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Valentin Fadeev

Pfaff equation

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Edited by Valentin Fadeev, Wednesday, 15 Sep 2010, 23:37

Moving on with Stepanov's book I have reached the subject equations which have the following form (3 variables):

sum with, 3 , summands cap p times d times x plus cap q times d times y plus cap r times d times z equals zero

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

Excercise 205:

open y times z minus z squared close times d times x minus x times z times d times y plus x times y times d times z equals zero

z times open y minus z close times d times x plus x times open y times d times z minus z times d times y close equals zero

Integrating factor: mu equals one divided by y squared

z divided by y times open one minus z divided by y close times d times x plus x times y times d times z minus z times d times y divided by y squared equals zero

z divided by y times open one minus z divided by y close times d times x plus x times d of z divided by y equals zero

d times x divided by x plus d of z divided by y divided by z divided by y times open one minus z divided by y close equals zero

z divided by y equals u

d times x divided by x plus d times u divided by u minus d times u divided by u minus one equals zero

x times u divided by u minus one equals cap c

x times z divided by z minus y equals cap c

Will we always be lucky to have 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:

normal cap phi of x comma y comma z equals cap c

or (locally, at least):

z equals phi of x comma y

Then

equation left hand side d times z equals right hand side normal partial differential times z divided by normal partial differential times x times d times x plus normal partial differential times z divided by normal partial differential times y times d times y

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

equation left hand side d times z equals right hand side negative cap p divided by cap r times d times x minus cap q divided by cap r times d times y

Comparing the coefficients:

equation sequence normal partial differential times z divided by normal partial differential times x equals negative cap p divided by cap r equals cap a

equation sequence normal partial differential times z divided by normal partial differential times y equals negative cap q divided by cap r equals cap b

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

Consider an infinitesimal shift along the manifold from open x comma y comma z close to open x plus d times x comma y close . Then z will take value:

equation left hand side z plus normal partial differential times z divided by normal partial differential times x times d times x equals right hand side z plus cap a times d times x

From here we move to the point with coordinates open x plus d times x comma y plus d times y close without leaving the manifold. New value of z:

equation left hand side sum with, 3 , summands z plus cap a times d times x plus normal partial differential divided by normal partial differential times y times open z plus cap a times d times x close times d times y equals right hand side sum with, 4 , summands z plus cap a times d times x plus cap b times d times y plus open cap a sub y plus cap a sub z times cap b close times d times x times d times y

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

sum with, 4 , summands z plus cap a times d times x plus cap b times d times y plus open cap b sub x plus cap b sub z times cap a close times d times x times d times y

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:

cap a sub y plus cap a sub z times cap b minus cap b sub x minus cap b sub z times cap a equals zero

negative normal partial differential divided by normal partial differential times y times open cap p divided by cap r close plus normal partial differential divided by normal partial differential times z times open cap p divided by cap r close times cap q divided by cap r minus cap q divided by cap r minus normal partial differential divided by normal partial differential times z times open cap q divided by cap r close times cap p divided by cap r equals zero

equation left hand side sum with, 3 , summands cap p times open normal partial differential times cap q divided by normal partial differential times z minus normal partial differential times cap r divided by normal partial differential times z close plus cap q times open normal partial differential times cap r divided by normal partial differential times x minus normal partial differential times cap p divided by normal partial differential times z close plus cap r times open normal partial differential times cap p divided by normal partial differential times y minus normal partial differential times cap q divided by normal partial differential times x close equals right hand side zero times open one close

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

1) First, equation open asterisk operator close   definitely points to some categories of vector analysis. Indeed, the factors of P, Q and R are the components of the rotor of the vector field cap v right arrow of cap p comma cap q comma cap r . Hence, the condition can be rewritten in a more compact form:

cap v right arrow times r times o times t times cap v right arrow equals zero

At first sight this should hold trivially for any cap v right arrow , for the rotor is by definition perpendicular to the plane defined by cap v right arrow and tangent normal partial differential times cap v right arrow . 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:

equation left hand side diamond operator equals right hand side normal partial differential divided by normal partial differential times x plus i times normal partial differential divided by normal partial differential times y

where equation left hand side i squared equals right hand side negative one .

Assuming again that the solution exists in the form z equals phi of x comma y and using the above shortcuts for partial derivatives we obtain:

equation left hand side prefix diamond operator of z equals right hand side cap a plus i times cap b

Now apply diamond operator super asterisk operator to both parts, where asterisk operator is complex conjugate:

equation sequence diamond operator super asterisk operator diamond operator z equals open normal partial differential divided by normal partial differential times x minus i times normal partial differential divided by normal partial differential times y close times open cap a plus i times cap b close equals open normal partial differential divided by normal partial differential times x times open cap a close plus normal partial differential divided by normal partial differential times y times open cap b close close minus i times open cap a sub y plus cap a sub z times cap b minus cap b sub x minus cap b sub z times cap a close

normal partial differential divided by normal partial differential times x times open cap a close stands for "full partial derivative" where dependance of z on x is taken into account. Replacing cap a and cap b with their values, we obtain:

equation left hand side diamond operator super asterisk operator diamond operator z equals right hand side normal cap delta times z minus i times open cap a sub y plus cap a sub z times cap b minus cap b sub x minus cap b sub z times cap a close

where normal cap delta is the Laplacian, On the other hand:

equation sequence diamond operator super asterisk operator postfix diamond operator equals open normal partial differential divided by normal partial differential times x plus i times normal partial differential divided by normal partial differential times y close times open normal partial differential divided by normal partial differential times x minus i times normal partial differential divided by normal partial differential times y close equals normal cap delta

Hence

equation sequence negative normal black letter cap i times diamond operator super asterisk operator diamond operator z equals cap a sub y plus cap a sub z times cap b minus cap b sub x minus cap b sub z times cap a equals zero

which gives the above integrability condition.

So this is another example of how recourse to complex values can reveal deep facts behind the otherwise unfamiliar looking expressions. And formulate them in a nice compact form as well.

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

sum with, 3 , summands d times x plus open y plus z close times d times y plus z times d times z

To solve it we rewrite it as follows:

sum with, 4 , summands d times x plus y times d times y plus z times d times z plus z times d times y equals zero

d times open x plus y squared plus z squared divided by two close plus z times d times y equals zero

Now let

x plus y squared plus z squared divided by two equals phi of y

where phi is an arbitrary function. Then

z equals negative phi super prime times open y close

These 2 relations give the general solution.

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