Math 2C03 - Differential Equations. Slides shown in class - Winter Laplace Transforms. March 4, 5, 9, 11, 12, 16,

Transcription

1 Math 2C03 - Differential Equations Slides shown in class - Winter 2015 Laplace Transforms March 4, 5, 9, 11, 12, 16,

2 Laplace Transform used to solve linear ODEs and systems of linear ODEs with constant coefficients especially useful if the RHS has a jump discontinuity involves ideas similar to using log and antilog OR differentiation and integration i.e., inverse process useful in the study of delay differential equations (an advanced topic beyond the scope of this course)

3 Laplace Transform Definition Assume that f (t) is a function defined for t 0. The Laplace transform of f (t) is the function F(s) = L {f } = 0 e st f (t) dt, s > a (1) provided the integral exists for all s > a for some a R. Convention: lower case used for function: g(t) upper case used for its Laplace transform: G(s), with s as the independent variable.

4 Linearity of the Laplace Transform Theorem (Linearity) Let f (t) and g(t) be functions that have a Laplace transform that exists for s > a, for some a R. Let α and β be any constants (real or complex). Denote, L{f (t)} = F(s) & L{g(t)} = G(s). Then, L{αf (t) + βg(t)} = αf(s) + βg(s), for s > a Proof: Follows because integration is linear. Can be used to find the Laplace transform of cos(kt) & of sin(kt).

5 Preliminary Definitions Definition (Piecewise Continuous) (a) f (t) is piecewise continuous on an interval [a, b], if [a, b] can be subdivided into a finite number of subintervals, and in each subinterval f (t) is continuous and has a finite left- and right-hand limit. (b) f (t) is piecewise continuous on [0, ), if it is piecewise continuous on [0, b] for every b > 0. No vertical asymptotes Only jump discontinuities.

6 Definition (Exponential Order) f (t) is of exponential order α R a t, if there exist positive constants K and M such that or equivalently, if f (t) Ke αt whenever t > M. f (t) lim t e αt = L, where L 0 is finite, f (t) is called O(e αt ) ( big Oh of e αt ) if L>0. f (t) is called o(e αt ), ( little oh of e αt ) if L=0.

7 Existence of the Laplace Transform Theorem Assume f (t) is piecewise continuous on [0, ) and of exponential order α as t. Then L{f } exists for s > α. Therefore, F(s) = 0 T e st f (t) dt = lim T 0 an improper integral, that converges for s > α. e st f (t) dt,

8 Inverse Laplace Transform Definition Given a function F(s), if there is a function f (t) that is continuous on [0, ) and satisfies L{f (t)} = F(s) ( ) then we say that f (t) is the inverse Laplace transform of F(s), and we write f (t) = L 1 {F(s)}. Note: If every function that satisfies ( ) is piecewise continuous, then one can choose any one of the functions as the inverse Laplace transform. Two piecewise continuous functions satisfying ( ) will only differ at a finite number of points.

9 Example (Inverse Laplace Transforms Several Examples) { 1 L 1 s { } s L 1 s 2 + k 2 } { } 1 = 1, L 1 = e kt, s k } = cos(kt), L 1 { k s 2 + k 2 = sin(kt). Proposition The inverse Laplace transform L 1 is a linear transformation, i.e. for any constants α, β, we have L 1 {α F(s) + βg(s)} = α L 1 {F(s)} + β L 1 {G(s)}

10 Laplace Transform of mth Derivative: f (m) = d m dt m f (t) Theorem Assume that f (t), f (t),... f (m) (t) are all continuous on [0, ) and of some exponential order. Let F(s) = L {f }. Then, { L f (m)} = s m F(s) s m 1 f (0) s m 2 f (0) f (m 1) (0). Therefore, L { f } = sf (s) f (0). L { f } = s 2 F(s) sf (0) f (0). L { f } = s 3 F(s) s 2 f (0) sf (0) f (0).

11 Shift Property or Translation in s Theorem (Translation on the s-axis) If k R and then L {f (t)} = F(s) { } L e kt f (t) = F(s k) Proof. L { e kt f (t) } = 0 e st e kt f (t) dt = 0 e (s k)t f (t) dt = F(s k).

12 Derivatives of Laplace Transforms Theorem (Derivatives of transforms) If F(s) = L {f (t)}, then L {t n f (t)} = ( 1) n d n F(s), n 0. dsn Therefore, It follows that, L {t f (t)} = F (s). { } L t 2 f (t) = F (s). L {t n } = L {t n 1} = ( 1) n d n ds n ( ) 1 = n! s s n+1

13 Methods to Compute L 1 {F (s)} 1 Directly from a table or memory in combination with properties of L 1, including linearity. 2 Completing the square and using the shift property: L{e kt f (t)} = F (s k) L 1 {F (s k)} = e kt f (t) = e kt L 1 F(s) 3 Partial fraction decomposition. 4 Convolution.

