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MATH 374 Lecture 26

MATH 374 Lecture 26. Chapter 9: An Introduction to Laplace Transforms. 9.1: The Laplace Transform. Definition: E is the class of continuous or piecewise continuous functions f(t), defined for at least t > 0, such that |f(t)| < Ae Bt for some A, B.

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MATH 374 Lecture 26

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  1. MATH 374 Lecture 26 Chapter 9: An Introduction to Laplace Transforms

  2. 9.1: The Laplace Transform • Definition:Eis the class of continuous or piecewise continuous functions f(t), defined for at least t > 0, such that |f(t)| < AeBt for some A, B. If f 2E , then we say f is of exponential type. • Note that not all functions are of exponential type. For example, f(t) = et2 is not of exponential type. 2

  3. The Laplace Transform • Definition:If f 2E, then the Laplace transform of f(t) is defined to be the function F(s) := s01 e-st f(t) dt , defined at least for all sufficiently large t. We will write F(s) = L{f(t)} = L{f}.

  4. Example 1: Laplace Transforms of Common Functions Note that for positive integer n, Gamma(n+1) = (n+1) = n!. Recall that (p+1) = s01 e-x xp dx. 4

  5. The Laplace Transform is Linear • Let F(s) and G(s) be the Laplace transforms of f(t) and g(t), respectively. Then L{a f(t)+b g(t)} = a L{f(t)}+ b L{g(t) = a F(s) + b G(s). • Example 2: Try applying this idea, using the table above, with h(t) = 4 tet – 3 cos t. L {h(t)} = L {4 tet – 3 cos t} = 4L {t et} – 3L {cos t} = 4(1/(s-1)2) – 3(s/(s2+1))

  6. Laplace Transforms of Derivatives • One of the main reasons that Laplace transforms are used is the fact that they turn differentiation into multiplication! • Theorem 9.1: Let f(t) and f '(t) be in E. Then L {f '(t)} = -f(0) + sL { f(t) } = -f(0)+ sF(s). Proof: Use integration by parts. 

  7. Laplace Transforms of Derivatives • Corollary 1:If f and its first n derivatives belong to E, then L {f(n)(t)}= - [sn-1f(0) + … + s f(n-2)(0) + f(n-1)(0)] + snL {f(t)} = - [sn-1f(0) + … + s f(n-2)(0) + f(n-1)(0)] + snF(s). • Note that if f(t) is such that f and all of its derivatives up to order n-1 are zero at t = 0, then Corollary 1 implies L {f(n)(t)} = snL {f(t)} = snF(s).

  8. Laplace Transforms of Derivatives • Corollary 2:Let T(D) be a differential operator with constant coefficients: T(D) = a0Dn + a1Dn-1 + ... + an. If f and all of its derivatives up to order n belong to E and f(0) = f '(0) = ... = f(n-1)(0) = 0, then L {T(D)f} = T(s) L{f} where T(s) = a0sn + a1sn-1 + ... + an.

  9. Inverse Laplace Transform • Definition: Given function F(s), the inverse Laplace transform of F(s) is the function f(t) such that L {f(t)} = F(s). Notation: L-1{F(s)} = f(t). • Since the Laplace transform is linear, it follows that the inverse Laplace transform is linear, i.e. L-1{c1F(s) + c2G(s)} = c1L-1 {F(s)} +c2L-1 {G(s)}. • To find inverse Laplace transforms, one can use tables such as those above. For example, L-1{1/(s-4)} = ? = e4t.

  10. Solving ODE’s via Laplace Transforms • Laplace transforms can be used to solve ordinary differential equations! • The way we do this is to take the Laplace transform of both sides of a differential equation involving unknown function y(t). • If initial conditions are given at t = 0, we can include this information by using Theorem 9.1 and its Corollaries. • The transformed equation can be solved algebraically for Y(s), the transform of y(t). • Then y(t) = L-1 {Y(s)}. • If no initial conditions at zero are given, this technique can still be used to find a particular solution to the ODE by assuming initial data taken at t = 0 is zero.

  11. Example 3: Solve y’’ + y = 3e-2t (1) using Laplace transforms. • Solution: First note that the auxiliary equation is m2 + 1 = 0 which has roots m = §i, so yc = c1sin t + c2cos t (2) is the solution of the associated homogeneous problem. • For a complete solution, we need to find a particular solution to (1).

  12. Example 3: Solve y’’ + y = 3e-2t (1) using Laplace transforms. • Assume yp(t) is a solution of (1); yp and its derivatives are in E; and yp(0) = yp’(0) = 0. • Let Y(s) = L {yp(t)}, which is still unknown at this point and substitute yp for y in (1). • Take the Laplace transform of both sides of (1): L{yp’’ + yp} = L {3e-2t} L{(D2+ 1)yp} = 3L {e-2t} ) (s2 + 1)Y(s) = 3(1/(s+2)) by Corollary 2. ) Y(s) = 3/((s2+1)(s+2)). (3)

  13. Y(s) = 3/((s2+1)(s+2)) (3) Example 3: Solve y’’ + y = 3e-2t (1) using Laplace transforms. • Hence, yp(t) is the function whose Laplace transform is given by (3). • To find yp(t), all we need to do is find the inverse Laplace transform of Y(s)! • How do we find L-1{Y(s)}? • A common procedure is to use partial fractions to rewrite the RHS of (3) and use tables such as the one above to find the inverse Laplace transforms of each of the functions in the partial fraction decomposition!

