Final Examination Solutions

Transcription

1 Math. 42 Fulling) December 22 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 + 2 u r 2 θ 2. Laplacian operator in spherical coordinates physicists notation ): 2 u = r 2 r 2 ) + r r Spherical harmonics satisfy [ sinθ ) + sinθ θ θ sin 2 θ Bessel s equation: 2 Z z 2 + z 2 Z z z Legendre s equation: sinθ Famous Green function integrals: r 2 sinθ 2 φ 2 sinθ ) + θ θ ] r 2 sin 2 θ 2 u 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). e ikx e k2 t dk = e x2 /4t, 2π 4πt 2π e ikx e k y dk = π Hyperbolic function identities: y x 2 +y 2. sinha±b) = sinhacoshb±coshasinhb, cosha±b) = coshacoshb±sinhasinhb.

2 42F-F2 Page 2. 3 pts.) Classify each equation as i) elliptic, hyperbolic, or parabolic, and ii) linear homogeneous, linear nonhomogeneous, or nonlinear. a) t + x 2 u x 2 = Parabolic, linear homogeneous. b) Elliptic, nonlinear. 2 u x 2 ++u2 ) 2 u y 2 = 2 u c) t 2 2 u x 2 +cos2t) = Hyperbolic, linear nonhomogeneous pts.) Solve Laplace s equation in a ball, 2 u = for r < R, R,θ,φ) = fθ,φ). r You may jump right to the answer if you know it. The spherical harmonic notation is strongly advised.) The general solution of Laplace s equation in spherical coordinates with regularity at the origin is ur,θ,φ) = l l= m= l c lm r l Y m l θ,φ). Impose the boundary condition fθ,φ) = l l= m= l c lm lr l Y m l θ,φ). Since the spherical harmonics are orthonormal, it follows that π c lm = l R l sinθdθ 2π dφy m l θ,φ) fθ,φ).

3 42F-F2 Page pts.) a) Solve by separation of variables or an equivalent transform technique: t 2 u = < x <, < t < ), x2,t) = < t < ), x ux,) = fx) < x < ). I will use the transform language since it s briefer. The infinite x interval and Neumann boundary condition suggest a Fourier cosine transform in x: So U t +ω2 U =, Uω,) = Fω). Uω,t) = Fω)e ω2 t. So far I didn t say how the transform is normalized; we need a net factor of you have Fω) = ux,t) = 2 π 2 π. So if you define cosωx)fx) dx, ) cosωx)e ω2 t Fω)dω. 2) b) Find the Green function that gives the solution to a) in the form ux,t) = Gx,z,t)fz)dz. There are two methods. Do evaluate the integral if your method leads to one.) Method : Substitute ) into 2) and rearrange: So we can identify Gx,z,t) = 2 π ux,t) = 2 π = 2π = 2π dz dωcosωx)cosωz)e ω2 t dω dωcosωx)cosωz)e ω2 t fz). [ dω e iωx+z) +e iωx+z) +e iωx z) +e iωx z)] e ω2 t [ e iωx+z) +e iωx z)] e ω2 t = 4πt [e x+z)2 /4t +e x z) 2 /4t ]. Method 2: We know that the first famous integral, with x replaced by x z, is the Green function for the analogous problem on the whole line. By the method of images, the Green function for our problem is obtained by adding the same function of z replaced by z.

4 42F-F2 Page pts.) Solve the heat equation in the region between two concentric spheres: t = 2 u for r < 2, ut,,θ,φ) =, ut,2,θ,φ) =, u,r,θ,φ) = fr,θ) independent of φ). Note that φ is the azimuthal angle, not the polar one.) Separate variables as u = Ψωr,θ,φ)e ω2t, arriving at ω 2 Ψ ω = 2 Ψ ω, the Laplacian in spherical coordinates being given on the first page of the test. The quickest way to proceed is to notice that since the data function in this problem is independent of φ, the relevant eigenfunctions will be, too; therefore, we can discard all the φ derivatives and get r 2 r r 2 U r ) + r 2 sinθ θ sinθ Ψ θ Multiply by r 2 and separate variables again: U = Rr)Θθ), d sinθ dθ ) +ll+)θ = sinθ dθ dθ ) d r 2 dr = ll+)r ω 2 r 2 R. dr dr So Θ = P l cosθ) and d 2 R dr r ) dr dr = ll ) r 2 ω2 R, ) = ω 2 U. whose solutions are j l ωr) and y l ωr), the spherical Bessel functions. We need to choose a linear combination of them that vanishes at r = ; will do. We also need R l ω,r) j l ωr)y l ω) y l ωr)j l ω) = R l ω,2) = j l 2ω)y l ω) y l 2ω)j l ω); this is the regular Sturm Liouville) eigenvalue condition to be solved for the allowed values ω j. The full solution now is ut,r,θ,φ) = C lj R l ω j,r)p l cosθ)e ω j 2 t. The coefficients are C ln = U lj,f U lj 2, j= l= U ljr,θ) = R l ω j,r)p l cosθ). The numerator and denominator are messy to write out; they both involve integrations of the type 2 r 2 dr π cosθdθ. There is no need to integrate over φ, or even to multiply by 2π because that will cancel. Remember, however, that the Legendre polynomials are not normalized to unit norm.

5 42F-F2 Page 5 5. Brief essays 3 pts.) Imagine that you are grading homework, or working in a help session, in this course. How would you explain to a student how [s]he is going wrong? a) The problem is the wave equation on the real line < x < ). The student has separated variables and arrived at mode functions of the form Aω cosωx)+b ω sinωx) ) C ω cosωt)+d ω sinωt) ). The solution must be a linear combination of normal modes, A ω cosωx)cosωt)+b ω sinωx)cosωt)+c ω cosωx)sinωt)+d ω sinωx)sinωt). In the product form, solutions for the coefficients that match the data may not exist and may not be unique. b) The problem is the heat equation on an interval of length π with data ux,) = fx), u,t) = T, uπ,t) =. The student has separated variables and arrived at the equations X n x) = n2 X n x), X n ) = T. You can t impose a nonhomogeneous boundary condition on the individual normal modes! When you add them, the condition will not be satisfied. In this problem you should find a steady-state solution first and subtract it off. c) The problem is Laplace s equation on the semiinfinite strip < x <, < y < L with boundary data The student has written down a sum ux,) =, ux,l) = fx), u,y) =. ux,y) = c n sin λ n x)sinh λ n y). n= What is λ n here? Because the data interval is infinite, the solution should be a transform integral), not a sum.