THE UNIVERSITY OF WESTERN ONTARIO. Applied Mathematics 375a Instructor: Matt Davison. Final Examination December 14, :00 12:00 a.m.

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1 THE UNIVERSITY OF WESTERN ONTARIO London Ontario Applied Mathematics 375a Instructor: Matt Davison Final Examination December 4, 22 9: 2: a.m. 3 HOURS Name: Stu. #: Notes: ) There are 8 question worth a total of marks and a bonus question worth marks. 2) NO calculators allowed. 3) Show all the steps in your calculation on the page with the question. (Rough work can be done on the back of the previous page.) 4) Do not remove any pages from the exam. Do not remove staples. There should be 2 pages in the booklet (including the cover page). FOR INSTRUCTOR S USE ONLY Bonus Total 2 2

2 AM375a Final Examination December 4, 22 2 []. Find the Laplace transform of t 2 sin(ωt) We have that: L [tf (t)] d ds F (s) L [ t 2 f (t) ] d ds ( d ) ds F (s) d2 ds2f (s) Where: F (s) L [f (t)] Now: So: L [sin ωt] ω s 2 + ω 2 L [ t 2 sin ωt ] d2 ω ds 2 s 2 + ω 2 ω d [ d ( s 2 + ω 2) ] ds ds ω d [ (2s) ( s 2 + ω 2) 2 ] ds 2ω d [s ( s 2 + ω 2) 2 ] ds [ (s 2ω 2 + ω 2) 2 ( + s ( 2) s 2 + ω 2) ] 3 (2s) 2ω [ 4s 2 (s 2 + ω 2 ) 3 s 2 ω 2] 2ω ( 3s 2 ω 2 (s 2 + ω 2 ) 3 )

3 AM375a Final Examination December 4, 22 3 [] 2. Find the inverse Laplace transform of F (s) s(s 2 + ) Soln. #: Use Partial Fractions. A s (s 2 + ) s + Bs + C s 2 + As 2 + As + Bs 2 + Cs s (s 2 + ) Therefore, A, B, C So: So we have: L s (s 2 + ) s s s 2 + [ s (s 2 L + ) s s s 2 + [ ] [ L L s cos t ] s s 2 + ]

4 AM375a Final Examination December 4, 22 4 Soln. #2: Convolution. We have that: L [ [ L s s 2 + ] ] sin t Thus: L s (s 2 + ) sin t t sin τ dτ cos τ t cos t cos cos t + Soln. #3 Recall that: [ ] t L f (τ) dτ L [ s F (s) ] s F (s) t f (τ) dτ So we have that: L s (s 2 + ) t L t [ sin τdτ cos t + ] dτ s 2 +

5 AM375a Final Examination December 4, 22 5 [] 3. Solve the Ordinary Differential Equation d 2 x dt + 3 dx 2 dt + 2x H(t 4), x(), (). dx dt Letting L [x (t)] X (s) and taking the Laplace transform of both sides we have: s 2 X + 3sX + 2X s s e 4s X ( s 2 + 3s + 2 ) s s e 4s X (s + ) (s + 2) X s s e 4s s (s + ) (s + 2) s (s + ) (s + 2) e 4s Using partial fractions we have: s (s + ) (s + 2) A s + B s + + C s + 2 A 2, B, C 2 So: s (s + ) (s + 2) 2s s (s + 2)

6 AM375a Final Examination December 4, 22 6 So: x (t) L L e 4s s (s + ) (s + 2) s (s + ) (s + 2) [ ] [ ] 2 L L + [ ] s s + 2 L s L e 4s + L e 4s s s + 2 L e 4s s e t + 2 e 2t 2 H (t 4) + e (t 4) H (t 4) 2 e 2(t 4) H (t 4) 2 [ H (t 4)] + e t [ e 4 H (t 4) ] 2 e 2t [ e 8 H (t 4) ]

7 AM375a Final Examination December 4, 22 7 [] 4. Use Laplace Transforms to solve the integral equation t f(t β) exp( aβ)dβ 2 ( )( ) exp(at) exp( at) sinh(at). Recognizing that the LHS is a convolution we have: f e αt sinh (at) Taking the Laplace transform of both sides gives: F (s) s + a a s 2 a 2 a (s + a) F (s) (s + a) (s a) a F (s) s a [ ] Therefore, f (t) al s a f (t) ae at

8 AM375a Final Examination December 4, 22 8 [] 5. Find the Fourier Series representation for the square wave function with period 2π: f(x), π x <, f(x), x π. f(x + 2nπ) f(x), n... 2,,,, 2,... f (x) is odd so there will only be sine terms in its expansion. The period is T 2π so we have that: f (x) ( ) 2πnx b n sin T where: n n b n sin (nx) b n 2 T T 2 T 2 f (x) sin (nx) dx But both f (x) and sin (nx) are odd. Thus, f (x) sin (nx) is even and this becomes: b n 2 2 2π π 2π 2 f (x) sin (nx) dx 2 sin (nx) dx π 2 nπ cos nx π 2 ( cos nπ) nπ if n is even if n is odd 4 nπ So: f (x) 4 π k sin [(2k + ) x] 2k +

