MATH 131P: PRACTICE FINAL SOLUTIONS DECEMBER 12, 2012

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1 MATH 3P: PRACTICE FINAL SOLUTIONS DECEMBER, This is a closed ook, closed notes, no calculators/computers exam. There are 6 prolems. Write your solutions to Prolems -3 in lue ook #, and your solutions to Prolems 4-6 in lue ook # to facilitate grading. You may solve the prolems in any order. Total score: 5 points. Prolem 4(iv is extra credit only; do it only if you are done with the rest of the exam. Prolem. ( points Solve the PDE and sketch the characteristics of the PDE. u x + xu y = u, u(, y = y, Solution. The characteristics going through the initial data curve {(, r : r R} satisfy the ODE x =, x(r, =, s y = x, y(r, = r. s Solving the x ODE gives x = s, thus the y ODE is y s = s, so y = s +r. Thus, s = x, r = y x is the change of coordinates from (x, y to (r, s. Finally the ODE along the characteristics is u s = u, u(, r = r, so u = r e s. Sustituting in r and s yields u = (y x e x. The characteristics are paraolae, y = x + r, with r fixed along the characteristic, and x varying. Prolem. ( points Consider the PDE (i u xx u xy u yy =, (ii u xx + 3u xy + 3u yy =. What is the type of these PDE? One of the PDE is hyperolic. Find its general solution. Solution. The PDE au xx + u xy + cu yy = is elliptic if 4ac >, hyperolic if 4ac <. Thus, (i is hyperolic since 4 ( < (, while (ii is elliptic since 4 3 > 3. Hence, we need to solve (i. To do so we factor x x y y = ( x y ( x + y. Since ( x y φ(x + y =, ( x + y ψ(x y =, we conclude that the general solution of the PDE is u(x, y = φ(x + y + ψ(x y with φ, ψ aritrary functions. One can also proceed y changing variales to new coordinates ξ, η; recall that one would want ( ξx ξ x =, ξ y ξ y otained y simply replacing each x-derivative in the PDE y a factor of ξ x, and each y-derivative η in the PDE y a factor of ξ y. x η y solves the same equation. This gives that ξ x = =, ξ y η x = + 8 =, η y so we can take ξ = x + y, η = x + y, which gives the same general solution (an aritrary function of ξ plus one of η as aove.

2 MATH 3P: PRACTICE FINAL SOLUTIONS DECEMBER, Prolem 3. Consider Laplace s equation on the rectangle R: u xx + u yy = on R = (, a x (, y with oundary conditions u(, y =, u(a, y = y, u y (x, =, u y (x, =. (i (7 points Interpret the PDE and the oundary conditions: if u is the steady state temperature on a metal plate, what is imposed at the various edges of the plate? Draw a picture. (ii (3 points Use separation of variales to solve the PDE. Find all coefficients in your expansion explicitly. Solution. The physical interpretation is that the edges y = and y = are insulated, and there is no heat flux through these edges, while the edge at x = is kept at temperature and the edge at x = a is kept at temperature u(a, y = y. Looking for separated solutions we write u(x, y = X(xY (y, impose the PDE and the homogeneous oundary conditions. This gives a constant, with This gives X Y + XY = X X = Y Y = λ, X( =, Y ( =, Y ( =. Y (y = cos( λy, λ = nπ, where n is an integer, so λ = nπ. Note that n > and n < give rise to the same Y, so we can restrict to n ; n = corresponds to Y (y =. Then the X equation gives, for n, X(x = sinh( ( nπx λ n x = sinh, while for n =, X(x = x. Taking linear cominations gives Sustituing in x = a yields u(x, y = A x + y = A a + ( nπx cos ( nπa cos.. ( nπa Thus, is the nth Fourier cosine coefficient of y on [, ] for n, while A a is the th Fourier cosine coefficient, so for n, while for n =, But so y cos ( nπa A a = = y cos dy y dy = y =. dy = y nπ sin sin dy nπ ( ( ( = cos = ( n, nπ nπ A = a, ( ( A n = ( ( n, n >, nπa sinh nπ

