4 Laplace transforms 4. Definition and basic properties The Laplace transform is a useful tool for solving differential equations, in particular initial value problems. It also provides an example of integral transforms, which play an important role in various branches of mathematics. It is named after Pierre-Simon Laplace (749-87). Definition 4. Let f : [, ) R. We define the Laplace transform of f as Sometimes we write L[f(t)] for f. f(s) = f(t)e st dt. (4.) Note that the integral defining the Laplace transform converges for s s provided f(t) Ke s t for some constant K. 4.. Examples (i) Let f(t). Then (ii) Let f(t) = e αt. Then f(s) = e st dt = s. (4.) for s > α. (iii) Let f(t) = sin(αt). Then f(s) = e (s α)t dt = s α (4.3) f(s) = = Im sin(αt)e st dt = Im e iαt e st dt e (s iα)t dt = Im s iα = Similarly one shows that if f(t) = cos(αt) then f(s) = s s + α. α s + α. (4.4) (iv) If f(t) = t then f(s) =. You should be able to prove (by induction) that if f(t) = tn s then f(s) = n! s n+ 4.. Simple properties of Laplace transforms It is easy to check that the Laplace transform is linear in the sense that L[f (t) + f (t)] = L[f (t)] + L[f (t)] and L[αf(t)] = αl[f(t)]. (4.5) 37
For us it is particularly important to work out the Laplace transform of the derivative of a function: and hence L[f (t)] = f (t)e st dt = f(t)e st t= t= + s f(t)e st dt It follows that and, by induction, L[f (t)] = s f(s) f(). (4.6) L[f (t)] = sl[f ](s) f () = s f(s) sf() f (), (4.7) L[f (n) (t)] = s n L[f] s n f() s n f ()... f (n ) (). (4.8) 4. Solution of initial value problems Consider the initial value problem Taking Laplace transforms we obtain y 3y + y = e 3t, y() =, y () =. (4.9) s ȳ(s) sy() y () 3(sȳ(s) y()) + ȳ(s) = s 3. (4.) Inserting the initial values and re-arranging terms yields and hence (s 3s + )ȳ(s) = s 3 + s 3 (4.) ȳ(s) = s 6s + (s )(s )(s 3). (4.) Our next goal is to invert the Laplace transform, i.e. to find the function y(t) whose Laplace transform is (4.). To achieve this we write the right hand side of (4.) in terms of partial fractions Then s 6s + (s )(s )(s 3) = A s + B s + C s 3. (4.3) s 6s + = A(s )(s 3) + B(s )(s 3) + C(s )(s ). (4.4) 38
Setting in turn s =, s = and s = 3 we find A = 5, B =, C =. (4.5) Thus ȳ(s) = 5 (s ) (s ) + (s 3). (4.6) Now we find y(t) by comparing (4.6) with (4.3). Using the linearity of the Laplace transform you can check that y(t) = 5 et e t + e3t (4.7) has the Laplace transform (4.6). One can show that that the Laplace transform of a function is (essentially) unique. It is thus possible to invert Laplace transforms by guessing or inspection. There a more systematic method based on contour integrals, but in many applications the method of inspection is the quickest. It is the one we will use in this course. To sum up, we have found that the Laplace transform translates the problem of solving a differential equation into an algebraic problem which even incorporates the initial values. The three steps in solving a differential equation using the method of Laplace transfroms are:. Apply the Laplace transform to the differential equation for y(t), using the initial conditions. Solve the resulting algebraic equation for ȳ(s). 3. Invert the Laplace transform to find y(t). Of these, the last step is often the trickiest. We therefore devote a special subsection to it. 4.3 Inverting Laplace transforms Finding the inverse Laplace transform of ȳ(s) by inspection is sometimes facilitated by expanding ȳ(s) in partial fractions. The example in the previous section contained the simplest sort of partial fraction. The following example is designed to remind you of some possible additional complications, such as repeated factors. Example: Find y(t) when ȳ(s) = s + s + s (s + ). Write s + s + s (s + ) = As + B s 39 + Cs + D s +
