MA 201, Mathematics III, July-November 2016, Laplace Transform

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1 MA 21, Mathematics III, July-November 216, Laplace Transform Lecture 18 Lecture 18 MA 21, PDE (216) 1 / 21

2 Laplace Transform Let F : [, ) R. If F(t) satisfies the following conditions: F(t) is piecewise continuous on the interval [,b] for each b >, There exist real constants M and real positive number a such that Definition (Laplace Transform) e at F(t) M, i.e. F(t) Me at for all t >, Then, the Laplace transform of F(t), denoted by L{F(t)} or f(s), is defined by L{F(t)} = f(s) = e st F(t) dt, (1) provided this integral exists. Here s is a positive real number or a complex number. Observe F(t)e st dt F(t) e st dt M e (s a)t dt = M (s a), s > a. The variable s also takes complex values, but for the time being, assume s to be real. Lecture 18 MA 21, PDE (216) 2 / 21

3 Derivative property of Laplace transform Suppose a differentiable function F(t) has Laplace transform f(s), we can obtain the Laplace transform of its derivative F (t): L{F (t)} = e st F (t) dt. Similarly if the function F(t) is twice, thrice,..., n-times differentiable, we can obtain the Laplace transforms of F (t),f (t),...,f (n) (t) respectively. The following theorems relate to the Laplace transforms of the derivatives of a function. These results are extremely helpful in solving differential equations of various orders. Lecture 18 MA 21, PDE (216) 3 / 21

4 Derivative property of Laplace transform If F(t) is a differentiable function of t and L{F(t)} = f(s), then L{F (t)} = F()+sf(s). (2) Proof: L{F (t)} = = e st F (t) dt [ e st F(t) ] +s e st F(t) dt = F()+sf(s), where F() is the value of F(t) at t =. This is the proof of the theorem. Lecture 18 MA 21, PDE (216) 4 / 21

5 Derivative property of Laplace transform If F(t) is a twice differentiable function of t and L{F(t)} = f(s), then L{F (t)} = s 2 f(s) sf() F (). (3) Proof: L{F (t)} = e st F (t) dt [ e = st F (t) ] +s e st F (t) dt [ se = F ()+ st F(t) ] +s2 e st F(t) dt = s 2 f(s) sf() F (), where F() and F () are the values of F(t) and F (t), respectively at t =. This is the proof of the theorem. Lecture 18 MA 21, PDE (216) 5 / 21

6 Derivative property of Laplace transform We can generalize the above results as follows: If F(t) is an n-times differentiable function of t, and L{F(t)} = f(s), then L{F (n) (t)} = s n f(s) s n 1 F() s n 2 F () sf (n 2) () F (n 1) (). (4) The proof can be completed very easily by induction. Example ODE1 (First order ODE): dx +3x =, x() = 1. dt By taking Laplace transform on both sides of the equation, L{ dx dt }+L{3x} = sl{x} x()+3l{x} = (s+3)l{x} = 1 L{x} = 1 s+3 Lecture 18 MA 21, PDE (216) 6 / 21

7 Laplace transform of some elementary functions 1 L{1} = 1 s, s >. 2 L{t} = 1 s2, s >. 3 L{t n } = n! sn+1, s >, n an integer or L{t n } = Γ(n+1) s n+1, n+1 >, n not an integer. 4 L{e at } = 1 s a, s > a. a 5 L{sinat} = s 2 +a2, s >. s 6 L{cosat} = s 2 +a2, s >. a 7 L{sinhat} = s 2 a2, s > a. s 8 L{coshat} = s 2 a2, s > a. Note: Gamma function Γ(α) is defined as an improper definite integral as Γ(α) = t α 1 e t dt, α > Lecture 18 MA 21, PDE (216) 7 / 21

8 Elementary properties of Laplace transform The Laplace transform has many interesting and useful properties, the most fundamental of which is linearity. If F 1 (t) and F 2 (t) are two functions whose Laplace transforms exist, then L{aF 1 (t)+bf 2 (t)} = al{f 1 (t)}+bl{f 2 (t)} (5) where a and b are arbitrary constants. Proof: L{aF 1 (t)+bf 2 (t)} = = = a (af 1 +bf 2 )e st dt (af 1 e st +bf 2 e st ) dt F 1 e st dt+b = al{f 1 (t)}+bl{f 2 (t)} F 2 e st dt Lecture 18 MA 21, PDE (216) 8 / 21

9 Elementary properties of Laplace transform (First Shifting ): If F(t) is a Laplace transformable function, i.e., L{F(t)} = f(s), then L{e at F(t)} = f(s a). (6) Proof: L{e at F(t)} = = = f(s a). e st {e at F(t)} dt e st e at F(t) dt e (s a)t F(t) dt This establishes the theorem. Lecture 18 MA 21, PDE (216) 9 / 21

10 Elementary properties of Laplace transform (Second Shifting ): { F(t a), t a If L{F(t)} = f(s) and G(t) =, t < a then L{G(t)} = e as f(s). (7) Proof: L{G(t)} = = = e st G(t) dt e st F(t a) dt a e s(a+u) F(u) du, by taking t a = u = e as e su F(u) du = e as f(s), which proves the theorem. Lecture 18 MA 21, PDE (216) 1 / 21

11 Some more properties Recall Heaviside s Unit Step Function, or simply the unit step function, is defined as {, t <, H(t) = 1, t. (8) Since H(t) is precisely the same as 1 for t, the Laplace transform of H(t) must be the same as the Laplace transform of 1, i.e., 1 s. If we shift the origin to t, {, t < t, H(t t ) = 1, t t. (9) The following results can be obtained very easily: L{H(t)} = 1 s. (1) L{H(t t )} = e st. (11) s Lecture 18 MA 21, PDE (216) 11 / 21

