Mathematical Physics Vol 1

Chapter 7. Partial differential equations

400

Problem 260

As a characteristic example of a partial differential equation of second order, which comes down to Bessel equation, let us observe the equation ∆ u + k 2 u = 0 , (7.303) on or inside of a circle (two variables) or, on or inside a cylinder (three variables). Let us find its solution. Solution Using the polar coordinates (two variables: r , ϕ ) the equation (7.303) can be written in the form 1 r ∂ ∂ r r ∂ u ∂ r + 1 r 2 ∂ 2 u ∂ϕ 2 + k 2 u = 0 . (7.304) and the equation (7.304) is decomposed to two ordinary differential equations 1 r d d r r d R d r + − λ r 2 + k 2 R = 0 , (7.306) Φ ′′ + λ Φ = 0 . (7.307) As we have shown previously, the equation (7.307) yields the dependence λ = n 2 . Let us now introduce the substitution x = kr , and equation (7.306) becomes the Bessel equation: 1 x d d x ( xy ′ )+ 1 − n 2 x 2 y = 0 , (7.308) R ( r )= y ( x )= y ( kr ) . (7.309) In case of radial symmetry ( n = 0) it comes down to the Bessel equation of zero order y ′′ + 1 x y ′ + y = 0 . (7.310) Assume, further, that the solution u can be represented in the form u = R ( r ) · Φ ( ϕ ) , (7.305)

Problem 261

Find the solution of equation

2 u

u tt = a

xx ,

that satisfies the boundary

u ( 0 , t )= 0 , u ( l , t )= 0 , 0 ≤ x ≤ 1