The Method of Laplace Transforms.

Transcription

1 The Method of Laplace Transforms. James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University May 25, 217

2 Outline 1 The Laplace Transform 2 Inverting the Laplace Transform 3 The Laplace Transform and ODE

3 Abstract This lecture is about the Laplace Transform method for linear ODE.

4 The Laplace Transform Recall what the improper integral f (t)dt means: Definition Let f be a Riemann Integrable on the interval [, L] for all L >. We define f (t)dt = lim L f (t)dt when this limit exists. Definition The Laplace Transform of the function f, L (f ) is the function whose value at s is given by L (f )(s) = e st f (t)dt when this improper integral exists. We want to apply the Laplace Transform to functions made up of linear combinations of functions in ker((d r) k, ker((d 2 + b 2 ) k, ker(d 2 2aD + a 2 + b 2 ) k.

5 The Laplace Transform The functions we will apply the Laplace Tranform to will thus have terms of the form f (t) = k a k t k + p b p t p e rt + q c q t q e a t (cos(b t) + sin(b t)) where all the summations are all finite. The first summation is for the root occurring k times, the second summation is for the root r occurring p times and the third summation is for the complex conjugate pair roots a ± bi occuring q times. Hence, the first summation is for terms in ker(d k ), the second summation is for terms in ker((d r) p ) and the third summation is for terms in ker((d 2 2aD + a 2 + b 2 ) q ). This form of f only uses the roots, r and a ± bi for simplicity. Of course, we can add additional roots to have f (t) = k c k t k + + roots a j ±b j i roots r i Q j P i p= b p,i t p e r i t f q,j t q e a j t (cos(b j t) + sin(b j t)) q=

6 The Laplace Transform Theorem If L (f ) exists for s > a, then L (tf (t))(s) = d L (f (t))(s) ds Proof Consider d ds L (f (t))(s) = d ds lim L e st f (t)dt We can show we can interchange the order of the differentiation and the limit on L to get

7 The Laplace Transform Proof d L (f )(s) = lim ds L d ds e st f (t)dt There is a rule for differentiating integrals of the form b(s) g(s, t)dt a(s) which is called Leibnitz s Rule. This is d ds b(s) a(s) g(s, t)dt = g(s, b(s))b (s) g(s, a(s))a (s) + b(s) a(s) g s dt Applying Leibnitz s Rule here is simple as a(s) = and b(s) = L. Thus, we have d ds e st f (t)dt = d L (f )(s = lim ds L te st f (t)dt = te st f (t)dt

8 The Laplace Transform Proof Thus, L (tf (t))(s) = d L (f (t))(s) ds Example The function f (t) = 1 has L (1)(s) = 1/s for s >. Solution L (1)(s) = lim L The limit is finite as long as s >. e st 1dt = lim L (1/s) e st L = lim L ( (1/s)e sl + 1/s) = 1/s

9 The Laplace Transform Theorem For s > and n 1, L (t n )(s) = n!/s n+1. Proof L (t 1)(s) = d ds L (1)(s) = d ds (1/s) = 1/s2 L (t t)(s) = d ds L (t)(s) = d ds (1/s2 ) = 2/s 3 L (t t 2 )(s) = d ds L (t2 )(s) = d ds (2/s3 ) = 3!/s 4 The pattern is now clear: L (t n )(s) = n!/s n+1.

10 The Laplace Transform Theorem For s > a, L (e (a+bi)t ) = 1/(s a i b)) and L (e (a bi)t ) = 1/(s a + i b)). Proof We do the first case. The other one is similar. L (e (a+bi )t ) = lim L e st e (a+bi )t dt = lim L e (s a i b)t dt = lim ( 1/(s a i b)) L e (s a i b)t L ( ) = lim ( 1/(s a i b)) e (s a i b)l + 1/(s a i b)) L = 1/(s a i b)

