1 x if 0 < x < 2, 1 x if 2 < x < 0.

Transcription

1 Problem The function f(x) = x, defined on the interval [0, 2], is to be extended to an odd function g with period 4. Sketch the graph of the function g on the interval [ 4, 4] and find the Fourier series of g. Possible Solution First we extend the function f to an odd function on the interval [ 2, 2] by setting g(x) := f( x) for 2 < x < 0. This means that g(x) = f( x) = ( ( x)) = x for 2 < x < 0. we have { x if 0 < x < 2, g(x) = x if 2 < x < 0. Next,we consider the 4-periodic extension of this function. In particular, we have for 2 < x < 4 that g(x) = g(x 4) = (x 4) = 3 x, and for 4 < x < 2 we have g(x) = g(x + 4) = (x + 4) = 3 x. the function g is defined on the interval where we need to sketch it as The sketch we obtain is: 3 x if 4 < x < 2, x if 2 < x < 0, g(x) = x if 0 < x < 2, 3 x if 2 < x < 4. y x One might note here that the function g is not only 4-periodic, but also 2-periodic. (Rather: Its fundamental period is 2). Because the function g is odd, the Fourier coefficients a n, n = 0,,..., are

2 all zero. Moreover, we obtain for the Fourier coefficients b n the expression b n = 2 2 = ( x) sin nπx 2 dx sin nπx 2 dx 2 = 2 (cos nπ ) nπ 0 x sin nπx 2 dx [ n 2 π 2 4 = 2 4 ( cos nπ) + cos nπ nπ nπ = 2 ( cos nπ + 2 cos nπ) nπ = 2 nπ ( + ( )n ) { 0 when n is odd = 4 nπ when n is even. the Fourier series we obtain is or simplified slightly, n= n= 4 2nπ sin nπx 2 2x nπx cos nπ 2 2nπx sin, 2 sin nπx. nπ ] x=2 (Note that it is also possible (and perfectly fine) to regard the function g as a 2-periodic function and compute the Fourier coefficients on the interval [, ] instead. This changes the integration interval to [0, ] and the factor in front of the integral to 2. The final series remains unchanged.) Problem 2 The 2π-periodic function f defined as f(x) = x 2 for π < x < π has the Fourier coefficients a 0 = π2 3, a n = ( ) n 4, n2 n =, 2,..., b n = 0, n =, 2,.... Use this information in order to compute the sum of the series Hint: Parseval s identity! n= n 4. x=0 2

3 Possible Solution We use Parseval s identity, which states that π f(x) 2 dx = 2a 2 ( 0 + a 2 π n + b 2 ) n. π n= In this case, this means that 2 π n 4 = π x 4 dx = 2π4 π π 5. Problem 3 n= n= n 4 = ( 2π 4 ) 6 5 2π4 = π Use the Laplace transform for solving the integro-differential equation with initial values Possible Solution y + y + y = t 2 t 0 y(τ)e t τ dτ y(0) =, y (0) = 0. We note that the integral on the right hand side of the equation is actually the convolution of the functions y and e t. Application of the Laplace transform to the equation thus yields s 2 Y sy(0) y (0) + sy y(0) + Y = 2 s 3 Y L(et ). Using that y(0) =, y (0) = 0, and L(e t ) = s, we obtain the equation s 2 Y s + sy + Y = 2 s 3 Y s. ( s 2 + s + + This can be simplified to From this we obtain that or Y = s ) Y = s s 3. s 3 s Y = s s 3. (s )(s + ) 2(s ) s 3 + s 6 Y = s s s 5 2 s 6. y(t) = t2 2 + t4 2 t

4 Problem 4 Perform three steps of fixed point iteration with starting value x 0 = 0 for solving the equation Show that the iteration converges. Possible solution x = + arctan x. 2 Let g(x) = + 2 arctan x. Then fixed point iteration is defined as x n+ = g(x n ). With x 0 = 0 we obtain We have and thus x =, x , x g (x) = 2( + x 2 ) 0 < g (x) 2 < for all x R. fixed point iteration converges for all starting values. Remark: It is not enough to show only that g (x) < for all x and by far not enough to show that g (x 0 ) <. The condition that there is a constant K < (here: K = /2) such that g (x) K for all x is crucial. However, the condition g (x) < for all x is enough to guarantee convergence, if one has already shown by other means (e.g. the mean value theorem) that the function g actually has a fixed point., an alternative solution to the problem would be: The function g(x) satisfies π/4 < g(x) < + π/4 for all x. Since g is continuous, there exists some π/4 < x < + π/4 such that g(x) = x (mean value theorem). g (x) = /(2( + x 2 )) < for all x, and thus the fixed point iteration converges. Problem 5 Use the trapezoid method with step length h = 0.4 for the computation of an approximation T of the integral I = Find an upper bound for the error I T. e x dx. 4

