x ct x + t , and the characteristics for the associated transport equation would be given by the solution of the ode dx dt = 1 4. ξ = x + t 4.

Transcription

1 . The solution is ( 2 e x+ct + e x ct) + 2c x+ct x ct sin(s)dx ( e x+ct + e x ct) + ( cos(x + ct) + cos(x ct)) 2 2c 2. To solve the PDE u xx 3u xt 4u tt =, you can first fact the differential operat to see ( x 4 ) ( t x + ) u =. t Now you can either change variables accding to the two different directions of characteristics that arise from the two facts of the operat, you can directly find a change of variable so that ξ = x 4 t η = x + t. To be consistant with the way we solved the wave equation in class, I will take the perspective of the characteristics. Thinking of the first fact, it looks like a transpt operat x 4 t, the characteristics f the associated transpt equation would be given by the solution of the ode dx dt = 4. The solution gives the characteristics x = t 4 + C, x + t 4 = C. Thus we could set ξ = x + t 4. Similarly, the other fact gives the transpt operat the solutions of dx dt =. Thus, the characteristics are x t = C, we can take. Thus, And we have η = x t x + t x = ξ ξ x + η η x = ξ + η. t = ξ ξ t + η η t = 4 Subbing into our facts f our PDE operat we have x 4 t = 5 η ξ η., it s characteristics are Thus the PDE transfms to x + t = 5 4 ξ u ξη =.

2 We can divide both sides by 25 4 then integrate with respect to η to get u ξ = f(ξ) f some function f. Following with integration in ξ, we have u = F(ξ) + G(η), where F(ξ) = f(ξ)dξ. Finally, changing variables back to x t, we get the general solution to the iginal PDE is u = F(x + t ) + G(x t). 4 Now if we apply the initial conditions, we have: u(x,) = F(x) + G(x) = φ(x) u t (x,) = 4 F (x) G (x) = ψ(x). Solving f F gives so Now solving f G gives F (x) = 4 5 F(x) = 4 5 φ(x) G (x) = 4 5 G(x) = 5 φ(x) 4 5 ( φ (x) + ψ(x) ) x ψ(x)ds. ( ) 4 φ (x) ψ(x) x ψ(s)ds. Finally, we can obtain the unique solution to the initial value problem ( 4 5 φ(x + t 4 ) + ) φ(x t) + 4 x+ t 4 ψ(s)ds The damped wave equation is given by u tt c 2 u xx + ru t =. Follow the basic outline of the proof of conservation of energy f the wave equation to show that f the damped wave equation, the total energy decreases over time if r >. Solution t 2 ρ(u t) 2 dx = ρu t u tt dx = ρu t (c 2 u xx ru t )dx = rρ (u t ) 2 dx [u t u x ρc 2 = rρ (u t ) 2 dx T 2 t (u x) 2 dx = rρ (u t ) 2 dx t 2 T(u x) 2 dx x t u tx u x dx

