Ordinary Differential Equations. Session 7

Transcription

1 Ordinary Differential Equations. Session 7 Dr. Marco A Roque Sol 11/16/2018

2 Laplace Transform Among the tools that are very useful for solving linear differential equations are integral transforms. An integral transform is a relation of the form F (s) = β α K(s, t)f (t)dt where K(s, t) is a given function, called the kernel of the transformation, and the limits of integration α and β are also given. It is possible that α = or β = or both. The relation, introduced above, transforms the function f into another function F, which is called the transform of f.

3 There are several integral transforms that are useful in applied mathematics, but we consider only the Laplace Transform ( ) (... Napoleon asked Laplace where God fit into his mathematical work Traite de mecanique celeste, and Laplace famously replied Sir, I have no need of that hypothesis... ). Laplace Transform Let f (t) be given for t 0. Then the Laplace transform of f, which we will denote by L {f (t)} = F (s), is defined by the equation L {f (t)} = F (s) = whenever this improper integral converges. 0 e st f (t)dt

4 The Laplace transform makes use of the kernel K(s, t) = e st. In particular for linear second order differential equations with constant coeficients is particular useful, since the solutions are based on the exponential function. The general idea in using the Laplace transform to solve a differential equation is as follows: 1. Use the relation L {f (t)} = F (s) to transform an initial value problem for an unknown function f in the t-domain (time domain) into an algebraic problem for F in the s-domain (frequency domain).

5 2. Solve this algebraic problem to find F. 3. Recover the desired function f from its transform F. This last step is known as inverting the transform (which in general involve complex integration) and denoted by L 1 {F (s)} (= lim ω 1 2πi σ+iω σ iω F (s)est ds). OBS The full power of Laplace Transform becomes available only when we regard F (s) as a function of a complex variable. However, for our purposes it will be enough to consider only real values for s. The Laplace transform F of a function f exists if f satisfies certain conditions:

6 Theorem 6.1 Suppose that 1. f is piecewise continuous on the interval 0 t A for any positive A. 2. f (t) Ke at when t M. In this inequality, K, a, and M are real constants, K and M necessarily positive. Then the Laplace transform L {f (t)} = F (s), defined by exists for s > a. L {f (t)} = F (s) = 0 e st f (t)dt

7 Remember that a function, f (t), is piecewise continuous on the interval α t β if the interval can be partitioned by a finite number of points α = t 0 < t 1 <... < t n = β so that 1. f is continuous on each open subinterval t i 1 < t < t i. 2. f approaches a finite limit as the endpoints of each subinterval are approached from within the subinterval.

8 Example 6.1 Find the Laplace transform for f (t) = 1, t 0 Solution L {1} = F (s) = L {f (t)} = F (s) = 0 0 e st dt = lim A e st f (t)dt e st s A = 1 0 s ; s > 0

9 Example 6.2 Find the Laplace transform for f (t) = e at, t 0 Solution L {e at } = F (s) = L {f (t)} = F (s) = lim A 0 e (s a)t s a 0 e at e st dt = A = 1 0 s a ; e st f (t)dt 0 e (s a)t dt = s > a

10 Example 6.3 Find the Laplace transform for 1 0 t < 1 f (t) = k t = 1 0 t > 1 where k is a constant. In engineering contexts f (t) often represents a unit pulse, perhaps of force or voltage. Solution L {f (t)} = F (s) = 0 e st s f (t)e st dt = 1 = 1 e s 0 s 1 0 e st dt =

11 Example 6.4 Find the Laplace transform for f (t) = sin(at), t 0 Solution L {f (t)} = F (s) = 0 e st f (t)dt = [ F (s) = lim e st cos(at) A A a s 0 a F (s) = 1 a s2 a 2 0 A 0 0 sin(at)e st dt = Int by Parts ] e st cos(at)dt = Int by Parts sin(at)e st dt = 1 a s2 a 2 F (s)

12 Hence, solving for F(s), we have a F (s) = s 2 + a 2 Now, the Laplace transform is a linear operator, that is, suppose that f 1 and f 2 are two functions whose Laplace transforms exist for s > a 1 and s > a 2, respectively. Then, for s > max{a 1, a 2 } L {c 1 f 1 + c 2 f 2 } = 0 e st {c 1 f 1 + c 2 f 2 }dt = c 1 e st f 1 dt + c 2 e st f 2 dt = c 1 L {f 1 } + c 2 L {f 2 } 0 Thus, we have 0 L {c 1 f 1 + c 2 f 2 } = c 1 L {f 1 } + c 2 L {f 2 }

