Linear partial differnetial equations (PDE's)


by Niles
Tags: differnetial, equations, linear, partial
Niles
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#1
Mar8-08, 09:30 AM
P: 1,863
1. The problem statement, all variables and given/known data
Please take a look at the example at the bottom (at eq. 18:18, at page 619):

http://books.google.dk/books?id=9p6s...l=da#PPA619,M1

Q1: In case (ii), why do they add g(x^2+y^2)?
Q2: Why do they not add it in case (i)?
Q3: In case (ii), if I chose f(p) = p + 1 instead of f(p) = 2p, then the solution would be u(x,y) = 1+ x^2 +y^2 + g(x^2 + y^2), right?

I hope you can help me; I find this really hard to understand, and I've spent hours trying to find out, but I find that the book is poorly written. They don't emphasize the important things at all, and the reader is left behind with so many questions.

Thanks in advance.
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Niles
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#2
Mar8-08, 12:25 PM
P: 1,863
The difference between (ii) and (i) is that in (i) it is a function, but in (ii) it is a single point. Although I am not sure about this?
Niles
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#3
Mar8-08, 03:33 PM
P: 1,863
Can you guys give me a hint? I still can't see the system in it.

Dick
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#4
Mar8-08, 04:29 PM
Sci Advisor
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Thanks
P: 25,170

Linear partial differnetial equations (PDE's)


The difference between i) and ii) is that in i) the boundary conditions are given along a line y=0 and in ii) they are specified only a a single point. In the first case that's enough information to specify a unique solution. In the second case you can add a general function that solves the homogeneous equation (g(x^2+y^2)) subject to only one condition. I don't see why they are messing around with splitting g into two parts. That is confusing. I would just write the solution as -3y+g(x^2+y^2) subject to the condition g(1)=2. Your other form for Q3 is fine, but you forgot to put the -3y in it.
Niles
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#5
Mar8-08, 04:47 PM
P: 1,863
Thank you for taking the time to help me.


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