Math 126 Final Exam Solutions

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1 Math 126 Final Exam Solutions 1. (a) Give an example of a linear homogeneous PE, a linear inhomogeneous PE, and a nonlinear PE. [3 points] Solution. Poisson s equation u = f is linear homogeneous when f and linear inhomogeneous when f. The PE u 2 = f is nonlinear. (b) State in words what it means for a PE to be well-posed. [4 points] Solution. A PE is well-posed if for any reasonable choice of initial and/or boundary data, there exists a unique solution of the PE, and this solution depends continuously on the initial/boundary data. You could also say the solution is stable under perturbations in the initial/boundary data. (c) Give an example of a PE that is not well-posed. [3 point] Solution. The backwards heat equation u t = u xx for t > is not well-posed (or ill-posed). Simpler examples are PE like (u x ) 2 = 1, which clearly does not have a solution. 2. Find the solution u(x, y) of the linear partial differential equation u x + yu y = that satisfies u(, y) = sin(y). Solution. The solution u is constant in the direction (1, y), so along curves y(x) with slope dy dx = y. Hence y(x) = Ce x and u(x, y) = f(ye x ). Since f(y) = u(, y) = sin(y) we find that u(x, y) = sin(ye x ). 3. (a) Write down d Alembert s formula for the solution of the one dimensional wave equation u tt = c 2 u xx with initial conditions u(x, ) = ϕ(x) and u t (x, ) = ψ(x). Solution. u(x, t) = 1 2 (ϕ(x ct) + ϕ(x + ct)) + 1 2c x+ct x ct ψ(s) ds. (b) Solve the wave equation u tt = c 2 u xx with initial conditions u(x, ) = x 2 and u t (x, ) = cos(x). 1

2 Solution. By d Alembert s formula we have u(x, t) = 1 2 ((x ct)2 + (x + ct) 2 ) + 1 2c x+ct x ct cos(s) ds = 1 2 (x2 2xct + c 2 t 2 + x 2 + 2xct + c 2 t 2 ) + 1 2c sin(s) x+ct = x 2 + c 2 t (sin(x + ct) sin(x ct)). 2c x ct 4. Find the solution u(x, t) of the heat equation u t = u xx on the halfline x > satisfying u(x, ) = 1 for x > and u(, t) = for t >. Solution. We use the method of odd reflection and define ϕ(x) = 1 for x > and ϕ(x) = 1 for x. Then u(x, t) = = x x x S(x y, t)ϕ(y) dy S(x y, t)(ϕ(y) + 1) dy S(x y, t) dy 1 S(z, t) dz 1 S(z, t) dz + 2 S(z, t) dz. S(z, t) dz 1 S(x y, t) dy 5. Suppose that u + u = f within a bounded open set R 3, and u = on the boundary. Show that u 2 dxdydz 1 f 2 dxdydz. 2 [1 points] [Hint: Multiply both sides of the PE by u and integrate over. Then apply Green s identity and make use of the inequality ab 1 2 a b2.] Solution. Following the hint we find that u u+u 2 dxdydz = uf dxdydz 1 f 2 dxdydz u 2 dxdydz.

3 Since u = on, it follows from Green s identity that u u dxdydz = u 2 dxdydz. Therefore we have u 2 dxdydz + u 2 dxdydz 1 f 2 dxdydz Simplifying we find that u 2 dxdydz + 1 u 2 dxdydz and the result immediately follows. f 2 dxdydz, u 2 dxdydz. 6. Suppose that u = within a bounded open set R 2. Show that max u = max u. That is, the length of u attains its maximum on. [Hint: Set v = u 2 = u 2 x + u 2 y, compute v, and use the maximum principle.] Solution. We compute and Likewise Therefore v x u x u xx + 2u y u xy, v xx u 2 xx + 2u x u xxx + 2u 2 xy + 2u y u xxy 2(u x u xxx + u y u xxy ). v yy 2(u x u xyy + u y u yyy ) v 2(u x ( u) x + u y ( u) y ) =. Hence v is subharmonic, and by the maximum principle, v attains its maximum on, which completes the proof. 7. A weak solution of Burger s equation has the form u t + uu x = for x R, t > u(x, t) = { x t+1, for x < s(t), for x > s(t), where s(t) is a shock curve (wave) starting at s() = 1. Find a formula for s(t). [Hint: Use the Rankine-Hugoniot condition to find an OE that s(t) satisfies.] 3

