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Mathematics LibreTexts

8.6E: Convolution (Exercises)

( \newcommand{\kernel}{\mathrm{null}\,}\)

Q8.6.1

1. Express the inverse transform as an integral.

  1. 1s2(s2+4)
  2. s(s+2)(s2+9)
  3. s(s2+4)(s2+9)
  4. s(s2+1)2
  5. 1s(sa)
  6. 1(s+1)(s2+2s+2)
  7. 1(s+1)2(s2+4s+5)
  8. 1(s1)3(s+2)2
  9. s1s2(s22s+2)
  10. s(s+3)(s2+4)(s2+6s+10)
  11. 1(s3)5s6
  12. 1(s1)3(s2+4)
  13. 1s2(s2)3
  14. 1s7(s2)6

2. Find the Laplace transform.

  1. t0sinaτcosb(tτ)dτ
  2. t0eτsina(tτ)dτ
  3. t0sinhaτcosha(tτ)dτ
  4. t0τ(tτ)sinωτcosω(tτ)dτ
  5. ett0sinωτcosω(tτ)dτ
  6. ett0τ2(tτ)eτdτ
  7. ett0eττcosω(tτ)dτ
  8. ett0e2τsinh(tτ)dτ
  9. t0τe2τsin2(tτ)dτ
  10. t0(tτ)3eτdτ
  11. t0τ6e(tτ)sin3(tτ)dτ
  12. t0τ2(tτ)3dτ
  13. t0(tτ)7eτsin2τdτ
  14. t0(tτ)4sin2τdτ

3. Find a formula for the solution of the initial value problem.

  1. y+3y+y=f(t),y(0)=0,y(0)=0
  2. y+4y=f(t),y(0)=0,y(0)=0
  3. y+2y+y=f(t),y(0)=0,y(0)=0
  4. y+k2y=f(t),y(0)=1,y(0)=1
  5. y+6y+9y=f(t),y(0)=0,y(0)=2
  6. y4y=f(t),y(0)=0,y(0)=3
  7. y5y+6y=f(t),y(0)=1,y(0)=3
  8. y+ω2y=f(t),y(0)=k0,y(0)=k1

4. Solve the integral equation.

  1. y(t)=tt0(tτ)y(τ)dτ
  2. y(t)=sint2t0cos(tτ)y(τ)dτ
  3. y(t)=1+2t0y(τ)cos(tτ)dτ
  4. y(t)=t+t0y(τ)e(tτ)dτ
  5. y(t)=t+t0y(τ)cos(tτ)dτ,y(0)=4
  6. y(t)=costsint+t0y(τ)sin(tτ)dτ

5. Use the convolution theorem to evaluate the integral.

  1. t0(tτ)7τ8dτ
  2. t0(tτ)13τ7dτ
  3. t0(tτ)6τ7dτ
  4. t0eτsin(tτ)dτ
  5. t0sinτcos2(tτ)dτ

6. Show that t0f(tτ)g(τ)dτ=t0f(τ)g(tτ)dτ by introducing the new variable of integration x=tτ in the first integral.

7. Use the convolution theorem to show that if f(t)F(s) then t0f(τ)dτF(s)s.

8. Show that if p(s)=as2+bs+c has distinct real zeros r1 and r2 then the solution of

ay+by+cy=f(t),y(0)=k0,y(0)=k1

is

y(t)=k0r2er1tr1er2tr2r1+k1er2ter1tr2r1+1a(r2r1)t0(er2τer1τ)f(tτ)dτ.

9. Show that if p(s)=as2+bs+c has a repeated real zero r1 then the solution of

ay+by+cy=f(t),y(0)=k0,y(0)=k1

is

y(t)=k0(1r1t)er1t+k1ter1t+1at0τer1τf(tτ)dτ.

10. Show that if p(s)=as2+bs+c has complex conjugate zeros λ±iω then the solution of

ay+by+cy=f(t),y(0)=k0,y(0)=k1

is

y(t)=eλt[k0(cosωtλωsinωt)+k1ωsinωt]+1aωt0eλtf(tτ)sinωτdτ.

11. Let w=L1(1as2+bs+c), where a,b, and c are constants and a0.

  1. Show that w is the solution of aw+bw+cw=0,w(0)=0,w(0)=1a.
  2. Let f be continuous on [0,) and define h(t)=t0w(tτ)f(τ)dτ. Use Leibniz’s rule for differentiating an integral with respect to a parameter to show that h is the solution of ah+bh+ch=f,h(0)=0,h(0)=0.
  3. Show that the function y in Equation 8.6.14 is the solution of Equation 8.6.13 provided that f is continuous on [0,); thus, it is not necessary to assume that f has a Laplace transform.

12. Consider the initial value problem

ay+by+cy=f(t),y(0)=0,y(0)=0,

where a,b, and c are constants, a0, and

f(t)={f0(t),0t<t1,f1(t),tt1.

Assume that f0 is continuous and of exponential order on [0,) and f1 is continuous and of exponential order on [t1,). Let

p(s)=as2+bs+c.

  1. Show that the Laplace transform of the solution of (A) is Y(s)=F0(s)+est1G(s)p(s) where g(t)=f1(t+t1)f0(t+t1).
  2. Let w be as in Exercise 8.6.11. Use Theorem 8.4.2 and the convolution theorem to show that the solution of (A) is y(t)=t0w(tτ)f0(τ)dτ+u(tt1)tt10w(tt1τ)g(τ)dτ for t>0.
  3. Henceforth, assume only that f0 is continuous on [0,) and f1 is continuous on [t1,). Use Exercise 8.6.11 (a) and (b) to show that y(t)=t0w(tτ)f0(τ)dτ+u(tt1)tt10w(tt1τ)g(τ)dτ for t>0, and y(t)=f(t)a+t0w(tτ)f0(τ)dτ+u(tt1)tt10w(tt1τ)g(τ)dτ for 0<t<t1 and t>t1. Also, show y satisfies the differential equation in (A) on(0,t1) and (t1,).
  4. Show that y and y are continuous on [0,).

13. Suppose

f(t)={f0(t),0t<t1,f1(t),t1t<t2,fk1(t),tk1t<tk,fk(t),ttk,

where fm is continuous on [tm,) for m=0,,k (let t0=0), and define

gm(t)=fm(t+tm)fm1(t+tm),m=1,,k.

Extend the results of Exercise 8.6.12 to show that the solution of

ay+by+cy=f(t),y(0)=0,y(0)=0

is

y(t)=t0w(tτ)f0(τ)dτ+km=1u(ttm)ttm0w(ttmτ)gm(τ)dτ.

14. Let {tm}m=0 be a sequence of points such that t0=0, tm+1>tm, and lim. For each nonegative integer m let f_m be continuous on [t_m,\infty), and let f be defined on [0,\infty) by

f(t)=f_m(t),\quad t_m\le t<t_{m+1}\quad m=0,1,2,...\nonumber

Let

g_m(t)=f_m(t+t_m)-f_{m-1}(t+t_m),\quad m=1,\dots,k.\nonumber

Extend the results of Exercise 8.6.13 to show that the solution of

ay''+by'+cy=f(t),\quad y(0)=0,\quad y'(0)=0\nonumber

is

y(t)=\int_0^t w(t-\tau)f_0(\tau)\,d\tau+\sum_{m=1}^\infty u(t-t_m) \int_0^{t-t_m}w(t-t_m-\tau)g_m(\tau) \,d\tau.\nonumber HINT: See Excercise 8.6.30


This page titled 8.6E: Convolution (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.

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