Some Review Problems

Transcription

1 Some Review Prolems Prolem 1. Show that the functions 1 x rational, f(x) = 1, x irrational and g(x) 3 are linearly independent on the interval (, ). Suppose, on the contrary, that f(x) and g(x) are linearly dependent on (, ). Since neither f nor g is identically equal to zero, this implies that f cg for some constant c 0. For x = 0 the equation f(0) = cg(0) gives us 1 = 3c, so that c = 1/3. On the other hand for x = π (an irrational numer) we have f(π) = cg(π), so that 1 = 3c and c = 1/3. Since 1/3 1/3, we otain a contradiction. Thus the assumption that f(x) and g(x) are linearly dependent on (, ) was false and hence f(x) and g(x) are linearly independent on (, ). Prolem. Find all the separated solutions of the system: u xx + u yy = 0, 0 < x < π, 0 < y < π, u(0, y) = u(π, y) = 0, 0 < y < π u(x, π) = 0, 0 < x < π. We need to look for separated solutions of the form u = X(x)Y (y), where X(x), Y (y) are some functions that are not identically equal to zero. Sustituting u = X(x)Y (y) in u xx + u yy = 0 we get X Y + Y X = 0 and hence X = Y = λ = const. X Y This gives us X + λx = 0 and Y λy = 0. The conditions u(0, y) = u(π, y) = 0 imply X(0) = X(π) = 0. Thus we get an eigenvalue prolem X + λx = 0 X(0) = X(π) = 0. 1

2 The eigenvalues of this prolem are λ n = n for n = 1,, 3,... and the corresponding eigenfunctions are X n = sin nx, n = 1,, 3,.... For each λ n = n we have the equation Y λ n Y = 0 which gives Y n Y = 0. The condition u(x, π) = 0 gives us X(x)Y (π) = 0, Y (π) = 0. The general solution of Y n Y = 0 is Y = A cosh ny + B sinh ny. The condition Y (π) = 0 gives us A cosh πn + B sinh πn = 0, cosh πn so that B = A. Now Y = A cosh ny + B sinh ny takes the form sinh πn cosh πn Y = A cosh ny + B sinh ny = A cosh ny A sinh ny = sinh πn A (sinh πn cosh ny cosh πn sinh ny) = sinh πn A sinh n(π y) sinh πn For the last equality we used the identity sinh(a ) = sinh a cosh cosh a sinh A with a = πn and = ny. Since is a constant not depending on sinh πn y, we can take Y = Y n = sinh n(π y). Thus the separated solutions are: u n (x, y) = X n Y n = sin nx sinh n(π y), n = 1,, 3,.... Prolem 3. Solve the system: y tt = 5y xx, 0 < x < 3, t > 0, y(0, t) = y(3, t) = 0, t > 0 y(x, 0) = 0, 0 < x < 3 y t (x, 0) = 10 sin πx, 0 < x < 3. This is Prolem B (for the wave equation) with L = 3 and a = 5 = 5 and g(x) = 10 sin πx. Therefore it has solution of the form y = B n sin πnat L πnx sin L = B n sin 5πnt 3 sin πnx 3, where B n = L πna n = 3 5πn n and g(x) n sin πnx is the Fourier 3 Sine Series of g(x).

3 We have g(x) = 10 sin π = n sin πnx 3, 0 < x < 3 and hence 6 = 10 and n = 0 for n 6. Thus B 6 = 3 5π6 6 = 30 and B n = 0, n 6. Hence y = 1 sin 10πt sin πx. π 3 30π = 1 π Prolem 4. An oject of mass m = 1kg is attached to a spring with Hooke s constant k = 4N/m and is acted on y a π-periodic force F (t) Newtons where F (t) = 1 for 0 < t < π and F (t) = 1 for π < t < 0. Determine whether or not pure resonance occurs. The displacement function x(t) satisfies the equation mx + kx = F (t), x + k m = F (t) m, x + 4x = F (t) k The natural frequency of the system is ω 0 = =. Let F (t) m n sin πnt = L n sin nt e the Fourier Series of the odd function F (t). Pure resonance occurs if there is n such that ω 0 = πn and L n 0. The condition ω 0 = = πn = n holds for n =. L We have = 1 π π π π F (t) sin t dt = since F (t) is odd π = F (t) sin t dt = sin t dt = π 0 π 0 = 1 π [cos t]π 0 = 0. Since = 0, pure resonance does not occur. Prolem 5. Let f(x) e a 6-periodic function such that f(x) = x x for 3 < x < 3. Let a n, n e the general Fourier series coefficients of f(x). Find a 0 + (a n cos 46πn 3 + n sin 46πn 3 )

