Suggested Solution to Assignment 5

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1 MATH 4 (5-6) prti diferenti equtions Suggested Soution to Assignment 5 Exercise 5.. () (b) A m = A m = = ( )m+ mπ x sin mπx dx = x mπ cos mπx + + 4( )m 4 m π. 4x cos mπx dx mπ x cos mπxdx = x mπ sin mπx 4x 4 sin mπxdx = ( )m mπ m π. 4. To find the Fourier series of the function f(x) = sin x, we first note tht this is n even function so tht it hs cos-series. If we integrte from to π nd mutipy the resut by, we cn tke the function sin x insted of sin x so tht Hence, we hve n = π = π sin x cos nxdx = f(x) = π 4 cos x ( π Substituting x = nd x = π, we hve 5. () From Pge.9, we hve Integrtion of both sides gives x x = sin xdx = 4 π. { 4 ( n )π n even n odd. cos 4x cos 6x ). 4n =. ( ) n 4n = π 4. m= m+ ( ) mπ mπx sin. = c + ( ) m mπx m cos. π m= The constnt of the integrtion is the missing coefficient (b) By setting x = gives or c = A = x dx = 6. = 6 + ( ) m m π, π = m= m= ( ) m+ m.

2 MATH 4 (5-6) prti diferenti equtions 8. The key point in the probem bove is to sove the foowing PDE probem. u t u xx =, u(x, ) = φ(x), u(, t) = u(, t) =, { φ(x) =, < x <, x, < x <. Through stndrd procedure of seprtion vribe method, we obtin where n = 9 φ(x) sin nπxdx = u(x, t) = n e n π t sin nπx, sin πn n π, so the soution T = u(x, t) + x. 9. From Section 4..7, we see tht the gener formu to wve eqution with Neu- mnn boundry condition is u(x, t) = (A + B t) + (A n cos nct + B n sin nct) cos nx, where φ(x) = A + A n cos nx, ψ(x) = B + ncb n cos nx. By further ccution, we hve B =, B = 4c nd the other coefficients re zero. Hence, the soution is u(x, t) = sin ct cos x t +. 4c Exercise 5.. Suppose α = p/q, where p, q re co-prime to ech other. Then is is not difficut to see tht S = qπ is period of the function. Suppose qπ = mt, where T is the minim period. Then cos x + cos αx = cos(x + T ) + cos(αx + αt ). Let x =, we hve the bove equity hods iff q/m, p/m re both integers. Therefore, m =. Hence, we finish the probem. 5. Let m = mπx φ(x) sin. Then we hve φ(x) = m= m sin mπx. 8. () If f is even, f( x) = f(x). Differentiting both sides gives f ( x) = f (x), so f ( x) = f (x), showing f is odd. If f is odd, f( x) = f(x). Differentiting both sides gives f ( x) = f (x), so f ( x) = f (x), showing f is even. (b) If f is even, consider f( x)dx = f(x)dx. Vi substitution, u = x, we hve f(u)du = f(x)dx. So if ignoring te constnt of integrtion, F ( x) = F (x), showing F is odd, where F is n ntiderivtive of f.simiry, for f odd, we hve f( x)dx = f(x)dx, so F ( x) = F (x), showing F is even.. () If φ is continuos on (, ), φ odd is continuous on (, ) if nd ony if im φ(x) =. x + (b) If φ(x) is differentibe on (, ), φ odd is differentibe on (, ) if nd ony if im x φ (x) exists, since + is n even function, so the ony thing to void is n infinite discontinuity t x =. φ odd

