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8.6E: Convolution (Exercises)

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    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

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


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

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

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


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

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

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


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

    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


    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

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


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

    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

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


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

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

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


    \[y(t)=\int_0^t w(t-\tau)f_0(\tau)\,d\tau+\sum_{m=1}^\infty u(t-t_m) \int_0^{t-t_m}w(t-t_m-\tau)g_m(\tau) \,d\tau.\nonumber\] 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; a detailed edit history is available upon request.