# 8.6.1: Convolution (Exercises)

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## Q8.6.1

1. Express the inverse transform as an integral.

1. $$\dfrac{1}{ s^2(s^2+4)}$$
2. $$\dfrac{s}{(s+2)(s^2+9)}$$
3. $$\dfrac{s}{(s^2+4)(s^2+9)}$$
4. $$\dfrac{s}{(s^2+1)^2}$$
5. $$\dfrac{1}{ s(s-a)}$$
6. $$\dfrac{1}{(s+1)(s^2+2s+2)}$$
7. $$\dfrac{1}{ (s+1)^2(s^2+4s+5)}$$
8. $$\dfrac{1}{(s-1)^3(s+2)^2}$$
9. $$\dfrac{s-1}{ s^2(s^2-2s+2)}$$
10. $$\dfrac{s(s+3)}{(s^2+4)(s^2+6s+10)}$$
11. $$\dfrac{1}{(s-3)^5s^6}$$
12. $$\dfrac{1}{(s-1)^3(s^2+4)}$$
13. $$\dfrac{1}{ s^2(s-2)^3}$$
14. $$\dfrac{1}{ s^7(s-2)^6}$$

2. Find the Laplace transform.

1. $$\displaystyle \int_0^t\sin a\tau\cos b(t-\tau)\, d\tau$$
2. $$\displaystyle \int_0^t e^\tau\sin a(t-\tau)\,d\tau$$
3. $$\displaystyle \int_0^t\sinh a\tau\cosh a(t-\tau)\,d\tau$$
4. $$\displaystyle \int_0^t\tau(t-\tau)\sin \omega\tau\cos\omega (t-\tau)\,d\tau$$
5. $$\displaystyle e^t\int_0^t\sin\omega\tau \cos\omega (t-\tau)\,d\tau$$
6. $$\displaystyle e^t\int_0^t\tau^2 (t-\tau)e^\tau\,d\tau$$
7. $$\displaystyle e^{-t}\int_0^t e^{-\tau}\tau\cos\omega (t-\tau)\,d\tau$$
8. $$\displaystyle e^t\int_0^t e^{2\tau}\sinh (t-\tau)\,d\tau$$
9. $$\displaystyle \int_0^t\tau e^{2\tau}\sin 2(t-\tau)\,d\tau$$
10. $$\displaystyle \int_0^t (t-\tau)^3 e^\tau\, d\tau$$
11. $$\displaystyle \int_0^t\tau^6 e^{-(t-\tau)}\sin 3(t-\tau)\,d\tau$$
12. $$\displaystyle \int_0^t\tau^2 (t-\tau)^3\, d\tau$$
13. $$\displaystyle \int_0^t (t-\tau)^7 e^{-\tau} \sin 2\tau\,d\tau$$
14. $$\displaystyle \int_0^t (t-\tau)^4\sin 2\tau\,d\tau$$

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

1. $$y''+3y'+y=f(t),\quad y(0)=0,\quad y'(0)=0$$
2. $$y''+4y=f(t),\quad y(0)=0,\quad y'(0)=0$$
3. $$y''+2y'+y=f(t),\quad y(0)=0,\quad y'(0)=0$$
4. $$y''+k^2y=f(t),\quad y(0)=1,\quad y'(0)=-1$$
5. $$y''+6y'+9y=f(t),\quad y(0)=0,\quad y'(0)=-2$$
6. $$y''-4y=f(t),\quad y(0)=0,\quad y'(0)=3$$
7. $$y''-5y'+6y=f(t),\quad y(0)=1,\quad y'(0)=3$$
8. $$y''+\omega^2y=f(t),\quad y(0)=k_0,\quad y'(0)=k_1$$

4. Solve the integral equation.

1. $$\displaystyle y(t)=t-\int_0^t (t-\tau) y(\tau)\,d\tau$$
2. $$\displaystyle y(t)=\sin t-2 \int_0^t\cos (t-\tau) y (\tau)\,d\tau$$
3. $$\displaystyle y(t)=1+2 \int_0^ty(\tau)\cos(t-\tau)\,d\tau$$
4. $$\displaystyle y(t)=t+\int_0^t y(\tau)e^{-(t-\tau)}\,d\tau$$
5. $$\displaystyle y'(t)=t+\int_0^t y(\tau)\cos (t-\tau)\,d\tau,\, y(0)=4$$
6. $$\displaystyle y(t)=\cos t-\sin t+ \int_0^t y(\tau)\sin (t-\tau)\,d\tau$$

5. Use the convolution theorem to evaluate the integral.

1. $$\displaystyle \int_0^t (t-\tau)^7\tau^8\, d\tau$$
2. $$\displaystyle \int_0^t(t-\tau)^{13}\tau^7\,d\tau$$
3. $$\displaystyle \int_0^t(t-\tau)^6\tau^7\, d\tau$$
4. $$\displaystyle \int_0^te^{-\tau}\sin(t-\tau)\,d\tau$$
5. $$\displaystyle \int_0^t\sin\tau\cos2(t-\tau)\,d\tau$$

6. Show that $$\displaystyle \int_0^tf(t-\tau)g(\tau)\,d\tau=\int_0^tf(\tau)g(t-\tau)\,d\tau$$ by introducing the new variable of integration $$x=t-\tau$$ in the first integral.

