MAE 200B Homework #3 Solutions University of California, Irvine Winter 2005

Transcription

1 Problem 1 (Haberman 5.3.2): Consider this equation: MAE 200B Homework #3 Solutions University of California, Irvine Winter 2005 a) ρ 2 u t = T 2 u u αu + β x2 t The term αu describes a force that is proportional to displacement, such as a spring or friction force. For such restoring forces α < 0. The term β u describes an aerodynamic damping term. This damping should reduce the t velocity so β < 0 b) u(x, t) = h(t)φ(x) Now we can use separation of variables to solve the equation. ρ(x)h φ = T 0 φ h + α(x)hφ + β(x)h φ ρ(x)h φ β(x)h φ = T 0 φ h + α(x)hφ hφ h ρ(x) h h β(x)h h = T φ 0 φ + α(x) h β(x) ρ(x) h h = T 0 φ ρ(x) φ + 1 ρ(x) α(x) For separation of variables to work the left side must only be a function of t and the right side must only be a function of x. Therefore for this to hold true β(x) = c where c = constant ρ(x) or β(x) = cρ(x). 1

2 c) If we look at the spatial equation: T 0 φ ρ(x) φ + 1 α(x) = λ2 ρ(x) T 0 φ + α(x)φ = λ 2 ρ(x)φ T 0 φ + ( α(x) + λ 2 ρ(x) ) φ = 0 [ ] d dφ T 0 + ( α(x) + λ 2 ρ(x) ) φ = 0 dx dx Now we can put this in Strum-Liouville form: [ d p(x) dφ ] + ( q(x) + λ 2 σ(x) ) φ = 0 dx dx Therefore, p(x) = T 0, q(x) = α(x), and σ(x) = ρ(x) Now we can solve the time equation with β(x) = cρ(x) : h h β(x) ρ(x) The roots for this equation come out to be: h h = λ2 h h ch h = λ2 h ch + λ 2 h = 0 r 1,2 = c 2 ± c 2 4 λ2 Case 1) c2 4 λ2 > 0 The solution to the time equation is: ( ) ( c 2 h(t) = C n e + c 2 4 λ2 c t 2 + Dn e c )t 2 4 λ2 Case 2) c2 4 λ2 = 0 2

3 h(t) = C n e ( c 2)t + D n te ( c 2)t Case 3) c2 4 λ2 < 0 h(t) = C n e c 2 t cos λ 2 c2 4 t + sin λ 2 c2 4 t Note c < 0 because β < 0 3

4 Problem 2 (Haberman 5.3.9): Consider this equation and boundary conditions: a) x 2 d2 φ dx 2 + xdφ dx + λφ = 0 φ(1) = 0 φ(b) = 0 If we divide through by 1 x we can put this equation into Strum-Liouville form: [ ] 1 x 2 d2 φ x dx + xdφ 2 dx + λφ = 0 x d2 φ dx 2 + dφ dx + 1 x λφ = 0 Notice that: [ d x dφ ] = x d2 φ dx dx dx + dφ 2 dx Therefore: [ d x dφ ] + 1 dx dx x λφ = 0 is in S.L. form where p(x) = x, q(x) = 0, and σ(x) = 1 x b) We can solve this problem using the Raleigh quotient λ = pφ dφ dx b a + [ b a p ( dφ dx b a φ2 σdx ) 2 qφ ] dx In our case a = 1, b = b, p(x) = x, q(x) = 0, and σ(x) = 1 x. Therefore: λ = dφ pφ dx b 1 + b 1 x ( ) dφ 2 dx dx b 1 1 x φ2 dx 4

5 Since φ(1) = φ(b) = 0, λ = b 1 x ( ) dφ 2 dx dx b 1 1 x φ2 dx For 1 x b, b 1 x ( ) 2 dφ dx 0 dx and b 1 1 x φ2 dx > 0 Thus λ 0 c) When solving the equidimentional or Cauchy-Euler equation we look for solutions in the form of φ = x r. Substituting this solution into our problem we get, x 2 r(r 1)x r 2 + xrx r 1 + λx r = 0 [r(r 1) + r + λ] x r = 0 r(r 1) + r + λ = 0 r 2 r + r + λ = 0 r 2 = λ r = ±i λ This gives us a the solution, φ = x ±i λ Since, x ±i λ = e ln(x±i λ ) φ = e ±i λln(x) 5

6 The general solution is then, φ = Ccos ( λln(x) ) + Bsin ( λln(x) ) Applying the first BC: Applying the second BC: Therefore, φ(1) = 0 = Ccos ( λln(1) ) + Bsin ( λln(1) ) λ n = C = 0 φ(b) = 0 = Bsin ( λln(b) ) λln(b) = nπ λ n = nπ ln(b) ( ) 2 nπ n = 1, 2, 3,... ln(b) ( ) φ(x) = Bsin λ n ln(x) d) By definition, φ n (x) = sin φ m (x) = sin ( ) nπ ln(b) ln(x) ( ) mπ ln(b) ln(x) < φ n, φ m >= σ(x) = 1 x b a φ n (x)φ m (x)σ(x)dx 6

