. 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. 25 4 u ξη =.
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) + 4 5 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. 5 5 3. 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
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.
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 2. 8. 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
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.