  14. Y(s) = 3/((s2+1)(s+2)) (3) Example 3: Solve y’’ + y = 3e-2t (1) using Laplace transforms. • Rewriting the RHS of (3) as: Y(s) = 3/((s2+1)(s+2)) = (As)/(s2+1) + B/(s2+1) + C/(s+2), we find that A = -3/5, B = 6/5, and C = 3/5. (This can be done algebraically by clearing out denominators, equating coefficients of like terms on each side of the resulting equation, and solving, or via Mathematica, using the Apart command!)

  15. Y(s) = -3/5(s/(s2+1) + 6/5(1/(s2+1) + 3/5(1/(s+2) Example 3: Solve y’’ + y = 3e-2t (1) using Laplace transforms. • Thus, a particular solution to (1) is: yp(t) = L-1{Y(s)} = L-1{-3/5(s/(s2+1) + 6/5(1/(s2+1) + 3/5(1/(s+2)} = -3/5L-1{s/(s2+1)} + 6/5L-1{1/(s2+1)} + 3/5L-1{1/(s+2)} = -3/5 cos t + 6/5 sin t + 3/5 e-2t (4) It follows from (2), (4) that the general solution to (1) is y(t) = c1 sin t + c2 cos t - 3/5 cos t + 6/5 sin t + 3/5 e-2t.

  16. y’’ + y = 3e-2t (1) Example 4: Using Laplace transforms, solve (1) with y(0) = 1;y’(0) = -1. (2). • Solution: As before, assume y(t) is a solution of (1), with y, y’ , and y’’ in E. • Take the Laplace transform of both sides of (1): L{y’’ + y} = L {3e-2t} This time, using Corollary 1, we see that ) – [s y(0) + y’(0)] + s2L {y} + L {y} = 3 L {e-2t} ) – [s∙1 + (-1)] + s2Y(s) + Y(s) = 3(1/(s+2)) ) – s + 1 + (s2+1)Y(s) = 3(1/(s+2)) ) Y(s) = (s-1)/(s2+1) + 3/((s2+1)(s+2)) (5)

  17. Y(s) = (s-1)/(s2+1) + 3/((s2+1)(s+2) (5) Example 4: Using Laplace transforms, solve (1) with y(0) = 1;y’(0) = -1. (2). • Splitting up the first term on the RHS of (5) and writing the second RHS term in its partial fraction decomposition (see Example 3), we have Y(s) = s/(s2+1) – 1/(s2+1) – 3/5(s/(s2+1)) + 6/5(1/(s2+1)) + 3/5(1/(s+2))

  18. Y(s) = (s-1)/(s2+1) + 3/((s2+1)(s+2) (5) Example 4: Using Laplace transforms, solve (1) with y(0) = 1;y’(0) = -1. (2). • Splitting up the first term on the RHS of (5) and writing the second RHS term in its partial fraction decomposition (see Example 3), we have Y(s) = s/(s2+1) – 1/(s2+1) – 3/5(s/(s2+1)) + 6/5(1/(s2+1)) + 3/5(1/(s+2)) = 2/5(s/(s2+1)) + 1/5(1/(s2+1)) + 3/5(1/(s+2)). (6)

  19. Y(s) = 2/5(s/(s2+1)) + 1/5(1/(s2+1)) + 3/5(1/(s+2)). (6) Example 4: Using Laplace transforms, solve (1) with y(0) = 1;y’(0) = -1. (2). • Taking the inverse Laplace transform of (6), we have: y(t) = L-1{Y(s)} = L-1{2/5(s/(s2+1)) + 1/5(1/(s2+1)) + 3/5(1/(s+2))} = 2/5 L-1{s/(s2+1)} + 1/5 L-1{1/(s2+1)} + 3/5 L-1{1/(s+2)} = 2/5 cos t + 1/5 sin t + 3/5 e-2t

  20. Laplace Transforms and Convolutions • Here is a useful property of the Laplace transform: • Theorem 9.2:Let f and g 2E, and le h = f*g (the convolution of f and g) be defined by h(t) = (f*g)(t) = s0t f(t-u) g(u) du. (7) Then L{h} = L{f} L{g}.

  21. Example 6 • Suppose L{y(t)} = 3/((s2 + 1)(s+2)). What is y(t)? • Solution:L{y(t)} = 3/((s2 + 1)(s+2)) = (3/(s2+1))(1/(s+2)) = L(3 sin t) L(e-2t) Therefore Theorem 9.2 implies that y(t) = (3 sin t)*(e-2t).

  22. Example 6 • Note: In Example 3 above, we used partial fractions to show that if L{y(t)} = 3/((s2 + 1)(s+2)), then y(t) = - 3/5 cos t + 6/5 sin t + 3/5 e-2t. Therefore, since we just found that y(t) = (3 sin t)*(e-2t), it follows that: (3 sin t)*(e-2t) = - 3/5 cos t + 6/5 sin t + 3/5 e-2t, i.e. s0t 3 sin (t – u) e-2u du = - 3/5 cos t + 6/5 sin t + 3/5 e-2t.

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