7 = b 1 ( ) ( ) b nπ mπ 1 < φ n, φ m >= sin 1 ln(b) ln(x) sin ln(b) ln(x) x dx [ ( ) ( )] 1 1 π π cos (n m)ln(x) cos (n + m)ln(x) dx 2 x ln(b) ln(b) If m n and we let ln(x) = u and 1 dx = du then: x If m = n, Therefore < φ n, φ m >= ln(b) 0 = 1 ln(b) sin ( 2 π [ ( ) ( )] 1 π π cos (n m)u cos (n + m)u du 2 ln(b) ln(b) (n m)u) sin ( n m π ln(b) < φ n, φ m >= = 1 ln(b) [0 + 0] = 0 2 π ln(b) 0 (n + m)u) n + m π ln(b) [ ( )] 1 π 1 + cos 2 ln(b) 2nu du = 1 2 ln(b) ln(b) 2 2πn [sin (2πn) sin (0)] = 1 2 ln(b) ln(b) 0 if n m < φ n, φ m >= 0 if n = m < φ n, φ m >= 1 2 ln(b) e) φ n = sin ( ) nπ ln(b) ln(x) = sin(nzπ) 1 < x < b z = ln(x) ln(b) As x goes from 1 to b, nz goes from 0 to n. Therefore φ n = sin(nzπ) has the following zeros for x in (1, b) 7

8 z k = k n x k = e k n ln(b) = b k n, k = 1, 2, 3,..., n 1 Problem 3: Consider the periodic Strum-Liouville problem: X (x) = ω 2 X(x) L < x < L X( L) = X(L) X ( L) = X (L) a) Plugging in the given solutions into the above equation yields: φ n = e iωnx ω n = nπ L φ n = iω n e iωnx φ n = ωne 2 iωnx φ n = ωnφ 2 n ok φ n = e iωnx φ n = iω n e iωnx φ n = ωne 2 iωnx φ n = ωnφ 2 n ok These solutions can be related to sin and cos by: e iβ = cos(β) + isin(β), e iβ = cos(β) isin(β) Therefore, cos(β) = 1 ( e iβ + e iβ), sin(β) = 1 ( e iβ e iβ) 2 2 β = ω n x 8

9 b) By definition, for φ n = e iωnx φ m = e iωmx < φ n, φ m >= < φ n, φ m >= L L L L φ n φ mdx = L L e i(ωm ωn)x dx = e iωnx e iωmx dx L L e i π L (m n)x dx { 2L m = n < φ n, φ m >= L i(m n)π ei π L (m n)x L L m n) { 2L m = n < φ n, φ m >= 0 m n) For the same m, < φ + m, φ m >= L L e im π L x e im π L x dx = < φ + m, φ m >= L 2π i2π ei L x L L = 0 Therefore, these eigenfunctions are orthogonal. c) L e i 2π L x L We can find the coefficients using the definition of the inner product. f(x) = + n= C n e i nπ L x Therefore, C n = nπ < f(x), ei L x > = 1 e i nπ L x 2 2L nπ < f(x), ei L x >= 1 L nπ i f(x)e L x dx 2L L 9

10 d) If f(x) is real, then C n = < f(x), e i nπ L x > 2L By plugging in the value of n into the equation generated above we can show that C n = 1 L ( n)π i f(x)e L x dx = 1 L 2L L 2L L C n = 1 L f(x)e 2L L (n)π i L x dx = C n f(x)e i (n)π L x dx 10

11 Problem 4: The governing equation is the 1D String Equation with a forcing function: 2 u t 2 = 2 u c2 + Asin(ωt) (0 < x < L) x2 BC: u(0, t) = 0 u x (L, t) = 0 IC: u(x, 0) = f(x) = u t (x, 0) = 0 u(x, 0) = f(x) { 3h L x (0 < x < L/3) h (L/3 < x < L) First we will solve this problem assuming that ω ω n = ck n. We need to solve this problem using the method of Eigenfunction Expansion. We start by looking at the related homogeneous problem which we have solved before. 2 v t = 2 v 2 c2 (0 < x < L) x 2 Using the method of separation of variables we can find the eigenfunctions of this related problem: φ n (x) = sin(k n x) k n = nπ 2L = (2n 1)π 2L This now can give us a general solution in terms of an eigenfunction expansion, u(x, t) = a n (t)φ n (x) Now we can expand the non-homogeneous part of the equation in terms of our eigenfunctions, 11

