Math Laplace transform

Transcription

1 1 Math Laplace transform Erik Kjær Pedersen November 22, 2005 The functions we have treated with the power series method are called analytical functions y = a k t k The differential equations we are most interested in are: Linear, second order differential equations with constant coeefficients 0

2 2 y + ay + by = r(t) The reason is that these equations show up in lots of mechanical and electrical systems. The term r(t) on the right hand side is usually given by some kind of outer force on a mechanical system or electromotive force in an electrical system. We would like to be able to handle more general functions on the right hand side such as { 0 for t < a r(t) = 1 for t > a because a function like that represents something we often do, namely Turn on the battery Another problem is how to represent the force given as Hitting with a hammer

3 3 This is represented by a function which is 0 except a very short interval where it is very large. Dirac s delta function, δ(t) is 0 everywhere except for t = 0, but at t = 0 it is so large that b a δ(t)dt = 1 whenever a < 0 and b > 0 This obviously does not make sense, but we can make functions approximating this behavior The Laplace transform is a way to deal with these problems Let f (t) be a function defined for t 0 F (s) = L(f ) = 0 e st f (t)dt This is not defined for all functions, but only for functions that do not grow too fast.

4 It is however defined for functions say of subexponential growth It turns out that there is a way to go back from F to get f back We have problems in t-space, solve them in s-space and go back to t-space. We now start investigating some of the properties of the Laplace transform Theorem L(af + bg) = al(f ) + bl(g) Here f and g are functions and a and b are constants. We will now develop a number of theorems to calculate the Laplace transform First shifting theorem: Replacing s by s a in the transform Theorem L(e at f (t)) = L(f )(s a) 4

5 or 5 e at f (t) = L 1 (F (s a)) Much more importantly, the Laplace transform of a derivative Theorem L(f ) = sl(f ) f (0) The proof is a small calculation L(f ) = 0 e st f (t)dt = [e st f (t)] 0 + s 0 e st f (t)dt. The limit when t has to be 0 so the result follows. So differentiating in t-space is (almost) just multiplication by s in s-space.

6 6 This theorem is very good for calculations. Obviously the constant function 0 has L(0) = 0. Letting f (t) = 1 we now get L(0) = sl(1) 1 This is because the derivative of the constant function 1 is 0, so L(1) = 1 s For the function f (t) = t using the same trick we get so L(t) = 1 s 2 Inductively using that L(1) = sl(t) 0

7 7 We obtain the formula nl(t n 1 ) = sl(t n ) 0 L(t n ) = n! s n+1 The first shifting theorem tells us that L(e at ) = 1 s a 1 Because s a is 1 s shifted by a. We thus have to compensate by multiplying 1 by e at

8 Let us find out about the transforms of cos(ωt) and sin(ωt). We use the following calculation where the left side is the derivatives 8 ωl(cos(ωt)) = L(sin(ωt)) 0 The last 0 is there because sin(0) = 0 Similarly ωl(sin(ωt)) = L(cos(ωt)) 1 In this case the last 1 is there because cos(0) = 1 From these two calculations we get that L(cos(ωt)) = s s 2 + ω 2 L(sin(ωt)) = The first shifting theorem now tells us that ω s 2 + ω 2

9 L(e at cos(ωt)) = How about Dirac s delta function s a (s a) 2 + ω 2 L(e at sin(ωt)) = 0 e st δ(t a)dt = e sa 9 ω (s a) 2 + ω 2 This is because the value of e st at t = a is e sa. So while Dirac s delta function may not quite make sense it has a perfectly good Laplace transform. What is the value of Laplace transform. Let us try an example where we already know the answer Example y + y = 1 y(0) = 0 y (0) = 1 We know the general solution is y = 1 + c 1 cos(t) + c 2 sin(t).

10 10 We can quickly determine c 1 = 1 and c 2 = 1 to get the solution y = 1 cos(t) + sin(t) Mechanically this represents an undampened spring with a constant force of 1 applied and a set of initial conditions. Let is try something more difficult. Apply the Laplace transform and denoting L(y) by Y We begin our calculation by and L(y ) = sy y(0) = sy L(y ) = sl(y ) y (0) = sl(y ) 1 = s 2 Y 1 Now apply L to the whole equation, to get

11 11 s 2 Y 1 + Y = 1 s because L(1) = 1 s so Y = s 1 + s 2 = s 2 s 1 + s s Continuing the calculation we see that L(y) = Y = L(sin(t)) L(cos(t)) + L(1) = L(1 + sin(t) cos(t)) Applying the inverse Laplace transform we luckily get the same result as before namely y = 1 cos(t) + sin(t)

12 12 Now let us hit the spring with a hammer at time t = 1 instead of applying a constant force of 1. The equation will now be y + y = δ(t 1) y(0) = 0 y (0) = 1 Our classical method to solve this does not work but apply the Laplace transform We get the same thing on the left side and the Laplace transform of the delta function on the right side s 2 Y 1 + Y = e s We can calculate the Laplace transform Y = L(y) Y = 1 + e s 1 + s 2

13 13 The question now becomes how to recover y from the Laplace transform Y. That sets the stage for the next theorem, the t-shifting theorem. Second shift theorem Assume we have a given function f (t), t 0. We now physically move the graph to the right to obtain a shifted function: { 0 for t < a g(t) = f (t a) for t a What happens to the Laplace transform Theorem L(g) = e as L(f )