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Complex Analysis

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humph:
The branch cut from the negative logarithm is because when you approach it from one side you get a different limit than from the other side. But note that , so it doesn't make any difference when you take an integer power.

/0:
o right, thanks

/0:
Just a few last minute questions xD

1. If you have a function like .


Is it ok if for and you choose different intervals of existence for and ?
i.e. can you rotate each of the branch cuts however you like?

And would this mean you have to deal with each logarithm as a separate entity? (i.e. anymore?)





2. Also, for , we take . The solutions take the left contour, but I'm wonder, why couldn't we just take the right contour and just ignore the pole at ?

Thanks

/0:
From the PDEs exam:

"

a) Obtain the general solution of the first order PDE:
 


b) If we prescribe on the upper portion of the hyperbola , , show that no solution exists unless is of a special form. Find this form and show that in such a case there are infinitely many solutions."


I tried solving this the way it's done in the lecture notes but it didn't work =/



,    ,   

I can't solve any of these straight away so the usual method is already looking shaky

But I notice that,



And,

, which means ,

Hence, , but is a function of so .

Still, this isn't the solution... and I couldn't do the next part of the question

How can you solve it?

humph:
Hah, this was one of my assignment questions a couple of years ago. I'll just post it up in full.


Let us parametrise the characteristic curve in the plane of this partial differential equation by , and let . As the chain rule implies that

it follows that we can write

Writing , we see that the latter two ordinary differential equations are equivalent to the vector-valued ordinary differential equation

The matrix above has eigenvalues with corresponding eigenvectors , and so the solution to this ordinary differential equation is given by

where . Thus we have that and . Substituting these values into the earlier expression for , we have that

which, by multiplying through by the integrating factor , can be rearranged to read

and so we can integrate to find that

or equivalently, by substituting back in , , and then multiplying through by ,

where depends on the chosen characteristic curve. These characteristic curves are the solutions to

which, by separating variables, we can solve to find that

where . Thus we have that for any arbitrary function , and so


With this solution of the partial differential equation, we therefore have that on the upper portion of the hyperbola , ,

Thus if is not of the form for some , then we cannot have that on the upper portion of the hyperbola , . If, on the other hand, is of this form, then there are infinitely many solutions of the partial differential equation satisfying this prescribed condition, as there are infinitely many functions satisfying ; for example, we could have for any .

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