12 g(x, t) = A n (t)φ n (x) = Asin(ωt) Since this is a periodic forcing with no spacial dependence we can easily find the coefficients for the expansion. A n (t) = L 0 g(x, t)φ n(x)dx L 0 φ2 n(x)dx = 2 L L 0 Asin(ωt)φ n (x)dx A n (t) = 2 L Asin(ωt) L 0 sin(k n x)dx = 2 Lk n Asin(ωt)(1 cos(k n L)) = 2A Lk n sin(ωt) Our goal now is to find the coefficient s a n (t) that will satisfy our equation. To find these coefficients we can plug our general solutions back into the original equation. Recall, 2 u t 2 = c2 2 u x 2 + Asin(ωt) and u(x, t) = a n (t)φ n (x) f(x, t) = A n (t)φ n (x) = Asin(ωt) φ n(x) = k 2 φ n (x) Differentiating our solution term by term yields, 2 u t 2 = 2 t 2 a n(t)φ n (x) = a n(t)φ n (x) c 2 2 u x 2 = c 2 2 u x 2 a n(t)φ n (x) = c 2 a n (t)φ n(x) = c 2 kna 2 n (t)φ n (x) Therefore, [ a n(t)φ n (x) = c 2 k 2 na n (t)φ n (x) + A n (t)φ n (x) ] 12

13 This reduces to a second order ODE in time, [ a n (t) = c 2 kna 2 n (t) + A n (t) ] Since we know A n (t) we can rewrite this equation, a n(t) + c 2 k 2 na n (t) = 2A Lk n sin(ωt) = sin(ωt) = 2A Lk n = 4A (2n 1)π Recall that this differential equation can be solved by finding the homogeneous solution and the particular solution. The particular solution in this case can be found using the method of undetermined coefficients. The homogeneous equation is an ODE which we have done many times. The solution is, a n,h(t) + c 2 k 2 na n,h (t) = 0 a n,h(t) = c 2 k 2 na n,h (t) a n,h (t) = C n cos(ck n t) + D n sin(ck n t) For the particular solution we can guess a solution in the form of a n,p (t) = E n sin(ωt) a n,p(t) + c 2 k 2 na n,p (t) = sin(ωt) ω 2 E n sin(ωt) + c 2 k 2 ne n sin(ωt) = sin(ωt) E n ( ω 2 + c 2 k 2 n) = E n = ( ω 2 + c 2 k 2 n) Therefore the solution for the temporal coefficients are, a n (t) = a n,h (t) + a n,p (t) = C n cos(ck n t) + D n sin(ck n t) + ( ω 2 + c 2 k 2 n) sin(ωt) That makes the whole solution so far, [ ] u(x, t) = C n cos(ck n t) + D n sin(ck n t) + ( ω 2 + c 2 kn) sin(ωt) sin(k 2 n x) 13

14 Now we can apply the the first IC, [ u t (x, t) = 0 = ck n C n sin(ck n t) + ck n D n cos(ck n t) + [ u t (x, 0) = 0 = D n ck n + Therefore, ] ( ω 2 + c 2 kn) ωcos(ωt) cos(k 2 n x) ] ( ω 2 + c 2 kn) ω sin(k 2 n x) D n = (ω 2 c 2 k 2 n)ck n ω Now we can apply the second IC to our new solution, [ u(x, t) = C n cos(ck n t) + (c 2 k 2 n + ω 2 )ck n ωsin(ck n t) + u(x, 0) = C n sin(k n x) = a n (0)sin(k n x) = f(x) ] (ω 2 c 2 kn) sin(ωt) sin(k 2 n x) As you can see we arrive at the same equation as we did in Homework #2 when trying to determine the coefficients for the Fourier Series. Notice that these coefficients are the temporal coefficients at the initial time. a n (0) = a n,h (0) + a n,p (0) = C n cos(ck n (0)) + D n sin(ck n (0)) + Therefore, ( ω 2 + c 2 k 2 n) sin(ω(0)) = C n thus, C n = 2 L L/3 0 C n = 2 L L 0 f(x)sin(k n x)dx 3h L xsin(k nx)dx + 2 L hsin(k n x)dx L L/3 Now we can integrate by parts or use a software package to find the solution for the coefficients. 24h 1)π C n = sin((2n ) ((2n 1)π)

