MATH 220: MIDTERM OCTOBER 29, 2015

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1 MATH 22: MIDTERM OCTOBER 29, 25 This is a closed book, closed notes, no electronic devices exam. There are 5 problems. Solve Problems -3 and one of Problems 4 and 5. Write your solutions to problems and 2 in blue book #, and your solutions to problems 3, 4 and 5 in blue book #2. Within each book, you may solve the problems in any order. Total score: points. Problem. (2 points) Solve the PDE xu x + u y = u, u(x, ) = e x, on as large of a subset of R 2 containing the x-axis as you can. Do all characteristics intersect the x-axis, and do so in a unique point, non-tangentially? What does this mean for the solutions of the initial value problem of the PDE? Sketch the characteristics. Solution. The PDE is semilinear. The associated vector field in R 2 is V (x, y) = (x, ), and the surface on which we prescribe the function values is Γ = {(r, ) : r R}. It can be readily seen that the vector field is non-characteristic, i.e., at no point (x, y) Γ the vector V (x, y) is tangent to Γ at (x, y). We write Γ(r) = (Γ x (r), Γ y (r)) = (r, ). The equations for the projected characteristics are and ẋ r (s) = x r (s), x r () = r, ẏ r (s) =, y r () =, v r (s) = v r (s), v r () = e r, where v r (s) = u(x r (s), y r (s)). It follows that x r (s) = re s, y r (s) = s, v r (s) = e r e s. To check whether we can perform the change of variables (r, s) (x, y) note that [ ] [ ] r x det r (s) r y r (s) e s = det s x r (s) s y r (s) re s = e s, for any (r, s), which implies the inverse map exists locally. We now address the question whether the characteristics reach every point in the plane. This is equivalent to finding the inverse map explicitly. Fix (x, y). We need to find (r, s) such

2 2 MATH 22: MIDTERM OCTOBER 29, y y x x Figure. Vector field and characteristics for the ODE in Problem. that x = re s, y = s, from where we deduce s = y and it follows that x = re y, r = xe y, and s = y. Since this is valid for any (x, y) R 2 we conclude for every point there is a characteristic curve that goes through it (see Fig. for a sketch of the characteristics). We conclude u(x, y) = e xe y e y. Problem 2. (25 points) Consider the (real-valued) variable coefficient wave equation on [, l] x [, ) t with Dirichlet boundary conditions: u tt = (c(x, t) 2 u x ) x, u(, t) = = u(l, t). where c > is allowed to depend on x and t, and there are constants c, c 2 > such that c c(x) c 2 for all x. Assume throughout that u is C 2. Let E(t) = 2 (u t (x, t) 2 + c(x, t) 2 u x (x, t) 2 ) dx. (i) Show that if c is independent of t then E is constant. (ii) Show that even if c depends on t, but t c C 3, E(t) e αt E() where α = 2C 3 /c. (Hint: consider d dt (e αt E(t)).) (iii) Show the solution of the variable coefficient wave equation (under the conditions mentioned above) with given initial condition is unique. Solution: We break down the solution by parts. (i) First note E is well-defined for t > as the integral is over a bounded domain, and the integrand is C 2. Since the boundary condition implies

