5.4: Discontinuous or Impulsive Terms

We now solve some more challenging ode’s with discontinuous or impulsive inhomogeneous terms.

Example \(\PageIndex{1}\)

Solve \(2\overset{..}{x}+\overset{.}{x}+2x=u_5(t)-u_{20}(t)\), with \(x(0)=\overset{.}{x}(0)=0\)

Solution

The inhomogeneous term here is a step-up, step-down function that is unity over the interval \((5, 20)\) and zero elsewhere. Taking the Laplace transform of the ode using Table 5.1.1, \[2\left(s^2X(s)-sx(0)-\overset{.}{x}(0)\right)+sX(s)-x(0)+2X(s)=\frac{e^{-5s}}{s}-\frac{e^{-20s}}{s}.\nonumber\]

Using the initial values and solving for \(X(s)\), we find \[X(s)=\frac{e^{-5s}-e^{-20s}}{s(2s^2+s+2)}.\nonumber\]

To determine the solution for \(x(t)\) we now need to find the inverse Laplace transform. The exponential functions can be dealt with using line 13 of Table 5.1.1. We write \[X(s)=(e^{-5s}-e^{-20s})H(s),\nonumber\] where \[H(s)=\frac{1}{s(2s^2+s+2)}.\nonumber\]

Then using line 13, we have \[\label{eq:1}x(t)=u_5(t)h(t-5)-u_{20}(t)h(t-20),\] where \(h(t)=\mathcal{L}^{-1}\{H(s)\}\).

To determine \(h(t)\) we need the partial fraction expansion of \(H(s)\). Since the discriminant of \(2s^2 + s + 2\) is negative, we have \[\frac{1}{s(2s^2+s+2)}=\frac{a}{s}+\frac{bs+c}{2s^2+s+2},\nonumber\] yielding the equation \[1=a(2s^2+s+2)+(bs+c);\nonumber\] or after equating powers of \(s\), \[2a+b=0,\quad a+c=0,\quad 2a=1,\nonumber\] yielding \(a=\frac{1}{2}\), \(b=-1\), and \(c=-\frac{1}{2}\). Therefore, \[\begin{aligned}H(s)&=\frac{1/2}{s}-\frac{s+\frac{1}{2}}{2s^2+s+2} \\ &=\frac{1}{2}\left(\frac{1}{s}-\frac{s+\frac{1}{2}}{s^2+\frac{1}{2}s+1}\right).\end{aligned}\]

Inspecting Table 5.1.1, the first term can be transformed using line 2, and the second term can be transformed using lines 8 and 9, provided we complete the square of the denominator and then massage the numerator. That is, first we complete the square:

\[s^2+\frac{1}{2}s+1=\left(s+\frac{1}{4}\right)^2+\frac{15}{16};\nonumber\] and next we write \[\frac{s+\frac{1}{2}}{s^2+\frac{1}{2}s+1}=\frac{\left(s+\frac{1}{4}\right)+\frac{1}{\sqrt{15}}\sqrt{\frac{15}{16}}}{\left(s+\frac{1}{4}\right)^2+\frac{15}{16}}.\nonumber\]

Therefore, the function \(H(s)\) can be written as \[H(s)=\frac{1}{2}\left(\frac{1}{s}-\frac{\left(s+\frac{1}{4}\right)}{\left(s+\frac{1}{4}\right)^2+\frac{15}{16}}-\left(\frac{1}{\sqrt{15}}\right)\frac{\sqrt{\frac{15}{16}}}{\left(s+\frac{1}{4}\right)^2+\frac{15}{16}}\right).\nonumber\]

The first term is transformed using line 2, the second term using line 9 and the third term using line 8. We finally obtain \[\label{eq:2}h(t)=\frac{1}{2}\left(1-e^{-t/4}\left(\cos (\sqrt{15}t/4)+\frac{1}{\sqrt{15}}\sin (\sqrt{15}t/4)\right)\right),\] which when combined with \(\eqref{eq:1}\) yields the rather complicated solution for \(x(t)\).