15 So our final solution is, [ u(x, t) = C n cos(ck n t) + (ω 2 c 2 k 2 n)ck n ωsin(ck n t) + ] ( ω 2 + c 2 kn) sin(ωt) sin(k 2 n x) Note that this solution only works if ω ck n for all n. If we now look at the case when ω = ω 2 = ck 2 we see a problem in our particular solution. In this case we need to come up with a different form of the solution for this part of the problem. For the particular solution if we guess a solution in the form of a n,p (t) = E n sin(ω 2 t) When n = 2, E n will not be defined. a n,p(t) + c 2 k 2 na n,p (t) = sin(ω 2 t) ω 2 2E n sin(ω 2 t) c 2 k 2 ne n sin(ω 2 t) = sin(ω 2 t) E n ( ω c 2 k 2 n) = E n = ( ω c 2 k 2 n) Therefore we look for a different particular solution in the form of a n,p (t) = E n tsin(ω 2 t) + F n tcos(ω 2 t), a n,p(t) = E n sin(ω 2 t) + ω 2 E n tcos(ω 2 t) + F n cos(ω 2 t) F n ω 2 tsin(ω 2 t) a n,p(t) = ω 2 E n cos(ω 2 t)+ω 2 E n cos(ω 2 t) E n ω 2 2tsin(ω 2 t) F n ω 2 sin(ω 2 t) F n ω 2 sin(ω 2 t) F n ω 2 2tcos(ω 2 t) a n,p(t) = 2ω 2 E n cos(ω 2 t) E n ω 2 2tsin(ω 2 t) F n 2ω 2 sin(ω 2 t) F n ω 2 2tcos(ω 2 t) a n,p(t) + c 2 k 2 na n,p (t) = sin(ω 2 t) 2ω 2 E n cos(ω 2 t) E n ω 2 2tsin(ω 2 t) F n 2ω 2 sin(ω 2 t) F n ω 2 2tcos(ω 2 t) c 2 k 2 n(e n tsin(ω 2 t)+f n tcos(ω 2 t)) = sin(ω 2 t) Now we can separate out the sin and cos terms to solve for the coefficients. E n ω 2 2tsin(ω 2 t) F n 2ω 2 sin(ω 2 t) c 2 k 2 ne n tsin(ω 2 t) = sin(ω 2 t) E n (ω 2 2t c 2 k 2 n) F n 2ω 2 = 15

16 2ω 2 E n cos(ω 2 t) F n ω 2 2tcos(ω 2 t) c 2 k 2 nf n tcos(ω 2 t) = 0 2ω 2 E n F n (ω 2 2t c 2 k 2 nt) = 0 Now when n = 2, F 2 2ω 2 = B 2 2ω 2 E 2 = 0 Therefore E 2 = 0 and F 2 = B 2 2ω 2, so our particular solution is, a 2,p (t) = B 2 2ω 2 tcos(ω 2 t) So our solution for this resonance case is, ( u(x, t) = C 2 cos(ck 2 t) + D 2 sin(ck 2 t) B ) 2 tcos(ω 2 t) sin(k 2 x)+ 2ω 2 [ ] C n cos(ck n t) + D n sin(ck n t) + ( ω 2 + c 2 kn) sin(ωt) sin(k 2 n x),n 2 Now we can apply the ICs, u t (x, t) = 0 =,n 2 [ u t (x, 0) = 0 = ( ck 2 C 2 sin(ck 2 t) + ck 2 D 2 cos(ck 2 t) B 2 2ω 2 cos(ω 2 t) + B 2 ck n C n sin(ck n t) + ck n D n cos(ck n t) + ( ω 2 + c 2 kn) ωcos(ωt) 2 ( ck 2 D 2 B ) 2 sin(k 2 x) + 2ω 2 ( ) B2 ck 2 D 2 sin(k 2 x) = 2ω 2,n 2 Now we can multiply both sides by sin(k n x), For n 2 D n ck n + ( ω 2 + c 2 k 2 n) ω = 2 L L 0,n 2 [ [ D n ck n + ) 2 tsin(ω 2t) sin(k 2 x)+ ] cos(k n x) ] D n ck n + ( ω 2 + c 2 kn) ω sin(k 2 n x) ] ( ω 2 + c 2 kn) ω sin(k 2 n x) ( ) B2 ck 2 D 2 sin(k 2 x)sin(k n x) 2ω 2 D n ck n + ( ω 2 + c 2 k 2 n) ω = 0 16

17 D n = ω ( ω 2 + c 2 kn)ck 2 n For n = 2 ( ) B2 L ck 2 D 2 2ω 2 2 = 0 B 2 2ω 2 = ck 2 D 2 B 2 2ω 2 ck 2 = D 2 = B 2 2ω 2 2 Now we can apply the second IC,,n 2 [ ( u(x, t) = C 2 cos(ck 2 t) + B 2 sin(ck 2ω2 2 2 t) B ) 2 tcos(ω 2 t) sin(k 2 x)+ 2ω 2 C n cos(ck n t) ] ωsin(ck ( ω 2 + c 2 kn)ck 2 n t) + n ( ω 2 + c 2 kn) sin(ωt) sin(k 2 n x) u(x, 0) = C 2 sin(k 2 x) + C n sin(k n x) = C n sin(k n x) = f(x),n 2 As you can see this solution will give you all the coefficients from a Fourier Sine Series expansion. Therefore, So our final solution is,,n 2 C n = 2 L L 0 f(x)sin(k n x)dx 24h 1)π C n = sin((2n ) ((2n 1)π) 2 6 ( u(x, t) = C 2 cos(ck 2 t) + B 2 sin(ck 2ω2 2 2 t) B ) 2 tcos(ω 2 t) sin(k 2 x)+ 2ω 2 [ ] C n cos(ck n t) + ωsin(ck ( ω 2 + c 2 kn)ck 2 n t) + n ( ω 2 + c 2 kn) sin(ωt) sin(k 2 n x) 17

