Edwards-Penney Laplace Theory, 2280 Chapter 7 Updated 2015 Section 7.1 ========= 7.1-18 Find L(f(t)) for f(t)=sin(3t)cos(3t) Trig identity: sin(2x) = 2 sin(x) cos(x) Then f(t)=sin(3t)cos(3t)=(1/2) sin(6t) Laplace table 7.1.2 is used to find the transform of sin(bt): L(sin(bt)) = b/(s^2+b^2) Then L(f(t)) = (1/2) L(sin(6t)) = (6/2)/(s^2+6^2) = 3/(s^2+36) Because all terms of f(t) are in the forward Laplace table, then this problem can be done without integrations of any kind. ================================================================ 7.1-22 Find L(f(t)) for f(t)=(sinh(3t))^2 ================================================================ Laplace table 7.1.2 is used to find the transform of f(t)=sinh^2(3t). The idea is to write sinh(w) = (exp(w) - exp(-w))/2 and then expand f(t) in terms of exponential functions. Because all terms of f(t) are in the forward Laplace table, then this problem can be done without integrations of any kind. ================================================================ 7.1-28 Find f(t) in the equation L(f(t))=F(s)=(3s+1)/(s^2+4) ================================================================ This problem uses the backward table, that is, use table 7.1.2 backwards or right-to-left. The details would look like L(f(t)) = F(s) 3s + 1 = -------- s^2 + 4 s 1 2 = 3 ----- + - ----- s^2+4 2 s^2+4 = 3 L(cos 2t) + 0.5 L(sin 2t) = L(3 cos 2t + 0.5 sin 2t) Lerch's Theorem (Theorem 3 in Section 7.1), or the inverse transform, finishes the problem. Section 7.2 ========= ================================================================ 7.2-6 x'' + 4x = cos(t), x(0)=x'(0)=0 ================================================================ See the web site slides for how to solve this problem and others like it. The landmark steps to verify are (s^2+4) L(x(t)) = L(cos t) s L(x(t)) = ----------------- (s^2 + 1)(s^2 + 4) As+B Cs+D = ----- + ----- Used partial fraction theory s^2+1 s^2+4 Heaviside's method manuscript = L(A cos(t) + B sin(t) + C cos(2t) + D sin(2t)/2) x(t) = A cos(t) + B sin(t) + C cos(2t) + D sin(2t)/2 by Lerch's theorem Then do partial fractions to find A,B,C,D. ================================================================ 7.2-14 ================================================================ x'' + 2x + 4y = 0, y'' + x + 2y = 0, x(0)=y(0)=0, x'(0)=y'(0)=-1 See Section 7.2 Example 3 for a similar problem. Here's what should appear in the details: 1. Transform each differential equation to get a 2x2 system for the unknowns L(x) and L(y). 2. Solve the system by Cramer's rule for L(x) and L(y). 3. Find x(t) from the formula for L(x), using the backward table, partial fractions, and the various Laplace rules. The last step should be Lerch's theorem or take an inverse transform. 4. Find y(t) from the formula for L(y), in the same way. There are web examples of the method in the Laplace theory manuscript. See also the Resolvent Method for second order systems, for an extension of these ideas to general systems of differential equations. ================================================================ 7.2-20 F(s) = L(f(t)) = (2s+1)/(s(s^2+9)), find f(t) ================================================================ It is possible to solve for f(t) in this example using Theorem 2 of section 7.2. You are advised to choose a more efficient partial fraction solution. Because the fraction in this problem has denominator s(s+3i)(s-3i), having distinct roots 0, 3i, -3i, Heaviside's coverup method is efficient to find the constants in the partial fraction expansion. 