Chapter 4 Laplace Transform It s time to stop guessing solutions and find a systematic way of finding solutions to non homogeneous linear ODEs. We define the Laplace transform of a function f in the following way. Definition 4.. The Laplace transform of f(t), written F(s), is given by (4.) L(f) = e st f(t)dt. That is, the Laplace transform acts on a function, f(t), integrates the t out, and creates function of s, which we denote F(s). Before we see why this is useful, we might want to know if the integral in the definition exists - it is after all an improper integral. We need another definition. Definition 4.2. We say a function f(t) is exponentially bounded or of exponential order if there exists non negative numbers a, k and M such that (4.2) f(t) ke at t M. We want to allow for interesting functions forcing an ODE. Definition 4.3. A function f is said to be piecewise continuous on a bounded interval if it has a finite number of discontinuities and the left and right limits exist at each discontinuity. It is said to be piecewise continuous on [, ] if it is piecewise continuous on every bounded subinterval I [, ]. And finally, Definition 4.4. A function f is said to be piecewise smooth if f and its derivative are piecewise continuous. Now we can say when the Laplace transform of a function exists. Theorem 4.5. Suppose f is piecewise continuous on [, ] and exponentially bounded. Then L(f) = F(s) exists for all s > a. 37
38 4. Laplace Transform We will not prove the theorem. However, the examples below will show why it is reasonable. Example 4.6. Compute L(). Solution. Let s apply Theorem 4.5 to see what to expect. The function f(t) = is exponentially bounded. Indeed, we can choosek =, a = M = in (4.2). Theorem 4.5 then implies the Laplace transform exists for all s > a =. To find the L() we compute it using the definition L() = e st dt = lim A A s e st = lim A s ( e As ). The limit of the exponential is zero as long as s >. Therefore, as expected. Example 4.7. Compute L(e at ). L() = s, s > Solution. Again we apply Theorem 4.5 to see what to expect. The function f(t) = e at is exponentially bounded if we choose k =, a = a, M = in (4.2). Theorem 4.5 then implies the Laplace transform exists for all s > a. To find the L(e at ) we compute it using the definition L(e at ) = e st e at dt = e (a s)t dt = lim A A a s e(a s)t = lim A s a ( ea(a s) ). We need the exponent to be negative forthe limit to exist. This is the case provided s > a. Therefore, L(e at ) = s > a. s a The next example is more difficult, but it shows the usefulness of the Laplace transform. Example 4.8. Compute L(f (t)), where f is a piecewise smooth exponentially bounded function. Solution. We are told there exists k, a, and M such that f(t) ke at for all t M. We use the definition of Laplace transform and integrate by parts A L(f (t)) = e st f (t)dt = lim e st f (t)dt A [ ] = lim e st f(t) A A + se st f(t)dt A [ = lim A e sa f(a) f()+ A se st f(t)dt ].
4. Laplace Transform 39 Since f is exponentially bounded lim A e sa f(a) lim A e sa ke aa = lim A ke(a s)a = provided s > a. Returning to our calculation [ L(f (t)) = lim A = s f()+ A e st f(t)dt f() = sl(f(t)) f(). se st f(t)dt This may look more useful if f is replaced with y. Then we have the formula L(y ) = sl(y) y(). That is, the Laplace transform turns a derivative y into an algebraic expression! Quite useful for solving ODEs! Indeed, we can easily compute L(y ). Using the formula for L(y ) we find L(y ) = sl(y ) y () = s 2 L(y) sy() y (). Table below provides the Laplace transform for many common functions. Example 4.9. Find the solution to by using Laplace transforms. y +3y +2y =, y() =, y () = Solution. We just take the Laplace transform of both sides. The left side of the ODE gives L(y +3y +2y) = L(y )+3L(y )+2L(y). It distributes this way because integrals behave this way, and the Laplace transform is an integral. Obviously L() =. Applying our formulas ( ) ( ) s 2 L(y) sy() y () +3 sl(y) y() +2L(y) =. Next we use the initial data and rearrange to find (s 2 +3s+2)L(y) = s+3, that is, s+3 L(y) = s 2 +3s+2. The Laplace transform turned an ODE into an algebraic problem. The only catch is that we need to find a function whose Laplace transform is (s+3)/(s 2 +3s+2). Table does not seem to help since this is not in the table. We have to modify the expression to use Table. He have using partial fractions s+3 s 2 +3s+2 = s+3 (s+)(s+2) = A s+ + B s+2. ]
4 4. Laplace Transform f(t) = L (F(s)) F(s) = L(f(t)). s s > 2. e at s a s > a 3. t n n! s n+ s > 4. t n at n! e (s a) n+ s > a 5. sin at a s 2 +a 2 s > 6. cos at s s 2 +a 2 s > 7. e at sinbt b (s a) 2 +b 2 8. e at cosbt s a (s a) 2 +b 2 s > a s > a 9. e at f(t) F(s a) s > a. u c (t)f(t) e cs L(f(t+c)). u c (t)f(t c) e cs L(f(t)) 2. δ(t c) e cs 3. t f(t τ)g(τ)dt F(s)G(s) 4. y (t) sl(y) y() 5. y (t) s 2 L(y) sy() y () Table. Laplace and inverse Laplace transforms for common functions. This implies s+3 = A(s+2)+B(s+), and A = and B = 2. Therefore, We may now use Table to find No guessing involved! L(y) = 2 s+ s+2. y(t) = 2e t e 2t.
