7.5: Constant Coefficient Equations with Piecewise Continuous Forcing Functions

We’ll now consider initial value problems of the form

$$\label{eq:8.5.1} ay''+by'+cy=f(t), \quad y(0)=k_0,\quad y'(0)=k_1,$$

where $a$, $b$, and $c$ are constants ($a\ne0$) and $f$ is piecewise continuous on $[0,\infty)$. Problems of this kind occur in situations where the input to a physical system undergoes instantaneous changes, as when a switch is turned on or off or the forces acting on the system change abruptly.

It can be shown (Exercises 8.5.23 and 8.5.24) that the differential equation in Equation $\ref{eq:8.5.1}$ has no solutions on an open interval that contains a jump discontinuity of $f$. Therefore we must define what we mean by a solution of Equation $\ref{eq:8.5.1}$ on $[0,\infty)$ in the case where $f$ has jump discontinuities. The next theorem motivates our definition. We omit the proof.

Theorem $\PageIndex{1}$

Suppose $a,b$, and $c$ are constants $(a\ne0),$ and $f$ is piecewise continuous on $[0,\infty).$ with jump discontinuities at $t_1,$ …, $t_n,$ where

$$0<t_1<\cdots<t_n. \nonumber$$

Let $k_0$ and $k_1$ be arbitrary real numbers. Then there is a unique function $y$ defined on $[0,\infty)$ with these properties:

1. $y(0)=k_0$ and $y'(0)=k_1$.
2. $y$ and $y'$ are continuous on $[0,\infty)$.
3. $y''$ is defined on every open subinterval of $[0,\infty)$ that does not contain any of the points $t_1,$ …, $t_n$, and $$ay''+by'+cy=f(t) \nonumber$$ on every such subinterval.
4. $y''$ has limits from the right and left at $t_1,$ …$,$ $t_n$.

We define the function $y$ of Theorem $\PageIndex{1}$ to be the solution of the initial value problem Equation $\ref{eq:8.5.1}$.

We begin by considering initial value problems of the form

$$\label{eq:8.5.2} ay''+by'+cy=\left\{\begin{array}{cl} f_0(t),&0\le t<t_1,\\[4pt]f_1(t),&t\ge t_1, \end{array}\right.\quad y(0)=k_0,\quad y'(0)=k_1,$$

where the forcing function has a single jump discontinuity at $t_1$.

Howto: Solve Constant Coefficient Equations with Piecewise Continuous Forcing Functions

We can solve Equation $\ref{eq:8.5.2}$ by the these steps:

• Step 1. Find the solution $y_0$ of the initial value problem $$ay''+by'+cy=f_0(t), \quad y(0)=k_0,\quad y'(0)=k_1. \nonumber$$
• Step 2. Compute $c_0=y_0(t_1)$ and $c_1=y_0'(t_1)$.
• Step 3. Find the solution $y_1$ of the initial value problem $$ay''+by'+cy=f_1(t), \quad y(t_1)=c_0,\quad y'(t_1)=c_1.\nonumber$$
• Step 4. Obtain the solution $y$ of Equation $\ref{eq:8.5.2}$ as $$y=\left\{\begin{array}{cl} y_0(t),&0\le t<t_1\\[4pt]y_1(t),&t\ge t_1. \end{array}\right.\nonumber$$

It is shown in Exercise 8.5.23 that $y'$ exists and is continuous at $t_1$. The next example illustrates this procedure.

Example $\PageIndex{1}$

Solve the initial value problem

$$\label{eq:7.5.3} y''+y=f(t), \quad y(0)=2,\; y'(0)=-1,$$

where

$$f(t)=\left\{\begin{array}{rl} 1,&0\le t< {\pi\over2},\\[4pt] -1,&t\ge {\pi\over2}. \end{array}\right. \nonumber$$

Figure $\PageIndex{1}$: Graph of Equation $\ref{eq:8.5.4}$

Solution

The initial value problem in Step 1 is

$$y''+y=1, \quad y(0)=2,\quad y'(0)=-1. \nonumber$$

We leave it to you to verify that its solution is

$$y_0=1+\cos t-\sin t. \nonumber$$

Doing Step 2 yields $y_0(\pi/2)=0$ and $y_0'(\pi/2)=-1$, so the second initial value problem is

$$y''+y=-1, \quad y\left({\pi\over2}\right)=0,\; y'\left({\pi\over 2}\right)=-1. \nonumber$$

We leave it to you to verify that the solution of this problem is

$$y_1=-1+\cos t+\sin t. \nonumber$$

Hence, the solution of Equation $\ref{eq:8.5.3}$ is

$$\label{eq:8.5.4} y=\left\{\begin{array}{rl} 1+\cos t-\sin t,&0\le t< {\pi\over2}, \\[4pt] -1+\cos t+\sin t,&t\ge {\pi\over2} \end{array}\right.$$

If $f_0$ and $f_1$ are defined on $[0,\infty)$, we can rewrite Equation $\ref{eq:8.5.2}$ as

$$ay''+by'+cy=f_0(t)+u(t-t_1)\left(f_1(t)-f_0(t)\right), \quad y(0)=k_0,\quad y'(0)=k_1, \nonumber$$

and apply the method of Laplace transforms. We’ll now solve the problem considered in Example [example:8.5.1} by this method.

