8.6E: Convolution (Exercises)
( \newcommand{\kernel}{\mathrm{null}\,}\)
Q8.6.1
1. Express the inverse transform as an integral.
- 1s2(s2+4)
- s(s+2)(s2+9)
- s(s2+4)(s2+9)
- s(s2+1)2
- 1s(s−a)
- 1(s+1)(s2+2s+2)
- 1(s+1)2(s2+4s+5)
- 1(s−1)3(s+2)2
- s−1s2(s2−2s+2)
- s(s+3)(s2+4)(s2+6s+10)
- 1(s−3)5s6
- 1(s−1)3(s2+4)
- 1s2(s−2)3
- 1s7(s−2)6
2. Find the Laplace transform.
- ∫t0sinaτcosb(t−τ)dτ
- ∫t0eτsina(t−τ)dτ
- ∫t0sinhaτcosha(t−τ)dτ
- ∫t0τ(t−τ)sinωτcosω(t−τ)dτ
- et∫t0sinωτcosω(t−τ)dτ
- et∫t0τ2(t−τ)eτdτ
- e−t∫t0e−ττcosω(t−τ)dτ
- et∫t0e2τsinh(t−τ)dτ
- ∫t0τe2τsin2(t−τ)dτ
- ∫t0(t−τ)3eτdτ
- ∫t0τ6e−(t−τ)sin3(t−τ)dτ
- ∫t0τ2(t−τ)3dτ
- ∫t0(t−τ)7e−τsin2τdτ
- ∫t0(t−τ)4sin2τdτ
3. Find a formula for the solution of the initial value problem.
- y″+3y′+y=f(t),y(0)=0,y′(0)=0
- y″+4y=f(t),y(0)=0,y′(0)=0
- y″+2y′+y=f(t),y(0)=0,y′(0)=0
- y″+k2y=f(t),y(0)=1,y′(0)=−1
- y″+6y′+9y=f(t),y(0)=0,y′(0)=−2
- y″−4y=f(t),y(0)=0,y′(0)=3
- y″−5y′+6y=f(t),y(0)=1,y′(0)=3
- y″+ω2y=f(t),y(0)=k0,y′(0)=k1
4. Solve the integral equation.
- y(t)=t−∫t0(t−τ)y(τ)dτ
- y(t)=sint−2∫t0cos(t−τ)y(τ)dτ
- y(t)=1+2∫t0y(τ)cos(t−τ)dτ
- y(t)=t+∫t0y(τ)e−(t−τ)dτ
- y′(t)=t+∫t0y(τ)cos(t−τ)dτ,y(0)=4
- y(t)=cost−sint+∫t0y(τ)sin(t−τ)dτ
5. Use the convolution theorem to evaluate the integral.
- ∫t0(t−τ)7τ8dτ
- ∫t0(t−τ)13τ7dτ
- ∫t0(t−τ)6τ7dτ
- ∫t0e−τsin(t−τ)dτ
- ∫t0sinτcos2(t−τ)dτ
6. Show that ∫t0f(t−τ)g(τ)dτ=∫t0f(τ)g(t−τ)dτ by introducing the new variable of integration x=t−τ in the first integral.
7. Use the convolution theorem to show that if f(t)↔F(s) then ∫t0f(τ)dτ↔F(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)=k0r2er1t−r1er2tr2−r1+k1er2t−er1tr2−r1+1a(r2−r1)∫t0(er2τ−er1τ)f(t−τ)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(1−r1t)er1t+k1ter1t+1a∫t0τer1τf(t−τ)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(cosωt−λωsinωt)+k1ωsinωt]+1aω∫t0eλtf(t−τ)sinωτdτ.
11. Let w=L−1(1as2+bs+c), where a,b, and c are constants and a≠0.
- Show that w is the solution of aw″+bw′+cw=0,w(0)=0,w′(0)=1a.
- Let f be continuous on [0,∞) and define h(t)=∫t0w(t−τ)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.
- 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, a≠0, and
f(t)={f0(t),0≤t<t1,f1(t),t≥t1.
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.
- Show that the Laplace transform of the solution of (A) is Y(s)=F0(s)+e−st1G(s)p(s) where g(t)=f1(t+t1)−f0(t+t1).
- 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(t−τ)f0(τ)dτ+u(t−t1)∫t−t10w(t−t1−τ)g(τ)dτ for t>0.
- 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′(t−τ)f0(τ)dτ+u(t−t1)∫t−t10w′(t−t1−τ)g(τ)dτ for t>0, and y″(t)=f(t)a+∫t0w″(t−τ)f0(τ)dτ+u(t−t1)∫t−t10w″(t−t1−τ)g(τ)dτ for 0<t<t1 and t>t1
. Also, show y satisfies the differential equation in (A) on(0,t1) and (t1,∞). - Show that y and y′ are continuous on [0,∞).
13. Suppose
f(t)={f0(t),0≤t<t1,f1(t),t1≤t<t2,⋮fk−1(t),tk−1≤t<tk,fk(t),t≥tk,
where fm is continuous on [tm,∞) for m=0,…,k (let t0=0), and define
gm(t)=fm(t+tm)−fm−1(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(t−τ)f0(τ)dτ+k∑m=1u(t−tm)∫t−tm0w(t−tm−τ)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