Math 5588 Final Exam Solutions Prof. Jeff Calder May 9, 2017 1. Find the function u : [0, 1] R that minimizes I(u) = subject to u(0) = 0 and u(1) = 1. 1 0 e u(x) u (x) + u (x) 2 dx, Solution. Since the Lagrangian L(x, z, p) = e z p + p 2 has no x-dependence, the Euler- Lagrange equation is which is L(u(x), u (x)) u (x)l p (u(x), u (x)) = 0 e u(x) u (x) + u (x) 2 u (x)(e u(x) 2u (x)) = 0 or u (x) 2 = 0. Therefore u is constant and u(x) = x due to boundary conditions. 1
2. [10 points] Find the entropy solution of Burger s equation u t + uu x = 0 with 1, if x < 1 u(x, 0) = f(x) = 0, if 1 < x < 1 1, if x > 1. Sketch the characteristics and write down u(x, t) explicitly. Solution. By the entropy and Rankine-Hugoniot conditions, shocks form at x = 1 and x = 1 with speeds dx/dt = 1/2 and dx/dt = 1/2, respectively. The solution therefore has the form 1, if x < 1 + t/2 u(x, t) = 0, if 1 + t/2 < x < 1 t/2 1, ifx > 1 t/2, for 0 < t < 2, and u(x, t) = { 1, if x < 0 1, if x > 0 for t > 2. The characteristics for 0 < t < 2 have speed u = 1 when x < 1 + t/2, speed u = 0 when 1 + t/2 < x < 1 t/2, and speed 1 when x > 1 t/2. When t > 2 the characteristics have speed 1 for x < 0 and speed 1 for x > 0. 2
3. [10 points] Let H(p) = e p. (a) [4 points] Show that q(log q 1), if q > 0 H (q) = 0, if q = 0., if q < 0. Solution. By definition H (q) = max p R {pq ep }. If q = 0 then the maximum (or supremum) is when p = and H (0). If q < 0 there is no upper bound on pq e p, since we can send p and pq e p +. Thus H (q) = for q < 0. When q > 0 the maximum occurs when q e p = 0 or p = log q, thus H (q) = q log q q. (b) [2 points] Is H convex? Is H superlinear? Solution. H is convex but not superlinear, since lim p H(p)/ p = 0. (c) [4 points] Compute H (p). Solution. By definition H (p) = max q R {pq H (q)}. The max cannot be attained when q < 0, since H (q) =. Therefore the max is attained for q 0. If q = 0 we haqve pq H (q) = 0. If the max is attained at q > 0 then p (H ) (q) = 0 or p log q = 0 and q = e p. Thus the maximum is 3
4. [10 points] Solve the wave equation u t u = 0 in n = 3 dimensions with initial conditions u(x, 0) = 0 and u t (x, 0) = 2x 2 1 x2 2 x2 3. Solution. We use Kirchoff s formula 1 u(x, t) = B(x, t) B(x,t) tg(y) + f(y) + f(y) (y x) ds(y). Here f = 0 and g(x) = 2x 2 1 x2 2 x2 3. Notice g = 0, so we can use the mean value formula to get u(x, t) = 1 4πt 2 B(x,t) tg(y) ds(y) = tg(x) = t(2x 2 1 x 2 2 x 2 3). 4
5. [10 points] Let R n and suppose that u(x, t) solves u t + u + H( u, 2 u) = 0 for x, t > 0, with initial condition u(x, 0) = f(x) for x, and boundary condition u(x, t) = 0 for x, t > 0. If H is (degenerate) elliptic, H(0, 0) = 0, and 0 f(x) 1 for all x, show that 0 u(x, t) e t for all x and t > 0. [Hint: Maximum principle] Solution. Notice that v(x, t) = e t solves v t + v + H( v, 2 v) = e t + e t + H(0, 0) = 0. Since v(x, 0) = 1 f(x) = u(x, 0) and v(x, t) 0 = u(x, t) for x, the maximum principle (or comparison principle) gives that u v. This is one direction. For the other direction, v(x, t) = 0 is also a solution of the PDE, so the maximum principle gives that u v = 0. 5
6. [10 points] Let R n and consider the functional I(u) = L(x, u, u, u) dx. Derive the Euler-Lagrange equation for I. [Hint: Write L = L(x, z, p, q) where x, z R, p R n and q R, and write L z and L q for the partial derivatives in z and q, and write p L for the gradient in p. Take variations of I in directions of compactly supported smooth test functions.] Solution. Let ϕ be smooth and compactly supported in. Then d dt I(u+tϕ) t=0 = L z (x, u, u, u)ϕ+ p L(x, u, u, u) ϕ+l q (x, u, u, u) ϕ dx. sing Green s formulas (integration by parts), any minimizer of I satisfies (L z (x, u, u, u) div p L(x, u, u, u) + L q (x, u, u, u))ϕ dx = 0. Since this holds for any test function ϕ, the vanishing lemma gives that L z (x, u, u, u) div p L(x, u, u, u) + L q (x, u, u, u) = 0, which is the Euler-Lagrange equation. We can also write this out: L z (x, u, u, u) n i=1 x i p L(x, u, u, u) + n 2 x 2 i=1 i L q (x, u, u, u) = 0. 6