1 L(f(t)) = -------- s(s^2+9) 1 = --------- s(s-3i)(s+3i) A B C = --- + --- + --- s s-3i s+3i A Ds+E = --- + ------ for some real constants A, D, E s s^2+9 = L(A + Dcos(3t)+Esin(3t)/3) Lerch's theorem then implies f(t) = A + Dcos(3t)+Esin(3t)/3 We are left to find constants A,D,E. The constant A=1/9 is found directly by Heaviside's coverup method. The other two D, E can be found by partially clearing the fraction s^2+9, which results in the equation 1 (s^2+9) A(s^2+9) ----------- = -------- + Ds+E s(s^2+9) s Setting s^2+9=0 (substitute s=either root of s^2+9=0) then implies 1 --- = Ds + E which implies 1 = Ds^2+Es s Because s^2+9=0, then 1=Ds^2+Es becomes 1=-9D+Es, which says that D=-1/9 and E=0. In summary, A=1/9, D=-1/9, E=0 and f(t) = 1/9 - (1/9)cos(3t) Maple Answer Check: inttrans[laplace](1/9 - (1/9)*cos(3*t),t,s); ================================================================ 7.2-22 F(s) = L(f(t)) = 1/((s(s^2-9)), find f(t) ================================================================ While it is possible to solve for f(t) in this example using Theorem 2 of section 7.2, you are advised to choose a more efficient partial fraction solution. Because the fraction in this problem has denominator s(s-3)(s+3), having distinct roots 0, 3, -3, Heavisides coverup method is efficient to find the partial fraction expansion. 1 L(f(t)) = --------- s(s-3)(s+3) A B C = --- + --- + --- s s-3 s+3 = A L(1) + B L(exp(3t)) + C L(exp(-3t)) = L(A + B exp(3t) + C exp(-3t)) Lerch's theorem says that the L-symbols cancel leaving f(t) = A + B exp(3t) + C exp(-3t) Now evaluate A, B, C by Heaviside's coverup method. Section 7.3 ========= ================================================================ 7.3-6 F(s) = L(f(t)) = (s-1)/(s+1)^3, find f(t) ================================================================ Details: s - 1 L(f(t)) = -------- (s+1)^3 u - 2 | [formally subst u=s+1] = ------ | u^3 | u ==> s+1 [prepare for shift] 1 2 | = ---- - --- | u^2 u^3 | u ==> s+1 1 2 | = ---- - --- | s^2 s^3 | s ==> s+1 [Write it as a real shift] = L(t - t^2) | s ==> s+1 = L((t-t^2)exp(-t)) by the first shifting theorem Finish with Lerch's theorem or inverse transform. ================================================================ 7.3-12 F(s) = L(f(t)) = (5s-6)/(s^2-3s), find f(t) ================================================================ 5s-6 L(f(t)) = ------- s(s-3) A B = --- + --- s s-3 = L(A+B exp(3t)) Then f(t)=A+B exp(3t) by Lerch's theorem. Find A,B using Heaviside's coverup method, A=2, B=3. Maple Answer Check: inttrans[laplace](2+3*exp(3*t),t,s); ================================================================ 7.3-16 F(s) = L(f(t)) = 1/(s^2+s-6)^2, find f(t) ================================================================ Details: Factor by inverse FOIL, s^2+s-6=(s+3)(s-2), then 1 L(f(t)) = --------------- (s+3)^2(s-2)^2 A B C D = --- + ------- + --- + ------- s+3 (s+3)^2 s-2 (s-2)^2 [ A B ]| [ C D ]| = [ - + ---]| + [ - + ---]| [ s s^2]| s ==> s+3 [ s s^2]| s ==> s-2 = L(A+Bt)| s ==> s+3 + L(C+Dt)| s ==> s-2 = L((A+Bt)exp(-3t)) + L((C+Dt)exp(2t)) Shifting theorem Combine into one L on the right, apply Lerch's theorem or inverse transforms to find f(t). Then evaluate A,B,C,D by partial