4 Solution. The flux for Burger s equation is F (u) = u 2 /2, and the Rankine-Hugoniot condition states that ṡ(t) = F (u l) F (u r ) u l u r = Since x = s(t) along the shock curve, We solve the OE as follows: ṡ(t) = 1 2 x 2 (t+1) 2 x t+1 = s(t) 2(t + 1). d ṡ(t) log(s(t)) = dt s(t) = 1 2(t + 1), x 2(t + 1). hence log(s(t)) = 1 2 log(t + 1) + C = log( t + 1) + C. Therefore s(t) = K t + 1 for K. Since s() = 1 we have K = 1 and s(t) = t Consider the following implicit scheme for the heat equation u n+1 j u n j t = un+1 j+1 2un+1 j + u n+1 j 1 x 2. Let s = t x 2. Prove that the scheme is unconditionally stable (that is, it is stable for all values of s). Solution. We look for a solution of the form u n j = λn k eij xk. Let s = t x 2 (λ n+1 k λ n k )eij xk = sλ n+1 k e ij xk (e i xk 2 + e i xk ). and write Simplifying we see that 1 1 λ k s(cos( xk) 1). Therefore 1 λ k = 1 2s(cos( xk) 1) 1 for all s. Therefore λ k 1, independent of s, which implies that the scheme is unconditionally stable. 9. Let f be a distribution and ψ be an infinitely differentiable function. We define the product ψf to be the distribution (ψf, ϕ) := (f, ψϕ) for all test functions ϕ. 4

5 (a) Show that e x δ(x) = δ(x) as distributions, where δ is the elta function. Solution. By definition (e x δ(x), ϕ) = (δ, e x ϕ) = e ϕ() = ϕ() = (δ, ϕ), for all test functions ϕ. Therefore e x δ(x) = δ(x). (b) Let f be a distribution and ψ an infinitely differentiable function. Show that the product rule (ψf) = ψf + ψ f holds in the distributional sense. Solution. By definition ((ψf), ϕ) = (ψf, ϕ ) = (f, ψϕ ) = (f, (ψϕ) ψ ϕ) for all test functions ϕ. Therefore in the distributional sense. (ψf) = ψf + ψ f = (f, (ψϕ) ) + (f, ψ ϕ) = (f, ψϕ) + (f, ψ ϕ) = (ψf, ϕ) + (ψ f, ϕ) = (ψf + ψ f, ϕ), 1. Let {α n } n=1 denote the positive solutions of the equation α tan(α) = 1. Find the solution u(x, y) of the boundary-value problem u =, < x < 1, y > u x (, y) =, y > u(1, y) + u x (1, y) =, y > u(x, ) = 1, < x < 1 that is bounded as y. [Hint: Use separation of variables and look for a series solution. You do not need to prove convergence of the series.] Solution. Let us look for a separated solution of the form Then Separating variables we have u(x, y) = X(x)Y (y). = u = X (x)y (y) + X(x)Y (y). X (x) X(x) = Y (y) Y (y) = λ 5

6 for some constant λ R. This gives us the pair of OE The boundary conditions give us and Therefore We check that X + λx = and Y λy =. (1) = u x (, y) = X ()Y (y) for all y > = u(1, y) + u x (1, y) = X(1)Y (y) + X (1)Y (y) for all y >. X () = and X(1) + X (1) =. (2) X(x)X (x) 1 = X(1)X (1) X()X () = X(1) 2. Therefore the eigenvalues are all nonnegative. Furthermore, if f and g satisfy the boundary conditions (2) then f (x)g(x) f(x)g (x) 1 = f (1)g(1) f(1)g (1) = f(1)g(1) + f(1)g(1) =. Therefore the boundary conditions are symmetric, which implies that eigenvectors with distinct eigenvalues are orthogonal. We first consider the zero eigenvalue λ =. Then X(x) = ax + b. The boundary conditions imply that a = and a + b + a =, hence b = as well. This gives the trivial eigenfunction X, so we ignore this. For λ > we can write λ = α 2 for some α >. Then the general solution of (1) is X(x) = A cos(αx) + B sin(αx) and Y (y) = Ce αy + e αy. Since we are looking for the bounded solution as y, and α >, we set C = to throw away the unbounded solution. We now need to select α to satisfy the boundary conditions (2). We have Therefore B =. We also have = X () = αa sin() + αb cos() = αb. = X(1) + X (1) = A cos(α) αa sin(α). Therefore cos(α) = α sin(α), or α tan(α) = 1. {α n } n=1. So we have discovered a family of separated solutions of the form u n (x, y) = e αny cos(α n x). Therefore, α is one of the numbers 6

7 We form the series solution u(x, y) = A n e αny cos(α n x). n=1 We need to select the coefficients A n so that the initial condition holds, i.e., 1 = u(x, ) = A n cos(α n x). n=1 Since the eigenfunctions are orthogonal, we can write cos(α n x) dx = A n cos 2 (α n x) dx. Therefore the coefficients of the series are the usual Fourier coefficients A n = cos(α nx) dx cos2 (α n x) dx. It is possible (but not necessary) to simplify the coefficients slightly. 7