FINAL EXAM, MATH 353 SUMMER I 2015

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1 FINAL EXAM, MATH 353 SUMMER I 25 9:am-2:pm, Thursday, June 25 I have neither given nor received any unauthorized help on this exam and I have conducted myself within the guidelines of the Duke Community Standard. Name: Signature: Instructions: This exam contains 2 pages and problems with a table of Laplace transform at the end. You have 8 minutes to answer all the questions. You may use a calculator or a review sheet, front and back, written in your own handwriting. Throughout the exam, show your work with clear reasoning and calculation. If you are using a theorem to draw some conclusions, quote the result. If you do not completely solve a problem, explain what you understand about it. No collaboration on this exam is allowed. Good luck! Problems Points Grade Total 2

2 . (2 points) Find the inverse Laplace transform f(t) = L {F (s)} of the following functions: s 2 () F (s) = s 2 + 2s + Solution. s 2 s 2 + 2s + = s 2 (s + ) = s + (s + ) (s + ) Thus, by the first shift theorem, f(t) = L {F (s)} = e t (cos 3t sin 3t). (2) F (s) = e s s 3 Solution. Note that L{t 2 } = 2! s 3, thus By the second shift theorem, we have L {s 3 } = 2 t2. f(t) = 2 u (t)(t ) 2. (3) F (s) = (s 2 4)(s 2 9) Solution. Note that (by partial fractions) (s 2 4)(s 2 9) = ( 5 s 2 9 ). s 2 4 Therefore, f(t) = sinh 2t + sinh 3t. 5 2

3 2. (2 points) Find the solution of the initial value problem y + 4y = 8 sinh(2t) + δ(t 2π) y() = y () =, where δ(t) is the Dirac delta function and sinh(t) is the hyperbolic sine function. Solution. Take the Laplace transform on both sides of the equation and note that the initial values are zero. We obtain (s 2 + 4)Y (s) = 6 s e 2πs. Thus, 6 Y (s) = (s 2 + 4)(s 2 4) + e 2πs s = 2 s s e 2πs s Now, taking the inverse Laplace transform, we have y(t) = sinh 2t sin 2t + 2 u 2π(t) sin 2(t 2π) = sinh 2t sin 2t + 2 u 2π(t) sin 2t. 3

4 3. (2 points) Consider the first order ordinary differential equation e x (x + ) + (ye y xe x ) dy dx =. () Find an integrating factor g(y) which makes this equation exact. Solution. The integrating factor g(y) should satisfy which is just Hence, and we can choose [g(y)e x (x + )] y = [g(y)(yey xe x )], x g (y)(e x (x + )) = g(y)( e x xe x ). g g =, g(y) = e y. (2) Find the (implicit) solution of the differential equation above with y() =. Solution. Multiplying the equation by e y, we have Thus, Furthermore, we need It follows that e x y (x + ) + (y xe x y ) dy dx =. F (x, y) = and we can choose h(y) = 2 y2. Therefore, the general solution is e x y (x + )dx + h(y) = e y xe x + h(y). y xe x y = F y (x, y) = e y xe x + h (y). h (y) = y, xe x y + 2 y2 = C, for some constant C. Using the given initial value, we have C = 2. We conclude that the solution of the initial value problem is xe x y + 2 y2 = 2. 4

5 4. (2 points) Find the general solution of the equation (x + )y xy y =, x > given that y(x) = e x is a solution. Your final answer could involve an indefinite integral. Solution. Use the method of Reduction of order: consider a solution of the form v(x)y(x) = v(x)e x and find v(x). We have which is equivalent to Letting z = v, we have Hence, We can choose and (x + )(v e x + 2v e x + ve x ) x(v e x + ve x ) ve x =, (x + )v + (x + 2)v =. z z = x + 2 x + = x +. ln z = x ln(x + ). z(x) = v(x) = e x x +, e x x + dx. Therefore, the general solution of the equation is y(x) = c e x + c 2 v(x)e x = e x ( c + c 2 e x ) x + dx. 5