3 MATH 3P: PRACTICE FINAL SOLUTIONS DECEMBER, 3 in u(x, y = A x + ( nπx cos. Prolem 4. Consider the wave equation u tt = c u xx on the half-line, i.e. on (, x R t, with homogeneous Neumann oundary condition u x (, t =, and with initial conditions u(x, = f(x and u t (x, = g(x for x. (i ( points Find u in terms of f, g. Make your answer as explicit and simplified as possile. (ii (6 points What are the time symmetric solutions u, i.e. solutions u such that u(x, t = u(x, t for all x, t? (iii ( points Suppose that c =, g, and f(x = for < x <, f(x = for x > and for < x <. Find u(, t explicitly for t = /, t =, t = 3/ and t =, and sketch the profiles. (iv (Bonus: 5 points Assume u is a solution of the wave equation with Neumann oundary condition, and on finite intervals [ T, T ], u vanishes for large x. The kinetic energy of the wave at time t is (up to a constant factor K(t = ut (x, t dx, while the potential energy is P (t = c u x (x, t dx. Let E(t = K(t + P (t e the total energy of the wave, and show that de dt =, hence E is constant (independent of t. You may use in any part of the prolem that if v solves v tt c v xx = on R x R t, then v(x, t = v(x ct, + v(x + ct, + c x+ct x ct v t (x, dx. Solution. Since the oundary condition is Neumann, we take even extensions of f and g across x =, i.e. we let f e (x = f(x for x, f e (x = f( x for x <, and similarly for g. Then the solution v of the wave equation on R R with initial data f e, g e, when restricted to x, gives us the solution u of the original prolem, so u(x, t = f e(x ct + f e (x + ct + c x+ct x ct g e (x dx. If x ct, then in all places where f e and g e are evaluated, they are equal to f, resp. g, so u(x, t = f(x ct + f(x + ct + c x+ct x ct g(x dx, x ct. On the other hand, if x < ct, then x ct <, so f e (x ct = f(ct x, and similarly g e (x = g( x when x <. We reak up the g e integral to the intervals [x ct, ] and [, x + ct], and use x ct g(x dx = ct x g(y dy (changing variales y = x to get u(x, t = f(ct x + f(x + ct + c x+ct g(x dx + c ct x For the solution to e time symmetric, we would need u t (x, = since g(x dx, x < ct. u(x, t u(x, t u t (x, = lim = t t in this case, i.e. we need g =. On the other hand, if g =, the solution of the extended prolem is v(x, t = fe(x+ct+fe(x ct, which is indeed time symmetric, for v(x, t = fe(x ct+fe(x+ct = v(x, t. Thus, the time symmetric solutions are exactly those given y g =. In particular, in the example, g =, so the solution is time symmetric, so the value at some t < is the same as at t >. In the example the solution is most easily computed most easily y considering v, and asking whether for a given (x, t, x ct and x + ct lie in the intervals [, ] or [, ], which is easily sketched in R y drawing the forward characteristics from the points,,, on the x-axis. When none of the ackward characteristics from (x, t hits the x-axis in one of these intervals, the solution is, when one does, the solution is, when oth do, the solution is. So at t = /, the solution in on [/, 5/], otherwise, at t =, the solution is on [, ] and [, 3], otherwise, at t = 3/, the solution is on [, /], / on [5/, 7/], otherwise, at t = the solution is on [, ] and [3, 4], otherwise. One should think of the value of the solution at the endpoints of these intervals is undefined, since the solution is not continuous there.

4 4 MATH 3P: PRACTICE FINAL SOLUTIONS DECEMBER, Consider de dt = (u t u tt + c u x u xt dx = (u t c u xx + c u x u xt dx, where the second step used the PDE. Now, if the integrand were the derivative in x of a function g which vanishes for large x, the fundamental theorem of calculus would express it as g(. But x (u t u x = u t u xx + u x u xt, so we can apply this with g = c u t u x, and note that g( = y the Neumann oundary condition. Thus, de dt =, and thus the energy is indeed conserved. Prolem 5. Consider Laplace s equation u = on the annular region A(, with inner and outer radii r = and r =, namely u =, u(r, θ = f(θ, u(r, θ = g(θ, θ [, π], f, g given functions. Recall that u = u rr + r u r + r u θθ in polar coordinates. (i (3 points If f(θ = a and g(θ = are constants, find u, and sketch the graph of u in each of the cases a >, a = and a <. What is the maximum value of u on the annulus? (ii ( points If f(θ = sin(θ and g(θ = sin(3θ, find u. Solution. Recall that, after separation of variales and keeping in mind that in θ the solution needs to e π-periodic, we get that the angular solution is Θ n (θ, a linear comination of cos(nθ and sin(nθ, n integer (and sine can e dropped if n =, while the radial solution is R n (θ, a linear comination of r n and r n if n >, a linear comination of and log r if n =. (These can all quickly e derived if one cannot recall the form of the solution. Thus, the general solution is given y A + C log r + r n (A n cos(nθ + B n sin(nθ + r n (C n cos(nθ + D n sin(nθ. where linear cominations of A n and C n give the cosine terms in the Fourier series of f and g, while those of B n and D n give the sine terms: A = π π f(θ dθ, A + C log = π A n + C n = π π n A n + n C n = π B n + D n = π π n B n + n D n = π π g(θ dθ, f(θ cos(nθ dθ, π g(θ cos(nθ dθ, f(θ sin(nθ dθ, π g(θ sin(nθ dθ. In particular, if oth f and g have, say, a vanishing nth sine Fourier coefficient, then B n and D n oth vanish, etc. Thus, when f, g are constants, all coefficients except A and C vanish, which are A = a, A + C log =, so C = log ( a. Thus, the solution in this case is u = a + ( a log r log. In particular, if a =, the solution is constant. Since u is a harmonic function, its maximum is attained on the oundary of the annulus (recall that this can e seen from u having no strict local maxima in the interior of the annulus, which follows from the Poisson integral formula y placing disks around a point inside the annulus, where it is max(a,, otained at r = if a >, at r = if > a, and throughout the domain if a =. This can also e checked directly from the explicit formula.

