MAT 372 K.T.D.D. Final Sınavın Çözümleri N. Course. Question 1 (Canonical Forms). Consider the second order partial differential equation

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1 OKAN ÜNİVERSİTESİ FEN EDEBİYAT FAKÜTESİ MATEMATİK BÖÜMÜ 1.5. MAT 7 K.T.D.D. Final Sınavın Çözümleri N. Course Question 1 (Canonical Forms). Consider the second order partial differential equation (sin x)u xx + (sin x)u xy + ( x)u yy = x. (1) (a) [1p] Calculate the discriminant (x, y) of (1). = B AC = (b) [p] Equation (1) is a hyperbolic PDE; parabolic PDE; elliptic PDE. (c) [p] Find the characteristic equation of (1). dy dx = B ± = cot x A (d) [5p] Find the characteristic curve(s) of (1). [HINT: sec z dz = log sec z + tan z + c, cot z dz = log sin z + c and ec z dz = log ec z + cot z + c] y = log sin x + c (sin x)u xx + +(sin x)u xy + ( x)u yy = x (1) (e) [15p] Find a canonical form for (1). [HINT: x and y MUST NOT appear in your answer, I only want to see u, ξ and η.] [HINT: z = 1 sin z.] etting ξ = y log sin x and η = y, we have ξ x = cot x ξ y = 1 ξ xx = 1 sin x ξ xy = ξ yy = η x = ξ y = 1 ξ xx = ξ xy = ξ yy = 1

2 Therefore Therefore A = B = C = + + x 1 = x ( ) 1 D = sin x sin = 1 x E = = F = G = x xu ηη + u ξ = x. Finally, we calculate that log sin x = y ξ = η ξ. Therefore sin x = e η ξ = x = 1 sin x = 1 e (η ξ) and x = sin 1 e η ξ. Hence, the answer 6 is ( 1 e (η ξ)) u ηη + u ξ = arcsin e η ξ. Question (Ortogonality). (a) [p] et f, g : [, β] R be piecewise continuous functions. Give the definition of the inner product of f and g on [, β]. [HINT: Repeating students should assume that the weighting function is w(x) 1 in other words; give the definition of the -inner product on [, β]. rd year students can ignore this comment.] f, g = β f(x)g(x) dx. (b) [6p] Show that the inner product satisfies the following conditions for all piecewise continuous functions f, g, h : [, β] R and for all λ, µ R: (a) f, f ; (b) f, g = g, f ; (c) λf + µg, h = λ f, h + µ g, h ; and (d) f, λg + µh = λ f, g + µ f, h. (a) Clearly f, f = β (f(x)) dx. (b) That f, g = β f(x)g(x) dx = β g(x)f(x) dx = g, f is trivial. 1 (c) λf + µg, h = β (λf(x) + µg(x))h(x) dx = λ β f(x)h(x) dx + µ β g(x)h(x) dx = λ f, h + µ g, h. (d) Follows immediately from (ii) and (iii). 1 (c) [p] Give the definition of an orthogonal system of functions on [, β]. The set of functions φ n } is called an orthogonal system on [, β] iff φ n, φ m = for all n m. et > and define φ n : [, ] R by φ n := nπx. ()

