Solutions to the Sample Problems for the Final Exam UCLA Math 135, Winter 2015, Professor David Levermore

Transcription

1 Solutions to the Sample Problems for the Final Exam UCLA Math 135, Winter 15, Professor David Levermore Every sample problem for the Midterm exam and every problem on the Midterm exam should be considered a sample problem for the Final exam. Between 3% and 5% of the Final exam will be on that material. Solutions to these problems have already been posted. Some Final exam problems will be drawn from the homework assignments. Here are some sample problems related to material covered since the Midterm exam. 1) Give a sequence of functions {f n } over [, 1] that converges to the function pointwise over [, 1] but that does not converge to in mean square over [, 1]. Give reasoning why your example works. Solution. One example is the sequence {f n } defined by f n x) = n xe nx. This sequence converges to the function pointwise over [, 1] because f n ) = for every n while if x, 1] then e x < 1, so that lim f nx) = x lim ne nx =. This sequence does not converge to in mean square over [, 1] because lim f n x) dx = lim = lim n 4n x e nx dx y e y dy = y e y dy = 1. Remark. There are many such examples. The easiest way to construct one is to first think of a sequence of nonnegative functions {g n } over [, 1] that converges to the function pointwise over [, 1] but that the area under their graphs over [, 1] does not converge to for example, that satisfies g n x) dx = 1, or lim g n x) dx = 1. Here the 1 on the right-hand sides above can be replaced by any positive number. The sequence {f n } is constructed from {g n } by setting f n x) = g n x). Then the sequence {f n } converges to the function pointwise over [, 1] because, by the continuity of the square root, for every x [, 1] we have lim f nx) = lim gn x) = lim g n x) = =. However, the sequence {f n } does not converge to in mean square over [, 1] because lim f n x) dx = lim g n x) dx = 1. The example used in the solution given above was constructed in this way from the sequence g n x) = 4n xe nx. Another example can be constructed in this way from the sequence g n x) = n x n 1 1 x). 1

2 ) Give a sequence of functions {f n } over [, 1] that converges to the function in mean square over [, 1] but such that for every x [, 1] the sequence of real numbers {f n x)} does not converge to the number. Give reasoning why your example works. Solution. Let {I n } be any sequence of intervals contained within [, 1] such that: I n as n ; for every x [, 1] we have x I n frequently. Here I n denotes the length of the interval. The second property means that for every x [, 1] and every m N there exits an n > m such that x I n. One such sequence of intevals is defined as follows. For every positive integer n there exists a unique nonnegative integer k n such that kn n < kn+1. Set [ n k n I n =, n + 1 ] kn. kn kn Clearly I n = 1/ kn as n. Moreover, for every x [, 1] we have x I n frequently because for every m N there exists k N such that k > m and we have k+1 1 n= k I n = [, 1], whereby for every x [.1] there exists n k > m such that x I n. Given any sequence of intervals {I n } with these properties, a sequence of functions {f n } over [, 1] can be constructed by setting f n x) = 1 In x), where 1 I denotes the indicator function of the interval I namely, the function defined by { 1 if x I, 1 I x) = otherwise. The sequence {f n } converges to in mean square over [, 1] because which implies that f n x) dx = lim 1 In x) dx = f n x) dx = lim I n =. 1 In x) dx = I n, For every x [, 1] the sequence of real numbers {f n x)} does not converge to the number because for every x [, 1] we have f n x) = 1 In x) = 1 frequently. 3) The Fourier series for e x over x [, π] has the form e x = 1a + an cosnx) + b n sinnx) ) for x, π). a) Calculate the a n and b n. n b) Evaluate lim 1) m a m. m=1

