2.14: Appendix I- Integrating Factors
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Suppose we have an inexact differential \mathchar′26dW=A∗idx∗i . Here I am adopting the ‘Einstein convention’ where we sum over repeated indices unless otherwise explicitly stated; A∗idx∗i=∑iA∗idx∗i. An integrating factor eL(→x) is a function which, when divided into \mathchar′26dF, yields an exact differential: dU=e−L\mathchar′26dW=∂U∂x∗idx∗i . Clearly we must have ∂2U∂x∗i∂x∗j=∂∂x∗i(e−LA∗j)=∂∂x∗j(e−LA∗i) . Applying the Leibniz rule and then multiplying by eL yields ∂A∗j∂x∗i−A∗j∂L∂x∗i=∂A∗i∂x∗j−A∗i∂L∂x∗j . If there are K independent variables {x∗1,…,x∗K}, then there are 12K(K−1) independent equations of the above form – one for each distinct (i,j) pair. These equations can be written compactly as Ω∗ijk∂L∂x∗k=F∗ij , where Ω∗ijk=A∗jδ∗ik−A∗iδ∗jk∑iF∗ij=∂A∗j∂x∗i−∂A∗i∂x∗j . Note that F∗ij is antisymmetric, and resembles a field strength tensor, and that Ω∗ijk=−Ω∗jik is antisymmetric in the first two indices (but is not totally antisymmetric in all three).
Can we solve these 12K(K−1) coupled equations to find an integrating factor L? In general the answer is no. However, when K=2 we can always find an integrating factor. To see why, let’s call x≡x∗1 and y≡x∗2. Consider now the ODE dydx=−A∗x(x,y)A∗y(x,y) . This equation can be integrated to yield a one-parameter set of integral curves, indexed by an initial condition. The equation for these curves may be written as U∗c(x,y)=0, where c labels the curves. Then along each curve we have 0=dU∗cdx=∂U∗x∂x+∂U∗c∂ydydx∑i=∂U∗c∂x−A∗xA∗y∂U∗c∂y . Thus, ∂U∗c∂xA∗y=∂U∗c∂yA∗x≡e−LA∗xA∗y . This equation defines the integrating factor L: L=−ln(1A∗x∂U∗c∂x)=−ln(1A∗y∂U∗c∂y) . We now have that A∗x=eL∂U∗c∂x,A∗y=eL∂U∗c∂y , and hence e−L\mathchar′26dW=∂U∗c∂xdx+∂U∗c∂ydy=dU∗c .