5 Possible solution In this case, the trapezoid rule yields T = 0.4 [ 2 e + e e e e e ] The error can be estimated by T I max x f (x) with f(x) = e x and thus f (x) = e x, implying that max f (x) = e. x T I e Problem 6a Given the equation 2 u t 2 2 u t + u = 2 u, 0 < x < π, t > 0, x2 find all solutions of the form u(x, t) = F (x)g(t) that satisfy the boundary conditions Possible solution u x (0, t) = 0 and u (π, t) = 0, t > 0. x Inserting F (x)g(t) into the PDE, we obtain which we can write as Dividing by F and G this becomes F G 2F Ġ + F G = F G, F ( G 2Ġ + G) = F G. G 2Ġ + G G = F F, which necessarily has to be a constant, say k R. we obtain the two ODEs F = kf, G 2Ġ + G = kg. Moreover, the boundary conditions imply that We now have three possibilities: F (0) = 0 and F (π) = 0. 5

6 k = p 2 > 0 with p > 0: Here The condition F (0) = 0 gives and thus F (x) = Ae px + Be px. 0 = pa pb A = B. Now the condition F (π) = 0 implies that or pae pπ pae pπ = 0 pa(e pπ e pπ ) = 0. Since e pπ > and e pπ <, this implies that we only have trivial solutions. k = 0: Here A = 0. F (x) = A + Bx. Now the condition F = 0 implies that B = 0. However, the function F (x) = A satisfies both boundary conditions, and thus we have found a non-trivial solution. Now we solve the equation for G, which, for k = 0 gives G 2Ġ + G = 0. The solutions of the equation are of the form we obtain the solutions k = p 2 < 0: Here G(t) = Ce t + Dte t. F (x)g(t) = Ce t + Dte t. F (x) = A cos(px) + B sin(px). The condition F (0) = 0 implies that 0 = F (0) = pa sin(0) + Bp cos(0) = Bp. B = 0 and F (x) = A cos(px). 6

7 Now the condition F (π) = 0 implies that Ap sin(pπ) = 0, and thus we obtain the non-trivial solutions F (x) = A cos(px) for p =, 2,.... Now we compute the solution of the ODE for G, which, with k = p 2 reads as G 2Ġ + G = p2 G. This equation has the solutions G(t) = e t( C cos(pt) + D sin(pt) ). we obtain the solutions (ignoring the unnecessary constant A) All in all, we have found the solutions and F (x)g(t) = e t( C cos(pt) + D sin(pt) ) cos(px). F (x)g(t) = Ce t + Dte t F (x)g(t) = e t( C cos(pt) + D sin(pt) ) cos(px) with p =, 2,.... Remark: It is perfectly fine and viable to solve the ODE for G using the Laplace transform: Denoting Y = L(G), the equation with p > 0 transforms to G 2Ġ + ( + p2 )G = 0 s 2 Y sg(0) Ġ(0) 2sY + 2G(0) + ( + p2 )Y = 0, which can be simplified as or (s 2 2s + ( + p 2 ))Y = sg(0) 2G(0) + Ġ(0) ((s ) 2 + p 2 )Y = sg(0) 2G(0) + Ġ(0). s Y = (s ) 2 + p 2 G(0) + (s ) 2 (Ġ(0) 2G(0)) + p2 s = (s ) 2 + p 2 G(0) + (s ) 2 (Ġ(0) G(0)). + p2 Applying the inverse Laplace transform, we obtain that G(t) = e t t G(0) cos(pt)g(0) + e sin(pt)ġ(0). p 7

