PDEs, Homework #2 Solutions. is even in x, if the initial data φ, ψ are even. Hint: show u( x, t) is also a solution.

Transcription

1 PDEs, Homework # Solutions 1 Show that the solution u(x, t) of the initial value problem u tt = c u xx, u(x, ) = φ(x), u t (x, ) = ψ(x) (WE) is even in x, if the initial data φ, ψ are even Hint: show u( x, t) is also a solution To show that w(x, t) = u( x, t) is also a solution, we note that w tt (x, t) = u tt ( x, t) = c u xx ( x, t) = c w xx (x, t) and that w(x, t) satisfies the initial conditions w(x, ) = u( x, ) = φ( x) = φ(x), w t (x, ) = u t ( x, ) = ψ( x) = ψ(x) Since the initial value problem has a unique solution, this implies u( x, t) Solve the Neumann problem for the wave equation on the half line That is, find the solution to (WE) when x > and the boundary condition u x (, t) = is imposed for all t Hint: argue as for the Dirichlet problem but use an even extension Extend the initial data φ, ψ to the whole real line in such a way that the extension is even Then the solution to the Cauchy problem on the whole real line is φ ext(x ct) + φ ext (x + ct) + 1 c ψ ext (s) ds and we need to express this in terms of φ, ψ When x > ct, we get the usual formula φ(x ct) + φ(x + ct) + 1 c ψ(s) ds because x ± ct are both positive When < x < ct, only x + ct is positive and so φ(ct x) + φ(x + ct) + 1 [ ] ψ( s) ds + ψ(s) ds c Making the substitution r = s in the first integral, we conclude that φ(ct x) + φ(ct + x) + 1 [ ct x ct+x ] ψ(r) dr + ψ(r) dr c

2 3 Find all solutions u = u(x, y) of the second-order equation u xx + 4u xy + 3u yy = First of all, let us factor the given PDE and write = ( x + 4 x y + 3 y)u = ( x + y )( x + 3 y )u If we can find variables v, w such that v = x + y and w = x + 3 y, then = v w u = w u = F 1 (w) = u = F (w) + F 3 (v) To actually find the variables v and w, we note that v = x v x + y v y, w = x w x + y w y by the chain rule, while v = x + y and w = x + 3 y by above This gives x v = y v = x w = 1, y w = 3 so we can let x = v + w and y = v + 3w Solving for v and w, we conclude that y x = w, 3x y = v = u = F (y x) + G(3x y) 4 Show that the solution to the wave equation (WE) need not remain bounded at all times, even though it is initially bounded Hint: take the initial data to be constant Suppose that φ(x) = ψ(x) = 1, for instance Then the corresponding solution is c ds = 1 + t by d Alembert s formula, so the solution does not remain bounded at all times 5 Solve the wave equation (WE) in the case that φ(x) = x and ψ(x) = x + 1 According to d Alembert s formula, the solution is given by (x + ct) + (x ct) + 1 c (s + 1) ds When it comes to the integral on the right hand side, one easily finds that (s + 1) ds = (x + ct) (x ct) + ct = cxt + ct Using this fact and a little bit of algebra, we conclude that x + (ct) + xt + t

3 6 Suppose that a < b and consider the Cauchy problem (WE) in the case that { } 1 if a x b φ(x) = ψ(x) = otherwise Compute the limit lim t u(x, t) for each fixed x R According to d Alembert s formula, the solution is given by φ(x ct) + φ(x + ct) + 1 c ψ(s) ds Since x R is fixed, we have x ± ct ± as t, and this implies lim t 1 c ψ(s) ds = 1 c b a ds = b a c 7 Solve the following non-homogeneous wave equation on the real line: u tt c u xx = t, u(x, ) = x, u t (x, ) = 1 According to Duhamel s formula, the solution is given by (x + ct) + (x ct) + 1 c ds + 1 c t x+c(t τ) x c(t τ) τ dy dτ When it comes to the rightmost integral, one easily finds that 1 c t τ x+c(t τ) x c(t τ) dy dτ = t τ(t τ) dτ = [ τ t τ ] 3 t = t3 3 τ= 6 Using this fact and simplifying the remaining terms, we conclude that x + (ct) + t + t3 6 8 Use the substitution v(x, t) = e λt u(x, t) to solve the initial value problem u tt u xx + λu t + λ u =, u(x, ) = φ(x), u t (x, ) = ψ(x) on the real line Hint: you should find that v satisfies the wave equation v tt = v xx etting v = e λt u, we have v xx = e λt u xx and also v t = λe λt u + e λt u t, hence v tt v xx = (λ e λt u + λe λt u t + e λt u tt ) e λt u xx = e λt (λ u + λu t + u tt u xx )