5 MATH 3P: PRACTICE FINAL SOLUTIONS DECEMBER, 5 In the second case considered all coefficients ut B, D, B 3 and D 3 are zero. Further, B + D =, 4B + 4 D =, so B = 5, D = 6 5, while B 3 + D 3 =, 8B D 3 =, so B 3 = 8 63, D 3 = 8 63, and thus u = sin(θ 5 ( r + 6r + 8 sin(3θ (r 3 r While the maximum of u is not part of the question, again since u is a harmonic function, its maximum is attained on the oundary of the annulus, where it is, otained where either r = and sin(θ = (so θ = π 4, 5π 4 or r = and sin(3θ = (so θ = π 6, 5π 6, 9π 6. Prolem 6. The purpose of this prolem is to solve u = f on R 3 when f( x vanishes for large x, more precisely to find the decaying solution u of this equation. (i (7 points Fourier transform the PDE to otain the solution u as an inverse Fourier transform. (ii (5 points Rewrite the inverse Fourier transform so that u is a convolution of f with a function which is the inverse Fourier transform of an explicit function. (iii (8 points Calculate the Fourier transform of the function g( x = x y considering g ɛ ( x = e ɛ x x first for ɛ >, computing the Fourier transform of g ɛ, and letting ɛ. (iv (5 points Write u as an explicit convolution without any (inverse Fourier transforms. Recall that the volume integral in R 3 takes the form π π h( x d x = h(r, θ, φ r sin θ dφ dθ dr R 3 in spherical coordinates (r, θ, φ where x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ. Solution. Fourier transform of the PDE gives ξ Fu = Ff, so u = F ( Ff ξ. ( Thus, if we let S( x e the inverse Fourier transform of, then ξ u = (π 3/ R 3 S( x yf( y d y. To compute the Fourier transform of g ɛ at ξ, we use spherical coordinates with axis of rotation given y ξ, so θ is the angle etween x and ξ, and so x ξ = x ξ cos θ. Correspondingly, the Fourier transform in question is (π 3/ times π π e i x ξ g ɛ ( x d x = e ir ξ cos(θ e ɛr r r sin θ dφ dθ dr R 3 = (π π e ir ξ cos(θ e ɛr r sin θ dθ dr r = (π ir ξ e ɛr e ir ξ cos(θ π dr = (π ξ ɛr i ξ (e ir e ir ξ ɛr dr = π ( e ir ξ ɛr i ξ i ξ ɛ eir ξ ɛr i ξ ɛ = π ( i ξ i ξ ɛ i ξ, ɛ so taking the limit as ɛ gives that the Fourier transform of g is (Fg( ξ = (π 3/ 4π ξ.

6 6 MATH 3P: PRACTICE FINAL SOLUTIONS DECEMBER, Correspondigly, as FS( ξ = ξ, we deduce that S( x = (π 3/ 4π x. Thus, the solution of Laplace s equation is u( x = 4π f( y d y. R x y 3 Correspondingly, one calls 4π x y a Green s function for the Laplacian on R3 ; it is the unique decaying Green s function (corresponding to u eing the unique decaying solution of the PDE.