3 MATH 4 (5-6) prti diferenti equtions (c) If φ is continuos on (, ), φ even is continuous on (, ) if nd ony if im φ(x) exists, since the ony x + thing to void is n infinite discontinuity t x =. (d) If φ(x) is differentibe on (, ), φ even is differentibe on (, ) if nd ony if im x φ (x) =, since + φ even is n odd function. Extr. u(, t) = u(, t) = tes us we cn do odd extension nd periodic extension with period. Thus define { sin (πx), x [n, n + ] φ(x) = sin (πx), x [n, n] { x( x), x [n, n + ] ψ(x) = x( + x), x [n, n] n =, ±, ±,... By d Aembert s formu,u(x, t) = [φ(x + t) + φ(x t)] + x+t 4 x t ψ(s)ds soves the probem. Exercise 5.. Since X() =, by the odd extension x( x) = X(x) for < x <, then X stisfies X + λx =, X ( ) = X () =. Hence, λ = [(n + )π] /, X n (x) = sin[(n + )πx/], n =,,,... Thus we botin the gener formu to this eqution u(x, t) = [A n cos (n + )πct n= + B n sin (n + )πct ] sin (n + )πx. By the boundry condition, we obtined tht B n re zero, whie A n ( ) n (n+ ) π. = sin (n+ )πx x dx = 5(). Let u(x, t) = X(x)T (t), then X (x) = λx(x), X() =, X () =. By Theorem, there is no negtive eigenvue. It is esy to check tht is not n eigenvue. Hence, there re ony positive eigenvues. Let λ = β, β >, then we hve X(x) = A cos βx + B sin βx. Hence the bounndry condtions impy A =, Bβ cos β =. So the eigenfunctions re β = (n + )π, n =,,,... X n (x) = sin (n + )πx, n =,,,...

4 MATH 4 (5-6) prti diferenti equtions 6. Let X (x) = λx(x), λ C, then X(x) = e λx. By the boundry condition X() = X(), we hve e λ =. Hence, λ n = nπi, X n (x) = e nπxi, n Z. Since, if m n, X n (x)x m (x)dx = e (n m)πxi dx =. Therefore, the eigenfunctions re orthogon on the interv (, ). 8. If nd then ( X X + X X ) b X () X () = X () X () =, X (b) + b X (b) = X (b) + b X (b) =, = X (b)x (b) + X (b)x (b) + X ()X () X ()X () = b X (b)x (b) X (b) b X (b) + X ()X () X () X () =. 9. For j =,, suppose tht X j (b) = αx j () + βx j() X j(b) = γx j () + δx j(). Then, (X X X X ) b = X (b)x (b) X (b)x (b) X ()X () + X ()X () = [γx () + δx ()][αx () + βx ()] [αx () + βx ()][γx () + δx ()] X ()X () + X ()X () = (αδ βγ )X ()X () + ( + βγ αδ)x ()X () = (αδ βγ )(X X ) x=. Therefore, the boundry conditions re symetric if nd ony if αδ βγ =.. By the divergence theorem, f g b = (f (x)g(x)) dx = f (x)g(x)dx = f (x)g(x) + f (x)g (x)dx, f (x)g (x)dx + f g b.. Substitute f(x) = X(x) = g(x) in the Green s first identity, we hve Since X = λx, so Therefore, we get λ since X. X (x)x(x)dx = λ X (x)dx + (X X) b. X (x)dx. 4

5 MATH 4 (5-6) prti diferenti equtions Exercise 5.4. The prti sum is given by S n = ( )n x n + x. () Obviousy for ny x fixed, S n. Thus it converges to pointwise. +x +x (b) Let x n = n, then xn e. Thus it does not converge uniformy. (c) It wi converge to S(x) = +x in the L sence since S n S dx = x 4n ( + x ) dx x 4n dx s n. 4n +. This is n esy consequence combined Theorem nd Theorem on Pge 4 nd Theorem 4 on Pge 5.. () For ny fixed point x, WLOG, we ssume x <. Then there is N such tht for n > N, x < n, which impies tht f n (x ). Thus f n (x) pointwisey. (b) Let x n = n, then f n(x n ) = γ n, which impies tht the convergence is not uniform. (c) By direct computtion, we hve f n(x)dx = n γndx + n + γndx = γ n n. For γ n = n, f n(x)dx = n s n. (d) By the computtion in (c), for γ n = n, fn(x)dx = n s n. 4. For odd n, For even n, Thus, for ny n, 4 + n dx = 4 n. n 4 + n dx = 4 n. n g n (x) L = s n. n 5

