2} dy.$$ This has $d \alpha = 0$, but you can't find a function so that $\alpha = df$.)

More generally, you can define a (partial) differential equation in $\mathbb R^n$ by taking $k$ differential 1-forms $\alpha_i, i=1\dots k$, and asking to find a surface $S$ with the property that every vector $v$ tangent to $S$ satisfies $\alpha_i(v) = 0$ for $i=1 \dots k$. The condition to find a solution is called the Frobenius integrability theorem and the special case when $k=1$ is the requirement that $d\alpha_1 = 0$. This is generally a topic you will see in a differential geometry class.

share|improve this answer
    
Thank you for this! –  Zack M. Davis Mar 20 '11 at 0:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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