8.6.1: Convolution (Exercises)

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

$$\begin{aligned} y(t)&=\; k_0\dfrac{r_2e^{r_1t}-r_1e^{r_2t}}{ r_2-r_1}+k_1\dfrac{e^{r_2t}-e^{r_1t} }{ r_2-r_1} \\[4pt] &+\dfrac{1}{ a(r_2-r_1)}\int_0^t(e^{r_2\tau}-e^{r_1\tau})f(t-\tau)\,d\tau.\end{aligned}\nonumber$$

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

$$\begin{aligned} y(t)&=\; e^{\lambda t}\left[k_0(\cos\omega t-\dfrac{\lambda}{\omega}\sin\omega t)+\dfrac{k_1}{\omega}\sin\omega t\right] \\[4pt] &+\dfrac{1}{ a\omega}\int_0^te^{\lambda t}f(t-\tau)\sin\omega\tau\, d\tau.\end{aligned}\nonumber$$

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

$$f(t)=\left\{\begin{array}{cc}f_0(t),&0\le t<t_1,\\[4pt] f_1(t),&t\ge t_1.\end{array}\right.\nonumber$$

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

$$f(t)=\left\{\begin{array}{cl} f_0(t),&0\le t < t_1,\\[4pt] f_1(t),&t_1\le t < t_2,\\[4pt] &\vdots\\[4pt] f_{k-1}(t),&t_{k-1}\le t < t_k,\\[4pt] f_k(t),&t\ge t_k, \end{array}\right.\nonumber$$

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