12 Elementary properties of Laplace transform (Change of Scale ): If L{F(t)} = f(s), then L{F(at)} = 1 a f ( s a ) a >. (12) Proof: L{F(at)} = = 1 a = 1 a = 1 a f ( s a e st F(at) dt e s u/a F(u) du, by taking at = u e ( s/a)u F(u) du ) which proves the theorem. Lecture 18 MA 21, PDE (216) 12 / 21

13 Elementary properties of Laplace transform If L{F(t)} = f(s), then L{tF(t)} = d ds f(s) and in general L{tn F(t)} = ( 1) n dn dsnf(s). (13) Proof: We know f(s) = L{F(t)} = e st F(t) dt. Differentiating w.r.t. s df ds = d e st F(t) dt ds = te st F(t) dt. Hence, L{tF(t)} = d ds f(s). Lecture 18 MA 21, PDE (216) 13 / 21

14 Elementary properties of Laplace transform Now assume that the result holds for n so that Differentiating the above w.r.t. s, L{t n F(t)} = ( 1) n dn ds nf(s) = t n+1 e st F(t) dt = ( 1) n dn+1 ds n+1f(s) e st (t n+1 n+1 dn+1 F(t)) dt = ( 1) ds n+1f(s) = L{t n+1 n+1 dn+1 F(t)} = ( 1) ds n+1f(s), which shows that it is true for all n and proves the theorem. Lecture 18 MA 21, PDE (216) 14 / 21

15 Some more properties If L{F(t)} = f(s), then { t } L F(u) du = f(s) s. (14) Proof: t Let G(t) = F(u) du. G() =, and G (t) = F(t). L{G (t)} = sl{g(t)} G() = sl{g(t)} i.e. sl{g(t)} = L{F(t)} or, L{G(t)} = f(s) { s t } L F(u) du = f(s) s which proves the theorem. Lecture 18 MA 21, PDE (216) 15 / 21

16 Some more properties If L{F(t)} = f(s), then { F(t) assuming that L t } = { F(t) L t } as s. f(u) du, (15) s Proof: Let G(t) = F(t) t so that F(t) = tg(t). Using the property L{tG(t)} = d ds L{G(t)}, f(s) = L{F(t)} = d ds L{G(t)} = d ds L { F(t) t Integrating both sides with respect to s from s to, [ { }] F(t) { } { } F(t) F(t) f(u) du = L = L = L, s t s t t s }. which proves the theorem. Lecture 18 MA 21, PDE (216) 16 / 21

17 Some more properties (Third Shifting ) If F(t) is a function of exponential order in t, then where f(s) is the Laplace transform of F(t). L{H(t t )F(t t )} = e st f(s), (16) Proof: L{H(t t )F(t t )} = = = e st H(t t )F(t t ) dt e st F(t t ) dt t e s(t+u) F(u) du = e st e su F(u) du = e st f(s), which proves the result. Lecture 18 MA 21, PDE (216) 17 / 21

18 Some more properties Example: Determine the Laplace transform of the sine function switched on at time t = 3. Solution: The function is { sint, t 3, S(t) =, t < 3. (17) We can use Heaviside function to write S(t) as S(t) = H(t 3)sint. sint = sin(t 3+3) = sin(t 3)cos3+cos(t 3)sin3. Taking Laplace transform L{S(t)} = L{H(t 3)sin(t 3)}cos3+L{H(t 3)cos(t 3)}sin3. Then using the third shifting theorem L{S(t)} = (cos3+ssin3)e 3s s Lecture 18 MA 21, PDE (216) 18 / 21

19 Inverse Laplace transform If F(t) has the Laplace transform f(s), i.e., L{F(t)} = f(s), then the inverse Laplace transform is defined by L 1 {f(s)} = F(t). (Linearity) The inverse Laplace transform is linear, i.e., L 1 {a 1 f 1 (s)±a 2 f 2 (s)} = a 1 L 1 {f 1 (s)}±a 2 L 1 {f 2 (s)}. (18) If L 1 {f(s)} = F(t), then L 1 {f(s a)} = e at F(t). (19) If L 1 {f(s)} = F(t), then { L 1 {e as F(t a), t a, f(s)} =, t < a. (2) Lecture 18 MA 21, PDE (216) 19 / 21

20 Inverse Laplace transform Example A: Find the inverse Laplace transform of the following: a s (1). s 2 a 2, (2). s 2 +a 2. It is very easy to find that the required functions are sinh at and cos at respectively. Example B: Determine L 1 { s 2 s 2 +4s+13 }. Solution B: s 2 s 2 +4s+13 = s 2 (s+2) = s+2 4 (s+2) Taking inverse F(t) = e 2t 3 [3cos3t 4sin3t]. Lecture 18 MA 21, PDE (216) 2 / 21

21 Inverse Laplace transform Example C: Determine L 1 { s 2 (s+3) 3 }. Solution C: Use the partial fractions to write s 2 (s+3) 3 = A s+3 + B (s+3) 2 + C (s+3) 3, Equating coefficients of various powers of s, we will ultimately get A = 1,B = 6,C = 9 so that s 2 (s+3) 3 = 1 s+3 6 (s+3) (s+3) 3. Taking inverse Laplace transform { } s L 1 2 (s+3) 3 = e 3t 6te 3t t2 e 3t. Lecture 18 MA 21, PDE (216) 21 / 21