and multiply by s (s + ) to obtain s + s + = As(s + ) + B(s + ) + (Cs + D)s = B + As + (B + D)s + (A + C)s 3. (4.8) Now compare coefficients of powers of s: s : B = s : A = s : B + D = D = s 3 : A + C = C = (4.9) Thus and hence ȳ(s) = s + s s s + + s + (4.) A second important tool is provided by the following Lemma 4. Let g(t) = e ct f(t). Then ḡ(s) = f(s c). Proof: This is a simple computation: ḡ(s) = y(t) = + t cos t + sin t (4.) The following example illustrate this result: L[cos(t)] = e (s c)t f(t)dt = f(s c). (4.) L[t] = s L[te t ] = (s ) s s + 4 L[et cos(t)] = (s ) (s ) + 4 More importantly we can use the lemma (4.) to invert Laplace transforms: Example: Find y(t) when (i) ȳ(s) = s + s + 3 (ii) ȳ(s) = s(s + s + 3). (i) Since ȳ(s) = (ii) Now use Thus we deduce (s + ) + = (s + ) + s(s + s + 3) we deduce y(t) = e t sin( t) = 3s s + (by partial fractions) 3 (s + s + 3) = 3s s + 3 (s + ) + 3 (s + ) +. y(t) = 3 3 e t cos( t) 3 e t sin( t). (4.3) 4
4.4 ODE s involving discontinuous functions The function defined by u c (t) = { if t < c if t c (4.4) is called the unit step function or Heaviside function (named after Oliver Heaviside, 859-95). The step function is useful in describing situations where an external effect (a force, a voltage) is suddenly switched on. Suppose for example that an external force f(t) acting on a spring is switched on at t = c. The force experienced by the spring is F (t) = { if t < c f(t) if t c (4.5) In terms of the step function we can simply write Similarly, if an external force is switched off at t = c, i.e. F (t) = u c (t)f(t). (4.6) F (t) = { f(t) if t < c if t c (4.7) 4
we can express the resulting function in terms of the step function F (t) = ( u c (t))f(t). (4.8) Since step functions arise naturally in a number of applications, we would like to be able to compute Laplace transforms of functions like (4.6) and (4.8). The following lemma is useful for that purpose: Lemma 4.3 L[u c (t)f(t c)] = e cs f(s). Proof: This follows again by direct computation. L[u c (t)f(t c)] = Now change variables to τ = t c. Then L[u c (t)f(t c)] = Corollary 4.4 L[u c (t)] = e cs s. u c (t)f(t c)e st dt = c f(t c)e st dt (4.9) f(τ)e s(τ+c) dτ = e cs f(τ)e sτ dτ = e cs f(s). (4.3) In order to illustrate the application of the lemma to situations like (4.6) and (4.8) consider the following examples: (i) Find the Laplace transform of y(t) = { if t < t if t (4.3) We rewrite y(t) as y(t) = u (t) t = u (t)(t ) + u (t). (4.3) Now we can apply the lemma (4.3) to the first term and its corollary to the second to find that ( ȳ(s) = e s s + ). (4.33) s (ii) If ȳ(s) = e s, find y(t). s 3 Recall L[t ] = s and hence write ȳ(s) as 3 e s L[ t ]. Then deduce from the lemma (4.3) that. y(t) = u (t) (t ) 4
(iii) Solve y + y + y = f(t), y() =, y () =, where { if t < 3 f(t) = t 3 if t 3 We note that f(t) = u 3 (t)(t 3) and take the Laplace transform of the equation. Using the given initial data we find Using (s + s + )ȳ(s) s = e 3s s ȳ(s) = s + (s + ) + e 3s s (s + ) (4.34) s + (s + ) + = = (s + ) (s + ) s + + (4.35) (s + ) and the partial fraction expansion s (s + ) = s + s + (s + ) + (s + ) (4.36) we have ȳ(s) = = s + + and deduce from the lemma (4.3) ( s + ) s + (s + ) + (s + ) (s + ) + e 3s s + + (s + ) + e 3s L[ + t + e t + te t ] (4.37) y(t) = e t + te t + u 3 (t)( + (t 3) + e (t 3) + (t 3)e (t 3) ) 4.5 The Dirac delta function = e t ( + t) + u 3 (t)( 5 + t + (t )e (t 3) ). (4.38) Consider the situation where a mass is subjected to an impulsive force, i.e. a very large force acting for a very short time. If a constant force F acts for a time t, the momentum of the mass is increased by P = F t. Now suppose we take t and F in such a way that P = F t remains constant. We say that the mass has been subjected to an impulse of size P. If the mass is subjected 43
