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Question: Kummer Function

Question:Kummer Function

Parham2016 70
kummersingularity
June 01 2016
0

 

Hi all,

 

It is believeble for me that the KummerU function is infinite or has a singular solution at r=0;

but without the command D(R) and dsolve ({er, D(R)(0)=0}, {R(r)}) that show the r=0 is a singular solution for the R(r), how could I prove it without using of these commands: for example is plotting of this function applicable??

while we have not the value of the  \lambda

````

restart

eq := diff(R(r), r, r)+(diff(R(r), r))/r+(-r^2+1)*lambda^2*R(r) = 0;

diff(diff(R(r), r), r)+(diff(R(r), r))/r+(-r^2+1)*lambda^2*R(r) = 0

 (1)

``

dsolve(eq);

R(r) = _C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+_C2*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2)

 (2)

``

dsolve({eq, (D(R))(0) = 0}, {R(r)})

R(r) = _C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)

 (3)

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R := proc (r) options operator, arrow; C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+_C2*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2) end proc

proc (r) options operator, arrow; C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+_C2*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2) end proc

 (4)

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D(R)

proc (r) options operator, arrow; -C1*lambda*r*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+2*C1*exp(-(1/2)*lambda*r^2)*((lambda*r^2-1/2-(1/4)*lambda)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+(1/2+(1/4)*lambda)*KummerM(-1/2-(1/4)*lambda, 1, lambda*r^2))/r-_C2*lambda*r*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2)+2*_C2*exp(-(1/2)*lambda*r^2)*((lambda*r^2-1/2-(1/4)*lambda)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2)-KummerU(-1/2-(1/4)*lambda, 1, lambda*r^2))/r end proc

 (5)

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