MATH 173: PRACTICE MIDTERM SOLUTIONS

Transcription

1 MATH 73: PACTICE MIDTEM SOLUTIONS This is a closed book, closed notes, no electronic devices exam. There are 5 problems. Solve all of them. Write your solutions to problems and in blue book #, and your solutions to problems 3, 4 and 5 in blue book #. Within each book, you may solve the problems in any order. Total score: points. Problem. ( points) Solve the PDE e y u x + u y = u, u(x, ) = x, for small y (i.e. in a neighborhood of the x axis). Solution. The characteristic ODEs are dx ds = ey, x(r, ) = r, dy =, y(r, ) =, ds dz ds = z, z(r, ) = r. From the ODE for y, y(r, s) = s, so the x ODE becomes dx ds = es, hence x(r, s) = e s + r. Finally the z ODE gives (for z ) z s = s, so r z(r, s) = s, and thus z = r s = r rs. Since s = y and r = x e s + = x e y +, x e y + u(x, y) = y(x e y + ). Problem. Consider the equation u t + u u x =, u(x, ) = φ(x), t. (i) (5 points) State the definition of u being a weak solution of this PDE. (ii) (9 points) Suppose c, and u(x, t) = if x < ct and u(x, t) = if x > ct. For what value of c does this give a weak solution of the PDE? Derive this directly from the definition. (iii) (6 points) Suppose the initial condition is φ(x) =, x <, x, < x <,, x >. Will the solution develop a shock in t >? Why/why not? Solution. We break down the solutions by parts. (i) First note the PDE corresponds to a conservation law with f(u) = u 3 /3. We say u is a weak solution of the PDE if (uψ t + f(u)ψ x ) dxdt + φ(x)ψ(x, ) dx =, x [, ) t

2 MATH 73: PACTICE MIDTEM SOLUTIONS for all ψ C ( ). (ii) We proceed in the same way as in Problem, Homework 3. Define ξ(t) = ct and Ω = {(x, t) : x < ξ(t)}, Ω + = {(x, t) : x > ξ(t)}, Γ = {(x, t) : x = ξ(t)}. By construction the restriction to u to Ω ± is constant. In particular { f(u) = 3 if (x, t) Ω, if (x, t) Ω +. Let φ C ( x (, ) t ). If u is a weak solution then = (uφ t + f(u)φ x ) dxdt ( = φ t + ) Ω 3 φ x dxdt = φν t ds + φν x ds Γ 3 Γ ( = ν t + ) 3 ν x φ ds, Γ where ν = (ν x, ν t ) is the outward-pointing unit normal vector to Ω. We can parameterize Γ using the function γ(s) = (cs, s). From this it follows that γ(s) = (c, ) and ν = (, c). Therefore ( = c + ) φ(cs, s) + c 3 ds ( = c + 3) + c φ(cs, s) ds, for any such φ. This is only possible if c = /3. Note this coincides with the ankine-hugoniot condition. (iii) This is a quasilinear equation for which we impose initial conditions on the surface Γ = {(r, ) : r }, and we parameterize it as Γ(r) = (r, ). We obtain the system of ODEs ṫ r (s) =, t r () =, ẋ r (s) = z r (s), x r () = r, ż r (s) =, z r () = φ(r), from where it readily follows that t r (s) = s, z r (s) = φ(r), and thus x r (s) = sφ(r) + r. Consequently x r (t) = tφ(r) + r. We know the solution will develop a shock if the characteristics intersect. Let r, r and r r. Without loss we can assume r < r and x r (t) x r (t) = r r + t(φ(r ) φ(r ) ) = (r r ) ( + t φ(r ) φ(r ) ). r r However, note that φ(r ) φ(r ) r r = r r r r φ (r)φ(r) dr,

