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Linear partial differnetial equations (PDE's)

  1. Mar 8, 2008 #1
    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):


    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.
    Last edited: Mar 8, 2008
  2. jcsd
  3. Mar 8, 2008 #2
    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?
  4. Mar 8, 2008 #3
    Can you guys give me a hint? I still can't see the system in it.
  5. Mar 8, 2008 #4


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    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.
  6. Mar 8, 2008 #5
    Thank you for taking the time to help me.
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