3 Solution a). The formulas for the Fourier coefficients are a n = 1 π cosnx)e x dx, b n = 1 π sinnx)e x dx. Primitives for these integrands may be computed with two applications of integration by parts. Alternatively, we can compute these primitives as particular solutions of the first-order nonhomogeneous linear differential equations with constant coefficients given by Dy 1 = cosnx)e x, Dy = sinnx)e x. These equations have characteristic polynomial pz) = z, which has root z =. Their forcings have degree d =, characteristic µ + iν = 1 + in, and multiplicity m =. They can be solved by either Key Identity Evaluations or Undetermined Coefficients. Because m + d = we need only the Key identity, which is De zx ) = ze zx. When this is evaluated at z = 1 + in we get Because D e 1+in)x) = 1 + in)e 1+in)x. ) e 1+in)x D = e 1+in)x = e x cosnx) + ie x sinnx), 1 + in we see that primitives of the integrands are ) ) e 1+in)x 1 in)e y 1 = Re = e x inx Re 1 + in 1 + n = ex 1 + n Re 1 in)cosnx) + i sinnx)) ) = ex ) cosnx) + n sinnx), 1 + n ) ) e 1+in)x 1 in)e y = Im = e x inx Im 1 + in 1 + n = ex 1 + n Im 1 in)cosnx) + i sinnx)) ) = ex ) sinnx) n cosnx). 1 + n Therefore the Fourier coefficients are given by a n = 1 π = 1) n 1 π b n = 1 π = 1) n n π cosnx)e x dx = 1 π e π e 1 + n, sinnx)e x dx = 1 π e π e 1 + n. e x 1 + n cosnx) + n sinnx) ) π e x 1 + n sinnx) n cosnx) ) π 3

4 4 Solution b). The π-periodic extension of e x will have a jump discontinuity at x = π + πm for every m Z where its value jumps from e π to e. Moreover, this extension is piecewise continuously differentiable. By the Dirichlet Convergence Theorem the Fourier Series will converge to the average of these values at each jump. In particular, at x = π we have e π + e = 1 a + an cosnπ) + b n sinnπ) ) = 1a + 1) n a n. Because we see that lim m=1 a = 1 π n 1) m a m = e x dx = 1 π ex π = 1 π 1) n a n = eπ + e 4) The sine series for x over x [, π] has the form x = e π e ), b n sinnx) for x [, π). a) Calculate the b n. n b) Evaluate lim b m sinm). m=1 eπ e π Solution a). The function x over [, π) has the odd extension x x over, π). The π-periodic extension of this odd extension is piecewise continuously differentiable. Its Fourier coefficients are b n = 1 π x x sinnx) dx = π x sinnx) dx. A primitive for this integrand may be computed with two applications of integration by parts. Alternatively, we can compute this primitive as a particular solution of the first-order nonhomogeneous linear differential equation with constant coefficients given by Dy = x sinnx). This equation has characteristic polynomial pz) = z, which has root z =. Its forcing has degree d =, characteristic µ + iν = in, and multiplicity m =. It can be solved by either Key Identity Evaluations or Undetermined Coefficients. Because m = and m+d = we need the Key identity and its first two derivatives with respect to z, which are D e zx) = z e zx, D x e zx) = z x e zx + e zx, D x e zx) = z x e zx + x e zx..

5 5 Evaluating these at z = in gives D e inx) = in e inx, D x e inx) = in x e inx + e inx, D x e inx) = in x e inx + x e inx. Multiplying the first relation by 1/in) and subtracting from the second yields D x e inx 1 ) in einx = in x e inx. Multiplying this fourth relation by /in) and subtracting from the third yields D x e inx in x einx ) n einx = in x e inx. Because x sinnx) = Imx e inx ), we divide the above relation by in and take its imaginary part to see that a primitive of the integrand is 1 y = Im in x e inx + n x einx ) in 3 einx = 1 n x cosnx) + n x sinnx) + n 3 cosnx). Therefore the Fourier coefficients are given by b n = x sinnx) dx π = 1n π x cosnx) + n x sinnx) + n ) π cosnx) 3 = 1)n π + 1)n 1 ). π n n 3 n 3 Solution b). The function x over [, π) has the odd extension x x over, π). The π-periodic extension of this odd extension is piecewise continuously differentiable. The Dirichlet Theorem implies that its Fourier series converges at every point of continuity of this extension. Because x = 1 is such a point of continuity, by plugging x = 1 into the given relation we see that 1 = b n sinn). Therefore lim m=1 n b m sinm) = 1.