Example $\PageIndex{2}$

Use the Laplace transform to solve the initial value problem

$$\label{eq:8.5.5} y''+y=f(t), \quad y(0)=2,\; y'(0)=-1,$$

where

$$f(t)=\left \{ \begin{array}{cl} \phantom{-}1,&0\le t< {\pi\over2},\\ -1,&t \ge {\pi\over2}. \end{array}\right. \nonumber$$

Solution

Here

$$f(t)=1-2u\left(t-{\pi\over2}\right), \nonumber$$

so Theorem $\PageIndex{1}$ (with $g(t)=1$) implies that

$${\cal L}(f)={1-2e^{-{\pi s/2}}\over s}. \nonumber$$

Therefore, transforming Equation $\ref{eq:8.5.5}$ yields

$$(s^2+1)Y(s)={1-2e^{-{\pi s/ 2}}\over s}-1+2s, \nonumber$$

so

$$\label{eq:8.5.6} Y(s)=(1-2e^{-{\pi s/ 2}}) G(s)+{2s-1\over s^2+1},$$

with

$$G(s)={1\over s(s^2+1)}. \nonumber$$

The form for the partial fraction expansion of $G$ is

$$\label{eq:8.5.7} {1\over s(s^2+1)}={A\over s}+{Bs+C\over s^2+1}.$$

Multiplying through by $s(s^2+1)$ yields

$$A(s^2+1)+(Bs+C)s=1, \nonumber$$

or

$$(A+B)s^2+Cs+A=1. \nonumber$$

Equating coefficients of like powers of $s$ on the two sides of this equation shows that $A=1$, $B=-A=-1$ and $C=0$. Hence, from Equation $\ref{eq:8.5.7}$,

$$G(s)={1\over s}-{s\over s^2+1}. \nonumber$$

Therefore

$$g(t)=1-\cos t. \nonumber$$

From this, Equation $\ref{eq:8.5.6}$, and Theorem 8.4.2,

$$y=1-\cos t-2u\left(t-{\pi\over2}\right)\left(1-\cos\left(t-{\pi \over2}\right)\right)+2\cos t-\sin t. \nonumber$$

Simplifying this (recalling that $\cos (t-\pi/2)=\sin t)$ yields

$$y=1+\cos t-\sin t-2u\left(t-{\pi\over2}\right)(1-\sin t), \nonumber$$

or

$$y=\left\{\begin{array}{cl}{1+\cos t-\sin t,}&{0\leq t<\frac{\pi }{2}}\\{-1+\cos t+\sin t,}&{t\geq \frac{\pi }{2}} \end{array} \right.\nonumber$$

which is the result obtained in Example $\PageIndex{1}$.

Note

It isn’t obvious that using the Laplace transform to solve Equation $\ref{eq:8.5.2}$ as we did in Example $\PageIndex{2}$ yields a function $y$ with the properties stated in Theorem $\PageIndex{1}$; that is, such that $y$ and $y'$ are continuous on $[0, ∞)$ and $y''$ has limits from the right and left at $t_{1}$. However, this is true if $f_{0}$ and $f_{1}$ are continuous and of exponential order on $[0, ∞)$. A proof is sketched in Exercises 8.6.11–8.6.13.

Example $\PageIndex{3}$

Solve the initial value problem

$$\label{eq:8.5.8} y''-y=f(t), \quad y(0)=-1,\; y'(0)=2,$$

where

$$f(t)=\left\{\begin{array}{cl} t,&0\le t<1,\\ 1,&t\ge 1. \end{array}\right.\nonumber$$

Solution

Here

$$f(t)=t-u(t-1)(t-1),\nonumber$$

so

$$\begin{align*} {\cal L}(f)&={\cal L}(t)-{\cal L}\left(u(t-1)(t-1)\right)\\[4pt] &=\cal L(t)-e^{-s}{\cal L}(t)\text{ (from Theorem 8.4.1})\\[4pt] &={1\over s^2}-{e^{-s}\over s^2}.\end{align*}\nonumber$$

Since transforming Equation $\ref{eq:8.5.8}$ yields

$$(s^2-1) Y(s)={\cal L}(f)+2-s,\nonumber$$

we see that

$$\label{eq:8.5.9} Y(s)=(1-e^{-s})H(s)+{2-s\over s^2-1},$$

where

$$H(s)={1\over s^2(s^2-1)}={1\over s^2-1}-{1\over s^2};\nonumber$$

therefore

$$\label{eq:8.5.10} h(t)=\sinh t-t.$$

Since

$${\cal L}^{-1}\left({2-s\over s^2-1}\right)=2\sinh t-\cosh t,\nonumber$$

we conclude from Equation $\ref{eq:8.5.9}$, Equation $\ref{eq:8.5.10}$, and Theorem $\PageIndex{1}$ that

$$y=\sinh t-t-u(t-1)\left(\sinh (t-1)-t+1\right)+2\sinh t- \cosh t,\nonumber$$

or

$$\label{eq:8.5.11} y=3\sinh t-\cosh t-t-u(t-1)\left(\sinh (t-1)-t+1\right)$$

We leave it to you to verify that $y$ and $y'$ are continuous and $y''$ has limits from the right and left at $t_1=1$.