7. [20 points: 4 for each part] Let f : R R be a probability density function, that is f 0 and f(x) dx = 1. Let us assume that R xf(x) dx = 0 (f has zero mean) and write σ 2 = x 2 f(x) dx for the variance. Also assume R x 3 f(x) dx <. (a) Show that R R 2π f(k) = 1 σ 2 2 k2 + O( k 3 ). [Hint: se the definition of f and the Taylor expansion e ikx = 1 + ( ikx) + ( ikx)2 2 + O( xk 3 ).] Solution. By the Taylor expansion and assumptions above 2π f(k) = f(x)e ikx dx = = f(x)(1 + ikx k2 x 2 + O( xk 3 )) dx 2 f(x) dx ik xf(x) dx k2 2 = 1 σ2 2 k2 + O( k 3 ). ) x 2 f(x) dx + O ( k 3 x 3 f(x) dx 7
(b) If X 1,..., X n are independent and identically distributed random variables with probability density f(x), then the sum S n := X 1 + + X n has probability density function g n (x) given by the n-fold convolution Show that 2πĝn (k) = g n = f f f f. }{{} n times ) n (1 σ2 2 k2 + O( k 3 ). Solution. By the convolution theorem iterated n times (or induction) ĝ n (k) = 2π f(k) F f f f f }{{} (k) (n 1) times = 2π f(k) 2 F f f f f }{{} (k) (n 2) times = (2π) 3/2 f(k) 3 F f f f f }{{} (k) (n 3) times =. = (2π) (n 1)/2 f(k) n 1 F(f)(k) = (2π) (n 1)/2 n f(k) = 1 ( ) n 1 σ2 2π 2 k2 + O( k 3 ). (c) Let h n be the probability density function for the normalized sum S n / n, that is, h n (x) = n g n ( nx). Show that 2π ĥ n (k) = (1 σ2 ( ) ) n 2n k2 + O k 3. n 3/2 Solution. By the scaling/dilating theorem for Fourier Transforms 2π ĥ n (k) = 2π ( 1 n n ĝ n (k/ ) n) = (1 σ2 ( ) ) n 2n k2 + O k 3. n 3/2 8
(d) Show that as n we have ĥ(k) := lim ĥ n (k) = 1 e σ2 2 k2. n 2π ( ) [Hint: You can neglect the error term O k 3 from part (c). se the identity n 3/2 lim (1 1 m m )m = 1 e.] Solution. Neglecting the error term we have ( ) n 1 ĥ(k) = lim ĥ n (k) = lim 1 σ2 n n 2π 2n k2. Set 1 m = σ2 2n k2 to get [( 1 ĥ(k) = lim 1 1 ) m ] σ 2 2 k2 = 1 e σ2 2 k2. m 2π m 2π (e) Find a formula for h(x). [The probability density h(x) is the limiting distribution of the normalized sums S n / n. You should get a familiar probability density; this is the celebrated Central Limit Theorem in probability.] Solution. We just need to invert the Fourier Transform. Recall that F(e x2 /2 ) = e k2 /2. Therefore and hence F(e x2 /2σ 2 ) = σe σ2 2 k2, h(x) = 1 2π F 1 (e σ2 2 k2 ) = 1 2πσ 2 e x2 /2σ 2. This is the normal (or Gaussian) distribution with mean zero and variance σ 2 (or standard deviation σ). The point of the central limit theorem is that you always get the Gaussian distribution in the limit as n, regardless of the distribution f you started with. The Gaussian is thus universal. 9
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Formula Sheet L(u(x), u (x)) u (x)l p (u(x), u (x)) = Constant L z (x, u(x), u (x)) d dx L p(x, u(x), u (x)) = 0. I(u) = L z (x, u, u) div ( p L(x, u, u)) = 0 u xi dx = uν i ds. u v dx = u v ν ds u v dx u v v u dx = u v ν v u ν ds v v dx = ν ds u div(f ) dx = u F ν ds u F dx. 1 F(u) = û(k) = (2π) n/2 u(x)e ik x dx. R n F 1 1 (û) = u(x) = (2π) n/2 û(k)e ik x dk. R n F(u xj ) = ik j û(k). F(u v) = (2π) n/2 û(k) v(k). û(k) 2 dk = u(x) 2 dx. R n R n F(e x 2 /2 ) = e k 2 /2. 1 u(x, t) = tg(y) + f(y) + f(y) (y x) ds(y). B(x, t) B(x,t) u(x, t) = 1 tf(y) + t f(y) (y x) + t 2 g(y) 2πt 2 dy. t 2 x y 2 B(x,t) kij s = (x i x l ) (x j x l ), kii s = x j x l 2 4Area(T s ) 4Area(T s ). dx dt = F (u l) F (u R ), u l > u R. u l u R { ( x y f(y) + tl u(x, t) = min y R n H (q) = max p Rn{p q H(p)}. t )}, L = H H(p, x) = min{f(x, a) p + r(x, a)}. a A 14