to an impulse of size at time t we may regard this as the limit lim t t (u t t(t) u t (t)) = lim t t (u t (t + t) u t (t)), (4.39) where the equality u t (t + t) = u t t(t) follows directly from the definition (4.4) of the Heaviside function. This limit is nothing but the derivative of the Heaviside function, for which we introduce a special name: Definition 4.5 We define the Dirac delta function as the derivative of the Heaviside function u t (t) with respect to t: δ(t t ) = d dt u t (t). The function δ(t t ) is named after the English physicist Paul Maurice Adrien Dirac who lived 9-984. (This is the first time that this course enters the twentieth century!) The Dirac delta function δ(t t ) is zero everywhere except at t where it is infinite. In terms of the delta function, the differential equation for a particle of mass m on a spring with spring constant k immersed in a fluid with damping constant r which is subject to an impulse P at time t = t is m dy dt + r dy dt + ky = P δ(t t ). (4.4) In order to solve equations of this form using Laplace transforms, we need to know the Laplace transform of the delta function. This will follow from the following important Theorem 4.6 For any continuous function f : R R, δ(t t )f(t) dt = f(t ). Proof: Going back to first principles and expressing the derivative of the step function in terms of the limit (4.39) we have δ(t t )f(t) dt = = lim t = lim t = lim t t (u t t(t) u t (t)) f(t) dt t ( ) f(t) dt f(t) dt t t t 44 t t t t f(t) dt = f(t ), (4.4)
where we used the continuity of f in the last line. As an immediate consequence we have Corollary 4.7 L[δ(t t )] = e st provided t. Proof: if t. L[δ(t t )] = e st δ(t t ) dt = e st Note in particular that L[δ(t)] =. Armed with this corollary we can tackle the following example. A mass of kg oscillates on a spring with spring constant N/m. Air resistance may be ignored. The mass is initially at rest in equilibrium and impulses of N sec and 3N sec are applied at times t = and t =. Describe the motion of the mass. The equation of motion is Taking Laplace transforms so that Thus y + y = δ(t) + 3δ(t ), y() = y () =. (4.4) (s + )ȳ(s) = + 3e s (4.43) ȳ(s) = s + + 3e s s + = L[sin t] + 3e s L[sin t]. (4.44) y(t) = sin t + 3u (t) sin(t ) = { sin t if t < sin t + 3 sin(t ) if t (4.45) 4.6 The convolution integral Definition 4.8 Let f, g : [, ) R. We define the convolution f g of the functions f and g as the function f g (t) = t f(t τ)g(τ)dτ. (4.46) As an elementary example, consider f(t) = sin t and g(t). Then f g (t) = t sin(t τ)dτ = t sin v dv = cos t, (4.47) where we have changed variables to v = t τ. A basic property of the convolution is that it is commutative: 45
Lemma 4.9 Let f, g : [, ) R. Then f g (t) = g f (t). (4.48) Proof: Changing integration variables from τ to v = t τ in the definition (4.46) we find f g (t) = t f(v)g(t v)dv = t g(t v)f(v)dv = g f (t). (4.49) A further remarkable property is the behaviour of the convolution under Laplace transforms. It is summed up in the convolution theorem: Theorem 4. For two functions f and g f g (s) = f(s)ḡ(s) (4.5) Thus the Laplace transform of a convolution of two function is the ordinary product of the Laplace transforms! Proof: f g (s) = = = (f g)(t)e st dt ( t ) f(t τ)g(τ)dτ e st dt ( ) g(τ) f(t τ)e st dt dτ (4.5) where we have changed the order of integration in the last step. Note that the limits are such that we integrate over the region τ t <. τ Next we change variables in the inner integral from t to v = t τ. Then ( ) f g (s) = g(τ) f(v)e s(v+τ) dv dτ = f(v)e sv dv g(τ)e sτ dτ = f(s)ḡ(s) (4.5) 46
This theorem is useful in a number of ways. It allows us to solve certain kinds of integral equations, involving a function and integrals of that function. Example: Find y(t) such that Taking Laplace transforms y(t) = 4t 3 t y(t τ) sin(τ)dτ. (4.53) ȳ(s) = 4 s 3ȳ(s) s + (4.54) we find ȳ(s) = 4(s + ) s (s + 4) = s + 3 s + 4 (4.55) and deduce that y(t) = t + 3 sin(t). (4.56) 47