6 5. (2 points) Use the series method to solve the initial value problem y + 3xy + e x y = 2x, y() =, y () =. You need only to calculate terms up to the fourth power of x. Note: The Taylor expansion of e x at x = is Solution. Let We have e x = + x + 2! x2 + 3! x3 + 4! x y(x) = a + a x + a 2 x 2 + a 3 x 3 + a 4 x y (x) = a + 2a 2 x + 3a 3 x 2 + 4a 4 x y (x) = 2a 2 + 6a 3 x + 2a 4 x Therefore, expanding all the functions in the equation in their Taylor series about x =, we obtain (2a 2 + 6a 3 x + 2a 4 x ) + 3x(a + 2a 2 x + 3a 3 x 2 + 4a 4 x )+ ( + x + 2! x2 + 3! x3 + 4! x4 +...)(a + a x + a 2 x 2 + a 3 x 3 + a 4 x ) = 2x. If we compare the coefficients of the terms, x, x 2 on both sides of the equation, we would obtain Note that the initial values are just Therefore, from the above equalities, 2a 2 + a =, 6a 3 + 3a + a + a = 2, 2a 4 + 6a 2 + a 2 + a + 2 a =. a =, a =. a 2 = 2, a 3 = 2 4a a 6 = 5 6, a 4 = a 2 a 7a 2 = 2 3. Therefore, the series solution is y(x) = x 2 x x3 + 3 x

7 6. (2 points) Show by calculation that the Fourier series for the function is f(x) = e x, π x < π, f(x) = f(x + 2π) g(x) = sinh π ( + 2 π n= ( ) n ) + n (cos nx n sin nx). 2 What is the value of g(π)? Now find the limit of the infinite series + n. 2 n= Solution. Note that the period is 2π. If we let g(x) = a 2 + n= (a n cos nx + b 2 sin nx), then a = π a n = π b n = π π π π π π π e x dx = π π ex = 2 sinh π, π π e x cos nxdx, e x sin nxdx, Using integration by parts, we could evaluate e x cos nxdx = e x cos nx+n e x sin nxdx, e x sin nxdx = e x sin nx n e x sin nxdx. If we let A = π π ex cos nxdx, B = π π ex sin nxdx, then the above equality tells us { A = (e π e π ) cos nπ + nb, Therefore, B = na. A = 2 2 cos nπ sinh π = + n2 + n 2 ( )n sinh π, 2 B = na = ( n) + n 2 ( )n sinh π. Plugging this in the expression of g(x) gives the Fourier expansion. By the Fourier convergence theorem and the periodicity of f(x), we know that g(π) = 2 (eπ+ + e π ) = 2 (e π + e π ) = cosh π. On the other hand, Hence, g(π) = sinh π ( + 2 π = sinh π ( + 2 π n= n= n= ( ) n ) + n 2 (cos nπ n sin nπ) ) + n 2. + n 2 = (π coth π ). 2 7

8 7. (2 points) Consider the initial-boundary value problem: u xx = u t sin x, < x < π, t > u(, t) = u(π, t) =, t > u(x, ) = 3 sin 2x + 6 sin 5x, x π () Find the steady state solution of this equation, i.e., a solution w(x, t) = w(x) (does not depend on t) which satisfies the equation and the boundary values. Solution. By the definition of steady state, we have that w(x) satisfies { w (x) = sin x, < x < π, w() = w(π) =. It is easy to see (or by integrating twice) that w(x) = sin x, < x < π. (2) Find the solution u(x, t) of the initial-boundary value problem. Hint: Which initial-boundary value problem does v(x, t) = u(x, t) w(x) satisfy? Solution. By the linearity of the equation, v(x, t) = u(x, t) w(x) satisfies the initialboundary value problem: v xx = v t, < x < π, t > v(, t) = v(π, t) =, t > v(x, ) = sin x + 3 sin 2x + 6 sin 5x, x π Note that this is just the heat equation with the temperature at both ends set to be zero. We know that the solution are in the form (α =, L = π) v(x, t) = b n sin nx e n2t. The initial value for v(x, t) is v(x, ) = b n sin nx = sin x + 3 sin 2x + 6 sin 5x. Therefore and We conclude that n= n= b =, b 2 = 3, b 5 = 6, b n = (n, 2, 5), v(x, t) = e t sin x + 3e 4t sin 2x + 6e 25t sin 5x. u(x, t) = v(x, t) + w(x) = ( e t ) sin x + 3e 4t sin 2x + 6e 25t sin 5x. 8