18 Now we can make the appropriate plots of the solutions for each of the cases. Case ω ω n The first plots show each of the first 5 terms of the Fourier Series and the associated sum for at 5 different times. For these plots, ω = 2πc L. 18

19 This next plot shows the time dependent solution at the center of the domain for each of the first five terms and the sum of those terms. Notice how the solution looks slightly different from the previous homework solution due to the effect of the forcing function. 19

20 These last plots are a 3D plot of the domain over the time of one period. If you would like to view the movie of these plots please refer to the Course Files on E3. 20

21 Case ω = ω 2 The first plots show each of the first 5 terms of the Fourier Series and the associated sum for at 5 different times. This next plot shows the time dependent solution at the center of the domain for each of the first five terms and the sum of those terms. Notice that at resonance we have a growing term that is unbounded. 21

22 22

23 These last plots are a 3D plot of the domain over the time of one period. If you would like to view the movie of these plots please refer to the Course Files on E3. 23

24 Problem 5: The governing equation is the 1D String Equation with a forcing function: BC: 2 u t + b u 2 t = 2 u c2 + Asin(ωt) (0 < x < L) x2 IC: u(0, t) = 0 u x (L, t) = 0 u(x, 0) = f(x) = u t (x, 0) = 0 u(x, 0) = f(x) { 3h L x (0 < x < L/3) h (L/3 < x < L) First we will solve this problem assuming that ω ω n = ck n. We need to solve this problem using the method of Eigenfunction Expansion. We start by looking at the related homogeneous problem which we have solved before. 2 v t 2 + b v t = c2 2 v x 2 (0 < x < L) Using the method of separation of variables we can find the eigenfunctions of this related problem: φ n (x) = sin(k n x) k n = nπ 2L = (2n 1)π 2L This now can give us a general solution in terms of an eigenfunction expansion, u(x, t) = a n (t)φ n (x) Now we can expand the non-homogeneous part of the equation in terms of our eigenfunctions, 24

25 g(x, t) = a n (t)φ n (x) = Asin(ωt) Since this is a periodic forcing with no spacial dependence we can easily find the coefficients for the expansion. a n (t) = 2 L Asin(ωt) L 0 sin(k n x)dx = 2A Lk n sin(ωt) Our goal now is to find the coefficient s a n (t) that will satisfy our equation. To find these coefficients we can plug our general solutions back into the original equation. Recall, and 2 u t 2 + b u t = c2 2 u x 2 + Asin(ωt) u(x, t) = a n (t)φ n (x) f(x, t) = a n (t)φ n (x) = Asin(ωt) φ n(x) = k 2 φ n (x) Differentiating our solution term by term yields, 2 u t 2 = 2 t 2 a n(t)φ n (x) = a n(t)φ n (x) c 2 2 u x 2 = u t = t a n(t)φ n (x) = a n(t)φ n (x) c 2 2 u x a n(t)φ 2 n (x) = c 2 a n (t)φ n(x) = c 2 kna 2 n (t)φ n (x) Therefore, [ a n(t)φ n (x) + ba n(t)φ n (x) = c 2 k 2 na n (t)φ n (x) + a n (t)φ n (x) ] 25

26 This reduces to a second order ODE in time, [ a n(t) + ba n(t) = c 2 k 2 na n (t) + a n (t) ] Since we know Ω n (t) we can rewrite this equation, a n(t) + ba n(t) + c 2 k 2 na n (t) = 2A Lk n sin(ωt) = sin(ωt) = 2A k n = 4A (2n 1)π Recall that this differential equation can be solved by finding the homogeneous solution and the particular solution. The particular solution in this case can be found using the method of undetermined coefficients. The homogeneous equation is an ODE which we have solved in the previous homework assignment. Recall that the solutions depend on the value of b, a n,h(t) + ba n(t) + c 2 k 2 na n,h (t) = 0 Now if ω n = ck n ζ = b 2ω n This equation has two roots r 1,2 = ζω n ± ω n ζ 2 1 = b 2 ± ω n ζ 2 1 Case #1 (Under-damped mode): ζ < 1 or b < 2ω n = 2n 1πc L r 1,2 = b 2 ± iω n 1 ζ 2 Therefore, the general solution is: a n,h (t) = e b 2 t [C n cos(β n t) + D n sin(β n t)] 26