6 MATH 4 (5-6) prti diferenti equtions 5. () We see tht A = terms re 4, pi dx = 4 nd A m = πx cos nd πx cos, π 4π cos mπx dx = mx cos 4πx. mπ sin. So, the first four nonzero (b) We cn express φ(x) = A + (A n cos nπx + B n sin nπx ). by Theorem 4 of Sectiion 4, since φ(x) nd its derivtive is piecewise continuous, so we get the fourier series wi converge to f(x) except t x =, whie the vue of this series t x = is. (c) By corory 7, we see tht it converge to φ(x) in L sense. (d) Put x =, we see tht the sine series vnish, it turns out to be tht φ() = π thus we obtin the sum of thee series is π. 6. The series is cos x = n sin nx. If n >, n = π cos x sin nxdx = + )x [cos(n + π n + cos(n )x n ] π = n( + ( )n ) (n. )π ( ) [ m ] m<,m n m If n =, =. The sum series is if x = π,, π. By Theorem 4 in Section 4, the sum series converges to cos x pointwisey in < x < π, nd to cos x for π < x <. 7. () Obviousy φ(x) is odd. Thus, its fu Fourier series is just the Sine Fourier series, i.e. where B n stisfies B n = (b) By (), the first three nonzero terms re (c) Since B n sin nπx, φ(x) sin nπxdx = nπ. π sin πx, π sin πx, sin πx. π φ(x) dx = ( x) dx, it cconverges in the men squre sense ccording to Corory 7. (d) Since φ(x) is continuous on (, ) except t the point x =. Therefore, Theorem 4 in Section 4 impies it converges pointwisey on (, ) expect t x =. (e) Since the series does not converge pointwisey, it does not converge uniformy. Exercise 5.6. () (Use the method of shifting the dt.) Let v(x, t) := u(x, t), then v soves v t = v xx, v x (, t) = v(, t) =, nd v(x, ) = x. By the method of sepertion of vribes, we hve v(x, t) = A n e (n+ ) π t cos[(n + )πx], n= cos mπ 6

7 MATH 4 (5-6) prti diferenti equtions where Hence, where A n is s before. (b). A n = ( ) n+ 4(n + ) π. u(x, t) = + A n e (n+ ) π t cos[(n + )πx], n=. In the cse j(t) = nd h(t) = e t, by () nd the initi condition u n () =, Therefore, u(x, t) = u n (t) = 5. It is esy to check tht et sin 5x + 5c soves Using the method of shifting the dt, we hve where nπk (λ n k + ) (et e λnkt ). nπk (λ n k + ) (et e λnkt ) sin nπx. v t t = c v xx + e t sin 5x, nd v(, t) = v(π, t) =. u(x, t) = et sin 5x + 5c + (A n cos(nct) + B n sin(nct)) sin(nx), A n = π sin 5x sin nx dx = + 5c sin 5x] sin nx dx + 5c B n = [sin x ncπ c n = = 5c( + 5c n = 5. ) otherwise So the formu of the soution cn be simpfied s u(x, t) = c sin ct sin x + + 5c + 5c n = 5 ; 5 otherwise (e t cos 5ct 5c sin 5ct ) sin 5x. 8. (Expnsion Method) Let u(x, t) = u n (t) sin nπx, u t (x, t) = v n (t) sin nπx, u x (x, t) = w n (t) sin nπx. 7

8 MATH 4 (5-6) prti diferenti equtions Then v n (t) = w n (t) = = u t nπx sin dx = du n dt, u nπx sin x dx = du n dt, ( nπ ) u(x, t) sin nπx = λ n u n (t) nπ ( ) n At, dx + (u x sin nπx nπ u cos nπx where λ n = (nπ/). Here we used the Green s second identity nd the boundry conditions. Hence, by the PDE u t = ku xx nd the initi condition u(x, ) =, we get ) du n dt = k[ λ n u n (t) nπ ( ) n At], u n () =. Hence, Therefore, where λ n = (nπ/). u(x, t) = u n (t) = ( ) n+ nπ A[ t λ n λ nk + e λnkt λ nk ]. ( ) n+ nπ A[ t λ n λ nk + e λnkt nπx λ ] sin, nk 8