9 8. (2 points) Find the solution u(x, t) of the following wave equation problem: u xx = u tt, < x < π, t > u(, t) = u(π, t) =, t > u(x, ) = 2 sin 3x, u t (x, ) = sin 4x, x π Solution. We consider instead the following two initial-boundary value problems u xx = u tt, < x < π, t > (I) u(, t) = u(π, t) =, t > u(x, ) = 2 sin 3x, u t (x, ) =, x π and u xx = u tt, < x < π, t > (II) u(, t) = u(π, t) =, t > u(x, ) =, u t (x, ) = sin 4x, x π From our discussion in class, solutions of (I) and (II) respectively take the form (L = π, c = ) u I (x, t) = d n sin nx cos nt, and u II (x, t) = n= k n sin nx sin nt. Now the non homogeneous parts of the initial conditions tell us u I (x, ) = d n sin nx = 2 sin 3x, Thus, u II t (x, ) = n= n= nk n sin nx = sin 4x. n= and the rest of the coefficients are all zero. We conclude that, by linearity, d 3 = 2, k 4 = 4, u(x, t) = u I (x, t) + u II (x, t) = 2 sin 3x cos 3t + sin 4x sin 4t. 4 9

10 9. (2 points) Find the solution u(r, θ) of the following Laplace equation in the semicircular region r <, < θ < π: u rr + u r r + u r 2 θθ =, u(r, ) = u(r, π) =, r < u(, θ) = f(θ), θ π, r <, < θ < π assuming that u(r, θ) is single-valued and bounded in the given region. Note: You are allowed to quote intermediate results in our solution of a similar problem with the region being circular. Solution. Same as in the case when the domain is circular, the separation of variable will give us two ODEs: { r 2 R + rr λr =, Θ + λθ =. Now the boundary conditions u(r, ) = u(r, π) = give us R(r)Θ() = R(r)Θ(π) =, which is just Θ() = Θ(π) =. Therefore, Θ(θ) satisfies the familiar two-point boundary value problem { Θ + λθ =, Θ() = Θ(π) =. We know that the eigenvalues and eigenfunctions of this problem are ( nπ ) 2 λ n = = n 2, Θ n (θ) = sin nx. (n =, 2, 3,...) π Correspondingly, R n (r) satisfies the Euler equation and its general solution takes the form r 2 R + rr n 2 R =, R n (r) = c r n + c 2 r n. As in the circular case, we must have c 2 = in order for the solution to be bounded, particularly as r +. Therefore, u n (r, θ) = R n (r)θ n (θ) = r n sin nθ. And by the principle of superposition, u(r, θ) = k n u n (r, θ) = n= k n r n sin nθ. Now, take into consideration of the non-homogeneous boundary value, we have f(θ) = u(, θ) = k n sin nθ, θ π. n= Therefore, the (sine) Fourier coefficients k n are This completes the solution. k n = 2 π π n= f(θ) sin nθdθ.

11 . (2 points) Consider the boundary value problem { [( + x 2 )y ] + y = λ( + x 2 )y y() y () =, y () + 2y() =. Let L be defined by L[y] = [( + x 2 )y ] + y. Prove that for any u(x), v(x) satisfying the boundary conditions above, L[u], v H = u, L[v] H, where u, v H denotes the Hermitian L 2 -inner product of u, v on the interval [, ]. Hint: You may use integration by parts twice. Solution. By the definition of L and the Hermitian L 2 -inner product, we have L[u], v H = = [( + x 2 )u ] vdx vd[( + x 2 )u ] = v( + x 2 )u = v( + x 2 )u v ( + x 2 )u dx v ( + x 2 )du(x) = v( + x 2 )u v ( + x 2 )u + [( + x 2 )v ] udx = v( + x 2 )u v ( + x 2 )u + u, L[v] H. Thus, to complete the proof, we need only to show v( + x 2 )u v ( + x 2 )u =. In fact, the boundary values (which are not separated) tell us Therefore, u () = u(), v () = v(), 2u() = u (), 2v() = v (). v( + x 2 )u v ( + x 2 )u =2v() u () v() u () 2v () u() + v () u() = v () u() v() u () + u () v() + v () u() =. This completes the proof.

12 Table of Laplace Transforms f(t) = L {F (s)} F (s) = L{f(t)} e at s a, t n (n >, integer) sin at cos at sinh at cosh at e at sin bt e at cos bt u c (t) s, s > s > a n! s, n+ s > a s 2 + a, 2 s > s s 2 + a, 2 s > a s 2 a, 2 s > s s 2 a, 2 s > b (s a) 2 + b, 2 s > a s a (s a) 2 + b 2, s > a e cs s, s > e ct f(t) F (s c) u c (t)f(t c) f(ct) e cs F (s) ( s ) c F, c > c t f(t τ)g(τ)dτ F (s)g(s) δ(t c) e cs f (n) (t) s n F (s) s n f()... f (n ) () t n f(t) ( ) n F (n) (s) 2