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18.034 - Honors Differential Equations, Spring 2007 Dr. Vera Mikyoung Hur LT1. LAPLACE TRANSFORM The Laplace transform of the function f , deﬁned for all real numbers t � 0, is the function F (s) deﬁned as � ∞ � b st (1) F (s) = L[f (t)](s) = e f (t)dt = lim est f (t)dt 0− a→0+ −a b→∞ provided that the limit exists for all sufﬁciently large s. The lower limit 0− ensures the inclusion of the unit impulse function (or the Dirac function)∗. The parameter s is in general considered to be complex. As an integral operator, the Laplace transform forms an important part of functional analysis. It is named in honor of Pierre-Simon Laplace, who used the transfrom in his work on probability theory. Functions of exponential type. A real- or complex-valued function f (t) is said to be of exponential type, and we write f ∈ E if there are positive constants A and B, depending on f such that: (a) f (t) is piecewise continuous on [0, ∞) and f (t) is bounded on the set of jump discontinuities where it is deﬁned, (b) |f (t)| � AeBt at all points t ∈ [0, ∞) where it is deﬁned. If f ∈ E and s > B then � ∞ � ∞ −st A e |f (t)|dt � e−st AeBt dt = . 0− 0− s−B This suggests that the integral in (1) exists for s > B, and thus, if f ∈ E then its Laplace transform 2 exists. (Note that et does not possess a Laplace transform.) Moreover, if f is continuous on [0, ∞) and f � ∈ E then f ∈ E and (2) L[f ](s) = sL[f ](s) − f (0−). This relation enables us to transform an initial value problem of a differential equation for y into an algebraic equation for Ly, which is much easier to solve. Once Lf is found, the inverse Laplace transform reverts back to the solution f to the differential equation. The theory takes its form from a symbolic method developed by the English engineer Oliver Heaviside. It necessitates the following justiﬁcations. Theorem 1. If y is a solution of p(D)y = f on [0, ∞) with f ∈ E then y and its derivative up to the order of p(D) are also in E. Its proof is left as an exercise. Theorem 2 (Uniqueness Theorem). If f and g are functions of class E and their Laplace transforms agree for all large s, then f (t) = g(t) at all points t � 0 where both functions are continuous. The proof depends on another theorem which is of independent interest. ∗ Kent H. Lundberg, Haynes R. Miller, and David L. Trumper, Initial conditions, generalized functions, and the Laplace transform, IEEE Control Systems magazine, 27 (2007) pp. 22-35. 1 Theorem 3. If q is continuous for 0 � x � 1, then � 1 xn q(x)dx = 0, for n = 0, 1, 2, . . . implies q(x) = 0 for 0 � x � 1. 0 Proof of Theorem 2. Given f and g as in the statement of Theorem 1, let u, v, w with their Laplace transforms U, V, W , respectively, be deﬁned by � t u(t) = f (t) − g(t), v(t) = u(τ )dτ, w(t) = e−ct v(t). 0 By assumption, then, U (s) = 0 for all large s, and our goal is to prove that u(t) = 0 at all points of continuity. Note that V (s) = U (s)/s = 0 for large s. Accordingly, if v(t) = 0 then by the fundamental theorem of calculus follows that u(t) = 0 at all points of continuity. Hence, we will work with the continuous function v rather than u. For the constant c large enough, by the shift theorem of the Laplace transform, W (s) = V (s + c) = 0 for all s > 0 (not only for large s). Since u ∈ E follows v ∈ E, and for large c, it yields (3) lim w(t) = 0. t→∞ By the change of vaiable x = e−t we obtain that � ∞ � 1 −st W (s) = e w(t)dt = xs−1 w(− ln x)dx. 0 0 Let � w(− ln x) for 0 < x � 1 q(x) = . 