4 According to the given PDE, this expression is zero and so v(x, t) satisfies v tt = v xx, v(x, ) = φ(x), v t (x, ) = λφ(x) + ψ(x) Solving this problem using d Alembert s formula with c = 1, we now find v(x, t) = φ(x + t) + φ(x t) + 1 x+t x t Thus, the solution to the original problem is given by [ e λt φ(x + t) + φ(x t) + x+t x t λφ(s) + ψ(s) ds ] λφ(s) + ψ(s) ds 9 Consider the wave equation with damping u tt c u xx + du t = on the real line Show that the energy is decreasing for all classical solutions of compact support, if d > Suppose u is a classical solution of compact support and consider the energy E(t) = 1 u t (x, t) + c u x (x, t) dx Since u is C by assumption, the integrand is C 1 and we have E (t) = = u t u tt + c u x u xt dx u t (c u xx du t ) dx + c u x u xt dx Integrating by parts and using the fact that u x vanishes at x = ±, we now get E (t) = = d u t (c u xx du t ) dx u t dx c u xx u t dx 1 Solve the Cauchy problem (WE) on the half line x > when φ(x) = ψ(x) = 1 and the Dirichlet condition u(, t) = is imposed for all t Is your solution a classical one? Hint: there are different formulas for the cases x > ct and x ct When x > ct, the solution is given by d Alembert s formula φ(x + ct) + φ(x ct) + 1 c When < x < ct, on the other hand, it is given by the formula φ(ct + x) φ(ct x) + 1 c ct+x ct x ψ(s) ds = 1 + t ψ(s) ds = x c Since these two expressions do not agree when x = ct, the solution is not continuous along the line x = ct, so it is certainly not a classical solution

5 11 Find the eigenfunctions and eigenvalues of x subject to Neumann boundary conditions on [, ] That is, find all nonzero functions F (x) and all λ R such that F (x) = λf (x), F () = F () = First of all, we multiply the given ODE by F (x) and then integrate to get λ F (x) dx = F (x)f (x) dx = F (x) dx Since the leftmost integral is positive, this implies λ If λ =, then we have F (x) = = F (x) = = F (x) = C If λ = m is positive, on the other hand, then we have F (x) = m F (x) = F (x) = C 1 sin(mx) + C cos(mx) In this case, the boundary condition F () = gives F (x) = mc 1 cos(mx) mc sin(mx) = = mc 1, while the boundary condition F () = gives = F () = mc sin(m) = m = kπ for some integer k In particular, we have m = kπ/ for some integer k, so ( kπx F (x) = C cos(mx) = C cos ), λ = m = ( ) kπ 1 Solve the wave equation u tt = 4u xx on the interval [, π] subject to the conditions u(x, ) = cos x, u t (x, ) = 1, u(, t) = = u(π, t) The solution of the Dirichlet problem for the wave equation u tt = c u xx on [, ] is n=1 ( a n sin nπct + b n cos nπct ) sin nπx, where the coefficients a n, b n are given by the formula a n = nπc sin nπx ψ(x) dx, b n = sin nπx φ(x) dx

6 In this case, we have c = and = π, so the coefficients a n are a n = 1 nπ sin(nx) dx = 1 cos(nπ) n π To compute the coefficients b n, one can integrate by parts to get b n = π sin(nx) cos x dx = n π and then integrate by parts again to arrive at b n = n π [ ] π cos(nx) cos x + n π cos(nx) sin x dx sin(nx) cos x dx This standard argument expresses b n in terms of b n, so it actually gives b n = n π [ ] cos(nπ) n b n = b n = n[cos(nπ) + 1] π(n 1) whenever n 1; the remaining case n = 1 can now be treated directly, as b 1 = π sin x cos x dx = 1 π [ (sin x) ] π =