Math 353 Lecture Notes Week 6 Laplace Transform: Fundamentals

Math 353 Lecture Notes Week 6 Laplace Transform: Fundamentals J. Wong (Fall 217) October 7, 217 What did we cover this week? Introduction to the Laplace transform Basic theory Domain and range of L Key properties linearity transform of derivatives Inversion Laplace transform of functions Some examples Initial value problems Using L to solve constant-coefficient, linear IVPs 1 A quick introduction We turn our attention now to transform methods, which will provide not just a tool for obtaining solutions, but a framework for gaining insight into structure. The idea is to define a transform operator L on functions, L : original space transformed space such that the ODE in the transformed space is much easier to solve. One class of transforms is an integral transform, which takes the form L[f(t)] = K(s, t)f(t) dt Ω 1

where Ω is some domain (usually (, ) or (, )) and K(s, t) is a function called the kernel of the transform. One of the two most important integral transforms is the Laplace transform L, which is defined according to the formula L[f(t)] = F (s) = e st f(t) dt, i.e. L takes a function f(t) as an input and outputs another function F (s) as defined above. Before developing the theory, here is a quick sketch of why the transform is useful for studying ODEs. The two key properties of L are (to be proven later): L is a linear operator L converts differentiation into multiplication by s (plus a little bit): L[f (t)] = sl[f] f(). Because of these two properties, linear, constant coefficient ODEs are converted into algebraic equations. An algebraic equation is much easier to work with than a differential equation! For example, let s solve = y y Let Y (s) = L[y(t)] be the Laplace transform of the solution. Applying L to the equation, we obtain the transformed equation Since L[] =, we get L[] = L[y ] L[y] = sy y() Y. = (s 1)Y y(), which is trivial to solve! The transformed solution to the ODE is then Y (s) = y() s 1. Here is the point at which we have to do actual work - the price of transforming the ODE is that we have to undo the transformation to get the desired solution, y(t). In this case, it is easy to show that L[e t ] = 1/(s 1), from which we can conclude that 2 Fundamentals y(t) = y()e t. Now we go through the basic theory. The treatment here is not be completely rigorous; some technical details are omitted in favor of getting to the key points. 2

2.1 Definition The Laplace transform F (s) of a function f(t) is given by L[f(t)] = F (s) = e st f(t) dt. (1) The transform, of course, is only defined when the integral above converges for some interval of s-values. 2.2 Domain/range of the Laplace transform Notice that the integral in (1) will be finite if f(t) does not grow too quickly. A function f(t) is of exponential type (or order) if there are constants a, K, and M such that f(t) Ke at for t > M. Informally, this means that f is bounded by an exponential as t. This bound ensures that the e st factor in the transform (1) can make the integrand decay as t. Note also that because f(t) is in an integral, it is okay for it to be continuous except isolated jump discontinuities (they don t affect the transform); we refer to such a functionas piecewise continuous 1 (asymptotes are not allowed; only finite jumps). Domain of the Laplace transform: Claim: If f is piecewise continuous and of exponential type with rate constant a, i.e. then L[f(t)] = F (s) is defined for s > a. f(t) Ke at for t > M, Proof. We need to show that e st f(t) dt is finite when s > a. Split the integral into (, M) and (M, ): e st f(t) dt = M e st f(t) dt + M e st f(t) dt. The first term is finite because f is piecewise continuous on [, M] and thus bounded. For the second term, if s > a then the exponential bound gives us that e st f(t) dt Ke (s a)t dt = K s a e (s a)m <. M The argument above is informal; technical details omitted 2. M 1 Formally, in any finite interval [a, b] there is a partition a = t 1 < t 2 < t n = b such that f is continuous on (t k, t k+1 ) for k =, n 1. 2 The argument as written is not rigorous because the improper integral M e st f(t) dt is not known to exist. To be correct, we must use a comparison test for integrals: If g, h are integrable functions and g(t) h(t) and then g(t) dt exists (and is finite). M M h(t) dt < 3

3 The key properties For the transform to be useful, it has to have a few key features: It must be easy to apply L to a linear ODE. L must transform derivatives in a nice way. L must be invertible (so we can go back after finding the solution in the transformed space). The Laplace transform has these properties, which we detail here. 3.1 Linearity The Laplace transform is a linear operator! Suppose f 1, f 2 are functions with Laplace transforms and c 1, c 2 be constants. Then L[c 1 f 1 + c 2 f 2 ] = e st (c 1 f 1 (t) + c 2 f 2 (t)) dt = c 1 e st f 1 (t) dt + c 2 e st f 2 (t) dt = c 1 L[f 1 ] + c 2 L[f 2 ]. The above assumes that s is large enough that L[f 1 ](s) and L[f 2 ](s) are defined. 3.2 Differentiation The other key property - which gives the Laplace transform its utility for solving ODEs, is that it acts in a nice way on derivatives. The formula is L[f (t)] = sl[f(t)] f(). Conceptually, it is essential to understand to that the above means derivatives in the original space multiplication in the transformed space up to the extra terms. That is, d dt (function) L s (transform) + (terms). L 1 To be precise, we have the following theorem. Theorem: Suppose f is of exponential order with rate constant a, i.e. f(t) Ke at for t M and that f is piecewise continuous. Then L[f (t)] exists for s > a and L[f (t)] = sl[f(t)] f(). 4