6 6 5) The cosine series for x over x [, π] is x = π 3 + 1) n 4 cosnx). n Use this fact to evaluate the following sums. Give your reasoning. 1 a) n 1) n cosn) b) n 1 c) n 4 Solution a). Because x is even over [, π], its π-periodic extension is continuous, with a piecewise continuous derivative. By the Dirichlet Theorem its Fourier series converges pointwise. Hence, by plugging x = π or x = ) into the given relation we see that π = π 3 + 1) n 4 cosnπ) = π n n. Therefore 1 n = 1 ) π π = π Solution b). Because x is even over [, π], its π-periodic extension is continuous, with a piecewise continuous derivative. By the Dirichlet Theorem its Fourier series converges pointwise. Hence, by plugging x = 1 or x = 1) into the given relation we see that 1 = π 3 + 1) n 4 cosn). n Therefore 1) n cosn) = 1 ) 1 π. n 4 3 Solution c). Because x is even over [, π], its π-periodic extension is continuous, with a piecewise continuous derivative. We can read off from the given relation that its Fourier coefficients are a = π 3, a n = 1) n 4 n, b n =, for n = 1,,. The Parseval equality states 1 π fx) dx = 1 a + ) a n + bn.

7 When fx) = x we have x ) dx = Hence, the Parseval equality becomes Therefore π 4 x 4 dx = 5 = 1 x 4 dx = x5 5 4π n. 4 π = π n = 1 ) π π4 = π ) Find the eigenvalues λ n and eigenfunctions y n x) that solve the eigenvalue problem y + λy = over [a, b] with ya) = yb) =. Solution. We break the analysis into three cases: λ >, λ =, and λ <. Let λ >. Set k = λ. Then the differential equation becomes y + k y =, which has the general solution yx) = c 1 coskx) + c sinkx). The boundary conditions then imply that = c 1 coska) + c sinka), = c 1 coskb) + c sinkb). This linear system has a nontrivial nonzero) solution if and only if ) coska) sinka) = det = sinkb) coska) coskb) sinka) = sin kb a) ). coskb) sinkb) This equation is satisfied if and only if kb a) = nπ for some positive integer n, by which we conclude that eigenvalues are given by ) nπ λ n = for n = 1,,. b a Nontrivial solutions of the linear system are c 1 = c sinka), c = c coska), for some c, whereby nontrivial solutions of the differential equation are yx) = c sinka) coskx) + c coska) sinkx) = c sin kx a) ). Therefore every eigenfunction assoicated with λ n is a multiple of ) nπx a) y n x) = sin. b a Let λ =. Then the differential equation becomes y =, which has the general solution yx) = c 1 + c x. The boundary conditions then imply that = c 1 + c a, = c 1 + c b.

8 8 This linear system has a nontrivial nonzero) solution if and only if ) 1 a = det = b a. 1 b Because b > a this equation is not satisfied, whereby λ = is not an eigenvalue. Let λ <. Set k = λ. Then the differential equation becomes y k y =, which has the general solution yx) = c 1 e kx + c e kx. The boundary conditions then imply that = c 1 e ka + c e ka, = c 1 e kb + c e kb. This linear system has a nontrivial nonzero) solution if and only if ) e ka e = det ka e kb e kb = e kb a) e kb a). Because k > and b > a, we see that e kb a) > 1 and e kb a) < 1, whereby e kb a) e kb a) >. Therefore this equation is not satisfied by any k >, which implies that λ < is never an eigenvalue. Therefore the eigenvalues λ n and eigenfunctions y n x) that solve the eigenvalue problem are ) nπ λ n =, y n x) = sin nπ x a ), for n = 1,,. b a b a Any nonzero multiple of y n x) is also an eigenfunction. 7) Find the eigenvalues λ n and eigenfunctions v n x) that solve the eigenvalue problem v + λv = over [, π] with v ) =, vπ) =. Solution. We break the analysis into three cases: λ >, λ =, and λ <. Let λ >. Set k = λ. Then the differential equation becomes v + k v =, which has the general solution which has derivative vx) = c 1 coskx) + c sinkx), v x) = c 1 k sinkx) + c k coskx). The boundary conditions then imply that = c 1 k sin) + c k cos) = c k, = c 1 coskπ) + c sinkπ). The first equation implies that c =, whereby the second implies that c 1 coskπ) =. This linear system has a nontrivial nonzero) solution if and only if coskπ) =. This condition is satisfied if and only if kπ = nπ + 1 π for some nonnegative integer n, by which we conclude that eigenvalues are given by λ n = n + 1 ) for n =, 1,. Moreover, every eigenfunction assoicated with λ n is a multiple of v n x) = cos n + 1 )x).

