PARTIAL DIFFERENTIAL EQUATIONS (MATH417) SOLUTIONS FOR THE FINAL EXAM

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1 PARTIAL DIFFERENTIAL EQUATIONS (MATH417) SOLUTIONS FOR THE FINAL EXAM Problem 1 (1 pts.) Classify each equation as linear homogeneous, linear inhomogeneous, or nonlinear: a) u = u x + x 3 b) u + u c) u x + u y = u 4 u y 4 x = 3u d) u x y = e x y Solution: (a) is a linear inhomogeneous equation; (b) is a linear homogeneous equation; (c) is a nonlinear (because of the right-hand side) equation; (d) is a linear inhomogeneous equation. Problem ( pts.) Let f(x) = x. (1) Find the Fourier series (complex form) of f(x) on the interval ( π, π). () Rewrite the Fourier series in the real form. (3) Sketch the function to which the Fourier series converges. (4) Use Parseval s equality to evaluate n=1 n 4. Solution: (1) The required series is f(x) n Z c ne inπx/l with c n = 1 π π π f(x)e inx dx. If n =, then c = 1 π x dx = 1 x 3 π π π 3 π π = π 3. If n, we can integrate by parts twice to get c n = ( 1)n n. We can also use the given formula x e iax dx = ( x ia + x a + i a 3 )eiax + c, a. According to this integral, c n = 1 π Thus π π x e inx dx = 1 π ( x in +x n i n 3 )e inx π π = 1 x π n e inx π π = e inπ + e inπ n. x π 3 + n,n Z 1 ( 1) n n e inx

2 SOLUTION FOR FINAL y x - Pi -Pi Pi Pi Figure 1. () Rewrite it in the real form, π 3 + Thus n,n Z ( 1) n n e inx = π 3 + ( 1) n n x π 3 + 4( 1) n n cos(nx). (e inx +e inx ) = π 3 + 4( 1) n n cos(nx) (3) Sketch the function to which the Fourier series converges. The series converges to the π-periodic function that coincides with f(x) for π < x < π. The sum is continuous and piecewise smooth hence the convergence is uniform. see Figure 1. (4) Use Parseval s equality to evaluate n=1 n 4. Recall that f(x) n Z c n e inx, c n = 1 π π π f(x)e inx dx Now Thus 1 π f dx = c n π π n Z 1 π f dx = 1 x 5 π π π 5 π π = π4 5, n Z c n = ( π 3 ) + ( ( 1)n n ) π 4 5 = π n 4

3 SOLUTION FOR FINAL 3 It follows that 1 n 4 = 1 8 (π4 5 π4 9 ) = π4 9. Problem 3 (4 pts.) a) By the method of your choice, solve the wave equation on the half-line u = u x < x <, < t <, u x b) Solve the problem (, t) = < t <, u(x, ) = x e x, u (x, ) =, < x <. u = u u x x < x <, < t <, (, t) = t < t <, u(x, ) = x e x, u (x, ) =, < x <. a) D Alembert s method: We use method of characteristic to give the solution here i)let f be the even extension of the function x e x (x > ), i.e. f(x) = x e x. Then we consider instead the problem on real line. ii) We know the general solution can be given by IC gives us that which means that we can set iii) thus we can know that u(x, t) = F (x t) + G(x + t). F (x) + G(x) = f(x), G (x) F (x) =, F (x) = f(x), G(x) = f(x). f(x + t) + f(x t) u(x, t) = = 1 [(x + t) e x+t + (x t) e x t ]. Fourier s method. In view of the BC, we use Fourier cosine transform to solve it. Recall that Then C[f](ω) = π f(x) cos ωxdx, f(x) = C[ df dx ] = f() + ωs[f], π S[ df dx ] = ωc[f]. C[f](ω) cos ωxdω. C[ d f dx ] = df df () + ωs[ π dx dx ] = df π dx () ω C[f].

4 4 SOLUTION FOR FINAL Apply these formula to the PDE, we get C[u](ω, t) = du π dx (, t)) ω C[u](ω, t) = ω C[u](ω, t). The ODE has general solution for ω, C[u](ω, t) = a(ω) cos ωt + b(ω) sin ωt. We also have the IC C[u](ω, t) C[u](ω, ) = C[f](ω), =. Thus the general solution to this ODE in t for fixed ω > is given by then Recall C[u](ω, t) = C[f](ω) cos ωt. cos ωt cos xω = cos(x t)ω + cos(x + t)ω = cos x t ω + cos x + t ω, u(x, t) = C 1 [C[u]](x, t) = C[f](ω) cos ωt cos xωdω = 1 [ C[f](ω) cos x t ωdω + = 1 [f( x t ) + f( x + t )] = 1 [(x + t) e x+t + (x t) e x t ]. C[f](ω) cos x + t ωdω] b) D Alembert s method: We use method of characteristic to give the solution. Let f(x) = x e x, x > i) We know the general solution can be given by IC gives us that which means that we can set u(x, t) = F (x t) + G(x + t). F (x) + G(x) = f(x), x > G (x) F (x) =, x > F (x) = G(x) = f(x) = x e x, x >. iii) To solve the problem, we still need to know F for negative variables, which comes from the BC. By BC, we see that then for x <, F ( t) + G (t) = t, t > F (x) = x G ( x), F (x) = c x + G( x)

