MATH 220 solution to homework 5 Problem. (i Define E(t = k(t + p(t = then E (t = 2 = 2 = 2 u t u tt + u x u xt dx u 2 t + u 2 x dx, u t u xx + u x u xt dx x [u tu x ] dx. Because f and g are compactly supported, we know u and its derivatives vanishes at x = (for fixed t, then E (t = 0, E(t = E(0, k(t + p(t = k(0 + p(0. (ii u(t, x = ( x+t g(x t + g(x + t + f(y dy, 2 x t k(t = u 2 t dx = ( g (x t + g (x + t + f(x t + f(x + t 2 dx, 4 p(t = u 2 x dx = (g (x t + g (x + t f(x t + f(x + t 2 dx, 4 k(t p(t = (( g (x t + g (x + t + f(x t + f(x + t 2 4 (g (x t + g (x + t f(x t + f(x + t 2 dx = (f(x + t + g (x + t(f(x t g (x t dx. Because f and g are compactly supported, there exists M > 0, such that supp(f ( M, M, supp(g ( M, M, supp(f + g ( M, M, supp(f g ( M, M.
(Here supp means the support of a function. Then t > M, x, f(x + t + g (x + t and f(x t g (x t cannot be both nonzero, i.e., (f(x + t + g (x + t(f(x t g (x t = 0, k(t p(t = 4 (f(x + t + g (x + t(f(x t g (x t dx = 0. If f and g are not compactly supported, the statement may not be true. k(t and p(t may not exist. Even if they exist, the statement may still be false. Counterexample: g(x = x e y2 dy, g (x > 0, x, f(x 0, k(t p(t = 4 k(t p(t < 0, t > 0. g (x + tg (x t dx < 0, t > 0, Problem 2. (i ĝ(ξ = = = = = g(xe 2πixξ dx f(x yh(ye 2πixξ dy dx f(x ye 2πi(x yξ h(ye 2πiyξ dy dx f(x ye 2πi(x yξ h(ye 2πiyξ dy d(x y f(ze 2πizξ dz h(ye 2πiyξ dy = ˆf(ξĥ(ξ. (ii Do Fourier transform on both sides, then û(t, ξ t = ˆK(ξû(t, ξ, û(t, ξ = û(0, ξe ˆK(ξt = ˆf(ξe ˆK(ξt. 2
For fixed t > 0, we have û(t, ξ dξ M(t ˆf(ξ dξ <, where M(t = sup e ˆK(ξt. (Since K(x is rapidly decaying, we know ˆK(ξ is bounded by ξ K L. So M(t is a well defined finite number for any t. Then u(t, x û(t, ξ dξ <, x, which means u(t, x is bounded (for fixed t > 0. We also have k u x k (t, ξ = (2πiξk ˆK(ξt ˆf(ξe, k u (t, x = (2πiξ k ˆK(ξt ˆf(ξe e 2πiξx dξ M(t xk (2πiξ k ˆf(ξ dξ <. The last step is because, the function (2πiξ k ˆf(ξ is in Schwartz class (since f is in Schwartz class, thus it s in L. So the definition for k u (t, x exists everywhere for any k, which means, xk for fixed t > 0, u is smooth. For the square of the L 2 norm of u(t, we have u(t, x 2 dx = û(t, ξ 2 dξ = ˆf(ξe ˆK(ξt 2 dξ M(t 2 ˆf 2 L = 2 M(t2 f 2 L 2, i.e., for fixed t, u(t, x 2 dx can be bounded by the initial data f(x and M(t. Furthermore, if ˆK(ξ 0, we have u(t, x 2 dx f 2 L 2. Problem 3. (i c 2 u t + p x = c 2 (φ t t + ( φ x x = c 2 φ tt φ xx = 0, p t + u x = ( φ x t + (φ t x = φ xt + φ tx = 0. (ii Do Fourier transform in x, we have: Aˆv t + 2πiξDˆv = 0, ˆv t = 2πiξA Dˆv. 3
Since A is symmetric positive definite and D is symmetric, we know A D can be diagonalized and its eigenvalues are all real, i.e., there exists a real diagonal matrix Λ and a real nonsingular matrix X such that A D = XΛX. This is because: Since A is S.P.D., we can write A as A = U A Λ A UA T where U A is orthogonal, and Λ A = diag(λ A,,..., Λ A,n is diagonal with every diagonal entry Λ A,i real positive. We define A /2 = U A Λ /2 A U A T where Λ/2 A is defined as diag(λ/2 A,,..., Λ/2 A,n, then A/2 is S.P.D. and A /2 A /2 = A. Similarly