4. Laplace Transform 4 Example 4.. Find the inverse Laplace transform of F(s) = 2s+ s 2 +4s+5, Solution. Just a matter of making it look like ones in the table. When the denominator does not factor, we complete the square. Recall to do this we take half the coefficient in front of the middle term, s, square it, then add and subtract it. Here s 2 +4s+5 = s 2 +4s+4 4+5 = (s+2) 2 +. Then F(s) = 2s+ s 2 +4s+5 = 2 (s+2) (s+2) 2 + 3 (s+2) 2 +. This is readily found in Table and f(t) = 2e 2t cost 3e 2t sint Homework 4. Find the inverse Laplace transform of the following.. F(s) = 3 s 2 +4 2. F(s) = 4 (s ) 3 3. F(s) = 2 s 2 +3s 4 4. F(s) = 3s s 2 s 6 5. F(s) = 2s+2 s 2 +2s+5 6. F(s) = 2s 3 s 2 4 7. F(s) = 2s+ s 2 2s+2 8. F(s) = 8s2 4s+2 s(s 2 +4) 9. F(s) = 2s s 2 +4s+5. F(s) = 2s 3 s 2 +2s+ Use the Laplace transform to solve the following ODEs. y y 6y = ; y() =, y () = 2. y 2y +2y = ; y() =, y () = 3. y +2y +5y = ; y() = 2, y () = 4. y IV y = ; y() =, y () =, y () =, y () = 5. y 2y +2y = cost; y() =, y () = Answers. f(t) = 3 2 sin2t 2. f(t) = 2t 2 e t 3. f(t) = 2 5 et 2 5 e 4t 4. f(t) = 9 5 e3t 6 5 e 2t 5. f(t) = 2e t cos2t 6. f(t) =?? 7. f(t) = 2e t cost+3e t sint 8. f(t) = 3+5cos2t 2sin2t 9. f(t) = 2e 2t cost+5e 2t sint. f(t) = 2e t cos3t 5 3 e t sin3t. y(t) = 5 (e3t +4e 2t ) 2. y(t) = e t sint 3. y(t) = 2e t cos2t+ 2 e t sin2t 4. y(t) = et +e t 2 5. y(t) = 5 (cost 2sint+4et cost 2e t sint)
42 4. Laplace Transform 4.2. Heaviside Function We can force ODEs with more interesting functions now that we have a more non guessing method for solving ODEs. Indeed, consider the Heaviside function given by { t < c (4.3) u c (t) = t c, where c >. We can think of the Heaviside function as a switch. It turns on at t = c. Thus the function u c (t) is a switch that turns off at t = c. 2 2.5.5 u c u c.5.5.5.5 2 t.5.5 2 t Figure. Left: Heaviside function with c =. Right: graph of u (t). Example 4.. Write the function in terms of Heaviside functions. f(t) = { t < 2π sint t 2π Solution. The function f starts out as the function until t = 2π. At t = 2π zero turns off and sint turns on. Thus f(t) = u 2π (t)+u 2π (t)sint = u 2π (t)sint. Keep in mind, u c (t) is a switch that turns on at t = c. A graph of f is given below. Example 4.2. Write the function, < t <, f(t) =, t < 2,, t 2, in terms of Heaviside functions.
4.2. Heaviside Function 43.5 f(t) -.5-5 5 2 t Figure 2. Graph of f(t) = u 2π (t)sint in Example 4.. Solution. The function f starts out as the function until t =. At t = zero turns off and turns on. At t = 2 one turns off turns on. Thus f(t) = u (t)+u (t) u 2 (t)+u 2 (t) = u (t) u 2 (t). A graph of this function is given below..5 f(t).5.5.5 2 2.5 3 t Figure 3. Graph of f(t) = u (t) u 2 (t) in Example 4.2. We need to compute the Laplace transform of the Heaviside function. In particular, the above examples imply we need to find the Laplace transform of functions in the form u c (t)f(t). Using the definition of Laplace transform, we find L(u c (t)f(t)) = = e st u c (t)f(t)dt = c e st f(t)dt e s(r+c) f(r+c)dr = e cs e sr f(r +c)dr = e cs L(f(t+c)).
