FirstEqn3 - Math 246, Fall 2009, Professor David Levermore...

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Unformatted text preview: Math 246, Fall 2009, Professor David Levermore 7. Exact Differential Forms and Integrating Factors Let us ask the following question. Given a first-order ordinary equation in the form (7.1) d y d x = f ( x,y ) , when do its solutions satisfy a relation of the form H ( x,y ) = c where c is an arbitrary constant ? Such an H ( x,y ) is called an integral of (7.1). This question is easily answered if we assume that all functions involved are as differntiable as we need. Suppose that such an H ( x,y ) exists, and that y = Y ( x ) is a solution of differential equation (7.1). Then H ( x,Y ( x )) = H ( x I ,Y ( x I )) , where x I is any point in the interval of definition of Y . By differentiating this equation with respect to x we find that x H ( x,Y ( x )) + Y ( x ) y H ( x,Y ( x )) = 0 . Therefore, wherever y H ( x,Y ( x )) negationslash = 0 we see that Y ( x ) = x H ( x,Y ( x )) y H ( x,Y ( x )) . For this to hold for every solution of (7.1), we must have d y d x = x H ( x,y ) y H ( x,y ) , or equivalently (7.2) f ( x,y ) = x H ( x,y ) y H ( x,y ) , wherever y H ( x,y ) negationslash = 0. The question then arises as to whether we can find an H ( x,y ) such that (7.2) holds for any given f ( x,y )? It turns out that this cannot always be done. In this section we explore how to seek such an H ( x,y ). 7.1. Exact Differential Forms. The starting point is to write equation (7.1) in a so-called differential form (7.3) M ( x,y ) d x + N ( x,y ) d y = 0 , where f ( x,y ) = M ( x,y ) N ( x,y ) . There is not a unique way to do this. Just pick one that looks natural. If you are lucky then there will exist a function H ( x,y ) such that (7.4) x H ( x,y ) = M ( x,y ) , y H ( x,y ) = N ( x,y ) . When this is the case the differential form (7.3) is said to be exact . 1 2 It turns out that there is test you can easily apply to find out if you are lucky. It de- rives from the fact that mixed partials commute namely, the fact that for any twice differentiable H ( x,y ) one has y ( x H ( x,y ) ) = x ( y H ( x,y ) ) . This fact implies that if (7.4) holds for some twice differentiable H ( x,y ) then M ( x,y ) and N ( x,y ) satisfy y M ( x,y ) = y ( x H ( x,y ) ) = x ( y H ( x,y ) ) = x N ( x,y ) . In other words, if the differential form (7.3) is exact then M ( x,y ) and N ( x,y ) satisfy (7.5) y M ( x,y ) = x N ( x,y ) . The remarkable fact is that the converse holds too. Namely, if the differential form (7.3) satisfies (7.5) for every ( x,y ) then it is exact i.e. there exists an H ( x,y ) such that (7.4) holds. Moreover, the problem of finding H ( x,y ) is reduced to evaluating two integrals. We illustrate this fact with examples....
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This note was uploaded on 01/24/2011 for the course MATH math246 taught by Professor Levermore during the Spring '10 term at Maryland.

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FirstEqn3 - Math 246, Fall 2009, Professor David Levermore...

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