Question #78893

Reduce the following PDE to a set of three ODEs by the method of separation variables.
1/r d/dr(r dv/dr)+1/r2 d2v/dϴ2+d2v/dz2+k2v=0

Expert's answer

Answer on Question #78893 – Math – Differential Equations

Question

Reduce the following PDE to a set of three ODEs by the method of separation of variables


1rr(r(Vr))+(1r2)(2Vθ2)+(2Vz2)+(k2)V=0\frac {1}{r} \frac {\partial}{\partial r} \left(r \left(\frac {\partial V}{\partial r}\right)\right) + \left(\frac {1}{r ^ {2}}\right) \left(\frac {\partial^ {2} V}{\partial \theta^ {2}}\right) + \left(\frac {\partial^ {2} V}{\partial z ^ {2}}\right) + (k ^ {2}) V = 0


Solution

Assume


V(r,θ,z)=R(r)Θ(θ)Z(z)V (r, \theta , z) = R (r) \Theta (\theta) Z (z)ΘZ(1r)ddr(r(dRdr))+RZ(1r2)(d2Θdθ2)+RΘ(d2Zdz2)+(k2)RΘZ=0\Theta Z \left(\frac {1}{r}\right) \frac {d}{d r} \left(r \left(\frac {d R}{d r}\right)\right) + R Z \left(\frac {1}{r ^ {2}}\right) \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right) + R \Theta \left(\frac {d ^ {2} Z}{d z ^ {2}}\right) + (k ^ {2}) R \Theta Z = 0


Divide through by VV :


1R(1r)ddr(r(dRdr))+1Θ(1r2)(d2Θdθ2)+1Z(d2Zdz2)+k2=0\frac {1}{R} \left(\frac {1}{r}\right) \frac {d}{d r} \left(r \left(\frac {d R}{d r}\right)\right) + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right) + \frac {1}{Z} \left(\frac {d ^ {2} Z}{d z ^ {2}}\right) + k ^ {2} = 01R(1r)dRdr+1Rd2Rdr2+1Θ(1r2)(d2Θdθ2)+k2=1Z(2Zz2)\frac {1}{R} \left(\frac {1}{r}\right) \frac {d R}{d r} + \frac {1}{R} \frac {d ^ {2} R}{d r ^ {2}} + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right) + k ^ {2} = - \frac {1}{Z} \left(\frac {\partial^ {2} Z}{\partial z ^ {2}}\right)


In the above equation the left-hand side depends on rr and θ\theta , while the right-hand side depends on zz . The only way these two members are going to be equal for all values of r,θr, \theta and zz is when both of them are equal to a constant. Let us define such a constant as l2-l^2 .

With this choice for the constant, we obtain:


d2Zdz2l2Z=0\frac {d ^ {2} Z}{d z ^ {2}} - l ^ {2} Z = 0


The general solution of this equation is:


Z(z)=a1elz+a2elzZ (z) = a _ {1} e ^ {l z} + a _ {2} e ^ {- l z}


Such a solution, when considering the specific boundary conditions, will allow Z(z)Z(z) to go to zero for zz going to ±\pm \infty , which makes physical sense. If we had given the constant a negative value, we would have had periodic trigonometric functions, which do not tend to zero for zz going to infinity.

Once sorted the zz -dependency, we need to take care of rr and θ\theta .


1R(1r)dRdr+1Rd2Rdr2+1Θ(1r2)(d2Θdθ2)=(k2+l2)\frac {1}{R} \left(\frac {1}{r}\right) \frac {d R}{d r} + \frac {1}{R} \frac {d ^ {2} R}{d r ^ {2}} + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right) = - (k ^ {2} + l ^ {2})rRdRdr+r2Rd2Rdr2+(k2+l2)r2=1Θ(d2Θdθ2)\frac {r}{R} \frac {d R}{d r} + \frac {r ^ {2}}{R} \frac {d ^ {2} R}{d r ^ {2}} + (k ^ {2} + l ^ {2}) r ^ {2} = - \frac {1}{\Theta} \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right)


Again we are in a situation where the only way a solution can be found for the above equation is when both members are equal to a constant. This time we select a positive constant, which we call m2m^2 . The equation for Θ\Theta becomes, then:


d2θdθ2+m2θ=0\frac {d ^ {2} \theta}{d \theta^ {2}} + m ^ {2} \theta = 0


Its general solution can be written as:


θ(θ)=b1sin(mθ)+b2cos(mθ)\theta (\theta) = b _ {1} \sin (m \theta) + b _ {2} \cos (m \theta)


This solution is well suited to describe the variation for an angular coordinate like θ\theta . Had we chosen to set both members of equation equal to a negative number, we would have ended up with exponential functions with a different value assigned to Θ(θ)\Theta(\theta) for each 360 degrees turn, a clear non-physical solution.

Last to be examined is the r-dependency. We have:


rRdRdr+r2Rd2Rdr2+(k2+l2)r2=m2\frac {r}{R} \frac {d R}{d r} + \frac {r ^ {2}}{R} \frac {d ^ {2} R}{d r ^ {2}} + (k ^ {2} + l ^ {2}) r ^ {2} = m ^ {2}r2d2Rdr2+rdRdr+((k2+l2)r2m2)R=0r ^ {2} \frac {d ^ {2} R}{d r ^ {2}} + r \frac {d R}{d r} + \left((k ^ {2} + l ^ {2}) r ^ {2} - m ^ {2}\right) R = 0


The equation ()(\ast) is a well-known equation of mathematical physics called parametric Bessel's equation. With a simple linear transformation of variable, x=(k2+l2)rx = (\sqrt{k^2 + l^2})r , equation ()(\ast) is readily changed into a Bessel's equation:


dRdr=dRdxdxdr=(k2+l2)R\frac {d R}{d r} = \frac {d R}{d x} \frac {d x}{d r} = \left(\sqrt {k ^ {2} + l ^ {2}}\right) R ^ {\prime}d2Rdr2=ddx((k2+l2)R)dxdr=(k2+l2)R\frac {d ^ {2} R}{d r ^ {2}} = \frac {d}{d x} \left(\left(\sqrt {k ^ {2} + l ^ {2}}\right) R ^ {\prime}\right) \frac {d x}{d r} = (k ^ {2} + l ^ {2}) R ^ {\prime \prime}x2k2+l2(k2+l2)R+xk2+l2(k2+l2)R+(x2m2)R=0\frac {x ^ {2}}{k ^ {2} + l ^ {2}} (k ^ {2} + l ^ {2}) R ^ {\prime \prime} + \frac {x}{\sqrt {k ^ {2} + l ^ {2}}} \left(\sqrt {k ^ {2} + l ^ {2}}\right) R ^ {\prime} + (x ^ {2} - m ^ {2}) R = 0x2R+xR+(x2m2)R=0x ^ {2} R ^ {\prime \prime} + x R ^ {\prime} + (x ^ {2} - m ^ {2}) R = 0


where RR'' and RR' indicate the first and the second derivatives with respect to xx . In what follows we will assume that mm is a real, non-negative number.

Linearly independent solutions are typically denoted Jm(x)J_{m}(x) (Bessel Functions) and Nm(x)N_{m}(x) (Neumann Functions).

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