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Mathematics LibreTexts

8.6.1: Convolution (Exercises)

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    30773
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    Q8.6.1

    1. Express the inverse transform as an integral.

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

    2. Find the Laplace transform.

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

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

    5. Use the convolution theorem to evaluate the integral.

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

    6. Show that

    \[\int_0^tf(t-\tau)g(\tau)\,d\tau=\int_0^tf(\tau)g(t-\tau)\,d\tau\nonumber \]

    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

    \[\int_0^tf(\tau)\,d\tau\leftrightarrow {F(s)\over s}.\nonumber \]

    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

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

    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

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

    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

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

    11. Let

    \[w={\cal L}^{-1}\left(1\over as^2+bs+c\right),\nonumber\]

    where \(a,b\), and \(c\) are constants and \(a\ne0\).

    1. Show that \(w\) is the solution of \[aw''+bw'+cw=0,\quad w(0)=0,\quad w'(0)={1\over a}.\nonumber\]
    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 \[Y(s)={F_0(s)+e^{-st_1}G(s)\over p(s)}\nonumber\] 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 \[y''(t)={f(t)\over a}+\int_0^t w''(t-\tau)f_0(\tau)\,d\tau+u(t-t_1)\int_0^{t-t_1} w''(t-t_1-\tau)g(\tau)\,d\tau\nonumber\] 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

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