Math512 PDE Homework 2

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1 Math51 PDE Homework October 11, 009 Exercise 1.3. Solve u = xu x +yu y +(u x+y y/ = 0 with initial conditon u(x, 0 = 1 x. Proof. In this case, we have F = xp + yq + (p + q / z = 0 and Γ parameterized as (s, 0, 1 s. Firstly we complete the initial curve Γ to a strip. φ and ψ should satisfy sφ + 0ψ + (φ + ψ / 1 s = 0 and s = φ + ψ0 which gives φ(s = s and ψ(s = ±1. Now the char equations are: dx dt = x + p; dy dt = y + q; dz dt = xp + p + yq + q dp dt = p + p = 0; dq dt = q + q = 0 We have that p = C 1 (s, q = C (s, and from the initial value φ(s = s and ψ(s = ±1, we have that p = s and q = ±1. Furthermore x = p = s, y = e t q = ±e t 1, and z = ±e t 1+s = ±y + 1 x. Exercise To generalize (6 to n variables, let u(x = u(x 1,..., x n and p j = u for j = 1,,..., n. Then consider F (x, u, p = 0 (a Find the char equation.[i.e. generalizations of (9,(31]. (b Find the conditions on the initial values of p.[i.e. generalizations of (31,(3]. Proof. (a We can rewrite F (x, u, p = 0, such taht F (x 1,..., x n, z, p 1,..., p n = 0, where p k = u x k, k = 1,,...n.. At (x 0, z 0, we have F (x 0, z 0, p = 0 which define p n as function of p 1,..., p n 1. From the family of tangent planes z z 0 = p i (x x i, we have dz = n i=1 p idx i. By taking / on the previous equation, we have 0 = dx i + p n dx n 0 = F pi + F pn p n 1

2 which give rise to dx i F pi = dxn F pn thus we have the generation of equation (9. dx i dt = F p i 1 i n (9 dz dt = dx k dt p k = p k F pk k=1 k=1 Now we derive dp i dt. By taking / of F = 0, we have 0 = F xi + F z p i + j F pj p j since p j = u ; p j = u equation becomes =. Furthermore, F pj j=1 which give rise to generation of (30. F pj p j = j=1 dx j dt = dx j. Then the last term of above dt = t dp i dt = F x i F z p i, 1 i n (30 (b Take f k (s, to be the paramterized form for x k,k = 1,,...n,. For the function φ k (s, k=1,,..n, the generation for (31, (3 is F (f 1 (s, f (s,..., f n (s, h(s, φ 1 (s,..., φ n (s = 0 (31 h (s = φ 1 (sf 1(s φ n (sf n(s Exercise The Hamilton-Jacobi equation u t + H( x u = 0, where H depends only on x u = (u x1,..., u xn, arises frequently in physics, such as geometric optics. Assume n = 1 here. (averify that condition (34 fails for u(x, t; a, b = ax th(a + b, yet this is a complete integral. (buse envelopes to generate a solution of u t + u x = 0 that is not linear in x and t. ( uxa u Proof. (afor any (a, b, det ta u xb u tb ( 1 H (a = det 0 0 = 0, which means that the condition (34 fails. But we can check, for any (a, b, that u(x, t; a, b is a solution of Hamilton Jocobi equation. u t = H(a; and H( x u = H(a, which gives u t + H( x u = 0. Next, we check if the parameters a and b are independent. u = u(x, y; a, b ( is mapping from ua u (a, b to (u, u x, u y, the Jocobian has full rank. Verify the condition: xa u ta u b u xb u tb

3 ( x th (a 1 H (a = has full rank. (b Take H(a = a,b = w(a = a and we use the fact that the general solution of u t + (u x is the envelope of u(x, t; a, b = ax ta + a to find the solution of the equation. Solve a u(x, t; a, b = x at + 1 = 0, we have a = x+1 ; substitute back to u(x, t; a, b, we t have that u(x, t = (x+1. 4t Exercise.1.1 Consider the initial value problem u zz = u u x +(u xy, u(x, y, 0 = x y, u z (x, y, 0 = sin x. Find the values of u xz, u yz, u zz, when z = 0. Proof. From u(x, y, 0 = x y, we know that u x (x, y, 0 = x y, u y (x, y, 0 = 1 and u xy (x, y, 0 = 0. And from u z (x, y, 0 = sin x, we know that u xz (x, y, 0 = u zx (x, y, 0 = cos x and u zy (x, y, 0 = u yz (x, y, 0 = 0. And u zz (x, y, 0 = u (x, y, 0u x (x, y, 0 + (u xy (x, y, 0 = (x y = (x y Exercise.1. Is the heat equation u t = ku xx in normal form for Cauchy data on the x-axis? On the t-axis? What form would be Cauchy data (3 take? Proof. The Heat equation u t = ku xx is normal form for Cauchy date on t-axis but not on x-axis. The Cauchy data form (3 will be u(0, t = g 1 (t and u x (0, t = g (t. Exercise.1.4 Find the Taylor series solution about x, y = 0 of the initial value problem u y = sin u x, u(x, 0 = πx/4. Proof. The Taylor expaonsion of u(x, y: u(x, y = j,k=0 x j y k u(0, 0 x j y k j!k! Since u(x, 0 = πx, we have that u(0, 0 = 0, u 4 x(0, 0 = π, u(k (0,0 = 0, with k, 4 x k u y (0, 0 = sin u x (0, 0 = sin π =, u 4 xy(0, 0 = (cos u x u xx (0, 0 = 0, u(k+j (0,0 = 0 with y k j + k. Thus we have that u(x, y = π 4 x + y Exercise..(b Find the general solution of y u xx yu xy + u yy = u x + 6y. 3

4 Proof. We know that a = y, b = y,c = 1. Then b 4ac = 4y 4y = 0. The Char curves are found by dy dx = y y = 1 y x + y = C And we choose another curve which is not tangent to the above one Which gives and { µ x = 1; η x = 0 µ y = y; η y = 1 plugging back to the equation, we have y = c µ(x, y = x + y η(x, y = y u x = u µ µ x = u µ ; u y = u µ y + u η And we have that u xx = u µµ ; u xy = u µµ y + u µη u yy = u µµ y + yu µη + u µη y + u ηη + u µ u ηη = 6η then we have that u(η, µ = η 3 +f(µη+g(µ, which is u(x, y = y 3 +f(x+ y y y+g(x+...4 Show that the minimal surface equation (1+u y u xx u yy u x u y u xy +(1+u x u yy = 0 is everywher elliptic. Proof. We have that a = (1 + u y, c = (1 + u x, and b = u x u y, which gives that b 4ac = 4u xu y 4(1 + u x (1 + u y = 4(u x + u y + 1 < 0 then we know that it is elliptic everwher...6 Reduce the form ( and solve the initial value problem. u t + ( u 1 x = ( 1, u(x, 0 = ( x 0 Proof. It is easy to get the eigenvalue λ 1, λ and eigenvector e 1, e of coefficient matrix. Thus we have that λ 1 = 1, λ = ; and e 1 = [ 1], e = [ 1 1] [ ] = [ ][ ][ ] 1 4

5 Thus the equation has the form v t +[ ]v x = [ which is v t + [ ]v x = [ which is 0 1 ], v(x, 0 = [ x x { v 1 t v 1 x = 0 with v 1 (x, 0 = x v t + v x = 1 with v (x, 0 = x. Then we have v = [ x t x 3t ] and u = [ ] 1 [ 1 1 ] where v = [ ] 1 u. ]v = [ x + 5t 4t ]. ], v(x, 0 = [ ] 1 [ x 0 ] 5

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