9 Let λ =. Then the differential equation becomes v =, which has the general solution vx) = c 1 + c x, whereby v x) = c. The boundary conditions then imply that = c, = c 1 + c π. The only solution of this system is c = c 1 =, which implies that λ = is not an eigenvalue. Let λ <. Set k = λ. Then the differential equation becomes v k v =, which has the general solution vx) = c 1 e kx + c e kx, whereby v x) = c 1 ke kx + c ke kx. The boundary conditions then imply that = c 1 ke + c ke, = c 1 e kπ + c e kπ. This linear system has a nontrivial nonzero) solution if and only if ) k k = det e kπ e kπ = ke kπ ke kπ. Because k >, we see that e kπ > 1 and e kπ < 1, whereby ke kπ e kπ ) <. Therefore this equation is not satisfied for any k >, which implies that λ < is never an eigenvalue. Therefore the eigenvalues λ n and eigenfunctions v n x) that solve the eigenvalue problem are λ n = n + 1 ), vn x) = cos n + 1 )x), for n =, 1,. Any nonzero multiple of v n x) is also an eigenfunction. 8) Find the solution wt, x) to the vibrating string problem w tt w xx = over x, π) with wt, ) = wt, π) =, w, x) = xπ x), w t, x) =. The solution can be expressed as an infinite series. Solution. Use separation of variables. Seek solutions of the partial differential equation in the form wt, x) = T t)sx) that also satisfies the homogeneous initial and boundary conditions. The homogeneous initial condition implies that T ) =. The homogeneous boundary conditions imply that S) = Sπ) =. The partial differential equation implies that T t)sx) T t)s x) =. By separating variables we see that there is a constant λ such that T t) T t) = S x) Sx) = λ. We see that Sx) satisfies the eigenvalue problem S x) + λsx) =, S) = Sπ) =, 9

10 1 which by problem 6 has the eigenvalues λ n and eigenfunctions S n x) given by λ n = n, S n x) = sinnx), for n = 1,,. For λ = n the associated T t) satisfies T t) + n T t) =, T ) =. A general solution of the differential equation is T t) = c 1 cosnt)+c sinnt). Because T t) = nc 1 sinnt) + nc cosnt), the initial condition T ) = implies that c =. Therefore we have found solutions of the partial differential equation in the form cosnt) sinnx) that also satisfy the homogeneous initial and boundary conditions. By taking a general superposition of these we obtain wt, x) = b n cosnt) sinnx). By setting t = in this relation we see that b n are the Fourier sine series coefficients of the initial data specifically, b n = π xπ x) sinnx) dx. A primitive for this integrand may be computed with two applications of integration by parts. Alternatively, we can compute this primitive as a particular solution of the first-order nonhomogeneous linear differential equation with constant coefficients given by Dy = πx x ) sinnx). This equation has characteristic polynomial pz) = z, which has root z =. Its forcing has degree d =, characteristic µ + iν = in, and multiplicity m =. It can be solved by either Key Identity Evaluations or Undetermined Coefficients. Because m = and m+d = we need the Key identity and its first two derivatives with respect to z, which are D e zx) = z e zx, D x e zx) = z x e zx + e zx, D x e zx) = z x e zx + x e zx. Evaluating these at z = in gives D e inx) = in e inx, D x e inx) = in x e inx + e inx, D x e inx) = in x e inx + x e inx. Multiplying the first relation by 1/in) and subtracting from the second yields D x e inx 1 ) in einx = in x e inx. Multiplying this fourth relation by /in) and subtracting from the third yields D x e inx in x einx ) n einx = in x e inx. Because πx x ) sinnx) = x πx) Reie inx ), we find a primitive from the last two relations. The details are omitted.