13 OBS Example 6.5 L 1 {d 1 F 1 + d 2 F 2 } = d 1 L 1 {F 1 } + d 2 L 1 {F 2 } Find the Laplace transform for f (t) = 5e 2t 3sin(4t), t 0 Solution L {5e 2t 3sin(4t)} = 5L {e 2t } 3L {sin(4t)} = L {5e 2t 3sin(4t)} = 5 s s ; s > 0 L 1 { 5 s s } = 5L 1 { 1 s + 2 } 3L 1 4 { s } = 5e 2t 3sin(4t)

14 To see how we can apply the method of Transform Laplace to solve linear differential equations with constant coefficients. We establish the following results. Theorem 6.2 Suppose that f is continuous and f is piecewise continuous on any interval 0 t A. Suppose further that there exist constants K, a, and M such that f (t) Ke at for t M. Then proof L {f } = sl {f } f (0) L {f, (t)} = 0 A e st f (t)dt = lim A e st f (t)dt = 0

15 lim A [ t1 0 e st f (t)dt + e st f (t)dt t2 t3 e st f (t)dt + t 1 t 2 ] t n 1 t n = Ae st f (t)dt and integrating by parts, we have s lim A {e st f (t) t e st f (t) t e st f (t) t 1 [ t1 0 t2 e st f (t)dt + e st f (t)dt t 1 = tn=a t n 1 tn=a + t n 1 e st f (t)dt ]} =

16 lim A [e sa f (A) f (0) + s In his way we obtain A 0 ] e st f (t)dt = s L {f } = sl {f } f (0) As a corollary, we have the following A 0 e st f (t)dt f (0) Corollary Suppose that the functions f, f,..., f (n 1) are continuous and that f (n) is piecewise continuous on any interval 0 t A.

17 Suppose further that there exist constants K, a, and M such that f (t) Ke at, f (t) Ke at,..., f (n 1) (t) Ke at for t M. Then L {f (n) (t)} exists for s > a and is given by L {f (n) (t)} = s n L {f (t)} s n 1 f (0) s n 2 f (0)... sf (n 2) (0) f (n 1) (0) We use the previous results to solve IVP s using Laplace Transform. It is most useful for problems involving nonhomogeneous differential equations. However, just to illustrate the method we will start with a homogeneus case

18 Example 6.6 Consider the IVP Solution y y 2y = 0; y(0) = 1, y (0) = 0 Using the traditional method we find that the general solution is y(x) = c 1 e t + c 2 e 2t and applying initial conditions we get c 1 = 2/3 and c 2 = 1/3. Hence, the particular solution is y(x) = 2 3 e t e2t

19 Now, let us solve the same problem by using the Laplace transform. We start off with the differential equation Applying the Laplace Transform y y 2y = 0 because of linearity L {y y 2y = 0} and using corollary L {y } L {y } 2L {y} = 0 s 2 L {y} sy(0) y (0) [sl {y} y(0)] 2L {y} = 0

20 Taking Y (s) = L {y} and applying initial coditions (y(0) = 1, y (0) = 0), we obtain s 2 Y (s) sy(0) y (0) [sy (s) y(0)] 2Y (s) = 0 = Y (s) = s 1 s 2 s 2 = s 1 (s 2)(s + 1) The above can be written, using partial fractions, as Y (s) = 1/3 s 2 + 2/3 s + 1 Now, applying the inverse Laplace transform

21 { 1/3 y(t) = L 1 {Y (s)} = L 1 s 2 + 2/3 } s + 1 y(t) = L 1 {Y (s)} = 1 { } 1 3 L { } 1 s 2 3 L 1 s + 1 but we know that L {e at } = 1 s a or equivalently L 1 { 1 s a } = eat y(t) = 1 3 e2t e t The same procedure can be applied to the general second order linear equation with constant coefficients ay + by + cy = f (t)

