Review Sol. of More Long Answer Questions

Transcription

1 Review Sol. of More Long Answer Questions 1. Solve the integro-differential equation t y (t) e t v y(v)dv = t; y()=. (1) Solution. The key is to recognize the convolution: t e t v y(v) dv = e t y. () Now taking the Laplace transform of the equation we reach which simplifies to Solving it, we get [s Y y()] L { e t} Y = L{t} (3) s Y s 1 Y = 1 +. (4) s Y (s)= s3 s +s 1 s (s. (5) s ) To take inverse Laplace transform we need to use partial fractions. a. Factorize the denominator: b. Write down which leads to c. Setting s = we get Setting s = we get Setting s = 1 we get s ( s s ) = s (s ) (s + 1); (6) s 3 s + s 1 s (s s ) = A s + B s + C s + D s + 1 s 3 s + s 1 = As(s )(s + 1)+B (s ) (s + 1) + Cs (s +1)+Ds (s ). (8) Finally comparing coefficients for s 3 : d. So we have 1 = B B = 1/; (9) 9 = 1 C C = 3/4; (1) 6= 3D D = ; (11) = A+C + D A = 3/4. (1) Y (s)= 3/4 s Taking inverse Laplace transform, we get (7) + 1/ s + 3/4 s + s +1. (13) y(t)= t et + e t. (14) Remark. Unfortunately there doesn t seem to be a quick way to check the solution. 1

2 Review Sol. of More Long Answer Questions. Find the solutions of the initial value problem (you can use all formulas derived in class) Solution. Transform the equation: The equation for Y is then y (t)+4 y(t)=sin(t)δ(t /), y()=, y ()=. (15) [s Y sy() y () ] +4Y = L{sin(t) δ(t /)} t = e st [sin(t)δ(t /)] dt t [ = e st sin(t) ] δ(t /)dt = [ e st sin(t) ] t=/ ( s + 4 ) Y =e s Y = e s = e s. (16) s +4. (17) To take the inverse Laplace transform, we have to first decide that the following formula should be used: Here a = / and F = 1 s +4. We easily get L 1{ e as F(s) } = u(t a) f(t a). (18) f(t)= 1 sin t f(t a)= 1 sin((t /)) = 1 sin(t). (19) So the final answer is y(t)=l 1 {Y }= 1 sin(t) u(t /). () Remark. We can check the solution using the fact that d u(t a) = δ(t a) and f(t) δ(t a) = dt f(a)δ(t a) to make sure the answer is correct: t < / y(t)= 1 sin(t) t > / (1) Note that y is in fact continuous at /. Thus { t < / y (t) = cos( t) t > / () Notice that y (t) jumped from to cos() =1 at t =/. In other words y (t)= cos(t) u(t /)=( cost 1)u(t /)+u(t /). (3) Note that the first term is continuous at /. Now taking derivative again: y (t) =δ(t /)+ { t < / sin(t) t > / (4) which clearly satisfies the equation.

3 3 3. Compute the Fourier series for { < x < f(x)= sin x < x < (5) The following identities might be useful for the solution of this problem: sin x cos y =.5(sin(x + y) + sin(x y)); (6) sin x sin y =.5(cos(x y) cos(x + y)). (7) Solution. As f is given on < x <, we have L (or T ) be in all the formulas. So the Fourier series should look like with So we compute a n = 1 a = 1 f(x) a + [a n cos(n x)+b n sin(n x)] (8) n=1 f(x) cos(n x)dx; f(x)dx = 1 b n = 1 f(x) sin(n x) dx. (9) x= sin xdx = 1 x= [ cosx] = 4 ; (3) a n = 1 f(x) cos(n x) dx = 1 sin x cos(n x)dx = 1 sin(n + 1) x sin(n 1)xdx = 1 [ cos(n +1)x cos(n 1)x + n +1 n 1 ]