11 The Laplace Transform Theorem For s >, L (cos(bt)) = s/(s 2 + b 2 ) and L (sin(bt)) = b/(s 2 + b 2 ) Proof Since cos(bt) = (1/2) ( e bi t + e bi t), L (cos(bt)) = (1/2)L (e bi t ) + (1/2)L (e bi t ) = (1/2) (1/(s i b) + 1/(s + i b)) = (1/2) ( 2s/(s 2 + b 2 ) ) = s/(s 2 + b 2 ) Since sin(bt) = (1/2i ) ( e bi t e bi t), L (sin(bt)) = (1/2i )L (e bi t ) (1/2i )L (e bi t ) = (1/2i ) (1/(s i b) 1/(s + i b)) = (1/2i ) ( 2bi /(s 2 + b 2 ) ) = b/(s 2 + b 2 )

12 The Laplace Transform Theorem If F (s) = L (f (t))(s) exists for s > c, then L (e at f (t))(s) = F (s a). Proof L (e at f (t))(s) = lim Hence, L (e at f (t))(s) = F (s a). L e st e at f (t)dt = lim e (s a)t f (t)dt = L (f )(s a) L

13 The Laplace Transform Theorem For s >, L (e at cos(bt)) = L (e at sin(bt)) = s a (s a) 2 + b 2 b (s a) 2 + b 2 L (t cos(bt)) = s (d/ds) s 2 + b 2 = s2 b 2 (s 2 + b 2 ) 2 L (t sin(bt)) = b (d/ds) (s 2 + b 2 = 2bs (s 2 + b 2 ) 2 L (te at cos(bt)) = L (te at sin(bt)) = (s a) 2 b 2 ((s a) 2 + b 2 ) 2 2b(s a) ((s a) 2 + b 2 ) 2

14 The Laplace Transform Homework Find L (cos(3t))(s) Find L (5 cos(3t) + 2 sin(4t))(s) Find L (2 + 3t)(s) Find L (e 3t )(s) Find L (4e 5t )(s) Find L (6e 2t cos(4t))(s) Find L (1te 2t )(s) Find L (1te 2t sin(3t))(s).

15 Inverting the Laplace Transform The inverse of the Laplace Transform is denoted by L 1. If F (s) = L (f (t))(s), then L 1 (F (s)) = f (t). Thus, ( ) n! t n = L 1 s n+1 ( ) ( ) cos(bt) = L 1 s (s 2 + b 2, sin(bt)) = L 1 b ) (s 2 + b 2 ) ( ) e at cos(bt) = L 1 s a (s a) 2 + b ( 2 ) e at sin(bt) = L 1 b (s a) 2 + b 2 ( s t cos(bt) = L 1 2 b 2 ) ( ) (s 2 + b 2 ) 2, t sin(bt) = L 1 2bs (s 2 + b 2 ) 2 ( (s a) te at cos(bt) = L 1 2 b 2 ) ((s a) 2 + b 2 ) ( 2 ) te at sin(bt) = L 1 2b(s a) ((s a) 2 + b 2 ) 2

16 Inverting the Laplace Transform Homework Find L 1 ( Find L 1 ( 3s + 5 s 2 s 4 s Find L 1 ( 7 s 2 +4 ). ). 3.4 Find L 1 ( ) 2 s Find L 1 ( 1 s 1 ) s. ( 4 ) 3.6 Find L 1 s (s 1) 2 +4 ). Hint: write numerator as (s 1) + 1 and split into two terms. 3.7 Find L 1 ( 6(s 2 4) (s 2 +4) 2 ). 3.8 Find L 1 ( 7 ((s 3) 2 9) ((s 3) 2 +9) 2 ).

