Last name: name: 1. (B A A B)dx = (A B) = (A B) nds. which implies that B Adx = A Bdx + (A B) nds.

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1 Last name: name: Notes, books, and calculators are not authorized. Show all your work in the blank space you are given on the exam sheet. Answers with no justification will not be graded. Question : Let be the curl operator acting on vector fields: i.e., let A = (A, A 2, A 3 ) : R 3 R 3 be a three-dimensional vector field over R 3, then A = ( 2 A 3 3 A 2, 3 A A 3, A 2 2 A ). Accept as a fact that (A B) = B A A B for all smooth vector fields A and B. Let Ω be a subset of R 3 with a smooth boundary Ω. Find an integration by parts formula for Ω B Adx. Using the divergence Theorem we infer that (B A A B)dx = which implies that Ω Ω B Adx = Ω Ω (A B) = Ω (A B) nds. A Bdx + (A B) nds. Ω Question 2: Let u, f : R R be two functions of class C. (a) Compute x f(u(x)). Using the chain rule we obtain where f denotes the derive of f. x f(u(x)) = f (u(x)) x u. (b) Let ψ : R R be functions of class C. Let F : R R be defined by F (v) = v 0 f (t)ψ (t)dt. Use (a) to compute x (F (u(x)) x (f(u(x)))ψ (u(x)). Using the chain rule we obtain x (F (u(x)) = F (u(x)) x u(x) = f (u(x))ψ (u(x)) x u(x) = x (f(u(x)))ψ (u(x)). This means that x (F (u(x)) = x (f(u(x)))ψ (u(x)). (c) Using the notation of (a) and (b), assume that u(± ) = 0 and compute x(f(u(x)))ψ (u(x))dx. Using (b) and u(± ) = 0 we have x (f(u(x)))ψ (u(x))dx = x (F (u(x)))dx = F (u(x)) + = F (0) F (0) = 0.

2 2 Mid-Term TEST, October, 6, 204 Question 3: Let φ be a smooth scalar field in R d, (d is the space dimension). (a) Prove that i () = i. Give all the details. Using the product rule, we have i () = d j φ i ( j φ) = i ( 2 j= d ( j φ) 2 ) = i ( 2 2 ) = i. j= (b) Let ɛ > 0. Show that ( 2 + ɛe i ) = Using the chain rule, we have +ɛ 2 i() where e i is the unit vector in the direction i. ( 2 + ɛe i ) = i ( 2 + ɛ) = ɛ i( 2 + ɛ) = 2 + ɛ i. Then using (a) we infer that ( 2 + ɛe i ) = 2 + ɛ i(). (c) Assume that φ(x) = 0 for all x R where R is a positive real number. Compute R d for all i =,..., d. Give all the details. Using (b) we have R d 2 + ɛ i()dx = ( 2 + ɛe i )dx, R d +ɛ i()dx, 2 where e i is the unit vector in the direction i. The divergence theorem together with the assumption that φ(x) = 0 for all x R implies that 2 + ɛ i()dx = 0. R d

3 Last name: name: 3 ( Question 4: Consider the vibrating beam equation tt u(x, t) + x 2 +cos(x) xx +x 2 xx u(x, t) ) = 0, u(x, 0) = f(x), t u(x, 0) = g(x), x (, + ), t > 0 with u(±, t) = 0, x u(±, t) = 0, xx u(±, t) = 0. Use the energy method to compute ([ tu(x, t)] 2 + x2 +cos(x) +x [ 2 xx u(x, t)] 2 )dx in terms of f and g. Give all the details. (Hint: test the equation with t u(x, t)). Using the hint we have 0 = ( x 2 ) + cos(x) ( tt u(x, t) t u(x, t) + xx + x 2 xx u(x, t) )dx Using the product rule, a t a = 2 ta 2 where a = t u(x, t), and integrating by parts two times (i.e., applying the fundamental theorem of calculus) we obtain 0 = = ( 2 t( t u(x, t)) 2 x ( x 2 + cos(x) ( t 2 ( tu(x, t)) 2 + We apply again the product rule a t a = 2 ta 2 where a = xx u(x, t), 0 = ( t 2 ( tu(x, t)) ) + x 2 xx u(x, t) t x u(x, t))dx ( x 2 ) + cos(x) + x 2 xx u(x, t) t xx u(x, t))dx. x 2 + cos(x) + x 2 t ( xx u(x, t)) 2 )dx. Switching the derivative with respect to t and the integration with respect to x, this finally gives In other words, 0 = 2 t ([ t u(x, t)] 2 + x2 + cos(x) + x 2 [ xx u(x, t)] 2 )dx = ([ t u(x, t)] 2 + x2 + cos(x) + x 2 [ xx u(x, t)] 2 )dx. (g(x) 2 + x2 + cos(x) + x 2 [ x f(x)] 2 )dx.