11 9) Find the solution ht, x) to the cooling rod problem h t = h xx h over x, π) with h x t, ) = h x t, π) =, h, x) = sinx). The solution can be expressed as an infinite series. Solution. Use separation of variables. Seek solutions of the partial differential equation in the form ht, x) = T t)sx) that also satisfies the homogeneous boundary conditions. The homogeneous boundary conditions imply that S ) = S π) =. The partial differential equation being implies that T t)sx) = T t)s x) T t)sx). By separating variables we see that there is a constant λ such that T t) T t) = S x) Sx) Sx) We see that Sx) satisfies the eigenvalue problem = λ 1. S x) + λsx) =, S ) = S π) =, It is clear that λ = is an eigenvalue with eigenfunction Sx) = 1. If Sx) is an eigenfunction that is not constant then its eigenvalue λ will be nonzero. In that case we set Y x) = S x) and we see that Y x) satisfies the eigenvalue problem Y x) + λy x) =, Y ) = Y π) =, which by problem 6 has the eigenvalues λ n and eigenfunctions Y n x) given by λ n = n, Y n x) = sinnx), for n = 1,,. Therefore the eigenvalues λ n and eigenfunctions S n x) are given by λ n = n, S n x) = cosnx), for n =, 1,. For λ = n the associated T t) satisfies T t) + n + 1)T t) =. A general solution of this equation is T t) = ce n +1)t. Therefore we have found solutions of the partial differential equation in the form e n +1)t cosnx) that also satisfy the homogeneous boundary conditions. By taking a general superposition of these we obtain ht, x) = 1a e t + a n e n +1)t cosnx). By setting t = in this relation we see that a n are the Fourier cosine series coefficients of the initial data. Because the initial condition is h, x) = sinx), we have a n = π Upon using the trigonometric identity cosnx) sinx) dx for n =, 1,. cosnx) sinx) = 1 sinn + 1)x) sinn 1)x) ), 11

12 1 we see that a 1 = and that for n 1 a n = 1 ) sinn + 1)x) sinn 1)x) dx π = 1 ) cosn 1)x) cosn + 1)x) π π n 1 n + 1 = 1 + 1) n. π n 1 Notice a n = for odd n. Therefore the solution is ht, x) = π e t )t π n 1 e n cosnx). n= even Remark. The definite integral giving a n can also be evaluated with two integrations by parts. Using the trigonometric identity is certainly faster! 1) Find the solution ux, y) to the boundary-value problem u xx + u yy = over, 4π), π) with u, y) = cosy), u4π, y) = ux, ) = ux, π) =. The solution can be expressed as an infinite series. Solution. Use separation of variables. Seek solutions of the partial differential equation in the form ux, y) = Xx)Y y) that also satisfies the homogeneous boundary conditions. The homogeneous boundary contitions imply that X4π) = and Y ) = Y π) =. The partial differential equation implies that X x)y y) + Xx)Y y) =. By separating variables we see that there is a constant λ such that X x) Xx) = Y y) Y y) = λ. We thereby see that Y y) satisfies the eigenvalue problem Y y) + λy y) =, Y ) = Y π) =, which by problem 6 has the eigenvalues λ n and eigenfunctions Y n y) given by λ n = n 4, Y ny) = sin 1 ny + π)), for n = 1,,. For λ = 1 4 n the associated Xx) satisfies X x) 1 4 n Xx) =, X4π) =. A general solution of this differential equation is Xx) = c 1 e 1 nx + c e 1 nx. The boundary condition at x = 4π implies that = c 1 e πn + c e πn. Setting c 1 = 1 eπn and c = 1 e πn, we obtain X n x) = sinh 1 n4π x)).

13 Therefore we have found solutions of the partial differential equation in the form X n x)y n y) that also satisfy the homogeneous boundary conditions. By taking a general superposition of these we obtain ux, y) = B n sinh 1 n4π x)) sin 1 ny + π)). By setting x = in this relation we see that B n are related to the Fourier sine series coefficients b n of the boundary data by b n = B n sinhπn), b n = 1 π cosy) sin 1 ny + π)) dy. By the trigonometric identity cosy) = cosy)), we have b n = 1 π π 1 + cosy)) sin 1 ny) dy. By using the trigonometric identity cosy) sin 1 ny) = 1 sin 1 n + 4)y) + sin 1 n 4)y)), we see that b n = for even n and that for odd n we have b n = 1 4π = 1 π = π π sin 1 ny) + sin 1 n + 4)y) + sin 1 n 4)y) dy cos 1 ny) n ) 1 1) n = 4 π n 8 n 3 16n + cos 1n + 4)y) n + 4 Finally, collecting the above results we have ux, y) = 4 n 8 π n 3 16n odd n 8 n 3 16n. sinh 1n4π x)) sinhπn) + cos 1 ) n 4)y) π n 4 sin 1 ny + π)). 11) Find the solution ur, θ) to the Dirichlet problem over the disk r R given by u rr + 1 r u r + 1 r u θθ =, ur, θ) = π θ for θ, π). The solution can be expressed as an infinite series. Solution. Use separation of variables. Seek solutions of the partial differential equation in the form ur, θ) = vr)wθ) where vr) is bounded as r and wθ) is π-periodic. The partial differential equation implies that v r)wθ) + 1 r v r)wθ) + 1 r vr)w θ) =. By separating variables we see that there is a constant λ such that r v r) + r v r) vr) = w θ) wθ) = λ. 13