27 Where β n = ω n 1 ζ 2 Case #2 (Critically damped mode): ζ = 1 or b = 2ω n = 2n 1πc L Therefore, the general solution is: r 1 = r 2 = b 2 a n,h (t) = C n e b 2 t + D n te b 2 t Case #3 (Over-damped mode): ζ > 1 or b > 2ω n = 2n 1πc L r 1,2 = b 2 ± ω n ζ 2 1 Therefore, the general solution is: a n,h (t) = C n e ( b 2 +ωn ζ 2 1)t + D n e ( b 2 ωn ζ 2 1)t a n,h (t) = C n cos(ck n t) + D n sin(ck n t) For the particular solution we can guess a solution in the form of a n,p (t) = E n sin(ωt) + F n cos(ωt). Note that since there is a damping term the form of our particular solution is different. a n,p(t) + ba n(t) + c 2 k 2 na n,p (t) = sin(ωt) ω 2 (E n sin(ωt)+f n cos(ωt))+b(e n ωcos(ωt) F n ωsin(ωt))+c 2 k 2 n(e n sin(ωt)+f n cos(ωt)) = sin(ωt) Like in the previous problem, we separate the sin(ωt) and cos(ωt) terms, ω 2 E n bf n ω + ω 2 ne n = E n (ω 2 n ω 2 ) bf n ω = F n (ω 2 n ω 2 ) + be n ω = 0 E n = ω2 ωn 2 F n bω 27

28 ω 2 c 2 kn 2 E n = (bω) 2 (ω 2 c 2 kn) 2 2 bω F n = (bω) 2 (ω 2 c 2 kn) 2 2 Therefore the particular solution for the temporal equation takes this form, ω 2 ωn 2 a n,p (t) = bω 2 (ω 2 ωn) sin(ωt) + B bω 2 2 n bω 2 (ω 2 ωn) cos(ωt) 2 2 Note that this solution works even when ω = ω n = ck n ωn 2 ωn 2 a n,p (t) = (bω n ) 2 (ωn 2 ωn) sin(ω bω n nt) + B 2 2 n (bω n ) 2 (ωn 2 ωn) cos(ω nt) 2 2 a n,p (t) = ω n cos(ω n t) This solution also conforms with theory because the resonance phenomena is 90 degrees out of phase with the forcing function. Now we can solve for the initial conditions for each of the cases. 1) Consider b < 2ω 1 = πc, Case #1 will be true for n = 1, 2, 3,... (all modes are underdamped) thus the solution L is: [ u n (x, t) = e b 2 t (C n cos(β n t) + D n sin(β n t)) sin(k n x)+ ω 2 ωn 2 ] (bω) 2 (ω 2 ωn) sin(ωt) + B bω 2 2 n (bω) 2 (ω 2 ωn) cos(ωt) sin(k 2 2 n x) For simplicity we will define, ω 2 ωn 2 P n = (bω) 2 (ω 2 ωn) 2 2 bω Q n = (bω) 2 (ω 2 ωn) 2 2 Therefore, [ u(x, t) = e b 2 t (C n cos(β n t) + D n sin(β n t)) + P n sin(ωt) + Q n cos(ωt) ] sin(k n x) 28

29 Now we apply the first IC: u t (x, t) = e b 2 [ t b ] 2 (C ncos(β n t) + D n sin(β n t)) + ( C n β n sin(β n t) + D n β n cos(β n t)) sin(k n x)+ [P n ωcos(ωt) + Q n ωsin(ωt)] sin(k n x) [ u t (x, 0) = b ] 2 (C n) + (D n β n ) + P n ω sin(k n x) = 0 Because of the uniqueness of the Fourier Series [ b ] 2 (C n) + (D n β n ) + P n ω = 0 Therefore, ( u(x, t) = [e b2 t C n cos(β n t) + D n = b 2β n (C n ) P nω β n ( b (C n ) P ) ) ] nω sin(β n t) + P n sin(ωt) + Q n cos(ωt) sin(k n x) 2β n β n Now we can apply the second IC: ( u(x, 0) = [e b2 (0) C n cos(β n (0)) + ( b (C n ) P ) ) ] nω sin(β n (0)) + P n sin(ω(0)) + Q n cos(ω(0)) sin( 2β n β n u(x, 0) = (C n + Q n ) sin(k n x) = f(x) Using superposition we see that the coefficients can be found using the Fourier series. C n = 2 L L/3 0 C n + Q n = 2 L L 0 f(x)sin(k n x)dx 3h L xsin(k nx)dx + 2 L hsin(k n x)dx + Q n L L/3 Again the coefficients can be solved using integration by parts or using a software program. 24h 1)π C n = sin((2n ) + Q ((2n 1)π) 2 n 6 29