4 4 Review Sol. of More Long Answer Questions [ ] = 1 ( 1)n ( 1)n 1 1 n +1 n 1 = 1 [ ] [ ( 1) n n ] n 1 = ( 1) n 1 1 n 1 n odd = 4. (31) (n n even 1) The above calculation is not legitimate for n = 1, we need to compute separately a 1 = 1 sin x cosxdx =. (3) We see that the general formula for a n is true for all n. On the other hand b n = 1 sin x sin (nx)dx = 1 [cos(n 1)x cos(n + 1) x] dx = 1 [ sin(n 1) x sin(n +1)x ] n 1 n + 1 =. (33) The above calculation is not legitimate for n = 1, we have to compute separately Summarizing, So finally we have b 1 = 1 sin x dx = 1 f(x) + n=1 [1 cosx] dx=1. (34) { 1 n = 1 b n = n > 1. (35) [a n cos(n x)+b n sin(n x)] (36) with a n, b n given by n odd a n = 4 (n 1) n even ; b 1 n = 1 n = n > 1 ; n = 1,, 3, (37) Remark. The calculation of b n shows that discussing n = 1 case separately is not just useless formality!

5 5 4. Find the first four non-zero coefficients in the power series expansion about x = for the solution to the problem y e x y =x, y() =1. (38) The following expansion may be useful for the solution of this problem e x = n= Solution. As the initial value is given at x =, we have x = and write x n n!. (39) y =a + a 1 x +a x + (4) (For problems asking for first four non-zero terms, the short-hand a n x n does not give us any advantage) and substitute into the equation [ a1 + a x + 3a 3 x + Now we compute the coefficients one by one. a = y() =1; x term: x 1 term: x term: So y up to first 4 nonzeroe terms is ] ] [1+x+ x + x3 6 + [a ] + a 1 x+a x + = x. (41) a 1 a = a 1 =1; (4) a a 1 a = 1 a = 3 ; (43) 3a 3 a a 1 a = a 3 =1; (44) y = 1+x+ 3 x + x 3 +. (45) Remark. A not-so-simple way to check the solution is to compute ( 1+x+ 3 ) x + x 3 (1 + x+ x + x3 6 + ) (1 +x+ 3 ) x + x 3 x (46) has no term below x 3 (because we obtain the coefficients by balancing up to x ).

6 6 Review Sol. of More Long Answer Questions 5. Use the method of separation of variables to find the formal solution of the problem u(x, t) = u(x, t) t x +e x ; <x<, t > (47) u(, t)= u(, t) = e ; t > (48) u(x, ) = e x, <x< (49) Note: You must show the calculation for obtaining the eigenvalues and eigenfunctions for the related boundary value problem. Solution. Notice that this problem is non-homogeneous and has nonzero source term (forcing) e x. So we have to pre-process first. Step. Preprocess. Find w(x) such that w xx + e x = ; w() =, w()= e. (5) Integrating the equation of w we get Now So we have w(x)=c 1 +C x e x. (51) w()= C 1 = C 1 = ; (5) w()= e C 1 + C e = e C =. (53) w(x)= e x. (54) Now set v(x, t)=u(x, t) w(x). The equation and boundary/initial conditions for v are v t = v xx (55) v(, t) =v(, t) =, (56) Step 1. Obtain the eigenvalue problem and solve it. We separate variable for v: v = X(x) T(t), to get v(x, ) = u(x, ) w(x) =. (57) X T = X T Consequently there is a constant λ such that T T X (t)= (x) (58) X X λx =; T λt =. (59) Now combining with the boundary conditions, we obtain the eigenvalue problem: X λx = ; X()=X() =. (6) We discuss the three cases: a. λ >. In this case X =C 1 e λ x +C e λ x and the boundary conditions enforce X()= C 1 + C = (61) X()= C 1 e λ + C e λ =. (6) The only C 1, C satisfying both are C 1 =C =. Thus no λ > can be an eigenvalue. b. λ=. In this case X = C 1 + C x and the boundary conditions enforce X() = C 1 = ; X() = C 1 + C = (63) so again necessarily C 1 = C =. Thus λ = is not an eigenvalue. ( ) ( ) c. λ <. In this case X = C 1 cos λ x + C sin λ x and the boundary conditions enforce X()= C 1 =; (64)