fractions. ================================================================ 7.3-22 F(s) = L(f(t)) = (2s^3-s^2)/(4s^2-4s+5)^2, find f(t) ================================================================ Begin by completing the square on the quadratic in the denominator: 4s^2 - 4s + 5 = 4(s-1/2)^2 +4 = 4[(s-1/2)^2 + 1) The shifting theorem suggests to replace s-1/2 by u, then try partial fractions. Details: L(f(t)) = F(s) 2s^3-s^2 = ---------------- (4(s-1/2)^2+4)^2 2(u+1/2)^3-(u+1/2)^2 | = --------------------- | (4u^2+4)^2 | u=s-1/2 [prepare for shift] (2u+1)(u+1/2)^2 | = --------------- | (4u^2+4)^2 | u=s-1/2 Au+B Cu+D | = ----- + --------- | u^2+1 (u^2+1)^2 | u=s-1/2 = L(Acos(t)+Bsin(t))+ | L(C(t/2)sin(t)+D(1/2)(sin(t)-t cos(t))) | s ==> s-1/2 Then Lerch's theorem implies f(t) = exp(t/2)(Acos(t)+Bsin(t))+C(t/2)sin(t)+D(1/2)(sin(t)-t cos(t))) The remainder of the problem is to find the constants A,B,C,D, using partial fraction theory. ================================================================ 7.3-28 x'' -6x' + 8x = 2, x(0)=x'(0)=0, solve for x(t) ================================================================ The transformed problem is (s^2-6s+8)L(x)=2/s. Then s^2-6s+8=(s-4)(s-2) implies 2 L(x(t)) = ----------- s(s^2-6s+8) 2 = ----------- s(s-4)(s-2) A B C = --- + ----- + ----- s (s-4) (s-2) = L(A + B exp(4t) + C exp(2t)) Lerch's theorem implies x(t) = A + B exp(4t) + C exp(2t) The rest of the problem finds the constants A,B,C by Heaviside's coverup method. Section 7.4 ========= ================================================================ 7.4-22 f(t) = (exp(t)-exp(-t))/t, find L(f(t)) ================================================================ The idea: tf(t) = exp(t)-exp(-t) L(tf(t)) = L(exp(t)-exp(-t)) (-d/ds)L(f(t)) = L(exp(t)-exp(-t)) The latter is a quadrature differential equation in s for L(f(t)). In more detail, The form of this quadrature equation is -dy/dx = 1/(x-1)-1/(x+1) where x=s, y=F(s)=L(f(t)) It is solved by the method of quadrature, that is, integrate across the equation to solve for y. The catch: there is a constant of integration C which has to be evaluated. Other than that, y=L(f(t)) has been found. The theorem lim L(f(t)) = 0 as s approaches infinity evaluates the constant C of integration. Then L(f(t)) has been determined. The book answer is correct. ================================================================ 7.4-28 F(s) = L(f(t)) = s/(s^2+1)^3, find f(t) ================================================================ Because F(s) = L(f(t)) as given is already a partial fraction, then the theory of partial fractions is of no immediate help. A simplistic plan is to use L((-t)g(t)) = (d/ds)L(g(t)) to work out formulas for the Laplace transforms of the functions t sin t, t cos t, t^2 cos t. You will find out that F(s) is a linear combination of these answers. Another way is to use the convolution theorem, although that involves integration which is not so easy. ================================================================ 7.4-36 Solve x''+4x=f(t), x(0)=x'(0)=0 using convolution f(t) stands for an arbitrary input. ================================================================ The answer is given in the problem, x(t) = int(f(t-u)k(u),u=0..t) where k(x)=(1/2) sin(2*x) The steps involve using the convolution theorem, on the last detail: L(x) = L(k) L(f) ===> x=convolution of k and f = convolution of f and k