14 Partial Fraction Decomposition Consider a rational function P(s) Q(s) where P(s) and Q(s) are polynomials with reals coefficients and: degree of P(s) < degree of Q(s) If not, use polynomial division, i.e. divide P(s) by Q(s). Factor & cancel common factors of P(s) & Q(s). For each linear term (s a) m, a R, in the denominator we include terms of the form A 1 s a + A 2 (s a) A m (s a) m. For each irreducible quadratic term (s α) 2 + β 2 ) p, α, β R, in the denominator we include terms of the form B 1 s + C 1 ((s α) 2 + β 2 ) + B 2 s + C 2 ((s α) 2 + β 2 ) B p s + C p ((s α) 2 + β 2 ) p.

15 Set P(s) Q(s) equal to the sum of all of these terms. Put over a common denominator. Equate numerators. To find the constants A i, B i & C i, Method I: Equate coefficients of s k, k = 0, 1, 2,..., n and solve the resulting system of equations. Method II: Evaluate both sides at the roots. If necessary differentiate both sides and then evaluate again at the roots, etc. until you obtain enough equations and unknowns to solve uniquely for the constants.

16 Finding Inverse Laplace Transforms after Partial Fraction Decomposition { } Use: shift L 1 {F(s a)} = e at f (t); L 1 (m 1)! s = t m 1 ; & m linearity: { } { } A A L 1 = A; L 1 = Ae at s s a { } L 1 Am s m = A mt m 1 L 1 { B1 s + C 1 (s α) 2 + β 2 (m 1)! ; L 1 } { } Am (s a) m = A me at t m 1 (m 1)! { } = L 1 B1 (s α) + B 1 α + C 1 (s α) 2 + β 2 ) sin(βt) β } = use CONVOLUTION = e αt B 1 cos(βt) + e αt ( αb1 + C 1 { L 1 B m s + C m ((s α) 2 + β 2 ) m

17 The Convolution Definition (Convolution) Let f (t) and g(t) be piecewise continuous for t 0. The convolution of f (t) and g(t), donoted (f g)(t), is (f g)(t) = t 0 f (t τ) g(τ) dτ. Properties f g = g f f (g + h) = (f g) + (f h) (f g) h = f (g h) f 0 = 0

18 The Convolution Theorem Theorem (Convolution theorem) If f (t) and g(t) are piecewise continuous and of exponential order for t 0, then so is (f g)(t) and, furthermore, L {f g} = F(s) G(s). where F(s) = L {f } and G(s) = L {g}.

19 Solving a Volterra Integral Equation Definition (Volterra integral equation) A Volterra integral equation involving the unknown function f (t) is an equation of the form t f (t) = g(t) + 0 f (τ) h(t τ) dτ, where g(t) and h(t) are given functions. This equation can be written f (t) = g(t) + (f h)(t), t 0, and can be solved by taking the Laplace transform of both sides.

20 Unit Step or Heaviside Function, u(t τ) place Transforms Definition For τ 0, the Heaviside { or unit step function, u(t τ) is 0, t < τ defined by u(t τ) = 1, t τ y t τ t piecewise continuous function (8.4.1) Figure Graph y = of u(t τ) u(t τ)

21 Theorem (Translation in t) Let F (s) = L{f (t)} and assume τ > 0. Then, L{u(t τ)f (t τ)} = e τs F(s). Proof. L {u(t τ) f (t τ)} = = e st f (t τ) dt τ }{{} v=t τ,dv=dt 0 e s(v+τ) f (v) dv = e sτ e sv f (v) dv 0 } {{ } F (s) = e τ s F(s).

22 L {u(t τ) f (t τ)} = e τ s F (s) Corollary 1 L 1 {e τs F(s)} = u(t τ)f (t τ) 2 L{u(t τ)} = e τs s 3 L{u(t τ)f (t)} = e τs L{f (t + τ)} Can express piecwise continuous functions using the step function. Useful to find the Laplace transform of piecewise continuous functions. Especially useful to solve IVPs with piecewise continuous functions on the RHS.

23 Table of Laplace Transforms, L{f (t)} = F (s) L{1} = 1 s L{e kt } = 1 s k L{sin(βt)} = β L{cos(βt)} = s s 2 +β 2 s 2 +β 2 L{t n } = n!, n 1 integer s n+1 L{e kt f (t)} = F (s k) L{t n f (t)} = ( 1) n F (n) (s) L{δ(t a)} = e as, a 0 L{f (n) (t)} = s n F(s) s n 1 f (0) s n 2 f (0) sf (n 2) (0) f (n 1) (0) L{u(t τ)f (t τ)} = e τs { F(s), τ 0, 0, t < τ, where u(t τ) = 1, t τ. is the unit step function or Heaviside function. If f (t + T ) = f (t) for all t, L{f (t)} = L{(f g)(t)} = L{ t o L{t r } = Γ(r+1) s r+1 T 0 e st f (t) dt 1 e st f (t v)g(v) dv} = F(s)G(s), r > 1 real, where Γ(t) = 0 e u u t 1 du, t > 0