22 obtaining Y (s) = (as + b)y(0) + ay (0) as 2 + bs + c + F (s) as 2 + bs + c ; F (s) = L {f (t)} The main difficulty that occurs in solving initial value problems by the transform method lies in the problem of determining the function y = y(t), corresponding to the inverse transform of Y (s). Since we will not deal with the formula for the inverse transform, because it requires complex integration, we will use a table of common Laplace Transform for basic functions (http: //

23 Example 6.7 Consider the IVP Solution y + y = sin(2t); y(0) = 2, y (0) = 1 Let us solve the problem by using the Laplace transform. We start off with the differential equation Applying the Laplace Transform y + y = sin(2t) L {y + y = sin(2t)}

24 because of linearity and using corollary L {y } + L {y} = L {sin(2t)} s 2 L {y} sy(0) y (0) + L {y} = 2 s taking Y (s) = L {y} and applying initial coditions (y(0) = 2, y (0) = 1), we obtain s 2 Y (s) 2s 1 + Y (s) = 2 s = Y (s) = 2s3 + s 2 + 8s + 6 (s 2 + 1)(s 2 + 4)

25 The above can be written, using partial fractions, as Y (s) = as + b s cs + d s = (as + b)(s2 + 4) + (cs + d)(s 2 + 1) (s 2 + 1)(s 2 = + 4) (a + c)s 3 + (b + d)s 2 + (4a + c)s + (4b + d) (s 2 + 1)(s 2 + 4) = 2s3 + s 2 + 8s + 6 (s 2 + 1)(s 2 + 4) Then, comparing coefficients of like powers of s, we have a + c = 2; b + d = 1; 4a + c = 0; 4b + d = 6;

26 Therefore, a = 2, b = 5/3, c = 0, d = 2/3 and Y (s) = Now, taking the inverse, we have 2s s /3 s /3 s y(t) = 2cos(t) sin(t) 1 3 sin(2t)

27 Example 6.8 Consider the IVP Solution y (4) y = 0; y(0) = 0, y (0) = 1, y (0) = 0, y (0) = 0 Let us solve the problem by using the Laplace transform. We start off with the differential equation Applying the Laplace Transform y (4) y = 0 L {y (4) y = 0}

28 because of linearity and using corollary L {y (4) } L {y} = 0 s 4 L {y} s 3 y(0) s 2 y (0) sy (0) y (0) Y (s) = 0 taking Y (s) = L {y} and applying initial coditions (y(0) = 0, y (0) = 1, y (0) = 0, y (0) = 0), we obtain s 4 L {y} s 2 y (0) Y (s) = 0 = Y (s) = s2 s 4 1 = s 2 (s 2 1)(s 2 + 1)

29 The above can be written, using partial fractions, as Y (s) = as + b s cs + d s = (as + b)(s2 + 1) + (cs + d)(s 2 1) (s 2 1)(s 2 = + 1) (a + c)s 3 + (b + d)s 2 + (a c)s + (b d) (s 2 1)(s 2 + 1) = s 2 (s 2 1)(s 2 + s) Then, comparing coefficients of like powers of s, we have a + c = 0; b + d = 1; a c = 0; b d = 0;

30 therefore, a = 0, b = 1/2, c = 0, and d = 1/2, and and take the inverse, we have Y (s) = 1/2 s /2 s y(t) = sinh(t) + sin(t) 2 = (et e t )/2 + sin(t) 2 = et 4 e t 4 + sin(t) 2

31 Example 6.9 The Laplace transforms of certain functions can be found conveniently from their Taylor series expansion. Using the Taylor series for sin(t) Let sin(t) = n=0 ( 1) n t 2n+1 (2n + 1)! f (t) = { sin(t) t t 0 1 t = 0

32 Find the Taylor series for f about t = 0. Assuming that the Laplace transform of this function can be computed term by term, determine L {f (t)} Solution For t 0 we have that f can written as f (t) = sin(t) t = n=0 ( 1) n t 2n+1 (2n + 1)!t = ( 1) n t 2n (2n + 1)! n=0 n=0 Applying the Laplace Transform { } ( 1) n t 2n L {f (t)} = L (2n + 1)!