3 MATH 22: MIDTERM OCTOBER 29, 25 3 E (t) = d 2 dt u t (, t) = u t (l, t) = for all t >, exchanging the derivative with respect to t and the integral shows that = 2 = = =. (u t (x, t) 2 + c(x) 2 u x (x, t) 2 ) dx d dt (u t(x, t) 2 + c(x) 2 u x (x, t) 2 ) dx (u t (x, t)u tt (x, t) + c(x) 2 u x (x, t)u xt (x, t)) dx (u tt (x, t) (c(x) 2 u x (x, t)) x ) }{{} = as u solves the PDE u t (x, t) dx + c(x) 2 u x (x, t)u t (x, t) x=l x= }{{} = by the boundary conditions (ii) Since c is bounded above, the same arguments as in (i) show E is welldefined. Since the boundary condition implies u t (, t) = u t (l, t) = for all t >, exchanging the derivative with respect to t and the integral shows that E (t) = d 2 dt = 2 (u t (x, t) 2 + c(x, t) 2 u x (x, t) 2 ) dx d dt (u t(x, t) 2 + c(x, t) 2 u x (x, t) 2 ) dx = (u t (x, t)u tt (x, t) + c(x, t) 2 u x (x, t)u xt (x, t)) dx + }{{} = = by the same computation performed in (i) c t (x, t)c(x, t)u x (x, t) 2 dx. Following the hint, define Ẽ(t) = e αt E(t). Then Ẽ (t) = αe αt E(t) + e αt E (t) = αe αt 2 = αe αt 2 c t (x, t)c(x, t)u x (x, t) 2 dx (u t (x, t) 2 + c(x, t) 2 u x (x, t) 2 ) dx + e αt c t (x, t)c(x, t)u x (x, t) 2 dx (u t (x, t) 2 + ( 2α c t (x, t) + c(x, t))c(x, t)u x (x, t) 2 ) dx. Since we want to show Ẽ is decreasing, we need to show the right-hand side is non-positive, i.e., the integral on the right-hand side is non-negative. Using the value of α = 2C 3 /c we can write 2 α c t(x, t) + c(x, t) c(x, t) 2 C 3 α = c(x, t) c, from where we conclude Ẽ(t). In particular this implies Ẽ(t) Ẽ() = E(t) eαt E().

4 4 MATH 22: MIDTERM OCTOBER 29, 25 There is an alternative way to prove the desired result. Both ways are valid, but we include both for completeness. Note that E (t) = C 3 c c t (x, t)c(x, t)u x (x, t) 2 dx 2C 3 c 2 αe(t), c c(x, t)u x (x, t) 2 dx c(x, t) 2 u x (x, t) 2 dx from where it follows that d dt log (E(t)) α = E(t) eαt E(), as E(t). This proves the desired inequality. (iii) If we fix the initial conditions, e.g., u(x, ) = φ(x) and u t (x, ) = ψ(x) for x (, l), then if u and u 2 are two solutions to the PDE, then w = u u 2 is a solution to the homogeneous problem w tt = (c(x, t) 2 w x ) x, w(, t) = w(l, t) =, w(x, ) = w t (x, ) =. In this case E() = 2 (w t (x, ) 2 + c(x, t) 2 w x (x, ) 2 ) dx =, and by (ii) we conclude E(t). Since E(t) it follows that E(t) for t and we have = E(t) = 2 2 (w t (x, t) 2 + c(x, t) 2 w x (x, t) 2 ) dx w t (x, t) 2 dx + 2 c2 w x (x, t) 2 dx. which implies w t (x, t) = w x (x, t) = for (x, t) [, l] x [, ) t as c >. Therefore w is a constant which has to be equal to zero by either the initial or boundary conditions. We conclude u = u 2 and the solution to the PDE with given initial conditions is unique. Continue to the next page!

5 MATH 22: MIDTERM OCTOBER 29, 25 5 Problem 3. (i) (8 points) Find the general C 2 solution of the PDE u xx + 5u xt + 6u tt =. (ii) (8 points) State what it means that a distribution u solves this PDE, and give an example of a discontinuous solution. Justify your answer. (Hint: if u j are distributions solving the PDE, and u j u in distributions, what can you say about u?) (iii) (8 points) Solve the initial value problem with initial condition u(x, x) = φ(x), u t (x, x) = ψ(x), with φ, ψ given C 2 functions. (iv) (6 points) If φ and ψ are C except at, where can you say for sure that u is C? Explain the geometric meaning. Solution: We break down the solution by parts. (i) We first determine the type of second-order operator. Note that [ ] 5/2 det = /2 6 4 = 4 <, and the PDE is hyperbolic. We can factor it as the product of two firstorder operators as (a x + b t )(c x + d t ) = ac xx + (ad + bc) xt + bd tt, from where it follows that ac =, ad + bc = 5, bd = 6. Choosing a = we see that c =, d + b = 5, and bd = 6. We conclude b = 2 and d = 3 and thus xx + 5 xt + 6 tt = ( x + 2 t )( x + 3 t ). As we have seen throughout the class, the general form of the C 2 solution is u(x, t) = f(2x t) + g(3x t), for some C 2 functions f and g. (ii) If u D (R 2 ) solves the PDE then φ C c (R 2 ) : ( xx u + 5 xt u + 6 tt u)(φ) =. Using the fact that D (R 2 ) is a linear space, the definition of the distributional derivative, and the fact that distributions are linear maps, we have that ( xx u + 5 xt u + 6 tt u)(φ) = xx u(φ) + 5 xt u(φ) + 6 tt u(φ) = u( xx φ) + 5u( xt φ) + 6u( tt φ) = u( xx φ + 5 xt φ + 6 tt φ). We conclude u D (R 2 ) solves the PDE if φ C c (R 2 ) : u( xx φ + 5 xt φ + 6 tt φ) =,

