8.5E: Constant Coefficient Equations with Piecewise Continuous Forcing Functions (Exercises)

Q8.5.1

In Exercises 8.5.1-8.5.20 use the Laplace transform to solve the initial value problem. Graph the solution for Exercise 8.5.6, 8.5.9, 8.5.13, and 8.5.19.

1. $y''+y=\left\{\begin{array}{cl} 3,& 0\le t<\pi,\\[4pt] 0,&t\ge\pi,\end{array}\right. \qquad y(0)=0, \quad y'(0)=0$

2. $y''+y=\left\{\begin{array}{cl} 3,&0\le t<4,\\[4pt]; 2t-5,&t > 4,\end{array}\right.\qquad y(0)=1,\quad y'(0)=0$

3. $y''-2y'= \left\{\begin{array}{cl} 4,&0\le t<1,\\[4pt] 6,&t\ge 1,\end{array}\right.\qquad y(0)=-6,\quad y'(0)=1$

4. $y''-y=\left\{\begin{array}{cl} e^{2t},&0\le t< 2,\\[4pt] 1,&t\ge 2,\end{array}\right.\qquad y(0)=3,\quad y'(0)=-1$

5. $y''-3y'+2y= \left\{\begin{array}{rl} 0,&0\le t<1,\\[4pt] 1,&1\le t<2,\\[4pt]-1,&t\ge 2, \end{array}\right.\qquad y(0)=-3,\quad y'(0)=1$

6. $y''+4y= \left\{\begin{array}{cl}|\sin t|,&0\le t<2\pi,\\[4pt] 0,&t\ge 2\pi,\end{array}\right.\qquad y(0)=-3,\quad y'(0)=1$

7. $y''-5y'+4y= \left\{\begin{array}{rl} 1,&0\le t<1\\[4pt] -1,&1\le t<2,\\[4pt] 0,&t\ge 2,\end{array}\right.\qquad y(0)=3,\quad y'(0)=-5$

8. $y''+9y=\left\{\begin{array}{ll}{\cos t,}&{0\leq t<\frac{3\pi }{2},}\\[4pt]{\sin t,}&{t\geq \frac{3\pi }{2},} \end{array} \right. \quad y(0)=0,\: y'(0)=0$

9. $y''+4y=\left\{\begin{array}{ll}{t,}&{0\leq t<\frac{\pi }{2},}\\[4pt]{\pi ,}&{t\geq \frac{\pi }{2},} \end{array} \right. \quad y(0)=0,\: y'(0)=0$

10. $y''+y=\left\{\begin{array}{cl}\phantom{-}t,&0\le t<\pi, \\[4pt]-t,&t\ge\pi ,\end{array}\right.\; y(0)=0,\; y'(0)=0$

11. $y''-3y'+2y=\left\{\begin{array}{cl} 0,&0\le t<2,\\[4pt]2t-4,&t\ge 2,\end{array}\right. ,\quad y(0)=0,\quad y'(0)=0$

12. $y''+y=\left\{\begin{array}{cl} t,&0\le t<2\pi,\\[4pt]-2t,&t\ge 2\pi,\end{array}\right.\quad y(0)=1,\quad y'(0)=2$

13. $y''+3y'+2y=\left\{\begin{array}{cl}\phantom{-}1,&0\le t<2,\\[4pt]-1,&t\ge 2,\end{array}\right.\; y(0)=0,\; y'(0)=0$

14. $y''-4y'+3y=\left\{\begin{array}{cl}-1,&0\le t<1,\\[4pt]\phantom{-}1,&t\ge 1,\end{array}\right.\; y(0)=0,\; y'(0)=0$

15. $y''+2y'+y=\left\{\begin{array}{cl} e^t,&0\le t<1,\\[4pt]e^t-1,&t\ge 1,\end{array}\right.\; y(0)=3,\; y'(0)=-1$

16. $y''+2y'+y=\left\{\begin{array}{cl} 4e^t,&0\le t<1,\\[4pt]0,&t\ge 1,\end{array}\right.\; y(0)=0,\; y'(0)=0$

17. $y''+3y'+2y=\left\{\begin{array}{cl} e^{-t},&0\le t<1,\\[4pt]0,&t\ge 1,\end{array}\right.\; y(0)=1,\; y'(0)=-1$

18. $y''-4y'+4y=\left\{\begin{array}{rl} e^{2t},&0\le t<2,\\[4pt]-e^{2t},&t\ge 2,\end{array}\right.\; y(0)=0,\; y'(0)=-1$

19. $y''=\left\{\begin{array}{cl}t^2,&0\le t<1,\\[4pt]-t,&1\le t<2,\\[4pt]t+1,&t\ge 2,\end{array}\right.\; y(0)=1,\; y'(0)=0$

20. $y''+2y'+2y=\left\{\begin{array}{rl}1,&0\le t<2\pi,\\[4pt]t,&2\pi\le t<3\pi,\\[4pt]-1,&t\ge 3\pi,\end{array}\right.\; y(0)=2,\quad y'(0)=-1$

Q8.5.2

21. Solve the initial value problem $$y''=f(t), \quad y(0)=0,\quad y'(0)=0,\nonumber$$ where $$f(t)=m+1,\quad m\le t<m+1,\quad m=0,1,2,\dots.\nonumber$$

22. Solve the given initial value problem and find a formula that does not involve step functions and represents $y$ on each interval of continuity of $f$.