The formula can be checked by using f(t)=1, in which case the initial value problem has solution 1/4 - 1/4*cos(2*t). The integral formula can be evaluated explicity to match the answer. ================================================================ Section 7.5 ========= ================================================================ 7.5-4 F(s) = L(f(t)) = (exp(-s)-exp(2-2s))/(s-1), find f(t) ================================================================ The factor s-1 suggests a shift, so we do it immediately to simplify the problem. L(f(t)) = F(s) exp(-s)-exp(2-2s) = ----------------- s-1 exp(-u-1)-exp(-2u)| [subst u=s-1 above] = ----------------- | u | u=s-1 exp(-s-1)-exp(-2s)| = ----------------- | s |s==>s-1 [write as a shift] = L(g(t))| s==>s-1 where exp(-s-1)-exp(-2s) L(g(t)) = ----------------- [work on inner fraction, no shift] s Then L(f(t)) = L(g(t))| s==>s-1 = L(exp(t)g(t)) [first shift theorem] Finally f(t) = exp(t)g(t) [Lerch's theorem] The plan moves to solving for g(t), which is a table lookup problem using Heaviside's unit function u(t)=0 for t<0, u(t)=1 otherwise. The basic table entries are L(u(t-a)) = exp(-as)/s [valid only for a>=0] L(h(t-a)u(t-a)) = exp(-as)L(h(t)) [any h(t), any a>=0] The table enties are memorized from the second one, because the first table entry is the special case h(t)=1. The details for g(t): exp(-s-1)-exp(-2s) L(g(t)) = ----------------- s exp(-s) exp(-2s) = exp(-1)------- - -------- s s = exp(-1)L(u(t-1)) - L(u(t-2)) [used L(u(t-a))=exp(-as)/s] = L(exp(-1)u(t-1)-u(t-2)) g(t) = exp(-1)u(t-1)-u(t-2) [by Lerch's theorem] Then the pieces are re-assmbled to finish the solution. We have from above, L(f(t)) = L(g(t))| s==>s-1 = L(exp(t)g(t)) [first shift theorem] f(t) = exp(t)g(t) [Lerch's theorem] f(t) = exp(t)exp(-1)u(t-1)-exp(t)u(t-2) f(t) = exp(t-1)u(t-1)-exp(t)u(t-2) === A SECOND SOLUTION ===== DIVIDE AND CONQUER. Let L(f1(t)) = exp(-s)/(s-1) and L(f2(t)) = -exp(2-2s)/(s-1).Then f =f1 + f2 is the answer. SOLVE for f1. Because exp(-s)/(s-1) must match exp(-as)L(h(t)), then exp(-s)=exp(-as) and a=1. Because 1/(s-1) must match L(h(t)), then L(exp(t))=L(h(t))) and h(t)=exp(t). Replace t by t-a to obtain h(t-a) = exp(t-a) = exp(t-1) [because a=1] Then L(f1)=exp(-s)/(s-1)=exp(-as)L(h(t)))=L(h(t-a)u(t-a)) implies L(f1)=L(exp(t-1)u(t-1)) and finally f1(t)=exp(t-1)u(t-1) [u(t)=unit step=Heaviside] The calculation for f2(t) is similar. MAPLE ANSWER CHECK. The code: with(inttrans): F:=exp(-s)/(s-1)-exp(2-2*s)/(s-1); invlaplace(F,s,t); # ans: exp(t)*(Heaviside(2-t)-1)+exp(t-1)*(-Heaviside(1-t)+1) # Mystery: why does this equal the answer reported above? # Most hand computation gets exp(t-1)u(t-1)-exp(t)u(t-2) laplace(exp(t-1)*Heaviside(t-1),t,s)-laplace(exp(t)*Heaviside(t-2),t,s); Answer to mystery: Heaviside(2-t)-1 == Heaviside(t-2) Heaviside(1-t)-1 == Heaviside(t-1) ================================================================ 7.5-14 f(t)=cos(Pi t) on 0 <= t <= 2, f(t)=0 otherwise, find L(f(t)) ================================================================ The function f(t) is recognized as a pulse, f(t) = cos(Pi t) pulse(t,0,2) = cos(PI t) (u(t) - u(t-2)) [u=unit step=Heaviside] To calculate the Laplace of f(t), we break it into two problems: f(t) = f1(t) - f2(t) f1(t) = cos(PI t) u(t) f2(t) = cos(PI t) u(t-2) The relevant Laplace table theorem is derived from the classic L(h(t-a)u(t-a))=exp(-as)L(h(t)) by