9 AM375a Final Examination December 4, 22 9 [2] 6. Solve the heat equation problem (with κ ): u t 2 u Your result from Q5 will be helpful. x, < x < π, t >, 2 u(, t) u(π, t), t >, u(x, ), x π. Use separation of variables. Let u (x, t) X (x) T (t). Substituting this into the PDE we get: XT Therefore, T From the equation for T we have: X T T X X λ T λt We need λ in order to avoid exponential runaway of the solution. So we will let λ µ 2. Solving the equation for T gives: Solving the equation for X gives: T (t) exp ( µ 2 t ) X + µ 2 X Therefore, X (x) a cos µx + b sin µx Now let s fit the boundary conditions to X. We have that: u (, t) X () Therefore, a cos (µ ) + b sin (µ ) a

10 AM375a Final Examination December 4, 22 And: So: u (π, t) X (π) b sin µπ Therefore, µ, 2,... u (x, t) n Now fit the final boundary condition: u (x, ) n b n sin nx exp ( n 2 t ) b n sin nx, x π So we must now expand f (x), x π using a sine series. Thus, we must use an odd extension of f (x) for the interval π x. Hence, the coefficients b n of the expansion will be those of the periodic function: f (x), π x, x π These coefficients were found in Q5 to be: b n, if n is even, if n is odd 4 nπ So we have that: u (x, t) 4 π 4 π n n odd k n sin nx exp ( n 2 t ) 2k + sin [(2k + ) x] exp [ (2k + ) 2 t ]

11 AM375a Final Examination December 4, 22 [] 7. Solve the wave equation problem on an infinite string: 2 u t 9 2 u 2 x, 2 u(x, ) sin(x), u t (x, ). This is an infinite string problem so we will use D Alembert s Solution. We know that a solution to the wave equation must be of the form: u (x, t) A (x ct) + B (x + ct) From the PDE we see that c 3. From the first boundary condition we have: u (x, ) A (x 3 ) + B (x + 3 ) sin x A (x) + B (x) sin x From the second boundary condition we have: Thus: u t (x, ) 3A (x 3 ) + 3B (x + 3 ) 3A (x) + 3B (x) A (x) B (x) Therefore, A (x) B (x) A (x) + B (x) sin x A (x) + A (x) sin x Therefore, A (x) B (x) 2 sin x So: u (x, t) 2 sin (x 3t) + sin (x + 3t) 2

12 AM375a Final Examination December 4, 22 2 [2] 8. Solve Laplace s equation on the disc r < a: where 2 Φ(r, θ), Φ(a, θ) f(θ), f(θ), θ < π, f(θ), π θ 2π. Your Fourier Series result from Q4 will be very useful here. Writing the Laplacian in polar coordinates we obtain the PDE: 2 Φ r + Φ 2 r r + 2 Φ r 2 θ 2 Use separation of variables. Let Φ (r, θ) R (r) Θ (θ). In order for Φ to yield real physical solutions we must impose that R () is finite and that Θ (θ) is periodic with period 2π. Substituting Φ RΘ into the PDE gives: R Θ + r R Θ + r 2RΘ r 2R R + rr R + Θ Θ Therefore, r 2R R + rr R Θ Θ λ So the equation for θ is: Θ + λθ This equation will yield solutions with period 2π only if λ n 2, n,, So we have: Θ + n 2 Θ Therefore, Θ (θ) The equation for R is: r 2 R + rr n 2 R Therefore, R (r) a cos nθ + b sin nθ n, 2,... n ar n + br n n, 2,... log r, n

13 AM375a Final Examination December 4, 22 3 So we must discard the solutions r n and log r. So we have that: Φ (r, θ) a n cos (nθ) r n + b n sin (nθ) r n Now let s fit the boundary condition. We have that: Φ (a, θ) a n a n cos (nθ) + b n a n sin (nθ) f (θ), θ 2π Thus, we must expand f (θ) using a Fourier series. From Q5, we see that the coefficients of this series are: a n n 4 b n nπ n odd n even Comparing these coefficients with those of Φ (a, θ) we see that our coefficients are: a n a n a n n n a n b n b n 4 nπ 4 a n nπ n odd n even n odd n even So the solution is: Φ (r, θ) 4 π n n odd k ( ) 4 r n nπ sin (nθ) a ( ) r 2k+ (2k + ) sin [(2k + ) θ] a

14 AM375a Final Examination December 4, 22 4 [] BONUS. Using Laplace Transforms and the Convolution Theorem, compute the integral: dx. x( x) The fact that [ ] π L t s will be very useful. We will consider the integral: and then find I (). I (t) t dx x (t x) Using the definition of a convolution we see that: I (t) [ t ] t [ ] t t Therefore, L [I (t)] L L L [I (t)] π s π s L [I (t)] π s ] [ Therefore, I (t) πl s I (t) π π Thus: I () dx x ( x) π