MA 201, Mathematics III, July-November 2018, Laplace Transform (Contd.)

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1 MA 201, Mathematics III, July-November 2018, Laplace Transform (Contd.) Lecture 19 Lecture 19 MA 201, PDE (2018) 1 / 24

2 Application of Laplace transform in solving ODEs ODEs with constant coefficients Example ODE1 (First order ODE): dx +3x = 0, x(0) = 1. dt By taking Laplace transform on both sides of the equation, L{ dx dt }+L{3x} = 0 sl{x} x(0)+3l{x} = 0 (s+3)l{x} = 1 L{x} = 1 s+3 Taking inverse transform x = e 3t. Lecture 19 MA 201, PDE (2018) 2 / 24

3 Application of Laplace transform in solving ODEs ODEs with constant coefficients Example ODE2 (Second order ODE): d 2 x dx +x = t, x(0) = 1, (0) = 2. dt2 dt By taking Laplace transform on both sides of the equation, L{ d2 x }+L{x} = L{t} dt2 s 2 L{x} sx(0) ẋ(0)+l{x} = 1/s 2 (s 2 +1)L{x} = 1 s 2 +s 2 1 L{x} = s 2 (s 2 +1) + s 2 s 2 +1 = 1 s 2 + s s s Taking inverse transform x = t+cost 3sint. Lecture 19 MA 201, PDE (2018) 3 / 24

4 Application of Laplace transform in solving ODEs ODEs with variable coefficients Example ODE3 (Second order ODE): t d2 x dt 2 +2(t 1)dx +(t 2)x = 0. dt By taking Laplace transform on both sides of the equation, L{t d2 x dt 2 }+2L{(t 1)dx dt }+L{(t 2)x} = 0 d ds L{ẍ} 2 d ds L{ẋ} 2L{ẋ} d ds L{x} 2L{x} = 0 which will ultimately lead to the differential equation: d 4s+4 L{x}+ ds s 2 +2s+1 L{x} = 3x 0 (s+1) 2, where x(0) = x 0 Lecture 19 MA 201, PDE (2018) 4 / 24

5 Application of Laplace transform in solving ODEs Giving us the solution (find the integrating factor as (s+1) 4 ) Taking inverse transform, L{x} = x 0 s+1 + C (s+1) 4. x = x 0 e t +C t3 6 e t. Lecture 19 MA 201, PDE (2018) 5 / 24

6 Application of Laplace transform in solving ODEs Simultaneous ODEs Example ODE4 (First order): } dx dt = 2x 3y, dy dt = y 2x, x(0) = 8, y(0) = 3. Taking Laplace transform on both sides of the first equation, (s 2)L{x}+3L{y} = 8. (1) Similarly, taking Laplace transform on both sides of the second equation, By application of Cramer s rule in (1) and (2), By taking the inverse transform, 2L{x}+(s 1)L{y} = 3. (2) L{x} = 5 s s 4, L{y} = 5 s+1 2 s 4. x(t) = 5e t +3e 4t, y(t) = 5e t 2e 4t. Lecture 19 MA 201, PDE (2018) 6 / 24

7 Convolution Example A: Find L 1 { s (s 2 +1)(s 2) } We can easily observe that the Laplace transforms of F(t) = e 2t and G(t) = cost, are, respectively, f(s) = 1 s s 2 and g(s) = s Can we have In other sense, whether L 1 {f(s)g(s)} = F(t)G(t)? L{F(t)G(t)} = {f(s)g(s)}? The result is true for a special product of functions, known as convolution The convolution of two given functions F(t) and G(t) is written as F G and is defined by the integral t F G = F(τ) G(t τ) dτ. (3) 0 Note that the commutative, associative and distributive properties hold true. Lecture 19 MA 201, PDE (2018) 7 / 24

8 Convolution Example B: t e at = t 0 t τe a(t τ) dτ = e at τe aτ dτ = 1 a 2(eat at 1). 0 Example C: sinat sinat = t 0 sinaτ sina(t τ)dτ = 1 2a (sinat atcosat). Lecture 19 MA 201, PDE (2018) 8 / 24