letting g(t)=h(t-a), then L(g(t)u(t-a)) = exp(-as)L(g(t+a)) [valid only for a >= 0] The symbol g(t+a) is a composite function, which is defined simply by g(t+a) = g(t) with t replaced by t+a | = g(t) | [a shift] | t ==> t+a This is why the result is called the second shifting theorem, it shifts on t instead of s. We can write it as L(g(t)u(t-a)) = exp(-as)L(g(t)|t==>t+a) We'll find the Laplace of f2(t) = cos(Pi t) u(t-2) and leave the Laplace of f1(t) as an exercise. L(f2(t)) = L(cos(Pi t) u(t-2)) = exp(-2s)L(cos(Pi t)| t==>t+2) [+2 comes from u(t-2)] = exp(-2s)L(cos(Pi t + 2 Pi)) = exp(-2s)L(cos(Pi t)) [because cos(A+B)=cos(A)cos(B)-sin(A)sin(B) and cos(2 Pi)=1, sin(2 Pi)=0] s = exp(-2s) -------- s^2+Pi^2 The book's answer is correct. In checking your work, observe that the second shifting theorem may not produce an answer with an exponential in the special case a=0, because exp(-0s)=1. ================================================================ 7.5-22 f(t) = t^3 for 1 <= t <= 2, f(t)=0 oherwise, find L(f(t)) ================================================================ First, write the function f(t) in terms of pulses and then unit steps: f(t) = t^3 pulse(t,1,2) = t^3(u(t,1)-u(t,2)) L(f) = L(t^3 u(t-1))-L(t^3 u(t-2)) L(t^3 u(t-1)) = L(g(t)u(t-1)) = exp(-s) L(g(t+1)) second shifting theorem = exp(-s) L((t+1)^3) [g(t)=t^3, g(t+1)=(t+1)^3] = exp(-s) L(t^3+3t^2+3t+1) L(t^3 u(t-2)) = L(g(t)u(t-2)) = exp(-2s) L(g(t+2)) second shifting theorem = exp(-2s) L((t+2)^3) [g(t)=t^3, g(t+2)=(t+2)^3] = exp(-2s) L(t^3+6t^2+12t+8) MAPLE ANS CHECK: #ans: 1/s^4*(-2*(4*s^3+6*s^2+6*s+3)*exp(-2*s) +(s^3+3*s^2+6*s+6)*exp(-s)) with(inttrans): laplace((t+1)^3,t,s); laplace((t+2)^3,t,s); laplace(t^3*Heaviside(t-1)-t^3*Heaviside(t-2),t,s); ================================================================ 7.5-28 f(t+2a)=f(t) [period=2a] with f(t)=f0(t) on 0<=t<2a, where f0(t)=t on 0<=t 2 ANSWER CHECK. # The maple code uses a special version of dsolve. de:=diff(x(t),t,t) + 2*diff(x(t),t) + x(t) = Dirac(t) - Dirac(t-2); ic:=x(0)=2,D(x)(0)=2; dsolve({de,ic},x(t),method=laplace); To get the answer in readable format, convert(%,piecewise,t); ================================================================ 7.6-22 Word problem, decoded below. ================================================================ The problem can be solved by considering this example: x'' + x = Dirac(t) + Dirac(t - 2 Pi) + Dirac(t - 4 Pi) x(0)=0, x'(0)=0 Answer check to the example using Maple code. de:=diff(x(t),t,t) + x(t) = Dirac(t) + Dirac(t - 2*Pi) + Dirac(t - 4*Pi); ic:=x(0)=0,D(x)(0)=0; dsolve({de,ic},x(t),method=laplace); convert(%,piecewise,t); The answer to 7.6-22, from maple code above. sin(t), t < 2*Pi undefined, t = 2*Pi x(t) = 2*sin(t), t < 4*Pi undefined, t = 4*Pi 3*sin(t), 4*Pi < t Solution steps to 7.6-22 using Laplace theory. (s^2+1)L(x) = 1 + exp(-2 Pi s) + exp(-4 Pi s) x = x1 + x2 + x3 L(x1) = 1/(s^2+1) = L(sin t) L(x2) = (1/(s^2+1))exp(-2 Pi s) = L(sin t) exp(-2 Pi s) = fill in using 2nd shifting theorem = L(sin(t-2 Pi)u(t-2 Pi)) L(x3) = (1/(s^2+1))exp(-4 Pi s) = L(sin t) exp(-4 Pi s) = fill in using 2nd shift theorem = L(sin(t-4 Pi)u(t-4 Pi)) Then x1, x2, x3 are found from Lerch's theorem [or inverse laplace]. Because the sine is 2Pi-periodic, all sine terms reduce to sin(t). There is an answer published in the book, which is correct. === end Chapter 7===