we define A /2 = U A Λ /2 A UA T = (A/2. Then A D = A /2 (A /2 DA /2 A /2. Note that A /2 DA /2 is symmetric, so it can be written as A /2 DA /2 = V ΛV T where Λ is a real diagonal matrix and V is orthogonal. Then A D = (A /2 V Λ(V T A /2. Set X = A /2 V and we have A D = XΛX. Then ˆv(t, ξ = exp( 2πiξtA D ˆv 0 (ξ = X exp( 2πiξtΛX ˆv 0 (ξ. Let w(t, x = X v(t, x, then ŵ(t, ξ = X ˆv(t, ξ, ŵ(t, ξ = exp( 2πiξtΛŵ 0 (ξ, w(t, x = j w 0,j (x λ j te j. where w 0 (x = (w 0,, w 0,2,..., w 0,n T = w(0, x = X v 0 (x, and e j is the j-th standard basis vector. Then we get v by v = Xw. The physical meaning is, the wave propagates along direction X j = X(:, j (the j-th eigenvector with speed λ j. (iii In this case we have ( v = (u, p, A = c 2 0, D = 0 ( 0, A D = 0 ( 0 c 2 0. Let X = ( c c, Λ = diag(c, c, then A D = XΛX, w 0 (x = ( ( c u0 (x, 2c c p 0 (x u(t, x = 2 (u 0(x ct + u 0 (x + ct + cp 0 (x ct cp 0 (x + ct. To get the D Alembert formula, we let φ(0, x = f(x, φ t (0, x = g(x, 4
then u 0 = g, p 0 = f, u(t, x = 2 (g(x ct + g(x + ct cf (x ct + cf (x + ct, t φ(t, x = f(x + u(s, x ds = ( f(x ct + f(x + ct + 2 c Thus we recovered the D Alembert formula. 0 x+ct x ct g(y dy. (iv E (t =2 v T Av t dx = 2 v T Dv x dx = (v T Dv x dx. Since the initial data of v decays rapidly at and the solution moves at a finite speed, we know v T Dv decays rapidly at, for any fixed t. then E (t = 0, E(t = E(0. For the wave equation, we have E(t = c 2 u2 + p 2 dx = c 2 φ2 t + φ 2 x dx. (v E (t = 2v T v t d(x, y = 2 v T (D v x + D 2 v y d(x, y = lim ( (vt D v + (vt D 2 v d(x, y B(0, x y = lim (v T D v, v T D 2 v n ds(x, y =0. B(0, The last step holds true because v decays rapidly at. 5
To solve the equation, do Fourier transform to (x, y, then ˆv t + 2πiξ D ˆv + 2πiξ 2 D 2ˆv = 0, ˆv(t, ξ = exp( 2πiξ td 2πiξ 2 td 2 ˆv 0 (ξ. For fixed ξ = (ξ, ξ 2, D(ξ = ξ D + ξ 2 D 2 can be diagonalized as then D(ξ = X(ξΛ(ξX (ξ, ˆv(t, ξ = X(ξ exp( 2πitΛ(ξX (ξ ˆv 0 (ξ. Thus, with the eigenvalues and the eigenvectors of D(ξ, the propagating operator exp( 2πiξ td 2πiξ 2 td 2 is diagonalized. It means that the eigenvalues and eigenvectors of D(ξ serve as the decoupling of the equation on the Fourier space. If we define ŵ(t, ξ := X (ξˆv(t, ξ, then ŵ,..., ŵ n are decoupled as follows ŵ j (t, ξ = exp( 2πiλ j (ξtŵ j (0, ξ. Problem 4. If 0 t <, then, x t, x/t, t x < 2t, u = 2, 2t x < t +, 0, t + x. If t < 4, then, x t, u = x/t, t x < 2 t, 0, 2 t x. If 4 t, then {, x 2 + t/2, u = 0, 2 + t/2 < x. Problem 5. (i Let w(t, x, y = u(t, x + t, y, then w t (t, x, y = u t (t, x + t, y + u x (t, x + t, y = u yy (t, x + t, y = w yy (t, x, y, t > 0, (x, y 2, w(0, x, y = u(0, x, y = f(x, y, (x, y 2. 6