44 4. Laplace Transform Example 4.3. Compute the Laplace transform of u 2π (t)sint. Solution. Using. in Table (just derived above) L(u 2π (t)sint) = e 2πs L(sin(t+2π)) = e 2πs L(sint) = e 2πs s 2 +. Example 4.4. Compute the Laplace transform of u 2 (t)(t t +t+). Solution. Using. in Table L(u 2 (t)(t 2 +t+) = e 2s L((t+2) 2 +(t+2)+)) = e 2s L(t 2 +5t+7) = e 2s ( 2 s 3 + 5 s 2 + 7 s Example 4.5. Compute the Laplace transform of t, t < 3, f(t) = (t 3), 3 t < 4,, t 4. Solution. Here The Laplace transform is ). f(t) = t u 3 (t)t+u 3 (t)(t 3) u 4 (t)(t 3)+u 4 (t). F(s) = s 2 e 3s L((t+3))+e 3s L((t+3) 3) e 4s L((t+4) 3)+ e 4s s = ( s 2 e 3s s 2 + 3 ( )+ e 3s s s 2 e 4s s 2 + )+ e 4s s s. Example 4.6. Compute the inverse Laplace transform of G(s) = (2s 3)e s s 2 +2s+. Solution. We need to make this look like in Table. We may write G(s) as 3e s G(s) = 2 (s+)e s (s+) 2 +3 2 5 3(s+) 2 +3 2 = e s F (s)+e s F 2 (s). Next we find the inverse of F (s) and F 2 (s) from the table. Here f (t) = 2e t cos3t, f 2 (t) = 5 3 e t sin3t. The inverse Laplace transform is g(t) = u (t)f (t )+u (t)f 2 (t) ( = u (t) 2e (t ) cos(3(t )) 5 ) 3 e (t ) sin(3(t )).
4.2. Heaviside Function 45 Example 4.7. Find the solution to the ODE { y t, t <, +y =, t. y() =, y () =. Solution. The right side is the same as ( t) u (t))( t). Taking the Laplace transform of both sides, or Set Then It follows that Here Hence, (s 2 L(y) s)+l(y) = s s 2 e s L( (t+)), L(y) = s s 2 + + s(s 2 +) s 2 (s 2 +) + e s s 2 (s 2 +), F(s) := G(s) := s(s 2 +) = s s (s 2 +) s 2 (s 2 +) = s 2 (s 2 +). L(y) = s s 2 + +F(s) G(s)+e s G(s). y(t) = cos(t)+f(t) g(t)+u (t)g(t ). f(t) = cost g(t) = t sint. y(t) = t+sint+u (t)(t sin(t )). Homework 4.2 Find the Laplace transform of the following. { < t < 2 < t < 3. f(t) =. f(t) = t < 2 t 2 2t+2 t t 2 { { < t < 2 cost < t < 2π 2. f(t) = (t 2) 2 4. f(t) = t 2 sint t 2π Find the inverse Laplace transform of the following. 5. F(s) = 3!e s (s 2) 4 6. F(s) = e 2s s 2 +s 2 7. F(s) = 2(s )e 2s s 2 2s+2 8. F(s) = 2e 2s s 2 4 9. F(s) = (s 2)e s s 2 4s+3. F(s) = e s +e 2s e 3s e 4s s
46 4. Laplace Transform Solve the following differential equations. {. y +y = f(t); y() =,y < t < 3π () = ; f(t) = t 3π { 2. y +3y +2y = f(t); y() =,y < t < () = ; f(t) = t 3. y +3y +2y = u 2 (t); y() =,y () = ; 4. y +y = f(t); y() =,y () = ; f(t) = 2. F(s) = 2e 2s /s 3 3. F(s) = e s (s 2 +2)/s 3 5. f(t) = u (t)(t ) 3 e 2(t ) 6. f(t) = 3 u 2(t)(e t 2 e 2(t 2) ) 7. f(t) = 2u 2 (t)e t 2 cos(t 2) 8. f(t) = u 2 (t)sinh2(t 2) 9.??. f(t) = u (t)+u 2 (t) u 3 (t) u 4 (t) Answers { t/2 < t < 6 3 t 6. y(t) = cost+sint u 3π (+cost) 2. y(t) = 2 (+e 2t ) e t u (t) ( 2 + 2 e 2(t ) e (t )) 3. y(t) = e t e 2t +u 2 (t) ( 2 e (t 2) + 2 e 2(t 2)) 4. y(t) = 2 (t+sint) 2 u 6(t)(t 6 sin(t 6)) 4.3. The Delta Function We have developed a sophisticated method of solving ODEs. So much so that we can think of forcing an ODE with a large force that acts over a short time. The physical phenomena we wish to describe might be a spark, or a baseball being hit by a bat. We could approximate such a force with the functions { /ǫ ǫ δ ǫ (t) = 2 < t < ǫ 2, otherwise where ǫ >. We think of ǫ as being a small positive number. Figure 4 below shows a graph of δ ǫ. Notice that the area under the curve for each ǫ > is one. We do not want an infinite velocity to result. We are interested ǫ small. In fact, we need to compute the limit as ǫ goes to zero if we want to think of the force as acting instantaneously. If we want to compute the Laplace transform of such a force, we will need to compute lim ǫ f(t)δ ǫ (t)dt for a continuous f(t). The limit is not hard to compute. Indeed, the function δ ǫ (t) is zero most of the time. If f continuous and ǫ is small, then f does not change