We briefly comment that it is also possible to solve this example without using the Laplace transform. The key idea is that both \(x\) and \(\overset{.}{x}\) are continuous functions of \(t\). Clearly from the form of the inhomogeneous term and the initial conditions, \(x(t) = 0\) for \(0 ≤ t ≤ 5\). We then solve the ode between \(5 ≤ t ≤ 20\) with the inhomogeneous term equal to unity and initial conditions \(x(5) =\overset{.}{x}(5) = 0\). This requires first finding the general homogeneous solution, next finding a particular solution, and then adding the homogeneous and particular solutions and solving for the two unknown constants. To simplify the algebra, note that the best ansatz to use to find the homogeneous solution is \(x(t) = e^{r(t−5)}\), and not \(x(t) = e^{rt}\). Finally, we solve the homogeneous ode for \(t ≥ 20\) using as boundary conditions the previously determined values \(x(20)\) and \(\overset{.}{x}(20)\), where we have made use of the continuity of \(x\) and \(\overset{.}{x}\). Here, the best ansatz to use is \(x(t) = e^{r(t−20)}\). The student may benefit by trying this as an exercise and attempting to obtain a final solution that agrees with the form given by \(\eqref{eq:1}\) and \(\eqref{eq:2}\).

Example \(\PageIndex{2}\)

Solve \(2\overset{..}{x}+\overset{.}{x}+2x=\delta (t-5)\) with \(x(0)=\overset{.}{x}(0)=0\)

Solution

Here the inhomogeneous term is an impulse at time \(t = 5\). Taking the Laplace transform of the ode using Table 5.1.1, and applying the initial conditions, \[(2s^2+s+2)X(s)=e^{-5s},\nonumber\] so that \[\begin{aligned}X(s)&=\frac{1}{2}e^{-5s}\left(\frac{1}{s^2+\frac{1}{2}s+1}\right) \\ &=\frac{1}{2}e^{-5s}\left(\frac{1}{\left(s+\frac{1}{4}\right)^2+\frac{15}{16}}\right) \\ &=\frac{1}{2}\sqrt{\frac{16}{15}}e^{-5s}\left(\frac{\sqrt{\frac{15}{16}}}{\left(s+\frac{1}{4}\right)^2+\frac{15}{16}}\right).\end{aligned}\]

The inverse Laplace transform may now be computed using lines 8 and 13 of Table 5.1.1:

\[\label{eq:3}x(t)=\frac{2}{\sqrt{15}}u_5(t)e^{-(t-5)/4}\sin (\sqrt{15}(t-5)/4).\]

It is interesting to solve this example without using a Laplace transform. Clearly, \(x(t) = 0\) up to the time of impulse at \(t = 5\). Furthermore, after the impulse the ode is homogeneous and can be solved with standard methods. The only difficulty is determining the initial conditions of the homogeneous ode at \(t = 5^+\).

When the inhomogeneous term is proportional to a delta-function, the solution \(x(t)\) is continuous across the delta-function singularity, but the derivative of the solution \(\overset{.}{x}(t)\) is discontinuous. If we integrate the second-order ode across the singularity at \(t = 5\) and consider \(\epsilon\to 0\), only the second derivative term of the left-hand-side survives, and \[\begin{aligned}2\int_{5-\epsilon}^{5+\epsilon}\overset{..}{x}dt&=\int_{5-\epsilon}^{5+\epsilon}\delta (t-5)dt \\ &=1.\end{aligned}\]

And as \(\epsilon\to 0\), we have \(\overset{.}{x}(5^+)− \overset{.}{x}(5^-) = 1/2\). Since \(\overset{.}{x}(5^-) = 0\), the appropriate initial conditions immediately after the impulse force are \(x(5^+) = 0\) and \(\overset{.}{x}(5^+) = 1/2\). This result can be confirmed using \(\eqref{eq:3}\).