3 MATH 73: PACTICE MIDTEM SOLUTIONS 3 5 t -5 5 x Figure. Characteristics for the ODE in Problem. as the integrand is non-negative. Therefore the characteristics do not intersect for any t > (see also Fig. for a sketch of the characteristics). Alternatively, using x r (s) = sφ(r) + r and t r (s) = s we see that [ ] [ ] r t det r (s) r x r (s) + sφ = det (r)φ(r) s t r (s) s x r (s) φ(r) = + sφ (r)φ(r), if s > as φ (r)φ(r). Problem 3. (i) ( points) Suppose that u D ( x,y). State the definition of u x, and show that this is consistent with the standard definition of partial derivatives if u is given by some f C ( ) (i.e. if u = ι f ). (ii) ( points) Suppose that u is given by a piecewise continuous function f on y, i.e. u(φ) = f(y)φ(x, y) dx dy for φ C c ( ). Show that u x =. Solution. The definition of u x is x u(φ) = u( x φ), φ C c ( ). If u = ι f then u( x φ) = ι f ( x φ) = f x φ dx dy = x fφ dx dy = ι xf (φ), so x ι f = ι xf indeed. In this case we have x u(φ) = u( x φ) = f(y) x φ(x, y) dx dy ( ) = f(y) x φ(x, y) dx dy = by the fundamental theorem of calculus, using that φ has compact support. Problem 4. (i) ( points) Find the general C solution of the PDE u xx 4u xt + 3u tt =. (ii) ( points) Solve the initial value problem with initial condition with φ, ψ given. u(x, x) = φ(x), u t (x, x) = ψ(x),

4 4 MATH 73: PACTICE MIDTEM SOLUTIONS Solution. We factor the operator as x 4 x t + 3 t = ( x t )( x 3 t ). The characteristics of x t are x + t constant, while those of x 3 t are 3x + t constant, so u(x, t) = f(x + t) + g(3x + t) certainly solves the PDE. That this is the general solution, would follow by a change of coordinates reducing it to the wave equation, in which case we have already shown the analogous formula. It can be also shown by solving the characteristic ODEs as in the practice exam solutions. We proceed to solve the initial value problem by substituting in the initial conditions: φ(x) = u(x, x) = f(3x) + g(5x), ψ(x) = u t (x, x) = f (3x) + g (5x). The first equation gives φ (x) = 3f (3x) + 5g (5x), so f (3x) = 5ψ(x) φ (x), g (5x) = φ (x) 3ψ(x) and thus Integrating yields f(s) = 3 φ(s/3) + 5 g(s) = 5 φ(s/5) 3 g (s) = φ (s/5) 3 ψ(s/5). s s f (s) = 5 ψ(s/3) φ (s/3) ψ(σ/3) dσ + A = 3 5 φ(s/3) + ψ(σ/5) dσ + B = 5 5 φ(s/5) s/3 s/5 Substituting into φ(x) = f(3x) + g(5x) gives B = A. Thus, u(x, t) = 3 φ((x + t)/3) φ((3x + t)/5) + (x+t)/3 (3x+t)/5 ψ(σ) dσ + A, ψ(σ) dσ + B. ψ(σ) dσ. Problem 5. ( points) Consider the damped wave equation on x t : u tt + a(x)u t = (c(x) u x ) x, where a, c > depending on x only, c a C, a a C function, and there are constants c, c > such that c c(x) c for all x. Suppose that u(x, ) and u t (x, ) vanish for x. Let E(t) = (u t (x, t) + c(x) u x (x, t) ) dx. Show that E is a decreasing (i.e. non-increasing) function of t, and use this to show that the solution of the damped wave equation with compactly supported initial data u(x, ) = φ(x), u t (x, ) = ψ(x), is unique. You may use that the damped wave equation also has finite propagation speed c. (This would be proved as in your homework problem.)

5 MATH 73: PACTICE MIDTEM SOLUTIONS 5 Solution. Because of the finite speed of propagation, all integrals considered below converge, and the last boundary term vanishes. We now compute de dt = (u t u tt + c u x u xt ) dx = (u t ( au t + (c u x ) x ) + c u x u xt ) dx = au t dx + (u t c u x ) x dx u t c u x + x= =, so E is indeed decreasing. Here we used that au t dx as a. Uniqueness for the initial value problem follows by considering the difference w = u () u () of two solutions solving the initial value problem with initial data (φ, ψ): indeed, w also solves the damped wave equation then (by linearity), with initial data, so the energy of w satisfies E() =, and thus E(t) E() = for all t >, so E(t) is identically. This gives w t (and w x ); the fundamental theorem of calculus using w(x, ) = then gives w, i.e. u () u ().