17 The Laplace Transform and ODE Theorem For a Linear ODE L(D), the solution to L(D)(x) = f (t) where f (t) is a linear combination of the kernel functions of ker((d r) k, ker((d 2 + b 2 ) k, and ker(d 2 2aD + a 2 + b 2 ) k, we know the solution x will also be a linear combination of those functions. Hence, L (x(t))(s) and its derivatives is well defined and L (x (t))(s) = sl (x)(s) x() L (x (t))(s) = s 2 L (x)(s) sx() x () Proof The well-defined nature of L (x)(s) and its derivatives follows from our discussions so far. To show the rest is a quick calculation:

18 The Laplace Transform and ODE Proof L (x (t))(s) = lim e st x (t)dt L ( = lim x(t)e st ) L L L se st x(t)dt Consider the first part of this limit. We know x(t) could have the form x(t) = k c k t k + + roots a j ±b j i roots r i Q j P i p= b p,i t p e r i t f q,j t q e a j t (cos(b j t) + sin(b j t)) q= Replace all the coefficients by their absolute values and overestimate all the sin and cos terms by their largest value 1. Then we have the overestimate

19 The Laplace Transform and ODE Proof x(t) k c k t k + roots r i P i p= b p,i t p e r i t + roots a j ±b j i Q j f q,j t q e a j t q= Let R = max{ r i } and A = max{ a j } and C = max{r, A}. Then we have x(t) k c k t k + roots r i P i p= b p,i t p e C t + roots a j ±b j i Q j q= f q,j t q e C t Now overestimate again by replacing the term k c k t k by ( k c k t k )e C t. Then, factoring out the common e Ct, we have x(t) k c k t k + roots r i P i p= b p,i t p + roots a j ±b j i Q j q= f q,j t q e C t

20 The Laplace Transform and ODE Proof The term in parenthesis in front of the e Ct is a polynomial in t which we will call p(t). This polynomial has positive coefficients. So we have x(t) p(t)e Ct Evaluating x(t)e st L, we get x(l)e sl x(). From our estimates above, we now see x(l) e sl p(l)ecl e sl when L when s > C. Hence, we can now calculate the limit on L to find L (x (t))(s) = = sl (x(t))(s) x() Applying the same sort of argument to x, we find L (x (t))(s) = sl (x (t))(s) x () = s 2 L (x(t)(s) sx() x ()

21 The Laplace Transform and ODE Example For x (t) 2x (t) 8x(t) = 6 + 8t. x() = 2, x () = 5. find L (x(t))(s) and then x(t) in terms of the inverse Laplace Transform. Solution Let L (x(t))(s) = X (s). Then L (x (t) 2x (t) 8x(t))(s) = L ( 6 + 8t)(s) ( s 2 X (s) sx() x () ) 2 (sx (s) x()) 8X (s) = 6 s + 8 s 2 Now use the ICs. Simplify ( s 2 X (s) + 2s 5 ) 2 (sx (s) + 2) 8X (s) = 6 s + 8 s 2 (s 2 2s 8)X (s) + 2s 5 4 = 6 s + 8 s 2

22 The Laplace Transform and ODE Solution Simplify some more: Now finish factoring: (s 2 2s 8)X (s) = 2s s + 8 s 2 = X (s) = 2s 9 s 2 2s 8 + X (s) = 6 s(s 2 2s 8) + 8 s 2 (s 2 2s 8) 2s 9 (s 4)(s + 2) + 6 s((s 4)(s + 2)) + 8 s 2 (s 4)(s + 2) Thus, ( ) ( ) x(t) = L 1 2s 9 + L 1 6 (s 4)(s + 2) s((s 4)(s + 2)) ( ) +L 1 8 s 2 (s 4)(s + 2)

23 The Laplace Transform and ODE Homework Calculate L (x (t) + 5x (t) + 6x(t))(s). This answer will have an x() and x () in it as they are not specified Calculate L (x (t) + 9x(t))(s). This answer will have an x() and x () in it as they are not specified Calculate L (x (t) + 6x (t) + 1x(t))(s). This answer will have an x() and x () in it as they are not specified For x (t) + 5x (t) + 6x(t) = 3 + 5t. x() = 2, x () = 4. find X (s) = L (x(t))(s) and find x(t) in terms of the inverse Laplace Transform.