So w behaves like heat equation in the plane (t, x 0, y for fixed x 0, then the solution is w(t, x, y = f(x, zg(t, y, z dz, where G(t, y, z = y z 2 exp ( is the dimension heat kernel. Then 4πt 4t u(t, x, y = w(t, x t, y = f(x t, zg(t, y, z dz. Since the solution w only smooth the initial data in the plane (t, x 0, y for fixed x 0, so discontinuity of the initial data or its derivatives can propagate if the discontinuity is in x. u is a shift of w so the property of the differentiability are the same with w. So, u s smoothness in x won t be better than f, still C 3, but in y it s infinitely differentiable. (ii Do Fourier transform in x and y, we have ˆv t ξ ˆv ξ2 = 4π 2 ξ 2 2 ˆv, t > 0, (ξ, ξ 2 2, ˆv(0, ξ, ξ 2 = ˆf(ξ, ξ 2. Let w(t, ξ, ξ 2 = ˆv(t, ξ, ξ 2 ξ t, then w t = 4π 2 (ξ 2 ξ t 2 w, t > 0, (ξ, ξ 2 2, w(0, ξ, ξ 2 = ˆf(ξ, ξ 2. Solve the equation we have ( w(t, ξ, ξ 2 = ˆf(ξ 4π 2 ((ξ 2 ξ t 3 ξ 3, ξ 2 exp 2. 3ξ Then ˆv(t, ξ, ξ 2 = w(t, ξ, ξ 2 + ξ t ( = ˆf(ξ 4π 2 (ξ2 3 (ξ 2 + ξ t 3, ξ 2 + ξ t exp 3ξ = ˆf(ξ, ξ 2 + ξ t exp ( π 2 t((2ξ 2 + tξ 2 + (tξ 2 /3. For fixed t > 0, the function g(t, ξ, ξ 2 := exp ( π 2 t((2ξ 2 + tξ 2 + (tξ 2 /3 is a Schwartz class function in (ξ, ξ 2. And ˆf(ξ, ξ 2 + ξ t M <, because f(x, y L ( 2. So, for any polynomial p(ξ, ξ 2, we have p(ξ, ξ 2 ( ˆf(ξ, ξ 2 + ξ tg(t, ξ, ξ 2 d(ξ, ξ 2 M p(ξ, ξ 2 g(t, ξ, ξ 2 d(ξ, ξ 2 <, so v(t, x, y is infinitely differentiable. 7
(iii For the two equations, we have the transport term and the diffusion term 2 x y 2. The diffusion term is the same for u and v, and the diffusion leads to a blurring in y direction, where the blurring comes from the Brownian motion in y direction. The blurring gives smoothness in y direction. The difference of u and v is that, the transport term for v is yv x, which has the coefficient y, while the one for u is just u x where the coefficient is a constant. The physical meaning of the coefficient is the transport speed. For the case of u, the speed in x is a constant, so the solution is only blurring in y and transporting in x with a constant speed, which implies that the unsmoothness in x direction will not be killed as time evolves. On the other hand, for v, since the transport speed in x depends on y, and the motion in y is the Brownian motion a stochastic process, thus the transportation speed is non-deterministic, and the solution will also be blurred in x direction, which gives more smoothness. One can also consider the equations after Fourier transform (the equations describing the frequency distributions of the solutions. For the first equation, the frequency doesn t always decay as t for some (ξ, ξ 2 sufficiently large, that is, we can always find some high frequency that does not decay with time. For the second equation, the frequency always decays rapidly as (ξ, ξ 2 for any t > 0, so the initial data is smoothed and the solution is infinitely differentiable for fixed t > 0. Problem 6. (i Suppose u(t, x = U(x ct, then cu (x + U(xU (x = εu (x, x. Integrate over, we have cu(x + U(x 2 /2 = εu (x + C. Let y +, we have C = 0, then cu(x + U(x 2 /2 = εu (x. Let x, we have c = /2, and U(x(U(x = 2εU (x. Solving the equation with boundary condition, we have U(x = + exp((x x 0 /(2ε, where x 0 is an undetermined parameter. Then u(t, x = U(x ct = + exp((x ct x 0 /(2ε. 8
(ii Similarly, if we assume u(t, x = U(x ct, we will have U(x(U(x = 2εU (x. Since 0 < U(x <, we have U (x < 0, which means U(x is concave. But a bounded concave function can only be a constant function, which leads to U (x 0, a contradiction. So such solution does not exist. 9