6 6 MATH 22: MIDTERM OCTOBER 29, 25 where the second-order operator acting on the argument of u is the same as in the original PDE. We claim h(x, t) = H(2x t) is a discontinuous solution to the PDE in the sense of distributions. To justify this is the case, consider a sequence {f k } in C 2 (R) with f k H pointwise, and satisfying f k, f k (x) = for x <, and f k (x) = for x > /k, for all k. Define h k (x, t) = f k (2x t). Clearly ι hk ι h in the sense of distributions. In fact, for π Cc (R 2 ) we have ι hk (φ) ι h (φ) f k (2x t) H(2x t) φ(x, t) dxdt = f k (y) H(y) φ(x, 2x y) dxdy = 2, /k /k ( f k (y) H(y) φ(x, 2x y) dx ) φ(x, 2x y) dx dy as k. Furthermore, by definition of the distributional derivative (in particular the fact it is consistent) we have ( xx + 5 xt + 6 tt )ι hk = ( x + 3 t )( x + 2 t )ι hk = ( x + 3 t )ι ( x+2 t)h k = ( x + 3 t )ι =, and ι hk solves the PDE in the sense of distributions for any k. Therefore for any fixed φ Cc (R 2 ) we have = lim ( xx + 5 xt + 6 tt )ι hk (φ) k = lim ι h k (( xx + 5 xt + 6 tt )φ) k = ι h (( xx + 5 xt + 6 tt )φ) = ( xx + 5 xt + 6 tt )ι h (φ), which justifies why h(x, t) = H(2x t) is a solution in the sense of distributions. (iii) Since the general solution to the PDE is u(x, t) = f(2x t) + g(3x t), we only need to find f, g C 2 (R) such that u satisfies the desired conditions. Since u t (x, t) = f (2x t) g (3x t), we obtain the system of equations φ(x) = f(x) + g(2x), ψ(x) = f (x) g (2x),

7 MATH 22: MIDTERM OCTOBER 29, 25 7 and by differentiating the first equation we obtain the system φ (x) = f (x) + 2g (2x), ψ(x) = f (x) g (2x). Adding both equations leads to φ (x) + ψ(x) = g (2x) from where it follows that x ( ) x ( ) ( ) g(x) g() = φ 2 y dy+ ψ 2 y dy = 2φ 2 x 2 x 2φ()+2 ψ(y) dy. Adding two times the second to the first leads to φ (x)+2ψ(x) = f (x) and f(x) f() = x Consequently φ (y) dy 2 x u(x, t) = f()+g() φ() φ(2x t)+2φ ψ(y) dy = φ(x) + φ() 2 ( 3 2 x ) 3 2 t +2 Note that g() + f() = φ() and thus ( 3 u(x, t) = φ(2x t) + 2φ 2 x ) 2 t x 2 t 2 x 2 t 2x t x ψ(y) dy 2 ψ(y) dy. ψ(y) dy. 2x t (iv) If φ, ψ are C except at the origin, then we can say for sure that the solution is C except at points (x, t) for which either 2x t = or 3x t =. Geometrically, this means that singularities propagate along the characteristics of the first-order operators. Solve one of Problems 4 and 5. Problem 4. Consider the equation u t + u 2 u x =, u(x, ) = φ(x), t. (i) (6 points) State the definition of u being a weak solution of this PDE. (ii) ( points) Suppose c R, 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) (9 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 =, R x [, ) t R for all ψ C (R 2 ). ψ(y) dy.