5 SOLUTION FOR FINAL 5 we would like to have F continuous at, which gives c = G( ) = G() =. Thus we can know that { x F (x) = e x x > x (ex 1) x < u(x, t) = { F (x t) + G(x + t) = 1 [(x + t) e x+t + (x t) e x t ] x t (x t) (e x t 1) + (x+t) e x t x < t Problem 4 (3 pts+5 bonus) Consider the regular Sturm-Liouville eigenvalue problem d dx (p(x)du ) + q(x)u + λσ(x)u =. dx We try to find the solution of form u(x) = A(x)e iλ1/ ϕ(x) for λ 1 with ϕ >. a)[1 pts] substitute to the DE, we may get [λb (x) + iλ 1/ B 1 (x) + B (x)]e iλ1/ϕ = determine B i (x). b)[5 pts] By letting B = B 1 =, solve for A = A (x) and ϕ. This will give the first approximation of the eigenfunction. c)[5 pts] In general, we can not have B = for the choice in ). Note that we have B = independent of A, the DE for A is iλ 1/ B 1 (x) + B (x) =. Instead, let A(x) = A (x) + λ 1/ A 1 (x) + λ 1 A (x) +. Comparing the coefficients of powers of λ, we can solve for A n (x). Use the described method to solve for A 1 in terms of A (together with p, q, σ). d)[1 pts] Consider the particular case p = q = σ = 1, u() = u x (π) =. By using the Rayleigh quotient, give an upper bound for λ 1, the smallest eigenvalue. e) (Bonus 5points) What is the precise value of λ 1? Solution: a) since u(x) = A(x)e iλ1/ ϕ(x), we have u (x) = A e iλ1/ϕ + iλ 1/ ϕ Ae iλ1/ ϕ u (x) = A e iλ1/ϕ + iλ 1/ ϕ A e iλ1/ϕ + iλ 1/ (ϕ A) e iλ1/ϕ λ(ϕ ) Ae iλ1/ ϕ Substitute in the equation we get d dx (p(x)du dx ) + q(x)u + λσ(x)u = pu + p u + qu + λσu =. λ(σ p(ϕ ) )Ae iλ1/ϕ +iλ 1/ [p(ϕ A +(ϕ A) )+p ϕ A]e iλ1/ϕ +(pa +p A +qa)e iλ1/ϕ = We see that B (x) = (σ p(ϕ ) )A B 1 (x) = p(ϕ A + (ϕ A) ) + p ϕ A = A pϕ + A(pϕ ) B (x) = pa + p A + qa b) B = tells us that x ϕ = (σ/p) 1/ ϕ(x) = (σ(t)/p(t)) 1/ dt

6 6 SOLUTION FOR FINAL B 1 = tells us that A A = 1 (pϕ ) pϕ A = c(pϕ ) 1/ = c(σp) 1/4 c) Fix the choice of ϕ, we want to have iλ 1/ B 1 (A) + B (A) =. Let A(x) = A (x) + λ 1/ A 1 (x) + O(λ 1 ), we have that is iλ 1/ B 1 (A ) + B (A ) + λ 1/ (iλ 1/ B 1 (A 1 ) + B (A 1 )) + O(λ 1 λ 1/ ) =, Let λ, we see that we must have B (A ) + ib 1 (A 1 ) = O(λ 1/ ) B (A ) + ib 1 (A 1 ) = i[a 1pϕ + 1 A 1(pϕ ) ] + pa + p A + qa = Integrating factor (pϕ ) 1/, we see Recall pϕ = (σp) 1/, [(pϕ ) 1/ A 1 ] = i (pϕ ) 1/ [qa + (pa ) ] A 1 = i (σp) 1/4 x (σp) 1/4 [qa + (pa ) ]dt d) RQ[u] = puu π + π [pu qu π ]dx π = [u u ]dx π u σdx u dx Based on the BCs u() = u x (π) =, we choose a trial function (without interior zeros) { x < x < π/ u(x) = π/ π/ < x < π then π/ dx π λ 1 RQ[u] = u dx π/ π = u dx π/ x dx + π/(π/) 1 = 3 π 1 e) In this case, the SL eigenvalue problem becomes d u dx + u + λu = which is just the standard problem with (p, q, σ) = (1,, 1) with eigenvalue λ + 1 d u + (λ + 1)u = dx For this problem, we know that λ + 1 >. u() = tells us that u(x) = sin(kx) u (π) = tells us that the first eigenfunction corresponding to k = 1. That is u 1 (x) = sin x/

7 SOLUTION FOR FINAL 7 and λ = (1/) λ 1 = 3/4