7 7 X()= ( C 1 cos λ ) + C sin ( λ We see that necessarily C 1 = C = unless λ is such that sin ) ( =. (65) ) λ =. Summarizing, we see that the eigenvalues satisfy ( ) sin λ = λ = integer λ =λ n = n, n = 1,, 3, (66) Thus n are eigenvalues with corresponding eigenfunctions Note that n range from 1 to. Step. Solve T n. The equation for T n is T n λ n T n = which becomes Step 3. Write down v(x, t)= n=1 X n = A n sin(n x), (67) T n + n T n = T n = B n e nt. (68) c n X n T n = n=1 Note that the constants A n, B n are absorbed in to c n. Setting t = we obtain v(x, )= n=1 c n e nt sin(n x). (69) c n sin(n x). (7) Step 4. Determine c n. Comparing the above with the given initial value, we have = n=1 c n sin(n x). (71) Therefore c n are coefficients of the Fourier Sine expansion of over < x <. Consequently c n = ( ) sin(n x)dx = 4 1 d(cos n x) n = 4 n cos(n x) = 4 [ ( 1) n 1 ] n 8 = n n odd. (7) n even Step 5. Post-processing. Recall that u =v +w with w = e x. We have with Remark. There is no simple way to check the solution. u(x, t) = e x + c n e nt sin(n x). (73) n=1 8 c n = n n odd. (74) n even

8 8 Review Sol. of More Long Answer Questions 6. Find the formal solution of Solution. We follow standard procedures. u t = 4 u x, < x <, t > (75) u(, t) = u(, t)=, t > (76) u(x, ) = x ( x), < x < (77) u (x, ) = <x <. (78) t Step 1. Separate variables and solve eigenvalue problem. Set u =X(x)T(t) we get thus there is λ such that T X =4X T The eigenvalue problem for X is then T T X (t)=4 (x) (79) X X λx = ; T 4λT =. (8) X λx =; X()=X()=. (81) As this eigenvalue problem is exactly the same as the one in the previous problem, instead of copy-paste, we simply omit the details here. n are eigenvalues with corresponding eigenfunctions Note that n range from 1 to. Step. Solve T n. The equation for T n is X n = A n sin(n x), (8) T n 4λ n T n = T n + 4n T n =. (83) The Characteristic equation is r + 4 n = r 1, = ±ni so Step 3. Write down with u(x, t)= n=1 T n = D n cos( n t)+e n sin( nt). (84) c n X n T n = n=1 [a n cos(nt) +b n sin(nt)] sin(nx). (85) u t (x, t)= n[b n cos( n t) a n sin(nt)] sin(n x). (86) n=1 Setting t = we reach u(x, ) = n=1 a n sin(n x); (87) u t (x, ) = nb n sin(n x). (88) n=1 Step 4. Determine the coefficients. Comparing with the given initial values we have x ( x)= n=1 a n sin(n x); = n=1 n b n sin(n x). (89) and conclude a n, b n are Fourier Sine series coefficients for x ( x) and respectively. Immediately we get b n =.

9 9 For a n, as the problem is given for <x< we have L (or T ) = and a n = x ( x) sin(n x)dx = x ( x)dcos(n x) n = [x ( x) cos(n x) n cos(n x)d ( x ( x) )] = ( ) x 3 x cos(n x) dx n = ( ) x 3 x n dsin(n x) = [ (x 3x ) n sin(n x) sin(n x)d ( x 3x )] = n ( 6 x) sin(n x)dx = n 3 ( 6x)dcos(n x) = [ n 3 ( 6x) cos(n x) ] cos(n x) d( 6x) = [ n 3 4 ( 1) n ] + 6 cos(n x)dx = [ n 3 4 ( 1) n + 6 ] n sin(n x) = [ 4 ( 1) n n 3 ] = 4 ( ( 1) n n ) = 4 n 3 n odd 1. n 3 n even So finally the solution is given by with u(x, t) = n=1 a n = [a n cos(nt)+b n sin(nt)] sin(n x). (9) 4 n 3 n odd 1 n 3 n even ; b n =. (91)