4 4 and Answers: We have and a 0 a 0 + (a n cos πn + n sin πn). + (a n cos 46πn 3 a 0 + n sin 46πn 3 ) = 6 + (a n cos πn + n sin πn) = 9. Prolem 6. Find the Fourier Sine Series of the function f(x) defined on the interval 0 < x < 3 as 1 0 < x, f(x) = 5, < x < 3. Prolem 7. Find the general solution of the following equation: y 6y + 9y = x e 3x + cos x. Note: This is a pretty long prolem! The general solution has the form y = y c + y 1 + y where y c is the general solution of the complimentary homogeneous prolem y 6y + 9y = 0, where y 1 is a particular solution of y 6y + 9y = x e 3x and where y is a particular solution of y 6y + 9y = cos x. The equation y 6y +9y = 0 has characteristic equation r 6r+9 = (r 3) = 0 and so y c = c 1 e 3x + c xe 3x. The Method of Undetermined coefficients tells us that we should look for y 1 in the form Therefore y 1 = x (A + Bx + Cx )e 3x = (Ax + Bx 3 + Cx 4 )e 3x. y 1 = (Ax + 3Bx + 4Cx 3 )e 3x + (3Ax + 3Bx 3 + 3Cx 4 )e 3x = = (Ax + (3A + 3B)x + (3B + 4C)x 3 + 3Cx 4 )e 3x

5 and y 1 = [A + (6A + 6B)x + (9B + 1C)x + 1Cx 3 )e 3x + (6Ax + (9A + 9B)x + + (9B + 1C)x 3 + 9Cx 4 ]e 3x = = [A + (1A + 6B)x + (9A + 9B + 1C)x + (9B + 4C)x 3 + 9Cx 4 ]e 3x. Sustituting this data in y 6y + 9y = x e 3x and re-grouping we get [A + (6A + 6B 1A)x + (9A + 9B + 1C 6(3A + 3B) + 9A)x + (9B + 4C 6(3B + 4C) + 9B)x 3 + (9C 18C + 9C)x 4 ]e 3x = x e 3x Hence A = 0, 6A + 6B = 0, 9B + 1C = 1, so that A = B = 0, C = 1/3. Thus y 1 = 1 3 x4 e 3x. By the Method of Undetermined coefficients we should look for y in the form y = E cos x + F sin x. Then y = E sin x + F cos x and y = E cos x F sin x. Sustituting this information in y 6y +9y = cos x we get ( E cos x F sin x) 6( E sin x+f cos x)+9(e cos x+f sin x) = cos x, and ( E 6F + 9E) cos x + ( F + 6E + 9F ) sin x = cos x. Hence 8E 6F = 1, 6E + 8F = 0 and therefore E = /5, F = 3/50 and y = cos x 3 sin x Thus the general solution of the equation y 6y +9y = x e 3x +cos x is y = c 1 e 3x + c xe 3x x4 e 3x + 5 cos x 3 sin x, 50 where c 1, c are aritrary constants. Prolem 8. Let y(x, t) e the solution of the system: y tt = 4y xx, 0 < x < 1, t > 0, y(0, t) = y(1, t) = 0, t > 0 y(x, 0) = x, 0 < x < 1 y t (x, 0) = 0, 0 < x < 1. Using d Alamert s solution find the precise value of y(1/, 10). Then find the formal infinite series solution of this system. 5

6 6 In this prolem a = 4 =, L = 1 and f(x) = x for 0 < x < 1. According to d Alamert s method we have y(x, t) = 1 [F (x + at) + F (x at)] where F (x) = f O (x) is the L-periodic odd extension of f(x). So y(1/, 10) = 1 [F (1/+ 10)+F (1/ 10)] = 1 [F (0.5)+F ( 19.5)]. The function F (x) is -periodic and therefore F (0.5) = F (1/ + 10) = F (1/) = f(1/) = (1/) = 1/4 F ( 19.5) = F (1/ 10) = F (1/) = f(1/) = (1/) = 1/4. Hence y(1/, 10) = 1 [ ] = 1 4. Prolem 9(1). Solve the following Dirichlet Prolem: (*) u xx + u yy = 0, 0 < x < a, 0 < y < u(x, 0) = u(x, ) = u(0, y) = 0, 0 < x < a, 0 < y < u(a, y) = g(y), 0 < y < 1) We first need to look for separated solutions of the form u = X(x)Y (y), satisfying the first two equations of (*), where X(x), Y (y) are some functions that are not identically equal to zero. Sustituting u = X(x)Y (y) in u xx + u yy = 0 we get X Y + Y X = 0 and hence X = Y X Y = λ = const. This gives us X λx = 0 and Y +λy = 0. The conditions u(x, 0) = u(x, ) = 0 imply Y (0) = Y () = 0. Thus we get an eigenvalue prolem Y + λy = 0 Y (0) = Y () = 0. The eigenvalues of this prolem are λ n = π n / for n = 1,, 3,... and the corresponding eigenfunctions are Y n = sin πny, n = 1,, 3,....