Proof. We sketch this only when f is continuous (the piecewise case is similar). Assume that s > a. We simply integrate by parts, being careful with the boundary terms: e st f (t) dt = e st f(t) t= = f() + s = f() + sl[f(t)]. For the second line, the boundary term at is zero since The above is not quite rigorous. 3 Higher order derivatives ( se st )f(t) dt e st f(t) dt e st f(t) Ke (s a)t as t. Applying the derivative formula n times, we get the transform of an n-th derivative. For example, L[f (t)] = sl[f (t)] f ()) = s(sl[f(t)] f()) f () = s 2 L[f(t)] sf() f () and so on. Thus an n-th derivative in the original space correspond to multiplications by s n in the transformed space (up to some polynomial in s). To be precise, we have: Theorem. If f (n) is piecewise continuous and there are constants a, K > and M such that f (j) (t) Ke at for t > M and j =, k 1 (i.e. the exponential bound holds for f and its derivatives except for the n-th) then L[f (n) (t)] = s n L[f(t)] s n 1 f() s n 2 f () sf (n 2) () f (n 1) (). 3.3 Inversion The Laplace transform can be inverted, i.e. there is an operator L 1 such that L 1 [L[f(t)]] = f(t) for all functions f(t) of exponential order. Unfortunately, the details (and definition of L 1 in general) require some complex analysis and are beyond the scope of this course. Thus we take the following result on faith: 3 We do not know in advance that the improper integral exists. It would be correct to instead show that b lim b e st f (t) dt. exists and equals what we want. But the careful version is the same aside from adding the limit. 5

Inverse of the Laplace transform: If F (s) is defined for s > a then there is a unique function f(t) such that L[f(t)] = F (s). That is, f(t) = L 1 [F (s)]. The inverse L 1 is defined on the set of functions that have no singularities for s > a. In practice, one typically computes L 1 [F (s)] by recognizing F (s) as the transform of some function by consulting a table or just being familiar with common transforms. 4 Inverses and transforms In this section we compute some common transforms and show strategies for computing the inverse transform of a function F (s). In doing so, we will need see a few new properties of the Laplace transform, as well as use some miscellaneous results from calculus. 4.1 Computing the Laplace transform We compute the inverse transform (typically) by generating a table of function-transform pairs; then given F (s) we decomposing it into simple parts and taking the inverse of each. Basic transforms In some cases, we simply integrate directly: Constant function: For the constant function f(t) = 1, Exponential: L[1] = L[e at ] = e st dt = 1 s, s >. e (s a)t dt = 1 s a, s > a. Complex exponential and sine/cosine: The Laplace transform can be applied to complex functions. In particular, L[e (a+bi)t 1 ] = s (a + bi) In particular (taking real parts) we have L[e at sin bt] = b (s a) 2 + b 2, L[eat cos bt] = s a (s a) 2 + b 2. 6

Polynomial: Let n be a positive integer. Then, using integration by parts, we can find the transform of t n in terms of the transform for t n 1 : L[t n ] = t n e st dt = 1 s tn e st t= = n s L[tn 1 ], s >. nt n 1 ( 1 s e st ) dt Note that both of the boundary terms in the integration by parts are zero (t n e st as t and is equal to zero at t = ). Iterating n times and using L[t ] = 1/s, we find that L[t n ] = n! s n+1. Power of t, non-integer: The transform for t p when p is not an integer does not always have a simple form. It is given in terms of a standard function used throughout mathematics, the gamma function. For real values, it is Γ(p) = e t t p 1 dt, p >. In particular, by the same sort of argument as above we have Γ(p + 1) = pγ(p) and Γ(n) = (n 1)! The Laplace transform of t p for p > is L[t p ] = Γ(p + 1) s p+1. (Check this! You ll need to rescale t in the integral to get the right form.) 4.2 Inverse transforms: some rules Not every Y (s) is going to be immediately recognizable as a standard transform. We sometimes need to manipulate it (and terms) into a standard form. If you see something like s 2 + 3 (s 1) 2 (2s 2) 2 you should think: how do I turn it into a sum of easy-to-invert functions? There are several tools for accomplishing this task. Some useful ones are listed below. Exploiting linearity: Sums can be inverted term by term; constant factors can be moved in/out of the transform.. L 1 [cf (s)] = cl 1 [F (s)], L 1 [F 1 (s) + + F n (s)] = L 1 [F 1 (s)] + + L 1 [F n (s)]. Rescale the variable: The argument to the function F (s) or f(t) can be scaled by a constant (positive) factor: f(ct) 1 F (s/c), c >. c 7