0 for x = 0 Equation (3) guarantees that q is continuous for 0 � x � 1. Since W (s) = 0 for all s � 0, we certainly have W (s) = 0 for s = 1, 2, 3, . . . . Then the assertion follows from Theorem 2. � The proof of Theorem 3 is found at the end of this note. Example 4. Here we consider the initial value problem y �� + 4y = 0, y(0) = 5, y � (0) = 6. Taking the transform gives s2 Ly − 5s − 6 + 4Ly = 0. Hence, 5s + 6 s 2 Ly = =5 2 +3 2 , s2 + 4 s +4 s +4 and y(t) = 5 cos 2t + 3 sin 2t. Further application of the Laplace transform to solve initial value problems of differential equa tions and useful properties of the Lapace transform, the Laplace transforms of frequently used functions are found in many standard textbooks on differential equations, for instance, [2]. The Laplace transform is particularly useful in dealing with discontinuous inputs (closing of a switch) and with periodic functions (rectiﬁed sine waves). A simple discontinuity at c has the left and right limits lim f (t) = f (c−), lim f (t) = f (c+) t→c− t→c+ but they are not equal. A function f is piecewise continuous if it is deﬁned and continuous except at ﬁnitely many points and all of its discontinuities are simple. 2 Theorem 5. If f � ∈ E and f is continuous except a simple discontinuity at a single value t = c > 0, then L[f � ](s) = sLf [s] − f (0+) − (f (c+) − f (c−))e−sc . Sketch of the proof. The assertion follows by adding � c � c −st � −st c e f (t)dt = e f (t)|0 + s f (t)e−st dt, 0 0 � ∞ � ∞ −st � −st ∞ e f (t) = e f (t)|c + s f (t)e−st dt. c c � The unit step function. The unit step function or the Heaviside function is deﬁned by � 0, t<0 (4) u(t) = 1, t � 0. One of the main uses of the unit step function is in connection with the shift theorem L[f (t − c)] = e−sc L[f (t)], c�0 with an additional condition that f (t) = 0. Note that e−cs (5) L[u(t − c)] = , c � 0. s Example 6. A rectiﬁed sine wave is f (t) = | sin t|. The function � sin t, 0�t<π f0 (t) = 0, π � t < 2π. agrees with f (t) on 0 � t < π and is zero elsewhere. It is easy to see that f0 (t) = u(t)f (t) − u(t − π)f (t). Taking the Laplace transform gives L[f0 ] = L[f ] − e−πs L[f ]. On the other hand, 1 + e−sπ L[f0 ](s) = . 1 + s2 Therefore, 1 1 + e−πs L[f ] = . 1 + s2 1 − e−πs We end this chapter by the proof of Theorem 3. � Proof of Theorem 3. Suppose q(x0 ) = 0 at some point in (0, 1). We may assume that q(x0 ) > 0. Continuity then gives positive constants �, δ such that |x − x0 | � 2δ implies q(x) � �. Let m be an arbitrary positive integer and deﬁne � 1 p(x) = 1 + 4δ 2 − (x − x0 )2 , Im = pm (x)q(x)dx. 0 By the binomial theorem pm is a polynomial in x, and hence Im = 0 under the hypothesis of Theorem 3. On the other hand Im = J1 + J2 + J3 , where J1 , J2 , J3 are integrals over the part of [0, 1] in which |x − x0 | < δ, δ � |x − x0 | � 2δ, |x − x0 | > 2δ, 3 respectively. On these three intervals we have p(x) � 1 + 3δ 2 , p(x) � 1, |p(x)| � 1 and also, if M is a sufﬁciently large constant, q(x) � �, q(x) � �, |q(x)| � M. It is easy to check that J1 � (1 + 3δ 2 )m δ�, J2 � 0, J3 � −M. The ﬁrst expression tends to inﬁnity as m → ∞, and hence IM → ∞. This contradicts that Im = 0. � R EFERENCES [1] Garret Birkhoff & Gian-Carlo Rota, Ordinary Differential Equations, 4th ed., New York: Wiley, 1978. [2] William Boyce & Richard DiPrima, Elementary differential equations and boundary value problems, 7th ed., New York; John Wiley, 2001. [3] Ray Redheffer & Dan Port Differential Equations, theory and applications, Jones and Barlett publishers, Boston, 1991. Problems. 1. If f (t) ∈ E, show that L[f (t)](s) → 0 as s → ∞. 2 √ 2 2. Showt that the Laplace transform of e−t /4 is πes erts. 3. (a) Let F (s) = L[f (t)](s). Suppose that f (t)/t has a limit as t → 0+. Prove that � ∞ L[f (t)/t](s) = F (u)du. s sin t (b) Use part (a) to ﬁnd the Laplace transform of f (t) = t . � � �� 4. Find L−1 ln s+a [t]. (Hint. F � (s) = −L[tf (t)].) s−a 5. Let f (t) = t2 /2 for 0 < t � 1 and f (t) = (t − 2)2 /2 for 1 � t � 2, and t(t) = 0 elsewhere. Sketch the graph of f (t). Find the Laplace transforms of f , f � and f �� . (Answers. 1 − e−2s 2e−s 1 − e−2s 2e−s 1 − e−2s L[f ] = − 2 , L[f � ] = − , L[f �� ] = .) s3 s s2 s s 6. (This problem requires knowledge of the geometric series for 1/(1 − r). ) Reason that L[f ] = L[u(t)f (t) + u(t − c)f (t − c) + u(t − 2c)f (t − 2c) + · · · ]. 1 − e−sc 4