Pfaff equation in the 3-dimensional case has the form:

$$Pdx+Qdy+Rdz=0$$

Where \(P\), \(Q\), \(R\) are sufficiently differentiable functions of \(x\), \(y\), \(z\).

This is Exercise 205 taken from the classic book by V.V. Stepanov:

$$\left(yz-z^{2}\right)dx-xzdy+xydz=0$$

$$z\left(y-z\right)dx+x\left(ydz-zdy\right)=0$$

Integrating factor: $$\mu=\frac{1}{y^2}$$

$$\frac{z}{y}\left(1-\frac{z}{y}\right)dx+x\frac{ydz-zdy}{y^{2}}=0$$

$$\frac{z}{y}\left(1-\frac{z}{y}\right)dx+xd\left(\frac{z}{y}\right)=0$$

$$\frac{dx}{x}+\frac{d\left(\frac{z}{y}\right)}{\frac{z}{y}\left(1-\frac{z}{y}\right)}=0$$

$$\frac{z}{y}=u$$

$$\frac{dx}{x}+\frac{du}{u}-\frac{du}{u-1}=0$$

$$\frac{dx}{x}+\frac{du}{u}-\frac{du}{u-1}=0$$

$$\frac{xu}{u-1}=C$$

$$\frac{xz}{z-y}=C$$

Will we always be lucky to be able to find an appropriate factor to cast the equation into a full differential form? The book gives a negative answer setting out a very specific condition on the coefficients.

Assume the equation does have a solution and this solution is a 2-dimensional manifold, i.e has the form:

$$\Phi(x,y,z)=C$$

or (locally, at least):

$$z=\phi(x,y)$$

Then

$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$

On the other hand, by virtue of the equation (assuming R does not vanish identically):

$$dz=-\frac{P}{R}dx-\frac{Q}{R}dy$$

Comparing the coefficients:

$$\frac{\partial z}{\partial x}=-\frac{P}{R}=A$$

$$\frac{\partial z}{\partial y}=-\frac{Q}{R}=B$$

This is an overdetermined system: one function, two equations, which generally does not have a solution. The integrability condition can be obtained by equating the mixed second derivatives, however I will quote the geometrical argument which may also shed light on another fact fact discussed later.

Consider an infinitesimal shift along the manifold from \((x,y,z)\) to \((x+dx,y)\). \(z\) will take value:

$$z+\frac{\partial z}{\partial x}dx=z+Adx$$

From this point we move to the point with coordinates $$(x+dx,y+dy)$$ without leaving the manifold. New value of z:

$$z+Adx+\frac{\partial}{\partial y}(z+Adx)dy=z+Adx+Bdy+(A_y+A_zB)dxdy$$

Similarly, if we first move along \(y\) and then along \(x\), we arrive at the following point:

$$z+Adx+Bdy+(B_x+B_zA)dxdy$$

Now we require that whatever route is chosen it leads to the same point on the manifold (up to the terms of the second order). This leads to the following equation:

$$A_y+A_zB-B_x-B_zA=0$$

$$-\frac{\partial}{\partial y}\left(\frac{P}{R}\right)+\frac{\partial}{\partial z}\left(\frac{P}{R}\right)\frac{Q}{R}-\frac{Q}{R}-\frac{\partial}{\partial z}\left(\frac{Q}{R}\right)\frac{P}{R}=0$$

$$P\left(\frac{\partial Q}{\partial z}-\frac{\partial R}{\partial z}\right)+Q\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)+R\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}\right)=0 \qquad (1)$$

Now that was the book, here are some thoughts about this theory.

1) First, equation \((*)\) definitely points to some categories of vector analysis. Indeed, the factors of P, Q and R are the components of the rotation of the vector field $$\vec{V}(P,Q,R)$$. Hence, the condition can be rewritten in a more compact form:

$$\vec{V}\operatorname{curl}\vec{V}=0$$

At first sight this should hold trivially for any \(\vec{V}\), for the rotation is by definition perpendicular to the plane defined by \(\vec{V}\) and tangent \(\partial \vec{V}\). However, this would only be true, if the solution were indeed an 2-dimensional manifold. If there is no such solution, then the whole derivation becomes invalid.

2) There is another reason why I prefer the geometric argument over comparing the mixed derivatives. The logic is very similar to that used to derive Cauchy-Riemann conditions for the analytic function. Remarkably enough, we can also apply complex formalism to the above problem. Consider the following operator:

$$\diamond = \frac{\partial}{\partial x}+i\frac{\partial}{\partial y}$$

Assuming again that the solution exists in the form $$z=\phi(x,y)$$ and using the above shortcuts for partial derivatives we obtain:

$$\diamond z = A+iB$$

Now apply \(\diamond^*\) to both parts, where \(*\) is complex conjugate:

$$\diamond^*\diamond z = \left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(A+iB\right)=\left(\frac{\partial}{\partial x} \{A \}+\frac{\partial}{\partial y} \{ B \} \right)-$$

$$-i\left(A_y+A_zB-B_x-B_zA\right)$$

$$-i\left(A_y+A_zB-B_x-B_zA\right)$$

\(\frac{\partial}{\partial x} \{A \}\) stands for "full partial derivative" where dependance of \(z\) on \(x\) is taken into account. Replacing \(A\) and \(B\) with their values, we obtain:

$$\diamond^*\diamond z =\Delta z - i\left(A_y+A_zB-B_x-B_zA\right)$$

where \(\Delta\) is the Laplacian.

On the other hand:

$$\diamond^*\diamond= \left( \frac{\partial}{\partial x}+i\frac{\partial}{\partial y} \right)\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)=\Delta$$

Hence
$$-\Im \diamond^*\diamond z =A_y+A_zB-B_x-B_zA=0$$

which gives the above integrability condition.

So this is another example of how recourse to complex values can reveal deep facts behind the obscure looking expressions.

Another fact which we implicitly assumed throughout the argument is the associability:

$$\diamond\left(\diamond^* z\right)=\left(\diamond\diamond^*\right)z$$

The fact that there exists examples which do not satisfy the above condition implies an interesting connection between algebraic properties of the operators and analytic properties of the solutions.

Another fact which we implicitly assumed throughout the argument is the associability:

$$\diamond\left(\diamond^* z\right)=\left(\diamond\diamond^*\right)z$$

The fact that there exists examples which do not satisfy the above condition implies an interesting connection between algebraic properties of the operators and analytic properties of the solutions.

In conclusion here is an example where integrability condition does not hold:

$$dx+(y+z)dy+zdz$$

To solve it we rewrite it as follows:

$$dx+ydy+zdz + zdy=0$$

$$d(x+\frac{y^2+z^2}{2})+zdy=0$$

Now let

$$x+\frac{y^2+z^2}{2}=\phi(y)$$

where \(\phi\) is an arbitrary function. Then

$$z=-\phi'(y)$$

These 2 relations give the general solution.

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