Final Examination Solutions

Transcription

1 Math. 42 Fulling) 4 December 25 Final Examination Solutions Calculators may be used for simple arithmetic operations only! Laplacian operator in polar coordinates: Some possibly useful information 2 u = 2 u r 2 + r r + r 2 θ 2. Laplacian operator in spherical coordinates physicists notation ): 2 u = r 2 r 2 ) + r r Spherical harmonics satisfy sinθ Bessel s equation: 2 Z z 2 + z 2 Z z z Legendre s equation: sinθ Airy s equation: sinθ ) + θ θ sin 2 θ Famous Green function integrals: r 2 sinθ 2 φ 2 sinθ ) + θ θ ] r 2 sin 2 θ φ 2. Y m l θ,φ) = ll+)y m l θ,φ). ) Z z + n2 z 2 Z = has solutions J n z) and Y n z). Z z + ll+) ) z 2 Z = has solutions j l z) and y l z). d sinθ dθ ) +ll+)θ = has a nice solution P dθ dθ l cosθ). 2 y zy = has solutions Aiz) and Biz). z2 e ikx e k2 t dk = e x2 /4t, 2π 4πt 2π e ikx e k y dk = π y x 2 +y 2.

2 42F-F5 Page 2. 3 pts.) Classify each equation as i) elliptic, hyperbolic, or parabolic, and ii) linear homogeneous, linear nonhomogeneous, or nonlinear. a) t 2 2 ) u 2 x 2 + +t 2 x 2 = x nonlinear, hyperbolic b) x u y 2 +e x y)2 = linear nonhomogeneous, elliptic c) t + x 2 u x 2 +x2 u = linear homogeneous, parabolic In any remaining problems that involve spherical coordinates, their range is always the whole sphere, < θ < π, π < φ π or equivalent). I reiterate that the meanings of θ and φ are reversed compared to Haberman s book pts.) Solve the heat equation in a ball, 2 u = for r < R, < t <, t ur,θ,φ,t) =, ur,θ,φ,) = fr,θ,φ). The spherical harmonic notation is strongly advised.) After separating the time variable we have to solve 2 u λ = λu λ with homogeneous Dirichlet condition on the sphere. The solutions are well known to be Yl m θ,φ)j l ω ln r), where j l is called a spherical Bessel function related to an ordinary Bessel function of order l + 2 ), the ω ln are the numbers for which j l ω ln R) =, and λ ln = ω 2 ln. Here ln is a pair of indices, not their product.) Then we can write the general solution l ur,θ,φ,t) = C lmn Yl m θ,φ)j l ω ln r)e λ lnt. Finally so fr,θ,φ) = l= m= ln= l l= m= ln= C lmn Y m l θ,φ)j l ω ln r), π π dφ π dθ R C lmn = drr2 sinθj l ω ln r)yl m θ,φ) fr,θ,φ) R. j lω ln r) 2 r 2 dr

3 42F-F5 Page pts.) a) Solve by separation of variables or an equivalent transform technique: x u = < x <, < y < ), y2,y) = < y < ), x ux,) = fx) < x < ). Weseethatwewillneedatransformonthehalf-lineof x. TheboundaryconditionindicatesaFourier cosine transform, since the cosines are the eigenfunctions that satisfy the Neumann condition. The infinite interval in y calls for decaying exponentials. So the expected form of the solution is Thus so ux,y) = fx) = Ak) = 2 π Ak)coskx)e ky dk. Ak) coskx) dk, fx)coskx)dx. b) Find the Green function that gives the solution to a) in the form ux,y) = Gx,z,y)fz)dz. There are two methods. Do evaluate the integral if your method leads to one.) Substituting the formula for A into the one for u, we get ux,y) = = 2 π fz)coskz)dzcoskx)e ky dk ] 2 dzfz) dk coskz)coskx)e ky. π So the object in the brackets is the Green function. Using coskz) = 2 eikz +e ikz ) etc., you can reduce this to integrals that can be evaluated by the second famous Green function formula. The result will be the same as the one I shall now get by the image method. Alternative method: Remember that the second famous Green function integral, with x replaced by x z, is the Green function for the similar problem with fz) given on the entire real line. For the half-line with the homogeneous Neumann condition at the end, the Green function is obtained by adding the effect of an image source at z instead of z : Gx,z,y) = π y x z) 2 +y 2 + y x+z) 2 +y 2 ].