8 Problem 6b Find the solution of the problem in item a) that additionally satisfies the initial conditions u(x, 0) = x 2, 0 < x < π, u (x, 0) = 0, 0 < x < π. t (You can use the Fourier series from problem 2.) Possible solution The general solution of the equation has the form u(x, t) = C 0 e t + D 0 te t + e t( C p cos(pt) + D p sin(pt) ) cos(px). p= The initial condition u(x, 0) = x 2 thus implies that x 2 = C 0 + C p cos(px) for 0 < x < π. p= the coefficients C 0 and C p are the Fourier-cosine coefficients of the function x 2. From problem 2 we thus obtain that C 0 = π2 3, C p = ( ) p 4, p =, 2,.... p2 In order to use the second initial condition, we compute u t (x, t) = C 0e t + D 0 (te t + e t ) + + e t( C p cos(pt) + D p sin(pt) ) cos(px) p= e t( pc p sin(pt) + pd p cos(pt) ) cos(px). p= Now the initial condition t u(x, 0) = 0 implies that and we see that 0 = C 0 + D 0 + u(x, t) = π2 3 et π2 3 tet + 4e t (C p + pd p ) cos(px). p= D 0 = C 0 D p = C p, p =, 2,.... p ( ( ) p cos(px) p 2 cos(pt) ) p 3 sin(pt). p= 8

9 Problem 7 We want to find a numerical solution of the partial differential equation with boundary conditions 2 2 u x u = 2y x, 0 < x <, 0 < y <, y2 u(0, y) = 0, u(, y) = + y, 0 y, u(x, 0) = x, u(, x) = 2x, 0 x. Set up a linear system for finding an approximation of the solution u in the points (x i, y i ) = (i h, j h) with h = /3. (It is not necessary to solve the system.) Possible solution The problem we are looking at y u = 2x u = xxu + yy u = 2y x u = + y u = x x The standard discretization of the second derivatives with finite differences gives us the equations 2 h 2 (u i,j 2u i,j + u i+,j ) + h 2 (u i,j 2u i,j + u i,j+ ) = 2y j x i, or, inserting h = /3, y j = j/3, x i = i/3, 8u i,j 54u i,j + 8u i+,j + 9u i,j + 9u i,j+ = 2j 3 i 3. Moreover we have the boundary conditions u,3 = 2 3, u 2,3 = 4 3, u 0,2 = 0, u 3,2 = 5 3, u 0, = 0, u 3, = 4 3, u,0 = 3, u 2,0 =

10 we obtain the following equations: At u : At u 2 : At u 2 : At u 22 : 54u + 8u u 2 = 3. 8u 54u u 22 = 0. 54u 2 + 8u u + 6 =. 8u 2 54u u = 2 3. Or, if we want to write the system in matrix form u u u 2 = u Problem 8 Let y(x) be the function that solves the ordinary differential equation with initial conditions y = cos(πx/2) y 2 y(0) =, y (0) = 0. Rewrite this second order equation as a system of first order differential equations, and use the improved Euler method with step length h = for approximating the value of y(x) in the point x = 2. Possible Solution We rewrite the equation as z = z 2, z (0) =, z 2 = cos(πx/2) z 2, z 2 (0) = 0. Or, z = F (x, z) with ( ) z F (x, z) = 2 cos(πx/2) z 2. Now the iterations for the improved Euler method with h = read as: 0

11 First step: ( ) ( 0 0 k = F (0, (, 0)) = = cos(0) 0) and ( ) 0 k 2 = F (, (, 0) + k ) = F (, (, 0)) = = cos(π/2) ( ) 0. z () = ( + 0) ( ) 2 (k + k 2 ) =. /2 Second step: ( ) /2 k = F (, (, /2)) = = cos(π/2) ( ) /2 and ( k 2 = F (2, (, /2)+( /2, )) = F (2, (/2, 3/2)) = ) 3/2 cos(π) (/2) 2 = ( ) 3/2. 5/4 z (2) = ( ) + /2 2 ( ) /2 + 2 ( ) ( ) 3/2 0 =. 5/4 3/8 The value we are interested in is z (2) 2 = 0. y(2) 0. Problem 9 Perform two iterations of the Gauss Seidel method for solving the linear system Use the starting value x (0) = (0, 0, 0). Possible Solution 4x x 2 x 3 = 4, x + 2x 2 =, x x 2 + 3x 3 = 3. The Gauß Seidel method for this problem reads as x (k+) = + 4 x(k) x(k) 3, x (k+) 2 = 2 2 x(k+), x (k+) 3 = + 3 x(k+) + 3 x(k+) 2.

12 , with x (0) = (0, 0, 0), we obtain x () =, x () 2 = 2 2 =, x () 3 = =, and x (2) = 4 4 = 2, 2 = = 3 4, 3 = = 3 2. x (2) x (2) 2