7. Use the convolution theorem to show that if $$f(t)\leftrightarrow F(s)$$ then $$\displaystyle \int_0^tf(\tau)\,d\tau\leftrightarrow \dfrac{F(s)}{ s}.$$

8. Show that if $$p(s)=as^2+bs+c$$ has distinct real zeros $$r_1$$ and $$r_2$$ then the solution of

$ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1\nonumber$

is

\begin{aligned} y(t)&=\; k_0\dfrac{r_2e^{r_1t}-r_1e^{r_2t}}{ r_2-r_1}+k_1\dfrac{e^{r_2t}-e^{r_1t} }{ r_2-r_1} \\ &+\dfrac{1}{ a(r_2-r_1)}\int_0^t(e^{r_2\tau}-e^{r_1\tau})f(t-\tau)\,d\tau.\end{aligned}\nonumber

9. Show that if $$p(s)=as^2+bs+c$$ has a repeated real zero $$r_1$$ then the solution of

$ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1\nonumber$

is

$y(t)=\; k_0(1-r_1t)e^{r_1t}+k_1te^{r_1t} +\dfrac{1}{ a}\int_0^t\tau e^{r_1\tau}f(t-\tau)\,d\tau.\nonumber$

10. Show that if $$p(s)=as^2+bs+c$$ has complex conjugate zeros $$\lambda\pm i\omega$$ then the solution of

$ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1\nonumber$

is

\begin{aligned} y(t)&=\; e^{\lambda t}\left[k_0(\cos\omega t-\dfrac{\lambda}{\omega}\sin\omega t)+\dfrac{k_1}{\omega}\sin\omega t\right] \\ &+\dfrac{1}{ a\omega}\int_0^te^{\lambda t}f(t-\tau)\sin\omega\tau\, d\tau.\end{aligned}\nonumber

11. Let $$\displaystyle w={\cal L}^{-1}\left(\dfrac{1}{ as^2+bs+c}\right),$$ where $$a,b$$, and $$c$$ are constants and $$a\ne0$$.

1. Show that $$w$$ is the solution of $aw''+bw'+cw=0,\quad w(0)=0,\quad w'(0)=\dfrac{1}{ a}.\nonumber$
2. Let $$f$$ be continuous on $$[0,\infty)$$ and define $h(t)=\int_0^t w(t-\tau)f(\tau)\,d\tau.\nonumber$ Use Leibniz’s rule for differentiating an integral with respect to a parameter to show that $$h$$ is the solution of $ah''+bh'+ch=f,\quad h(0)=0,\quad h'(0)=0.\nonumber$
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 $$[0,\infty)$$; thus, it is not necessary to assume that $$f$$ has a Laplace transform.

12. Consider the initial value problem

$ay''+by'+cy=f(t),\quad y(0)=0,\quad y'(0)=0, \tag{A}$

where $$a,b$$, and $$c$$ are constants, $$a\ne0$$, and

$f(t)=\left\{\begin{array}{cc}f_0(t),&0\le t<t_1,\\ f_1(t),&t\ge t_1.\end{array}\right.\nonumber$

Assume that $$f_0$$ is continuous and of exponential order on $$[0,\infty)$$ and $$f_1$$ is continuous and of exponential order on $$[t_1,\infty)$$. Let

$p(s)=as^2+bs+c.\nonumber$

1. Show that the Laplace transform of the solution of (A) is $Y(s)=\dfrac{F_0(s)+e^{-st_1}G(s)}{ p(s)}\nonumber$ where $$g(t)=f_1(t+t_1)-f_0(t+t_1)$$.
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 $y(t)=\int_0^t w(t-\tau)f_0(\tau)\,d\tau+u(t-t_1)\int_0^{t-t_1} w(t-t_1-\tau)g(\tau)\,d\tau\nonumber$ for $$t>0$$.
3. Henceforth, assume only that $$f_0$$ is continuous on $$[0,\infty)$$ and $$f_1$$ is continuous on $$[t_1,\infty)$$. Use Exercise 8.6.11 (a) and (b) to show that $y'(t)=\int_0^t w'(t-\tau)f_0(\tau)\,d\tau+u(t-t_1)\int_0^{t-t_1} w'(t-t_1-\tau)g(\tau)\,d\tau\nonumber$ for $$t>0$$, and $y''(t)=\dfrac{f(t)}{a}+\int_0^t w''(t-\tau)f_0(\tau)\,d\tau+u(t-t_1)\int_0^{t-t_1} w''(t-t_1-\tau)g(\tau)\,d\tau\nonumber$ for $$0<t<t_{1}$$ and $$t>t_{1}$$. Also, show $$y$$ satisfies the differential equation in (A) on$$(0,t_1)$$ and $$(t_1,\infty)$$.
4. Show that $$y$$ and $$y'$$ are continuous on $$[0,\infty)$$.

13. Suppose

$f(t)=\left\{\begin{array}{cl} f_0(t),&0\le t < t_1,\\ f_1(t),&t_1\le t < t_2,\\ &\vdots\\ f_{k-1}(t),&t_{k-1}\le t < t_k,\\ f_k(t),&t\ge t_k, \end{array}\right.\nonumber$

where $$f_m$$ is continuous on $$[t_m,\infty)$$ for $$m=0,\dots,k$$ (let $$t_0=0$$), and define

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

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

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

is

$y(t)=\int_0^t w(t-\tau)f_0(\tau)\,d\tau+\sum_{m=1}^ku(t-t_m) \int_0^{t-t_m}w(t-t_m-\tau)g_m(\tau)\,d\tau.\nonumber$

14. Let $$\{t_m\}_{m=0}^\infty$$ be a sequence of points such that $$t_0=0$$, $$t_{m+1}>t_m$$, and $$\lim_{m\to\infty}t_m=\infty$$. 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 8.6.1: Convolution (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.