14 14 We see the wθ) satisfies the eigenvalue problem w θ) + λwθ) =, wθ + π) = wθ). We consider the cases λ >, λ =, and λ <. Let λ >. Set k = λ >. Then a general solution of the differential equation is wθ) = c 1 coskθ) + c sinkθ). This solution is periodic with period π/k. This is π-periodic if and only if there must exist a positive integer n such that nπ/k = π. Therefore an eigenvalue is λ = n for some positive integer n with eigenfunctions wθ) = c 1 cosnθ) + c sinnθ). Let λ =. Then a general solution of the differential equation is wθ) = c 1 + c θ. This is π-periodic if and only if c =. Therefore an eigenvalue is λ = with eigenfunctions wθ) = c. Let λ <. Set k = λ >. Then a general solution of the differential equation is wθ) = c 1 e kθ + c e kθ. This is not periodic. Therefore there are no negative eigenvalues. Because λ = n for some nonnegative integer n, we see that vr) must be a bounded solution of the differential equation r v r) + rv r) n vr) =, over [, R]. We consider the cases n = and n > separately. If n = then the differential equation for vr) becomes rv r) + v r) =. If we think of this as a first-order homogeneous linear equation for v r) then we find that v r) = c 1 /r, whereby vr) = c 1 logr) + c. The requirement that this solution be bounded as r implies that c 1 =. If n > then we seek solutions of the differential equation in the form vr) = r k, whereby = r v r) + r v r) n vr) = kk 1)r k + kr k n r k = k n )r k. By taking k = ±n we obtain the solutions vr) = r n and vr) = r n. A general solution has the form vr) = c 1 r n + c r n. The requirement that this solution be bounded as r implies that c =. Therefore we have found solutions of the partial differential equation in the forms r n cosnθ) and r n sinnθ). By taking a general superposition of these we obtain ur, θ) = A + An r n cosnθ) + B n r n sinnθ) ). At r = R we have ur, θ) = A + An R n cosnθ) + B n R n sinnθ) ). Therefore, by the boundary condition at r = R we have A = 1 a, A n R n = a n, B n R n = b n, for n = 1,,,

15 we see that px)µx)y px)µx) ) y ) + qx)µx)y ) = µx)fx). where a n = 1 π π θ ) cosnθ) dθ, b n = 1 π π θ ) sinnθ) dθ. Because π θ ) is an even function of θ over, π), we see that Moreover, we have a = π a n = π a n = π π θ ) cosnθ) dθ, b n =. π θ ) dθ = π π θ 1 3 θ3) π = 4 3 π. θ cosnθ) dθ = 1) n 4 n for n = 1,,. Here the value of a n for n = 1,,, can be read off from the information given in problem 5, which saves us from computing it. Finally, upon collecting the above results we obtain ur, θ) = π 3 1) n 4 n r n R n cosnθ). 1) Let px) > be twice continuously differentiable over [, 1]. Let qx) be continuously differentiable over [, 1]. Let rx) and fx) be continuous over over [, 1]. Show that the problem of solving the second-order nonhomogeneous linear differential equation px)y + qx)y + rx)y = fx), reduces to the problem of solving two first-order nonhomogeneous linear differential equations if we know one positive solution µx) of the second-order homogeneous linear differential equation px)µ ) qx)µ ) + rx)µ =. Solution. Multiplying the first differential equation by µx) and the second by y we obtain px)µx)y + qx)µx)y + rx)µx)y = µx)fx), px)µx) ) y qx)µx) ) y + rx)µx)y =. Upon subtracting the bottom equation from the top one we find that Because px)µx)y px)µx) ) y + qx)µx)y + qx)µx) ) y = µx)fx). px)µx)y px)µx) ) y = px)µx)y px)µx) ) y ), qx)µx)y + qx)µx) ) y = qx)µx)y ), 15