4.3. The Delta Function 47 5 4 3 δ(τ) 2-2 - 2 t Figure 4. Graph of δ ǫ(t) for various ǫ. much on the small interval ǫ 2 < t < ǫ 2. In fact, it is close to f(). Thus the integral is approximately lim ǫ More generally (4.4) lim f(t)δ ǫ (t)dt lim ǫ ǫ = f()lim = f(). f()δ ǫ (t)dt ǫ f(t c)δ ǫ (t)dt = f(c). δ ǫ (t)dt = f()lim ǫ We cannot move the limit inside the integral since lim t δ ǫ (t) does not exist. However, formally we set (4.5) lim t δ ǫ (t) = δ(t) and we write f(t c)δ(t)dt = f(c) with the understanding that (4.5) really means (4.4). The function δ(t) is called the Delta function. The Laplace transform of the Delta function is easy to compute. Indeed, L(δ(t c)) = = e cs. Example 4.8. Find an anti derivative of δ(t ). e st δ(t c)dt Solution We just need to solve y = δ(t ) y() =.
48 4. Laplace Transform Taking the Laplace transform of both sides, we find That is, and L(y) y() = e s. L(y) = e s s, y(t) = u (t). Example 4.9. Find an two anti derivatives of δ(t ). Solution We just need to solve y = δ(t ) y() =, y () =. Taking the Laplace transform of both sides, we find That is, and s 2 L(y) sy() sy () = e s. L(y) = e s s 2, y(t) = u (t)(t ). There is a pattern here. Figure 5 plots the Delta function and its anti derivatives. Does the pattern make sense to you? Example 4.2. Find the solution to y 4y +3y = δ(t ), y() =, y () =. Solution. Taking the Laplace transform of both sides, we find Set Then L(y) = e s (s 3)(s ) + (s 3)(s ). F(s) = (s 3)(s ) = 2 s 3 2 s. f(t) = 2 e3t 2 et, and y = u (t)f(t )+f(t) or y = u (t) ( e 3(t ) e t ) + ( e 3t e t). 2 2
4.3. The Delta Function 49 e+7 8e+6 6e+6 δ(τ) 4e+6 2e+6.5.5 2 τ.4.2 u (t).8.6.4.2.5.5 2 t.4.2 y=u (t)(t-).8.6.4.2.5.5 2 t Figure 5. Graph of (top) δ(t ); its anti derivative (middle) u (t ); and it second anti derivative (bottom) u (t)(t ). Homework 4.3 Solve the following differential equations.. y +2y +2y = δ(t π); y() =,y () = 2. y +4y = δ(t π) δ(t 2π); y() =,y () = 3. y +4y +3y = +δ(t ); y() =,y () =. 4. y +y = δ(t 2π)cost; y() =,y () = Answers. y(t) = e t cost+e t sint+u π (t)e (t π) sin(t π) 2. y(t) = 2 u π(t)sin2(t π) 2 u 2π(t)sin2(t 2π) 4. y(t) = sint+u 2π (t)sint
5 4. Laplace Transform 4.4. The Convolution There is a way to avoid the use of partial fractions to compute the inverse Laplace transform. Suppose F(s) = L(f(t)), G(s) = L(g(t)). Set H(s) = F(s)G(s). It is too bad the inverse Laplace transform of H(s) is not f(t)g(t). However, it s not hard to show That is, L (F(s)G(s)) = = t t = f g. f(τ)g(t τ)dτ f(t τ)g(τ)dτ (4.6) h(t) = L (F(s)g(s)) = f g. Equation (4.6) is called the convolution of f and g. Example 4.2. Find the inverse Laplace transform of H(s) = s 2 (s 2 +). Solution. Set Then and F(s) = s2, G(s) = s 2 +. f(s) = t, L (H(s)) = t g(s) = sint, τ sin(t τ)dτ = τ cos(t τ) = t+sin(t τ) = t sint. t t Alternatively we could use partial fractions and s 2 (s 2 +) = s 2 s 2 +. Then the inverse is easy and the same as above. cos(t τ)dτ
4.4. The Convolution 5 Homework 4.3 Solve the following differential equation.. y +y = g(t); y() =,y () =. y(t) = t sin(t τ)g(τ)dτ +sint Answer