# Exact Differential Equation

1. Dec 5, 2009

### abcdefg10645

We can use Green's theorem to understand why the Exact Differential Equation satisfy the conditions it should have .....

How about a DE for more than two variables ?

Eg.dF=P(x,y,z,w)dx+Q(x,y,z,w)dy+R(x,y,z,w)dz+S(x,y,z,w)dw

IF the equation above is an Exact Differential Equation , what condition it would satisfy ?

2. Dec 5, 2009

### HallsofIvy

Staff Emeritus
An equation of the form f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz= 0 is "exact" if and only if there exist some F(x,y,z) such that
$$dF= \frac{\partial F}{\p artial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}= f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz$$

That means that we must have
$$\frac{\partial F}{\partial x}= f(x,y,z)$$
$$\frac{\partial F}{\partial y}= g(x,y,z)$$
and
$$\frac{\partial F}{\partial z}= h(x,y,z)$$

So, as long as those functions are continuous, we must have the "mixed partials" equal:
$$\frac{\partial^2 F}{\partial x\partial y}= \frac{\partial f}{\partial y}= \frac{\partial g}{\partial x}= \frac{\partial^2 F}{\partial y\partial x}$$
etc.

Since you mention "Green's theorem" (for two variables) you might want to look at $\nabla\times f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ and Stoke's theorem.

Last edited: Dec 5, 2009
3. Dec 5, 2009

### abcdefg10645

What u mentioned contains three variables ~and I've seen such a case in the book I read...

Suppose $$\vec{F}$$=P(x,y,z)$$\hat{i}$$+Q(x,y,z)$$\hat{j}$$+R(x,y,z)$$\hat{k}$$

dT=P(x,y,z)dx+Q(x,y,z)dy+R(x,y,z)dz is exact if only if $$\vec{F}$$$$\bullet$$($$\nabla$$$$\times$$$$\vec{F}$$)=0

I'm now requesting for the case which contains more than three variables~