4 4 Mid-Term TEST, October, 6, 204 Question 5: Let k, f : [, +] R be such that k(x) = 3, f(x) = 6 if x [, 0] and k(x) =, f(x) = 2 if x (0, ]. Consider the boundary value problem x (k(x) x T (x)) = f(x) with T () = and x T () =. (a) What should be the interface conditions at x = 0 for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x = 0. Let T denote the solution on [, 0] and T + the solution on [0, +]. One should have T (0) = T + (0) and k (0) x T (0) = k + (0) x T + (0), where k (0) = 3 and k + (0) =. (b) Solve the problem, i.e., find T (x), x [, +]. Give all the details. On [, 0] we have k (x) = 3 and f (x) = 6 which implies 3 xx T (x) = 6. This in turn implies T (x) = x 2 + ax + b. The Dirichlet condition at x = implies that T () = = a + b. This gives a = b and T (x) = x 2 + bx + b. We proceed similarly on [0, +] and we obtain xx T (x) = 2, which implies that T + (x) = x 2 + cx + d. Neumann condition at x = implies T + () = = 2 + c. This gives c = 3 and T (x) = x 2 + 3x + d. The interface conditions T (0) = T + (0) and k (0) x T (0) = k + (0) x T + (0) give b = d and 3b = 3, respectively. In conclusion b =, d = and { x 2 + x + if x [, 0], T (x) = x 2 + 3x + if x [0, ]. The

5 Last name: name: 5 Question 6: (a) Compute the coefficients of the sine series of f(x) = x for x [0, +π]. (Recall that by definition SS(f)(x) = + m= b m sin(mx) with b m = 2 π f(x) sin(mx)dx.) The definition of SS(f)(x) implies that π 0 b m = 2 π π 0 = 2 m ()m+. x sin(mx)dx = 2 π π 0 m cos(mx)dx + 2 π [ x m cos(mx)]π 0 As a result SS(f)(x) = + m= 2 m ()m+ sin(mx). (b) For which values of x in [0, +π] does the sine series coincide with f(x)? (Explain). The sine series coincides with the function f(x) over the entire interval [0, +π) since f(0) = 0 and f is smooth over [0, +π). The series does not coincide with f(+π) since f(+π) 0. (c) The sine series of x 2 over [0, +π] is SS(x 2 )(x) = + sine series of h(x) = x(π x). (Hint: use (a)) m= ( 4 Let h(x) = x(π x). Note that by linearity of the sine series we have as a result b m (h) = πb m (x) b m (x 2 ), i.e., SS(h)(x) = SS(πx)(x) SS(x 2 )(x), m 3 π (()m ) + 2π m ()m+ ) sin(mx). Compute the In conclusion b m (h) = π 2 4 m ()m+ ( m 3 π (()m ) + 2π m ()m+ ) = 4 m 3 π ( + ()m+ ). SS(h)(x) = m= 4 m 3 π ( + ()m+ ) sin(mx). (d) Compute the cosine series of the function g(x) := π 2x defined over [0, +π]. (Hint: x (x(π x)) = π 2x.) Observe that h(0) = h(π) = 0; as a result the sine series of h is continuous at 0 and +π. This in turn implies that it is the legitimate to differentiate the sine series of h term by term to obtain the cosine series of h (x) = g(x). In other words, 4 CS(g)(x) = x SS(h)(x) = m 2 π ( + ()m+ ) cos(mx). m= (e) Compute the sine series of h(x) = sin(x) for x [0, +π]. Obviously SS(h)(x) = sin(x), x R.

6 6 Mid-Term TEST, October, 6, 204 Question 7: Using cylindrical coordinates and the method of separation of variables, solve the equation, r r(r r u)+ r 2 θθ u = 0, inside the domain D = {θ [0, 3 2 π], r [0, 3]}, subject to the boundary conditions θu(r, 0) = 0, u(r, 3 2π) = 0, u(3, θ) = 9 cos(θ). (Give all the details of all the steps.) () We set u(r, θ) = φ(θ)g(r). This means φ = λφ, with φ (0) = 0 and φ( 3 d 2π) = 0, and r dr (r d dr g(r)) = λg(r). (2) The usual energy argument applied to the two-point boundary value problem φ = λφ, φ (0) = 0, φ( 3 π) = 0, 2 implies that λ is non-negative. If λ = 0, then φ(θ) = c + c 2 θ and the boundary conditions imply c = c 2 = 0, i.e., φ = 0, which in turns gives u = 0 and this solution is incompatible with the boundary condition u(3, θ) = 9 sin(2θ). Hence λ > 0 and φ(θ) = c cos( λθ) + c 2 sin( λθ). (3) The boundary condition φ (0) = 0 implies c 2 = 0. The boundary condition φ( 3 2 π) = 0 implies that cos( λ 3 2π) = 0, i.e., λ 3 2 π = (2n + ) π 2 with n N. This means λ = 3 (2n + ), n = 0,, 2,.... (4) From class we know that g(r) is of the form r α, α 0. The equality r d dr (r d dr rα ) = λr α gives α 2 = λ. The condition α 0 implies 3 (2n + ) = α = λ. The boundary condition at r = 3 gives 9 cos(θ) = c 3 3 (2n+) cos( 3 (2n + )θ) for all θ [0, 3 2 π]. This implies n = and c = 3. (5) Finally, the solution to the problem is u(r, θ) = 3r cos(θ).