Example $\PageIndex{4}$

Solve the initial value problem

$$\label{eq:8.5.12} y''+y=f(t), \quad y(0)=0,\; y'(0)=0,$$

where

$$f(t)=\left\{\begin{array}{cl}{0,}&{0\leq t<\frac{\pi }{4}}\\{\cos 2t}&{\frac{\pi }{4}\leq t< \pi }\\{0,}&{t\geq \pi } \end{array} \right.\nonumber$$

Solution

Here

$$f(t)=u(t-\pi/4)\cos2t-u(t-\pi)\cos2t, \nonumber$$

so

$$\begin{align*} {\cal L}(f)&={\cal L}\left(u(t-\pi/4)\cos2t\right)-{\cal L}\left( u(t-\pi)\cos2t\right)\\[4pt] &=e^{-{\pi s/4}}{\cal L}\left(\cos2(t+\pi/4)\right)-e^{-\pi s} {\cal L}\left(\cos2(t+\pi)\right)\\[4pt] &=-e^{-{\pi s/4}}{\cal L}(\sin2t)-e^{-\pi s} {\cal L}(\cos2t)\\[4pt] &=-{2e^{-{\pi s/ 4}}\over s^2+4}-{se^{-\pi s}\over s^2+4}.\end{align*}\nonumber$$

Since transforming Equation $\ref{eq:8.5.12}$ yields

$$(s^2+1)Y(s)={\cal L}(f),\nonumber$$

we see that

$$\label{eq:8.5.13} Y(s)=e^{-{\pi s/ 4}} H_1(s)+e^{-\pi s} H_2(s),$$

where

$$\label{eq:8.5.14} H_1(s)=-{2\over (s^2+1)(s^2+4)}\quad\mbox{ and }\quad H_2(s)=-{s \over (s^2+1)(s^2+4)}.$$

To simplify the required partial fraction expansions, we first write

$${1\over (x+1)(x+4)}={1\over3}\left[{1\over x+1}-{1\over x+4}\right].\nonumber$$

Setting $x=s^2$ and substituting the result in Equation $\ref{eq:8.5.14}$ yields

$$H_1(s)=-{2\over3}\left[{1\over s^2+1}-{1\over s^2+4}\right] \quad\mbox{ and }\quad H_2(s)=-{1\over3}\left[{s\over s^2+1}-{s\over s^2+4}\right].\nonumber$$

The inverse transforms are

$$h_1(t)=-{2\over3}\sin t+{1\over3}\sin2t \quad\mbox{ and }\; h_2(t)=-{1\over3}\cos t+{1\over3}\cos2t.\nonumber$$

From Equation $\ref{eq:8.5.13}$ and Theorem 8.4.2,

$$\label{eq:8.5.15} y=u\left(t-{\pi\over4}\right) h_1\left(t-{\pi\over4}\right)+ u(t-\pi) h_2(t-\pi).$$

Since

$$\begin{aligned} h_1\left(t-{\pi\over4}\right)&=-{2\over3}\sin\left(t-{\pi\over 4}\right)+{1\over3}\sin2\left(t-{\pi\over4}\right)\\ &=-{\sqrt{2}\over3} (\sin t-\cos t)-{1\over3}\cos2t\end{aligned}\nonumber$$

and

$$\begin{align*} h_2(t-\pi)&=-{1\over3}\cos (t-\pi)+{1\over3}\cos2(t-\pi)\\ &={1\over3}\cos t+{1\over3}\cos2t,\end{align*}\nonumber$$

Equation $\ref{eq:8.5.15}$ can be rewritten as

$$y=-{1\over3}u\left(t-{\pi\over4}\right)\left(\sqrt{2}(\sin t-\cos t)+\cos2t\right) + {1\over3} u(t-\pi) (\cos t+\cos2t)\nonumber$$

or

$$\label{eq:8.5.16} \left\{\begin{array}{cl}{0,}&{0\leq t< \frac{\pi }{4}}\\{-\frac{\sqrt{2}}{3}(\sin t-\cos t)-\frac{1}{3}\cos 2t,}&{\frac{\pi }{4}\leq t<\pi ,}\\{-\frac{\sqrt{2}}{3}\sin t+\frac{1+\sqrt{2}}{3}\cos t,}&{t\geq\pi } \end{array} \right.$$

We leave it to you to verify that $y$ and $y'$ are continuous and $y''$ has limits from the right and left at $t_1=\pi/4$ and $t_2=\pi$ (Figure $\PageIndex{2}$).

Figure $\PageIndex{2}$: Graph of Equation $\ref{eq:8.5.16}$