30 Therefore the final solution for this case is: ( u(x, t) = [e b2 t C n cos(β n t) + ( b (C n ) P ) ) ] nω sin(β n t) + P n sin(ωt) + Q n cos(ωt) sin(k n x) 2β n β n 2) Consider b = 2ω 1 = πc L < 2ω n n = 2, 3,... In this case mode 1 is critically damped and all other modes are under-damped. u 1 (x, t) = ( C 1 e b 2 t + D 1 te b 2 t + P 1 sin(ωt) + Q 1 cos(ωt) ) sin(k 1 x) u n (x, t) = e b 2 t (C n cos(β n t) + D n sin(β n t) + P n sin(ωt) + Q n cos(ωt)) sin(k n x) n = 2, 3,... Therefore the general solution is: u(x, t) = ( C 1 e b 2 t + D 1 te b 2 t + P 1 sin(ωt) + Q 1 cos(ωt) ) sin(k 1 x)+ e b 2 t [(C n cos(β n t) + D n sin(β n t)) + P n sin(ωt) + Q n cos(ωt)] sin(k n x) n=2 First we can find the coefficients when n = 1. u(x, t) = ( C 1 e b 2 t + D 1 te b 2 t + P 1 sin(ωt) + Q 1 cos(ωt) ) sin(k 1 x) Applying the first B.C. u t (x, t) = ( b 2 C 1e b 2 t + D 1 e b 2 t + D 1 t b ) 2 e b 2 t + P 1 ωcos(ωt) Q 1 ωsin(ωt) sin(k 1 x) u t (x, 0) = [ ] b 2 C 1 + D 1 + P 1 ω sin(k 1 x) = 0 b 2 C 1 + D 1 + P 1 ω = 0 D 1 = b 2 C 1 P 1 ω Therefore, ( ( u(x, t) = C 1 e b b 2 t + 2 P 1 ω C 1 ) ) te b 2 t + P 1 sin(ωt) + Q 1 cos(ωt) sin(k 1 x)+ e b 2 t [(C n cos(β n t) + D n sin(β n t)) + P n sin(ωt) + Q n cos(ωt)] sin(k n x) n=2 30

31 For n = 2, 3,... we can apply the first IC as in the precious problem and find that, So we can rewrite our solution as: D n = ( ( u(x, t) = C 1 e b b 2 t + 2 P 1 ω C 1 ( [C n e b2 t cos(β n t) + n=2 Now we can apply the second IC: ) b (C n ) P nω 2β n β n te b 2 t + P 1 sin(ωt) + Q ) 1 cos(ωt) sin(k 1 x)+ C 1 C 1 ( b P ) ) ] nω sin(β n t) + P n sin(ωt) + Q n cos(ωt) sin(k n x) 2β n C n β n u(x, 0) = (C 1 + Q 1 ) sin(k 1 x) + (C n + Q n ) sin(k n x) = f(x) Using superposition we see that the coefficients can be found using the Fourier series. n=2 C n = 2 L L/3 0 C n + Q n = 2 L L 0 f(x)sin(k n x)dx 3h L xsin(k nx)dx + 2 L hsin(k n x)dx Q n L L/3 Again the coefficients can be solved using integration by parts or using a software program. 24h 1)π C n = (1 + Q n ) sin((2n ) Q ((2n 1)π) 2 n 6 Therefore the final solution for this case is: ( ( u(x, t) = C 1 e b b 2 t + 2 P 1 ω C 1 ( [C n e b2 t cos(β n t) + n=2 ) ) te b 2 t + P 1 sin(ωt) + Q 1 cos(ωt) sin(k 1 x)+ ( b P ) ) ] nω sin(β n t) + P n sin(ωt) + Q n cos(ωt) sin(k n x) 2β n C n β n 3) Consider b = 2πc = 4ω L 1. We know that ω 1 = 2πc, ω L 2 = 3 2πc = 3ω L 1, ω 3 = 5 2πc = 5ω L 1. Thus 2ω 1 < b < 2ω 2 <... Therefore the first mode is over-damped and the rest are still under-damped. 31