4 42F-F5 Page pts.) Solve Laplace s equation in the region between two concentric spheres: 2 u = for < r < 3,,θ,φ) =, r u3,θ,φ) = fθ) independent of φ). Note that φ is the azimuthal angle, not the polar one.) Again we have spherical harmonics, but this time the radial solution involves r l and r l+), not Bessel functions. Which linear combination satisfies the homogeneous boundary condition? Rr) = Ar l +Br l+), = R ) = la l+)b. Of course we can only determine the ratio at this point, so take A =, B = l l+. ur,θ,φ) = l l= m= l C lm Yl m θ, φ) r l + l l + r l+)]. But because there is no φ dependence, only m = contributes: ur,θ,φ) = l= C l Yl θ,φ) r l + l l+ r l+)]. This could also be written in terms of P l cosθ), but the spherical harmonic notation automatically takes care of the normalization factor.) Then fθ) = l= C l Yl θ,φ) 3 l + l l+ 3 l+)]. The spherical harmonic here is actually independent of φ and is some multiple of P l cosθ). Also, it is real.) So C l = 2π π dθ sinθy l θ,φ)fθ) 3 l + l l+ 3 l+).

5 42F-F5 Page pts.) Using Fourier series, solve this modified wave equation on a circle with periodic boundary conditions): t 2 +tu = 2 u x 2 < t <, π < x π), ux,) = fx), x,) = gx). t If you don t know the solutions for the time dependence, give them convenient names and proceed.) Since this one involves a nonstandard operator, it is prudent to go through the whole separation of variables from the beginning. Let u sep = Xx)Tt). Then XT +txt = X T, so T +tt T = X X = λ. The spatial angular!) equation is X = λx, and because of the periodicity we must have λ = n 2 for integer n) and X n = e inx n =,±,±2,...) or, alternatively, X n = sinnx) n =,2,...) or cosnx) n =,,...). I will use the exponential form because it calls for less typing. The time equation is T + tt = n 2 T. It is second-order, linear, homogeneous, so it has a two-dimensional vector space of solutions. In other words, any solution is a linear combination of two basis solutions. Furthermore, the equation is nonsingular, so each solution can be characterized by its initial data at t =. In this problem the initial time was chosen to be instead of just to keep the coefficient t from changing sign in the region of interest, but that turns out not to be very important.) Let s introduce the notation p and q for the solutions with the data p) =, p ) =, q) =, q ) =. These functions also depend on λ = n 2, so they should carry a subscript n ; but p n and p n are the same thing, and the same for q. Now we write the general solution. For each λ there are four independent solutions, each of which needs a yet-to-be-determined coefficient: ux,t) = n= An p n t)e inx +B n q n t)e inx]. If you use the trigonometric basis, you must literally write four terms: ux,t) = an p n t)cosnx)+b n q n t)cosnx)+c n p n t)sinnx)+d n q n t)sinnx) ]. n=

6 42F-F5 Page 6 Recall that writing the most general solutions of the X equation and the T equation and multiplying them does not give the correct linear) construction.) Determine the coefficients by imposing the initial conditions: Therefore, fx) = n= A n e inx, gx) = n= B n e inx. A n = π e inx fx)dx, B n = π e inx gx)dx. 2π π 2π π Finally, can we say anything about what a n and b n are? Try introducing a new variable τ = t+n 2 ). This transforms the T equation to T τt = where the primes now indicate derivatives with respect to τ but the second derivative is the same for either variable). This is Airy s equation. So we know that p n t) and q n t) are certain linear combinations of the Airy fuctions Ai t n 2 ) and Bi t n 2 ). To find the coefficients in these linear combinations we would need to use a table or a computer program to find the values of the Airy functions and their derivatives at all the points z = n 2 +).