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:
\(y(0)=k_0\) and \(y'(0)=k_1\).
\(y\) and \(y'\) are continuous on \([0,\infty)\).
\(y''\) 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.
\(y''\) 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
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.
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
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.
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}