33 because of linearity L {f (t)} = n=0 ( 1) n (2n + 1)! L {t2n } and using the table of Laplace Transforms, L {t m } = m! s m+1 L {f (t)} = L {f (t)} = n=0 ( 1) n (2n)! (2n + 1)! s 2n+1 n=0 ( 1) n (2n)! (2n + 1)!s 2n+1 = ( 1) n (2n)! (2n + 1)(2n)!s 2n+1 n=0

34 L {f (t)} = but, if we remember Therefore, we have n=0 ( 1) n (2n + 1)s 2n+1 = tan 1 (x) = n=0 n=0 ( 1) n x 2n+1 (2n + 1) L {f (t)} = tan 1 (1/s) ( 1) n (1/s) 2n+1 (2n + 1)

35 Laplace Transform To deal effectively with functions having jump discontinuities, it is very helpful to introduce a function known as the unit step function or Heaviside function. This function will be denoted by u c and is defined by { 0 t < c u c (t) = 1 t c

36 Laplace Transform The step can also be negative. For instance { 1 t < c u(t) = (1 u c (t)) = 0 t c

37 Laplace Transform In fact, any piecewise-defined function can be written as a linear combination of u c (t) s functions. For instance consider the function 2 0 t < t < 2 f (t) = 2 2 t < t

38 Laplace Transform We start with the function f 1 (t) = 2u 0, which agrees with f (t) on [0, 1). To produce the jump down of one unit at t = 1, we add u 1 (t) to f 1 (t), obtaining f 2 (t) = 2u 0 u 1 (t), which agrees with f (t) on [1, 2). The jump of one unit at t = 2 corresponds to adding u 2 (t), which gives f 3 (t) = 2u 0 u 1 (t) + u 2 (t). Thus we obtain f (t) = f 3 (t) = 2u 0 u 1 (t) + u 2 (t) The Laplace transform of u c for c 0 is easily determined: L {u c } = 0 e st u c dt = c e st dt = e cs, s > 0 s

39 Laplace Transform For a given function f defined for t 0, we will often want to consider the related function g defined by g(t) = u c f (t c) which represents a translation of f a distance c in the positive t direction.

40 Laplace Transform Theorem 6.3 If F (s) = L {f (t)} exists for 0 a < s, and if c is a positive constant, then L {u c f (t c)} = e cs L {f (t)} = e cs F (s), Conversely, if f (t) = L 1 {F (s)}, then L 1 {e cs F (s)} = u c f (t c)

41 Laplace Transform Example 6.10 If the function f is defined { sin(t) 0 t < π/4 f (t) = sin(t) + cos(t π/4) π/4 t find L {f (t)}.

42 Laplace Transform Solution Note that f (t) = sint + g(t), where Thus { 0 0 t < π/4 g(t) = cos(t π/4) π/4 t g(t) = u π/4 cos(t π/4)

43 Laplace Transform and L {f (t)} = L {sin(t)} + L {u π/4 cos(t π/4)} = L {sin(t)} + e πs/4 L {cos(t)} and using the table of Laplace Transforms L {f (t)} = 1 s s e πs/4 s = 1 + se πs/4 s 2 + 1

44 Laplace Transform Let s consider the following theorem Theorem 6.4 If F (s) = L {f (t)} exists for 0 a < s, and if c is a constant, then L {e ct f (t)} = F (s c), Conversely, if f (t) = L 1 {F (s)}, then s > a + c L 1 {F (s c)} = e ct f (t)

45 Laplace Transform Example 6.11 Find the inverse transform of Solution H(s) = 1 s 2 4s + 5 First of all the polynomial s 2 4s + 5, has complex roots. By completing the square in the denominator, we can write 1 H(s) = (s 2) 2 = F (s 2) + 1 where F (s) = (s 2 + 1) 1. L 1 {F (s)} = sin(t). It follows from the previous theorem that h(t) = L 1 {H(s)} = L 1 {F (s 2)} = e 2t sin(t)

46 Laplace Transform Shift in the time domain (t domain) If F (s) = L {f (t)} exists for 0 a < s, and if c is a positive constant, then L {u c f (t c)} = e cs F (s) L 1 {e cs F (s)} = u c f (t c) Shift in the frequency domain (s domain) If F (s) = L {f (t)} exists for 0 a < s, and if c is a constant, then L {e ct f (t)} = F (s c), s > a + c L 1 {F (s c)} = e ct f (t)