9 Convolution Theorem If F(t) and G(t) are two functions of exponential order and given L 1 {f(s)} = F(t) and L 1 {g(s)} = G(t), then t L 1 {f(s)g(s)} = F(τ) G(t τ) dτ = F G, (4) 0 where is the convolution operator defined in (3) above. Example A: We need to find L 1 { s (s 2 +1)(s 2) } Now, use convolution theorem to have { } L 1 s (s 2 = L 1 {f(s)g(s)} = F(t) G(t) +1)(s 2) t = e 2t cos(t) = e 2τ cos(t τ) dτ 0 = 2 5 e2t (sint 2cost). Lecture 19 MA 201, PDE (2018) 9 / 24

10 Convolution Proof: By definition L{F(t) G(t)} = t e st F(τ)G(t τ) dτ dt. 0 0 The domain of this repeated integral takes the form of a wedge in the t,τ-plane. Write L{F(t) G(t)} = t 0 0 e st F(τ)G(t τ) dτ dt. Integrating with respect to t first L{F(t) G(t)} = = 0 e st F(τ)G(t τ) dτ dt τ { } F(τ) e st G(t τ) dt 0 τ dτ. Lecture 19 MA 201, PDE (2018) 10 / 24

11 Convolution In the inner integral above, put u = t τ so that it can be written as τ e st G(t τ) dt = 0 = e sτ g(s). e s(u+τ) G(u) du Therefore L{F(t) G(t)} = 0 = f(s)g(s), F(τ)e sτ g(s) dτ which gives us the following desired result: L 1 {f(s)g(s)} = t 0 F(τ) G(t τ) dτ = F G. Lecture 19 MA 201, PDE (2018) 11 / 24

12 Inverse Laplace transform by method of residue If s is considered as complex, then the inverse Laplace transform F(t) of f(s) is given by F(t) = 1 2πi a+i a i f(s)e st ds, where the integration is along Re(s) = a is a vertical line. We can use the method of residues to find inverse Laplace transform. Lecture 19 MA 201, PDE (2018) 12 / 24

13 Inverse Laplace transform by method of residue The following theorem is to be used for evaluating inverse Laplace transform. Theorem If the Laplace transform f(s) of F(t) is an analytic function of s, except at a finite number of singular points-the poles-each of which lies to the left of the vertical line Re(s) = a and if sf(s) is bounded as s approaches infinity through the half-plane Re(s) a, then L 1 {f(s)} = 1 2πi a+i a i f(s)e st ds = Residues of f(s)e st at all poles = R 1 +R 2 +R 3 + (5) where { lim s ak {(s a k )f(s)e st }, k = 1,2,...,n; a k simple pole R k = 1 d lim r s ak r! ds {(s a r k ) r+1 f(s)e st }, a k multiple pole of order r +1. (6) Lecture 19 MA 201, PDE (2018) 13 / 24

14 Inverse Laplace transform by method of residue Example D: Evaluate { L 1 s 2 } s+3 s 3 +6s s+6 Solution D: We can see that lim s sf(s) = 1 which is bounded. The poles are found to be s = 1, 2, 3 which are all simple poles. In order to evaluate the inverse Laplace transform using (6), we need to calculate the residues of the function due to the poles. We get R 1 = (residue at s = 1) = 5 2 e t, R 2 = (residue at s = 2) = 9e 2t, R 3 = (residue at s = 3) = 15 2 e 3t. Therefore { L 1 s 2 } s+3 s 3 +6s 2 = 5 +11s+6 2 e t 9e 2t e 3t. Lecture 19 MA 201, PDE (2018) 14 / 24

15 Inverse Laplace transform by method of residue Example E: Evaluate { } L 1 1 (s+1)(s 2) 2. Solution E: Here s = 1 is a simple pole whereas s = 2 is a double pole. R 1 = (residue at s = 1) = e t 9, R 2 = (residue at s = 2) = ( t )e2t Therefore { } L 1 1 (s+1)(s 2) 2 = e t 9 +(t )e2t. Lecture 19 MA 201, PDE (2018) 15 / 24

16 Application of Laplace transform in solving PDEs Definition The Laplace transform of a function u(x,t) with respect to t is defined as L{u(x,t)} = u(x,s) = The Laplace transforms of the partial derivatives t u(x,t), x u(x,t), 2 2 t 2u(x,t), x2u(x,t) are as follows: 0 e st u(x,t) dt. (7) { } L t u(x,t) { } L x u(x,t) { } 2 L t 2 u(x,t) { } 2 L x 2 u(x,t) = sū(x,s) u(x,0), (8) = d ū(x,s), (9) dx = s 2 ū(x,s) su(x,0) u t (x,0), (10) = d2 ū(x,s). (11) dx2 Lecture 19 MA 201, PDE (2018) 16 / 24

