... A1= 0, A2= 4 3, A3= A 4 = ··· = 0Ψ(θ, t) = −13P0(cos θ) + 4 3P2(cos θ) exp−6a2R2tΨ(θ, t) = −13+2 cos2θ −23exp−6a2R2tSolution 37. 34 Since we have ... inhomogeneous partial differential equation, we will expand the solution in a series of eigenfunctions inx for which the coefficients are functions of t. The solution for u has the form,u(x, t) ... areJ0j0,nrRj0,nJ1(j0,n)∼2Rπj0,nrcosj0,nrR−π 4 j0,n2πj0,ncosj0,n−3π 4 ∼Rr(n − 1 /4) πcos(n−1 /4) πrR−π 4 cos ((n − 1)π).1809From this solution, we see that the critical current...