For example, L 1 [e as ] = 1 a L 1 [e s ]. Translation (in s), exponential factors: Multiplying by an exponential corresponds to translation in the transformed space. To be precise, e ct f(t) F (s c), c >. Derivatives in s: A dual property to the rule for f (t). A derivative in the transformed space corresponds to multiplication by ( t) in the original space: ( t) n f(t) F (n) (s) For example, L[te t ] = d ds ( ) 1 = s 1 1 (s 1) 2. 5 Solving equations We are now ready to use the Laplace transform to solve linear, constant coefficient IVPS (of any order). The procedure is as follows: Apply the Laplace Transform to the ODE to obtain a transformed equation for Y (s) = L[y(t)]. This part is straightforward. Use the formula to transform L[y (n) (t)] For an IVP, plug in y(), y () etc. Solve the (algebraic) equation for Y (s) (the simplest part) Decompose Y (s) into a sum of functions that are easy to invert (typically functions that can be recognized as the transform of some known function). Consult standard results or derive new ones to figure calculate each term of L 1 [Y (s)] (there are tables for this step; see the book or the formula sheet on Sakai). It is worth noting, and we will see later, that the end goal is not always just to obtain a solution y(t). If we are interested in understanding the behavior of solutions, the transformed setting can be the right place to do analysis. Thus the last step (the inversion) is not always important. A few examples will do best to illustrate the process. 8

Example 1: y 2y + y =, y() = 1, y () =. Let Y (s) = L[y(t)]. Take the transform of the ODE, then apply the initial conditions: = s 2 Y sy() y () 2(sY y()) + Y = s 2 Y s 2sY + 2 + Y = (s 2 2s + 1)Y s + 2 = (s 1) 2 Y s + 2. The solution in the transformed domain is therefore Now we write Y (s) = s 2 (s 1) 2. Y (s) = s 2 (s 1) 2 = s 1 1 (s 1) 2 = 1 s 1 1 (s 1) 2. We recognize the first term as L[e t ]. The second term is the derivative of a function we know how to invert: 1 (s 1) = d ( ) 1 2 ds s 1 Thus we can use the fact that to obtain L 1 [1/(s 1) 2 ] = ( t)e t. Thus L[( t) n f(t)] = F (n) (s) L 1 [Y (s)] = L 1 1 [ s 1 ] + 1 L 1 [ (s 1) ] 2 = e t + ( t)e t = (1 t)e t. 9

Example 2: y + y = sin ωt, y() =, y () = 1, ω ±1. Take the transform (using the standard result for sine ): Apply the initial condition and obtain so we get Now use partial fractions: which gives s 2 Y sy() y () + Y = (s 2 + 1)Y = 1 + ω s 2 + ω 2. ω s 2 + ω 2. Y = 1 s 2 + 1 + ω (s 2 + 1)(s 2 + ω 2 ). 1 (s 2 + 1)(s 2 + ω 2 ) = A s 2 + 1 + B s 2 + ω 2 1 = As 2 + Aω 2 + Bs 2 + B so A + B = and Aω 2 + B = 1, solved by A = 1/(ω 2 1) and B = 1/(ω 2 1). Thus Y, after using partial fractions, is Y = 1 ( ) s 2 + 1 + ω 1 (ω 2 1) s 2 + 1 1. s 2 + ω 2 Now we recognize that L[sin t] = 1 s 2 + 1, L[sin ωt] = ω s 2 + ω 2, and use the known transforms to invert each term of Y : Y = ω2 + ω 1 ω 2 1 sin t 1 sin ωt. ω 2 1 1

Example 3: The method works for any linear constant-coefficient ODE (of any order). For example, we can use it to solve the fourth-order equation y (4) 5y + 4y =, y() = 1, y () =, y () = 3, y () =. Take the Laplace transform and use the initial conditions: Solve for Y to obtain Y (s) = (s 4 Y s 3 3s) 5(s 2 Y s) + 4Y =. s3 2s s 4 5s 2 + 1 = s 3 2s (s 1)(s + 1)(s 2)(s + 2). Notice that the denominator just the characteristic polynomial of the ODE (which will be true in general). We can invert Y (s) by using partial fractions to write it in the form Y (s) = 1/6 s 1 + 1/6 s + 1 + 1/3 s 2 + 1/3 s + 2. (The calculation here is tedious but straightforward). Each term can be inverted using L[e at ] = 1/(s a) to obtain the solution y(t) = 1 6 (et + e t ) + 1 3 (e2t + e 2t ). 11