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Linear partial differnetial equations (PDE's) 
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#1
Mar808, 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. 


#2
Mar808, 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?



#3
Mar808, 03:33 PM

P: 1,863

Can you guys give me a hint? I still can't see the system in it.



#4
Mar808, 04:29 PM

Sci Advisor
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Thanks
P: 25,251

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.



#5
Mar808, 04:47 PM

P: 1,863

Thank you for taking the time to help me.



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