7 Last name: name: 7 Question 8: Let p, q : [, +] R be smooth functions. Assume that p(x) 0 and q(x) q 0 for all x [, +], where q 0 R. Consider the eigenvalue problem x (p(x) x φ(x)) + q(x)φ(x) = λφ(x), supplemented with the boundary conditions x φ() = 0 and x φ() = 2φ(). (a) Prove that it is necessary that λ q 0 for a non-zero (smooth) solution, φ, to exist. (Hint: q 0 φ2 (x)dx q(x)φ2 (x)dx.) As usual we use the energy method. Let (φ, λ) be an eigenpair, then ( x (p(x) x φ(x))φ(x) + q(x)φ 2 (x))dx = λ After integration by parts and using the boundary conditions, we obtain λ φ 2 (x)dx = = which, using the hint, can also be re-written Then Assume that φ is non-zero, then φ 2 (x)dx. (p(x) x φ(x) x φ(x) + q(x)φ 2 (x))dx 2p(x) x φ(x)φ(x) + (p(x) x φ(x) x φ(x) + q(x)φ 2 (x))dx + 2p()φ() 2 (p(x) x φ(x) x φ(x) + q 0 φ 2 (x))dx + 2p()φ() 2 λ 2p()φ() 2 + λ q 0 p(x)( x φ(x)) 2 dx (λ q 0 ) p(x)( xφ(x)) 2 dx + 2p()φ() 2 0, φ2 (x)dx φ 2 (x)dx. which proves that it is necessary that λ q 0 for a non-zero (smooth) solution to exist. φ 2 (x)dx. (b) Assume that p(x) p 0 > 0 for all x [, +] where p 0 R +. Show that λ = q 0 cannot be an eigenvalue, i.e., prove that φ = 0 if λ = q 0. (Hint: p 0 ψ2 (x)dx p(x)ψ2 (x)dx.) Assume that λ = q 0 is an eigenvalue. Then the above computation shows that p 0 ( x φ(x)) 2 dx p(x)( x φ(x)) 2 dx = 0, which means that ( xφ(x)) 2 dx = 0 since p 0 > 0. As a result x φ = 0, i.e., φ(x) = c where c is a constant. The boundary condition φ() = 2φ() implies that c = 0. In conclusion φ = 0 if λ = q 0, thereby proving that (φ, q 0 ) is not an eigenpair.

8 8 Mid-Term TEST, October, 6, 204 Question 9: Use the Fourier transform technique to solve t u(x, t) xx u(x, t)+cos(t) x u(x, t)+(+2t)u(x, t) = 0, x R, t > 0, with u(x, 0) = u 0 (x). (Hint: use the definition F(f)(ω) := + 2π f(x)eiωx dx), the result F(e αx2 )(ω) = 4πα e ω2 4α, the convolution theorem and the shift lemma: F(f(x β))(ω) = F(f)(ω)e iωβ. Go slowly and give all the details.) Applying the Fourier transform to the equation gives This can also be re-written as follows: t F(u)(ω, t) + ω 2 F(u)(ω, t) + cos(t)( iω)f(u)(ω, t) + ( + 2t)F(u)(ω, t) = 0 t F(u)(ω, t) F(u)(ω, t) = ω 2 + iω cos(t) ( + 2t). Then applying the fundamental theorem of calculus between 0 and t, we obtain log(f(u)(ω, t)) log(f(u)(ω, 0)) = ω 2 t + iω sin(t) (t + t 2 ). This implies F(u)(ω, t) = F(u 0 )(ω)e ω2t e iω sin(t) e (t+t2). Using the result F(e αx2 )(ω) = 4πα e ω2 4α Then setting g = u 0 e x2 4t This finally gives u(x, t) = where α = 4t, this implies that π F(u)(ω, t) = t F(u 0)(ω)F(e x 2 4t )(ω)e iω sin(t) e (t+t2 ) π = t F(u 0 e x 2 4t )(ω)e iω sin(t) e (t+t2). the convolution theorem followed by the shift lemma gives F(u)(ω, t) = π F(g(x 2) sin(t)))(ω)e (t+t. 2π t g(x 2) sin(t))e (t+t = e ) (t+t2 4πt 4πt u 0 (y)e (x sin(t) y)2 4t dy.