Discontinuous forcing · Differential Equations

Intuition

With the step function, a forcing that switches on and off becomes one formula, and the transform handles it without solving on each piece separately and matching the pieces up. The solution comes out as a sum of delayed responses, each switched on when its piece of forcing starts. Even when the forcing jumps, the solution of a second-order equation stays smooth: it is the acceleration that jumps, not the position or the velocity.

Pushing a swing for a few seconds and letting go. The push switches on and off, but the swing never jumps: it gathers speed while the push lasts, and swings on freely afterwards.

$y''+y=u(t-1)-u(t-4)$ from rest: a push of size $1$ from $t=1$ to $t=4$ . The response $u(t-1)\bigl(1-\cos(t-1)\bigr)-u(t-4)\bigl(1-\cos(t-4)\bigr)$ is smooth at both switches, and after the push the spring swings freely.

One formula for a switched problem

To solve $y''+py'+qy=g(t)$ with switched on and off: write with steps, transform it — each piece switched on at gains a factor — solve for , and invert each term, keeping its exponential as a delay. If is the response from rest to a unit step switched on at , the response to one switched on at is delayed by . The answer holds for all at once, and and are continuous at every switch: the jump of appears only in .

Y=\frac{e^{-cs}}{s\,(s^{2}+ps+q)}\ \Longrightarrow\ y=u(t-c)\,h(t-c)

Switched problems

$y''+y=u(t-c)$ from rest: $\frac{1}{s (s^{2} + 1)} = \frac{1}{}$ , so and .

A delayed switch gives a delayed response

\begin{aligned}&y''+py'+qy=u(t-c)\ \text{from rest}\\&\Longrightarrow\ y=u(t-c)\,h(t-c)\end{aligned}

Transform both problems. From rest, the transform of each solution is the transform of its forcing divided by the characteristic polynomial. The delayed step’s transform is the undelayed one times e to the minus cs, so the same factor multiplies the solution’s transform, and by the second shifting rule that factor is a delay.

Proof steps

\mathcal{L}\{h\}=\frac{1}{s\,(s^{2}+ps+q)}

The step response from rest: the step at 0 has transform 1/s.

\mathcal{L}\{y\}=\frac{e^{-cs}}{s\,(s^{2}+ps+q)}=e^{-cs}\,\mathcal{L}\{h\}

The delayed step has transform e to the minus cs over s.

y=u(t-c)\,h(t-c)

By the second shifting rule, the factor is a delay.

Applications

Practice

Switched Forcing, One Formula

Write the forcing with steps; each piece switched on at c brings a factor e to the minus cs into the transform.

y''+y=u(t-2):\quad(s^{2}+1)Y=\frac{e^{-2s}}{s}

Try it

$y''+4y=u(t-1)$ from rest. What is $Y$ ?