16 16 Hence, y solves the first-order nonhomogeneous linear differential equation px)µx)y px)µx) ) y + qx)µx)y = µx)fx) dx. Because we have assumed that px) and µx) are positive and continuous over [, 1], this equation can be put into normal form ) px)µx) y y + qx) px)µx) px) y = 1 µx)fx) dx. px)µx) Therefore finding an explicit solution reduces to finding three primitives: one to calculate the right-hand side above, and two to solve this nonhomogeneous linear differential equation. 13) Let qx) and rx) be continuous over over [a, b]. Let y = ux) and y = vx) be solutions of the boundary-value problem y + qx)y + rx)y =, ya) + y a) =, yb) y b) =. Show that ux) and vx) must be linearly dependent. Solution. Because ux) and vx) both satisfy the boundary condition at x = a, we see that ua) + u a) =, va) + v a) =. These equations can be put into the matrix form ) ua) u 1 va) v a)) =, 1 ) from which it follows that ) ) ua) u = det a) ua) va) va) v = det a) u a) v = W [u, v]a), a) where W [u, v]x) denotes the Wronskian of u and v. Because W [u, v]a) = and because u and v satisfy the same second-order homogeneous linear differential equation over [a, b], the Abel Theorem implies that W [u.v]x) = over [a, b]. Moreover, because W [u, v]x) = over [a, b] and because u and v satisfy the same second-order homogeneous linear differential equation over [a, b], the functions ux) and vx) must be linearly dependent. Remark. The argument can also be begun by using the boundary condition at x = b to infer that W [u, v]b) =. 14) Let R > 1. Consider the Sturm-Liouville problem r [r r u] + λr u = over [1, R] with u1) = ur) =. a) Find the eigenvalues λ n and eigenfunctions u n r) for this problem. Hint: Let vr) = rur) and consider the transformed problem for vr).) b) Derive the orthogonality relation satisfies by u m r) and u n r) when m n.

17 Solution a). Let ur) = vr)/r. Then whereby r u = rv r v r, = r [r r u] + λr u = r [r r v v] + λrv = r rr v + r v r v + λrv = r rr v + λv ). Therefore v satisfies the Sturm-Liouville problem rr v + λv = over [1, R] with v1) = vr) =. The Sturm-Liouville problem for v is the eigenvalue problem for y of Problem 6 with a = 1 and b = R. The solution will not be repeated here. The result was that the eigenvalues λ n and eigenfunctions v n r) are λ n = nπ R 1 ), v n r) = sin nπ r 1 R 1 ), for n = 1,,. Therefore the eigenvalues λ n and eigenfunctions u n r) for the given Sturm-Liouville problem for u are ) nπ λ n =, u n r) = 1 R 1 r sin nπ r 1 ), for n = 1,,. R 1 Any nonzero multiple of u n r) is also an eigenfunction. Solution b). Let u m r) and u n r) be eigenfunctions with m n. Because an r appears in the λ term of the Sturm-Loiuville problem, these eigenfunctions must satisfy the orthogonality relation R 1 u m r) u n r) r dr =. Remark. For Sturm-Loiuville problems in the general form x [px) x u] + λqx)u + rx)u = over [a, b] with ua) = ub) =, any eigenfunctions u m x) and u n x) that correspond to different eigenvalues satisfy the orthogonality relation b a u m x) u n x) qx) dx =. This is the only fact you need to know to answer part b. 15) Find a first-order ordinary differential equation satisfied by an extreme point of the functional F [u] = 1 u x 11 4 u ) dx that also satisfies the boundary conditions u) = and u1) = 1. Solution. Because fu, u x ) does not depend upon x, the Euler equation can be integrated once to obtain the first-order equation u x ux fu, u x ) fu, u x ) = c. 17