32 ( ) u 1 (x, t) = C 1 e ( b 2 +ω 1 ζ 2 1)t + D 1 e ( b 2 ω 1 ζ 2 1)t + P 1 sin(ωt) + Q 1 cos(ωt) sin(k 1 x) u n (x, t) = [ e b 2 t (C n cos(β n t) + D n sin(β n t)) + P n sin(ωt) + Q n cos(ωt) ] sin(k n x) n = 2, 3,... Therefore the general solution is: u(x, t) = ( ) C 1 e ( b 2 +ω 1 ζ 2 1)t + D 1 e ( b 2 ω 1 ζ 2 1)t + P 1 sin(ωt) + Q 1 cos(ωt) sin(k 1 x)+ [ e b 2 t (C n cos(β n t) + D n sin(β n t)) + P n sin(ωt) + Q n cos(ωt) ] sin(k n x) n=2 First we can find the coefficients when n = 1. u(x, t) = ( ) C 1 e ( b 2 +ω 1 ζ 2 1)t + D 1 e ( b 2 ω 1 ζ 2 1)t + P 1 sin(ωt) + Q 1 cos(ωt) sin(k 1 x) Applying the first B.C. [( u t (x, t) = b ) ( 2 + ω 1 ζ 2 1 C 1 e ( b 2 +ω 1 ζ 2 1)t + b ) ] 2 ω 1 ζ 2 1 D 1 e ( b 2 ω 1 ζ 2 1)t sin(k 1 x)+ + (P 1 ωcos(ωt) Q 1 ωsin(ωt)) sin(k 1 x) [( u t (x, 0) = b ) ( 2 + ω 1 ζ 2 1 C 1 + b ) ] 2 ω 1 ζ 2 1 D 1 + P 1 ω sin(k 1 x) = 0 Therefore, ( b 2 + ω 1 ) ( ζ 2 1 C 1 + b ) 2 ω 1 ζ 2 1 D 1 + P 1 ω = 0 D 1 = b + ω 2 1 ζ2 1 b ω 2 1 ζ2 1 C 1 P 1 ω ( u(x, t) = C 1 (e ( b 2 +ω 1 ζ b 2 1)t 2 + ω 1 ζ2 1 b ω 2 1 ζ2 1 P 1 ω C 1 + (P 1 sin(ωt) + Q 1 cos(ωt)) sink 1 x ) ) e ( b 2 ω 1 ζ 2 1)t sin(k 1 x)+ [ e b 2 t (C n cos(β n t) + D n sin(β n t)) + P n sin(ωt) + Q n cos(ωt) ] sin(k n x) n=2 For n = 2, 3,... 32

33 This was the same as in the previous case, therefore: So we can rewrite our solution as: D n = b (C n ) P nω 2β n β n ( u(x, t) = C 1 (e ( b 2 +ω 1 ζ b 2 1)t 2 + ω 1 ζ2 1 b ω 2 1 ζ2 1 P 1 ω C 1 ) ) e ( b 2 ω 1 ζ 2 1)t sin(k 1 x)+ + (P 1 sin(ωt) + Q 1 cos(ωt)) sink 1 x ( ( [C n e b2 b t cos(β n t) + P ) ) ] nω sin(β n t) + P n sin(ωt) + Q n cos(ωt) sin(k n x) 2β n C n β n n=2 Now we can apply the second IC: ( b 2 u(x, 0) = C 1 (1 + ω 1 ζ2 1 b ω 2 1 ζ2 1 P ) 1 ω C 1 + Q ) 1 sin(k 1 x) + C 1 (C n + Q n ) sin(k n x) n=2 The solution for the coefficients deliver the same integral as in the previous cases with the addition that the first coefficient has a over-damped factor in front of it. So for the coefficients we find that: 1 C 1 = 1 b 2 +ω 1 b 2 ω 1 ζ 2 1 ζ h π 2 sin(π 6 ) Q 1 24h 1)π C n = sin((2n ) Q ((2n 1)π) 2 n n = 2, 3,... 6 Therefore the final solution for this case is: ( u(x, t) = C 1 (e ( b 2 +ω 1 ζ b 2 1)t 2 + ω 1 ζ2 1 b ω 2 1 ζ2 1 P 1 ω C 1 ) ) e ( b 2 ω 1 ζ 2 1)t sin(k 1 x)+ + (P 1 sin(ωt) + Q 1 cos(ωt)) sink 1 x ( ( [C n e b2 b t cos(β n t) + P ) ) ] nω sin(β n t) + P n sin(ωt) + Q n cos(ωt) sin(k n x) 2β n C n β n n=2 We can now examine the plots for each of these cases and compare it to the case of no damping in Problem #4. 33

34 1): The first case is known as an under-damped case. The first plots show each of the first 5 terms of the Fourier Series and the associated sum for at 5 different times. 34

35 This next plot shows the time dependent solution at the center of the domain for each of the first five terms and the sum of those terms. Notice how the solution looks slightly different from the previous homework solution due to the affect of the forcing function. 35

36 These last plots are a 3D plot of the domain over the time of one period. If you would like to view the movie of these plots please refer to the Course Files on E3. 36

37 2) This case is known as an critically damped case. The first plots show each of the first 5 terms of the Fourier Series and the associated sum for at 5 different times. This next plot shows the time dependent solution at the center of the domain for each of the first five terms and the sum of those terms. 37

38 38

39 These last plots are a 3D plot of the domain over the time of one period. If you would like to view the movie of these plots please refer to the Course Files on E3. You can see the oscillatory behavior if you look at a zoomed in graph later in time. 39

40 3) The last case is known as an over-damped solution. This next plot shows the time dependent solution at the center of the domain for each of the first five terms and the sum of those terms. 40

41 41

42 These last plots are a 3D plot of the domain over the time of one period. If you would like to view the movie of these plots please refer to the Course Files on E3. 42

43 After a long time the transient solutions will die out leaving behind only the forced reponse. u(x, t) = (P n cos(ωt) + Q n sin(ωt))sin(k n t) Therefore the amplitude of oscillation is R n (ω) = P 2 n + Q 2 n The following plot shows R n for various values of n. 43