3 Thus, we can see that KE t we can rearrange this to see that = rρ (KE + PE) t (u t ) 2 dx PE t = rρ (u t ) 2 dx. Thus, the total energy of the wave is decreasing over time f the damped wave equation. 4. Part (a): Since the minimum value of u(x,) = 4x( x) on the interval [, ] is since the maximum value of u(x,) is (which occurs at x = /2), also since u(, t) = = u(, t), then by the strong maximum minimum principles f the diffusion equation we have that < u(x, t) < f any < x < t >. Part(c): t u 2 dx = 2uu t dx = u(ku xx )dx = kuu x k (u x ) 2 dx = k (u x ) 2 dx < Thus, since the time derivative of the integral is negative, it is a strictly decreasing function over time. (Note we do not get that the final integral above can equal zero because we cannot have u x (x, t) = f all x t, as this would imply that u is constant in x f(t). This in turn would imply, since u t = u xx = that C. The only way u = C is possible is if u = to satisfy the boundary conditions. But this doesn t agree with our initial data.) 5. Part (a): Verifying the solution: u x = 2t 2x, u xx = 2 u t = 2x, thus u t = xu xx. The critical points of u(x, t) occur when both u x = u t =, so that t = x x =, which implies that t = x =. At (, ), the concavity in the x-direction is u xx = 2 in the t-direction is u tt =, so that the only kind of extremum the point can be is a local maximum if it s an extremum at all. We get no extrema on the interi of the region. Checking the boundary pieces, we see that on x = 2, u( 2, t) = 4t 4 which has it s maximum value at t = gives a max value of u( 2, ) =. On x = 2, we have u(2, t) = 4t 4 which has it s maximum value at t =, giving a max value of 4. On t =, u(x,) = x 2 which has it s max value at x = the max value is. Finally, on t =, we see u(x,) = 2x x 2 whose maximum occurs at x = giving a maximum value of u(, ) =. Comparing the values of u over the boundary at the critical point we see that the absolute maximum value of u is it occurs at the point (x, t) = (, ). Note that this point is NOT on x = 2, x = 2 t =, so that the statement of the maximum principle is not true f this equation. Part(b): Again letting v(x, t) = u(x, t) + ǫx 2 differentiating v with the same operat as the PDE, we have: v t xv xx = u t xu xx 2kǫ = 2kǫ Let M be the max value of u on the boundary edges x = 2, x = 2 t =. Then on those same edges, v(x, t) M + ǫl 2. If v has an interi maximum at (x, t ), then v t (x, t ) = v xx (x, t ). Thus, v t xv xx = xv xx, but because x could be positive negative, we cannot say anything about the sign of this expression! Hence we cannot arrive at the same contradiction we saw in the regular diffusion equation case.

4 6. Let w = u v. Then w t kw xx = (u v) t k(u v) xx = w(, t), w(l, t) w(x,). By the (strong) maximum principle applied to w (we can apply it to w because it s a solution of the heat equation), we see that w(x, t) < f all x in (, l) t >. Hence, u v < f all x in (, l) t >, u < v f all x in (, l) t >. 7. e (x y)2 / φ(y)dy = 3 e (x y)2 / dy+ e (x y)2 / dy Again, letting p = x y, we have dp = dy subbing in, we have 3 x/ e p2 dp Rearranging in der to use the err function, this becomes 3 e p2 dp 3 x/ e p2 dp x/ x/ e p2 dp. e p2 dp e p2 dp 3 2 erf( ) 3 2 erf(x/ ) + 2 erf(x/ ) 2 erf(). Since erf( ) =, by symmetry erf() =, we have 2 erf(x/ ). Note that as t, u(x, t) tends to the steady state If u(x,) = e 3x, then e (x y)2 / e 3y dy We can begin to rearrange our terms so that we have a power in the fm of a perfect square within the integr: completing the square gives e x2 / e( x2 +( x 6kt) 2 )/ e9kt+3x e ( 2xy+y2 2kty)/ dy e (( 2x 2kt)y+y2 +( x 6kt) 2 )/ dy e (y x 6kt)2 / dy Now we can make the substitution p = y x 6kt, we have dp = dy, e9kt+3x e 9kt+3x. e p2 dp

5 9. We make the change of variables e bt v(x, t), so the u t = be bt v + e bt v t u xx = e bt v xx. Subbing into the PDE, we have be bt v + e bt v t ke bt v xx + be bt v = e bt (v t kv xx ) = Since e bt is nowhere zero, we can divide by it get v t kv xx =. Thus, since u(x,) = e v(x,) = v(x,) = φ(x) v(x, t) = e bt e (x y)2 / φ(y)dy e (x y)2 / φ(y)dy is the solution to the diffusion equation with constant dissipation.. Now we make the change of variable y = x V t so that we are tracking space by a moving frame of reference (our position y depends on time) keep time measured in the same way so that our new time variable ˆt is ˆt = t. Then u x = u y y x +uˆtˆt x = u y u t = u y y t +uˆtˆt t = V u y +uˆt. Subbing into the PDE, we have uˆt ku yy = Thus, since we also know that when t =, y = x ˆt =, then u(y, ) = φ(y), u(y, ˆt) = 4kˆt e (y s)2 /4kˆt φ(s)ds the solution of the diffusion equation with transpt over the whole real line is: e (x V t s)2 / φ(s)ds.