18 18 For fu, u x ) = 1u x 11 4 u ) this becomes u x u x 1u x u ) = c, which reduces to ux = c 11 u ). 16) Let a, b, c) R 3 with a + b + c. Maximize fx, y, z) = ax + by + cz subject to the constraint x + y + z = 1. Solution. By the method of Lagrange multipliers, we seek critical points of This leads to the equations Hx, y, z, λ) = ax + by + cz λ 1 x + y + z 1). = x Hx, y, z, λ) = a λx, = y Hx, y, z, λ) = b λy, = z Hx, y, z, λ) = c λz, = λ Hx, y, z, λ) = 1 x + y + z 1). The first three equations imply a + b + c = λ x + y + z ). Because we have assumed a + b + c, we see that λ and that x + y + z. Hence, the first three equations yield x = a λ, y = b λ, z = c λ. When this result is plugged into the last equation, we obtain a + b + c 1 =, λ which yields λ = ± a + b + c. Therefore the critical points of Hx, y, z, λ) are x, y, z, λ) where a x = ± a + b + c, y = ± b a + b + c, z = ± c a + b + c, λ = ± a + b + c. Therefore the critical points of fx, y, z) = ax + by + cz subject to the constraint x + y + z = 1 are The corresponding critical values are a, b, c) x, y, z) = ± a + b + c. fx, y, z) = ± a + b + c. Therefore the maximum of fx, y, z) = ax + by + cz subject to the constraint x + y + z = 1 is a + b + c. Remark. The maximizer is the critical point x, y, z) = a, b, c) a + b + c.

19 The minimum of fx, y, z) = ax + by + cz subject to the constraint x + y + z = 1 is a + b + c and the minimizer is the critical point a, b, c) x, y, z) = a + b + c ) Let ux) be continuously differentiable over [, π] with u) = uπ) =. Minimize subject to the constraint E[u] = u x) dx ux) dx = π. Solution. By the method of Lagrange multipliers, we seek critical points of ) 1 F λ [u] = u x) dx λ 1 ux) dx π = The Euler equations for this problem are d dx u + λu =, 1 u x) λ 1 ux) 1 ) dx. ux) dx = π. Recalling the boundary conditions u) = uπ) =, this shows that every critical point must be a solution of the eigenvalue problem u + λu =, u) = uπ) =. This is problem 6 with a = and b = π. Its eigenvalues λ n and eigenfunctions u n x) are given by λ n = n, u n x) = sinnx), for n = 1,,, where the is chosen so that u n x) dx = sinnx) dx = 1 cosnx) dx = π. These eigenfunctions and their negatives are thereby the critical points of E[u] subject to the constraint. A direct calculation shows that E[±u n ] = u nx) dx = n cosnx) dx = n 1 + cosnx) dx = n π. Therefore π is the minimum of E[u] subject to the constraint, which is attained at the minimizers ± sinx). Remark. Strictly speaking, there are gaps in the above argument. However, it is at the level of rigor we have been using in the course. The answer we arrived at is correct, namely, that π is the minimum of E[u] subject to the constraint. Moreover, it is also true that each eigenfunction ±u n is a critical point of E[u] with critical value n π. This shows a beautiful relationship between eigenvalue problems and variational problems that generalizes in many directions.

20 18) Let xt), yt)) be a continuously differentiable curve in R parametrized by t [, 1] with x), y)) = 1, ) and x1), y1)) = 1, ). Minimize F [y] = yt)x t) dt subject to the constraint x t) + y t) dt = π 3. Solution. By the method of Lagrange multipliers, we seek critical points of H λ [x, y] = yt)x t) dt λ x t) + y t) dt π ). 3 The Euler equations associated with xt) and yt) are ) = d x t) yt) λ, dt x t) + y t) ) = x t) λ d x t). dt x t) + y t) These may each be integrated in t to obtain yt) c 1 λ = x t) x t) + y t), xt) c λ = x t) x t) + y t). Therefore xt) c ) + yt) c 1) = 1. λ λ The boundary conditions x), y)) = 1, ) and x1), y1)) = 1, ) imply that 1 + c ) + c 1) 1 c ) = 1, + c 1) = 1, λ λ λ λ which implies that c = and λ = c 1 ) + 1. Therefore xt), yt) lies on a circle x + y c) = 1 + c for some c R. This is a circle with center at, c) and radius 1 + c. The solution is an arc on this circle of length π/3 that connects the points 1, ) and 1, ). Because the distance between these points is and the length of the arc is π/3, which is just a bit larger than, the center of the circle must lie above the x-axis i.e. we have c >. The half-angle of the arc will have sine 1/ 1 + c, so c is determined by setting ) 1 + c sin 1 1 = π 1 + c 3. This has solution c = 3, which implies that the circle has radius and the angle of the arc is π/3. A picture then shows that the minimum of F [x, y] is 3 π. 3 Remark. There will be nothing this hard on the final exam! Good Luck!