8 8 MATH 22: MIDTERM OCTOBER 29, 25 (ii) We proceed in the same way as in Problem 4, Homework 2. 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 (R 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 2 ds ( = c + 3) + c 2 φ(cs, s) ds, for any such φ. This is only possible if c = /3. Note this coincides with the Rankine-Hugoniot condition. (iii) This is a quasilinear equation for which we impose initial conditions on the surface Γ = {(r, ) : 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) 2, 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) 2 + r. Consequently x r (t) = tφ(r) 2 + r. We know the solution will develop a shock if the characteristics intersect. Let r, r 2 R and r r 2. Without loss we can assume r < r 2 and x r2 (t) x r (t) = r 2 r + t(φ(r 2 ) 2 φ(r ) 2 ) = (r 2 r ) ( + t φ(r 2) 2 φ(r ) 2 ). r 2 r

9 MATH 22: MIDTERM OCTOBER 29, t -5 5 x Figure 2. Characteristics for the ODE in Problem 4. However, note that φ(r 2 ) 2 φ(r ) 2 r 2 r = 2 r 2 r r2 r φ (r)φ(r) dr, as the integrand is non-negative. Therefore the characteristics do not intersect for any t > (see also Fig. 2 for a sketch of the characteristics). Alternatively, using x r (s) = sφ(r) 2 + r and t r (s) = s we see that [ ] [ ] r t det r (s) r x r (s) + 2sφ = det (r)φ(r) s t r (s) s x r (s) φ(r) 2 = + 2sφ (r)φ(r), Problem 5. if s > as φ (r)φ(r). (i) (5 points) State the definition of a function f being Schwartz: f S(R n ). (ii) (7 points) Show that for f Schwartz, the Fourier transform of D j f = i jf is ξ j (Ff)(ξ). (iii) (7 points) On R, use the Fourier transform (by taking the F.T. of both sides) to solve the ODE d dx u + u = 2 e x. You may leave your answer as the inverse Fourier transform of a function. You may use that the Fourier transform of f(x) = e ax2, a >, is (π/a) /2 e ξ2 /(4a). (iv) (6 points) Is the solution of the ODE obtained by the Fourier transform unique? How do you reconcile this with the fact that the ODE existence and uniqueness theorem guarantees at least local solvability and uniqueness to the initial value problem, where one also imposes u() = c, with c C arbitrary? (Hint: what are the solutions of the homogeneous ODE?) Solution: We break down the solutions by parts.

10 MATH 22: MIDTERM OCTOBER 29, 25 (i) A function f C (R n ) is in S(R n ) if sup x R n x α β f(x) <, for any multiindices α, β. (ii) Fix f S(R n ) and define g(x) = f(x)e ix ξ. Then D j g(x) = D j f(x)e ix ξ ξ j f(x)e ix ξ, is absolutely integrable. Then = D j g(x) dx = D j f(x)e ix ξ dx ξ j f(x)e ix ξ dx, from where we conclude (FD j f)(ξ) = ξ j (Ff)(ξ). (iii) By taking Fourier transform on both sides we have iξ(fu)(ξ) + (Fu)(ξ) = π /2 e ξ2 /4 (Fu)(ξ) = 2 + iξ e ξ /4, which shows Fu is in fact in S(R). Therefore we can take inverse Fourier transform to obtain u(x) = e ξ 2 /4 2π + iξ eixξ dξ, and conclude that u S(R). (iv) Using the Fourier transform we obtain a unique solution which is also Schwartz. Note this leaves no degrees of freedom to find a solution, not even locally, for a given initial condition. However, note that the solution of the ODE for given initial conditions is the sum of a solution to the homogeneous problem plus a particular solution to the inhomogeneous problem. Note the solutions to the homogeneous problem are du dx + u = u(x) = Ae x, for some A R. It can be seen that u is not Schwartz, as it does not decay as x. Therefore taking Fourier transform allows us only to find a particular solution to the inhomogeneous problem when the inhomogeneous term is a Schwartz function. To understand why this is the case, if we were to solve the homogeneous ODE we would have (iξ + )(Fu)(ξ) =, implying that (Fu)(ξ) = for all ξ.

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