1. $y''+y=f(t), \quad y(0)=0,\quad y'(0)=0$;
   $f(t)=m+1,\quad m\pi\le t<(m+1)\pi,\quad m=0,1,2,\dots$.
2. $y''+y=f(t), \quad y(0)=0,\quad y'(0)=0$;
   $f(t)=(m+1)t, \quad 2m\pi\le t<2(m+1)\pi,\quad m=0,1,2,\dots$ HINT: You'll need the formula $$1+2+\cdots+m={m(m+1)\over2}.\nonumber$$
3. $y''+y=f(t), \quad y(0)=0,\quad y'(0)=0$;
   $f(t)=(-1)^m,\quad m\pi\le t<(m+1)\pi,\quad m=0,1,2,\dots.$
4. $y''-y=f(t), \quad y(0)=0,\quad y'(0)=0$;
   $f(t)=m+1,\quad m\le t<(m+1),\quad m=0,1,2,\dots.$
   HINT: You will need the formula $$1+r+...+r^{m}=\frac{1-r^{m+1}}{1-r}(r\neq 1).\nonumber$$
5. $y''+2y'+2y=f(t), \quad y(0)=0,\quad y'(0)=0$;
   $f(t)=(m+1)(\sin t+2\cos t),\quad 2m\pi\le t<2(m+1)\pi,\quad m=0,1,2,\dots.$
   (See the hint in d.)
6. $y''-3y'+2y=f(t), \quad y(0)=0,\quad y'(0)=0$;
7. $f(t)=m+1,\quad m\le t<m+1,\quad m=0,1,2,\dots.$
   (See the hints in b and d.)

23.

1. Let $g$ be continuous on $(\alpha,\beta)$ and differentiable on the $(\alpha,t_0)$ and $(t_0,\beta)$. Suppose $A=\lim_{t\to t_0-}g'(t)$ and $B=\lim_{t\to t_0+}g'(t)$ both exist. Use the mean value theorem to show that $$\lim_{t\to t_0-}{g(t)-g(t_0)\over t-t_0}=A\quad\mbox{ and }\quad \lim_{t\to t_0+}{g(t)-g(t_0)\over t-t_0}=B.\nonumber$$
2. Conclude from (a) that $g'(t_0)$ exists and $g'$ is continuous at $t_0$ if $A=B$.
3. Conclude from (a) that if $g$ is differentiable on $(\alpha,\beta)$ then $g'$ can’t have a jump discontinuity on $(\alpha,\beta)$.

24.

1. Let $a$, $b$, and $c$ be constants, with $a\ne0$. Let $f$ be piecewise continuous on an interval $(\alpha,\beta)$, with a single jump discontinuity at a point $t_0$ in $(\alpha,\beta)$. Suppose $y$ and $y'$ are continuous on $(\alpha,\beta)$ and $y''$ on $(\alpha,t_0)$ and $(t_0,\beta)$. Suppose also that $$ay''+by'+cy=f(t) \tag{A}$$ on $(\alpha,t_0)$ and $(t_0,\beta)$. Show that $$y''(t_0+)-y''(t_0-)={f(t_0+)-f(t_0-)\over a}\ne0.\nonumber$$
2. Use (a) and Exercise 8.5.23c to show that (A) does not have solutions on any interval $(\alpha,\beta)$ that contains a jump discontinuity of $f$.

25. Suppose $P_0,P_1$, and $P_2$ are continuous and $P_0$ has no zeros on an open interval $(a,b)$, and that $F$ has a jump discontinuity at a point $t_0$ in $(a,b)$. Show that the differential equation $$P_0(t)y''+P_1(t)y'+P_2(t)y=F(t)\nonumber$$ has no solutions on $(a,b)$. HINT: Generalize the result of Exercise 8.5.24 and use Exercise 8.5.23c.

26. Let $0=t_0<t_1<\cdots <t_n$. Suppose $f_m$ is continuous on $[t_m,\infty)$ for $m=1,\dots,n$. Let $$f(t)= \left\{\begin{array}{cl} f_m(t),&t_m\le t< t_{m+1},\quad m=1,\dots,n-1,\\[4pt] f_n(t),&t\ge t_n. \end{array}\right.\nonumber$$ Show that the solution of

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

as defined following Theorem 8.5.1, is given by

$$y=\left\{\begin{array}{cl} z_0(t),&0\le t<t_1,\\[4pt] z_0(t)+ z_1(t),&t_1\le t<t_2,\\[4pt] &\vdots\\[4pt] z_0+\cdots+z_{n-1}(t),&t_{n-1}\le t<t_n,\\[4pt] z_0+\cdots+ z_n(t),&t\ge t_n, \end{array}\right.\nonumber$$

where $z_0$ is the solution of

$$az''+bz'+cz=f_0(t), \quad z(0)=k_0,\quad z'(0)=k_1\nonumber$$

and $z_m$ is the solution of

$$az''+bz'+cz=f_m(t)-f_{m-1}(t), \quad z(t_m)=0,\quad z'(t_m)=0\nonumber$$

for $m=1,\dots,n$.