3 MAT 7 K.T.D.D. Final Sınavın Çözümleri (d) [1p] Show that φ n } n=1 is an orthogonal system on [, ]. [HINT: (A + B) + (A B) =? and (A + B) (A B) =?] For n m, we calculate that φ n, φ m = = = 1 [ =. nπx mπx dx 1 (n + m)πx + 1 (n m)πx dx (n + m)πx (n m)πx sin + sin (n + m)π (n m)π ] Question (Integration of Fourier Series). et > and define f : [, ] R by f(x) 1. (a) [1p] Calculate the Fourier Sine Series of f on [, ]. [HINT: Fourier Sine Series means that you should have a n = for all n =, 1,,,...] [HINT: The formulae on page are for a function g : [, ] R, so don t just copy them blindly: Do you remember how we changed the formulae to calculate the Fourier Sine/Cosine series of a function g : [, ] R?] b k = f(x) sin kπx dx = sin kπx dx = = ( 1 ( 1) k ) 5 kπ Therefore 1 ( ) 1 ( 1) k sin kπx π k = π k=1 [ ] kπx kπ 1 kπx sin k 5 (b) [1p] By integrating your answer to (a), show that x 1 kπx π k π + ( 1 π π π + 1 ) [HINT: x 1 dt = [t]x = x] x x π = π = π = π 1 kπt sin k dt k π kπt x kπx k π + π 1 kπx k π + π k π ( 1 π π + 1 ). (c) [5p] Given that the Fourier Cosine Series for x is where a = x dx, show that x a + a n nπx n= = π 8. ()

4 And an easy problem to finish with: Since a = that = a = π Rearranging gives the required formula. 1 x dx = ( 1 π π + 1 [ 1 x] ). = 1, we can see Question (Separation of Variables). Consider the heat equation on a rod of length : u t = ku xx u x (, t) = < x <, < t u(, t) = u(x, ) = h(x). () (a) [5p] If u(x, t) = X(x)T (t), show that X and T satisfy for some constant λ R. X + λx = and T + kλt = = T (t) kt (t) Since XT u t = ku xx = kx T, we have that X (x) X(x). The left-hand side is a function only of x; the right-hand side is a function only of t. Therefore both sides must be equal to a constant; equal to λ say T kt = λ = T + kλt = 1. (b) [p] What boundary conditions does X satisfy?. Then X X = λ = X + λx = and First note that = u x (, t) = X ()T (t) and = u, t) = X()T (t). Since we don t want T (t) = t, we must have... optional X () = 1.5 X() =. 1.5 (c) [1p] By considering the cases λ <, λ = and λ > separately, find all the eigenvalues and eigenfunctions of X + λx = subject to the boundary conditions that you wrote in part (b). CASE 1: λ <. The solution of X + λx = is X(x) = Ae λx + Be λx. Then = X () = A λe B λe = A = B; and = X() = A(e λ + e λ ) = A = = B =. There are no eigenvalues and no non-trivial eigenfunctions in this case. CASE : λ =. The solution of X = is X(x) = Ax + B. Then = X () = A and = X() = A + B = A = = B. There are no eigenvalues and no non-trivial eigenfunctions in this case. CASE : λ >. The solution of X + λx = is X(x) = A λx + B sin λx. So = X () = A λ sin λ + B λ λ = B λ = B = ; and = X() = A λ. Since we don t want A =, we must have that λ =. So λ = (n 1 )π, n = 1,,,.... So ( ) (n 1 λ n = )π are eigenvalues, with eigenfunctions Xn (x) = (n 1 )πx. 6

5 MAT 7 K.T.D.D. Final Sınavın Çözümleri (d) [5p] Find the general solution of u t = ku xx < x <, < t u x (, t) = u(, t) =. The solution of T n + kλ n T n = is T n (t) = a n e kλnt u(x, t) = n=1. So ( ) a n e k (n 1 )π t (n 1 )πx for some constants a n. Question 5 (Fourier Transforms). [5p] Use the Fourier Transform to solve u t = ku xxx, < x <, < t <, u(x, ) = f(x). (5) [HINT: You may give your answer as a double integral, or as a convolution of functions.] Taking Fourier Transforms, the problem becomes U t = (iω) ku = ikω U U(ω, ) = F (x). 5 which has solution U(ω, t) = F (ω)e ikωt. 5 If we define G(ω) = e ikωt, then we have U(ω, t) = F (ω)g(ω, t). So u(x, t) = g(x, t) f(x) 5 where g(x, t) = F 1 [G](x, t) = e ikωt e iωx dω = e i(kω t ωx) dω. 5 Therefore u(x, t) = g(x, t) f(x) = 1 f(ξ)g(x ξ, t) dξ. 5 π 5