17 Application of Laplace transform in solving PDEs Example PDE1: (First order) Find a bounded solution of the following problem u x = 2 u t +u subject to u(x,0) = 6 e 3x. Solution PDE1: Taking Laplace transform on both sides of the given PDE and using the initial condition, After finding the integrating factor, dū dx (2s+1)ū = 12e 3x. ū(x,s) = 6 s+2 e 3x +C e (2s+1)x. Lecture 19 MA 201, PDE (2018) 17 / 24

18 Application of Laplace transform in solving PDEs u(x,t) should be bounded when x. Hence its Laplace transform ū(x,s) should also be bounded as s and we take C = 0. ū(x,s) = 6 s+2 e 3x. Taking the inverse transform u(x,t) = 6e (2t+3x). Lecture 19 MA 201, PDE (2018) 18 / 24

19 Application of Laplace transform in solving PDEs Example PDE2: (Second order) Consider the following one-dimensional heat conduction equation subject to the following conditions: u t = 2 u x2, 0 < x < 1, t > 0 u(0,t) = 1, u(1,t) = 1, t > 0; u(x,0) = 1+sinπx, 0 < x < 1. Solution PDE2: Taking Laplace transform on both sides and applying the given initial condition, d 2 ū(x,s) sū(x,s) = (1+sinπx). dx2 The complementary function and the particular integral of the above equation can be derived as ū c (x,s) = Ae sx +Be sx, ū p (x,s) = 1 s + sinπx s+π 2, Lecture 19 MA 201, PDE (2018) 19 / 24

20 Application of Laplace transform in solving PDEs ū(x,s): ū(x,s) = Ae sx +Be sx + 1 s + sinπx s+π2. (12) Convert the boundary conditions in u(x, t) to boundary conditions in ū(x, s): u(0,t) = 1 ū(0,s) = 1 s, u(1,t) = 1 ū(1,s) = 1 s. Using these in (12) 1 s 1 s = A+B + 1 A+B = 0, s = Ae s +Be s + 1 Ae s +Be s = 0. s Lecture 19 MA 201, PDE (2018) 20 / 24

21 Application of Laplace transform in solving PDEs Both these conditions together imply A = 0 = B. Solution is obtained by taking the inverse ū(x,s) = 1 s + sinπx s+π2. (13) u(x,t) = 1+e π2t sinπx. (14) Lecture 19 MA 201, PDE (2018) 21 / 24

22 Application of Laplace transform in solving PDEs Example PDE3: (Second order) ( πx ) U tt = c 2 U xx +sin sin(σt), 0 < x < L, t > 0, L U(0,t) = 0, U(L,t) = 0, t > 0, U(x,0) = 0, U t (x,0) = 0, 0 < x < L. Taking Laplace transform on the equation, it gets reduced to d 2 s2 dx2u(x,s) c2u(x,s) = σsin(πx/l) c 2 (s 2 +σ 2 ) the solution of which can be obtained as which can be simplified to u(x,s) = A(s)e s c x +B(s)e s c x + σ c 2 u(x,s) = A(s)e s c x +B(s)e s c x + σsin(πx/l) ( c2 π 2 L 2 σ 2 ) sin(πx/l) (s 2 +σ 2 )( s2 c 2 + π2 L 2 ) ( 1 s 2 +σ 2 1 s 2 + c2 π 2 L 2 Lecture 19 MA 201, PDE (2018) 22 / 24 )

23 Application of Laplace transform in solving PDEs Taking transforms on the boundary conditions Reducing u(x, s) simply to u(x,s) = L 2 σ c 2 π 2 σ 2 L 2 u(0,s) = 0, u(l,s) = 0 A(s) = 0 = B(s) ( 1 s 2 +σ 2 1 s 2 + c2 π 2 L 2 ) sin(πx/l) Inverting That is U(x,t) = U(x,t) = L 2 [ σ 1 c 2 π 2 σ 2 L 2 σ sinσt L ] cπ sin(πct L ) sin(πx/l) L 2 [ c 2 π 2 σ 2 L 2 sinσt Lσ ] cπ sin(πct L ) sin(πx/l) Lecture 19 MA 201, PDE (2018) 23 / 24

24 Best wishes for rest of the course. Keep in mind Hard work has no substitute. Lecture 19 MA 201, PDE (2018) 24 / 24