7 For each λ n = π n / we have the equation X λ n X = 0 which gives X π n / X = 0. The condition u(0, y) = 0 gives us X(0)Y (y) = 0, so that X(0) = 0. The general solution of X π n / X = 0 is X = A cosh πnx + B sinh πnx. The condition X(0) = 0 gives us A 1B 0 = 0, so that A = 0. Thus X = B sinh πnx and we can take B = 1 giving X n = sinh πnx. Thus the separated solutions are: u n (x, y) = X n Y n = sinh πnx sin πny, n = 1,, 3,.... ) We now look for the formal solution u(x, y) of the Dirichlet prolem (*) in the form ( ) u = c n u n = c n sinh πnx The condition u(a, y) = g(y), 0 < y < yields: c n sinh πna sin πny sin πny. = g(y), 0 < y <. From here we conclude that c n sinh πna = n, where n is the Fourier Sine Series coefficient for g(y), 0 < y <. Therefore the solution of the Dirichlet prolem (*) is given y ( ) where c n = sinh πna 0 g(y) sin πny dy, n = 1,, 3,... 7 Prolem 9(). Solve the following Dirichlet Prolem: (**) u xx + u yy = 0, 0 < x < a, y > 0 u(0, y) = u(a, y) = 0, 0 < x < a, 0 < y < u(x, y) ounded as y, u(x, 0) = f(x), 0 < x < a 1) We first need to look for separated solutions of the form u = X(x)Y (y), satisfying the first three equations of (**), where X(x), Y (y) are some functions that are not identically equal to zero.

8 8 Sustituting u = X(x)Y (y) in u xx + u yy = 0 we get X Y + Y X = 0 and hence X = Y = λ = const. X Y This gives us X + λx = 0 and Y λy = 0. The conditions u(0, y) = u(a, y) = 0 imply X(0) = X(a) = 0. Thus we get an eigenvalue prolem X + λx = 0 X(0) = X(a) = 0. The eigenvalues of this prolem are λ n = π n /a for n = 1,, 3,... and the corresponding eigenfunctions are Y n = sin πnx, n = 1,, 3,.... a For each λ n = π n /a we have the equation Y λ n Y = 0 which gives Y π n /a Y = 0. The general solution of this equation is Y = A exp( πny πny ) + B exp( ). The condition u(x, y) = X(x)Y (y) a a ounded as y implies that Y (y) is ounded as y. This gives us A = 0. Thus Y = B exp( πny ). We can take the constant B a to e B = 1 which yields Y n = exp( πny ). a Thus the separated solutions are: u n (x, y) = X n Y n = exp( πny πnx ) sin, n = 1,, 3,.... a a ) We now look for the formal solution u(x, y) of the Dirichlet prolem (**) in the form ( ) u = c n u n = c n exp( πny πnx ) sin a a. The condition u(x, 0) = f(x), 0 < x < a yields: c n exp( πny πnx ) sin a a = f(x), 0 < x < a. From here we conclude that c n exp( πn0) = c a n = n, where n is the Fourier Sine Series coefficient for f(x), 0 < x < a. Therefore the solution of the Dirichlet prolem (*) is given y ( ) where c n = a f(x) sin πnx dy, n = 1,, 3,... a a 0 Prolem 10. What does it mean for the functions f(x), g(x) to e orthogonal on an interval [a, ]?

9 For each pair of the functions elow determine if they are orthogonal on the interval indicated: (1) the functions f(x) = cos(3πx), g(x) = sin(4πx), on the interval [ π/, π/]; () the functions f(x) = cos(3πx), g(x) = sin(4πx), on the interval [ 4π, 4π]; (3) the function f(x) = sin(x) and the function sin(x), 0 x π, g(x) = 0, π x < 0 on the interval [ π, π]. Prolem 11 Consider the equation y + 9y = x sin(3x). Then according to the Method of Undetermined Coefficients a particular solution y p can e found in the form: (a) y p = x(a sin(3x) + B cos(3x)); () y p = x (A sin(3x) + B cos(3x)); (c) y p = x (Ax + B) sin(3x); (d) None of the aove. Answer: (d) The correct form for y p is y p = x[(ax + B) sin(3x) + (Cx + D) cos(3x)]. 9 Prolem 1 Let f(t) = t + 5 for 0 < t < π and let n e the Fourier Sine Series Coefficients for f(t). Then: (a) We have 1 = n n cos(nt) for each t (, ). () We have 1 = n n cos(nt) for each t (0, π). (c) We have 6 n sin(n). (d) None of the aove. Answer: (d) Note that term-wise differentiation is not applicale here since the odd extension f O (t) of f(t) is not a continuous function on (, ). Prolem 13. Let f(t) = 3t t 4 for 0 < t <. Let a n, n = 0, 1,,..., e the coefficients of the Fourier Cosine Series of f(t). Find the precise values of the following expressions: (1) a n; () a n cos( 5πn); (3) πn a n sin( πn).

