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    I'm trying to find a function A(r) that fits the following conditions:

     A(0) = 0
     A(0<r<1) > 0
     A(1) = 0

    And if B(r) = \frac{1}{r}\left(\frac{\partial}  {\partial r} r A(r)\right):

     B(0) = 1
     B(1) = 0
     B(0<r<1) > 0

    (i.e. A(r) has to be zero at r = 0 and r = 1, and B(r) has to have a maximum at r = 0 and has to be 0 at r = 1, and both functions must be greater than or equal to zero for r between 0 and 1).

    Is this possible? So far I've been using a function based on \cos^2(\frac{\pi r}{2 r_0}) as B(r) which satisfies all of the above apart from A(1)=0 but I really need to find something that fulfils this condition too.
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    (Original post by Plagioclase)
    I'm trying to find a function A(r) that fits the following conditions:

     A(0) = 0
     A(0<r<1) > 0
     A(1) = 0

    And if B(r) = \frac{1}{r}\left(\frac{\partial}  {\partial r} r A(r)\right):

     B(0) = 1
     B(1) = 0
     B(0<r<1) > 0

    (i.e. A(r) has to be zero at r = 0 and r = 1, and B(r) has to have a maximum at r = 0 and has to be 0 at r = 1, and both functions must be greater than or equal to zero for r between 0 and 1).

    Is this possible? So far I've been using a function based on \cos^2(\frac{\pi r}{2 r_0}) as B(r) which satisfies all of the above apart from A(1)=0 but I really need to find something that fulfils this condition too.
    Here's a hint: take your differential expression for B in terms of A and invert it by integration to get A in terms of B. Then note that A(1) is an integral of stuff that is strictly poisitive and therefore cannot be zero.
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    (Original post by Gregorius)
    Here's a hint: take your differential expression for B in terms of A and invert it by integration to get A in terms of B. Then note that A(1) is an integral of stuff that is strictly poisitive and therefore cannot be zero.
    Ugh silly me, thanks for that!
 
 
 
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