10 10 [This was done in class on Wednesday, May 5] Prolem 14. Let f(t) e a 6-periodic function defined as t, if 0 < t < 3, f(t) = 3t, if 3 < t < 6, 10, if t = 0, 3, 6. Let a n, n e the general Fourier series coefficients for f(t). Then: (a) We have n = 0 for all n 1. () We have a n = 0 for all n 0. (c) We have 0 = a 0 + n=0 a n. (d) None of the aove. Answer: (d) Note that for 3 < t < 0 we have 3 < t+6 < 6 and f(t) = f(t+6) = 3(t + 6). Prolem 15. Solve the following heat equation prolem: y t = 5y xx, 0 < x < 3, t > 0, y(0, t) = y(3, t) = 0, t > 0 y(x, 0) = 4, 0 < x < 3 We have k = 5, L = 3. By Theorem 1 in Ch. 9.5 the solution is u(x, t) = n exp( n π kt ) sin πnx L L = n exp( 5n π t ) sin πnx 9 3, where n are the Fourier Sine Series Coefficients of the function f(x) = 4, 0 < x < 3. We have Hence n = sin πnx 3 dx = 8 3 πnx [ cos 3 πn 3 ]3 0 = 8 8 [ cos(πn) + 1] = πn πn [ ( 1)n + 1] = Prolem 16 18, πn n odd, 0, n even u(x, t) = 18 π t πn exp( 5n ) sin πnx 9 3 n odd

11 An oject of mass m = kg is attached to a spring with a spring constant k = 18 N/m and is acted upon y an odd periodic force F (t) Newtons. In each of the following cases determine whether or not pure resonance occurs: (1) The function F (t) has a period of π seconds and is defined as: 8, 0 < t < π, F (t) = 8, π < t < 0. () The function F (t) has a period of seconds and is defined as: 8, 0 < t < 1, F (t) = 8, 1 < t < 0. (3) The function F (t) has a period of π seconds and has the Fourier Series: F (t) = ( 1) n sin nt. n even k The natural frequency of the system is ω 0 = = 18/ = 3. m (1) The function F (t) is odd π-periodic. Therefore its Fourier Series is: F (t) n 4 n sin nt. Since nt = 3t = ω 0 t for n = 3, we need to check if 3 = 0. We have: 3 = 1 π π π π F (t) sin(3t) dt = π π 0 π 0 F (t) sin(3t) dt = 8 sin(3t) dt = 16 3π [ cos(3t)]π 0 = [ cos(3π) + 1] = [ ( 1) + 1] = 3π 3π 3π 0. Thus 3 0 and hence pure resonance does occur. () The function F (t) is odd -periodic. Therefore its Fourier Series is: F (t) n sin(πnt). 11

12 1 Recall that ω 0 = 3. Since π is irrational, we have πn 3 for n = 1,,.... Therefore pure resonance does not occur. (3) The function F (t) has a period of π seconds and has the Fourier Series: F (t) = ( 1) n sin nt. n even Since n = 3 is odd, the coefficient corresponding to sin 3t in the aove decomposition of F (t) is equal to 0. Therefore pure resonance does not occur. Some Trigonometric and hyperolic trigonometric formulas n 4 sin A cos B = 1 [sin(a + B) + sin(a B)] cos A cos B = 1 [cos(a + B) + cos(a B)] sin A sin B = 1 [cos(a B) cos(a + B)] sin A = 1 cos A, cos A = 1 + cos A. cosh(x) = ex + e x, sinh(x) = ex e x cosh( x) = cosh(x), sinh( x) = sinh(x), cosh(0) = 1, sinh(0) = 0, cosh (x) = sinh(x), cosh x sinh x = 1, sinh (x) = cosh (x), sinh(x + y) = sinh(x) cosh(y) + cosh(x) sinh(y), sinh(x y) = sinh(x) cosh(y) cosh(x) sinh(y).