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Discontinuous Forcing Functions

Differential Equations · Section 6.4 · generated practice set

IVPs whose forcing switches on or off, solved in one pass instead of interval by interval.

Practice this set → Fresh numbers on every attempt. No account needed.

Method

This is 6.2 and 6.3 used together: an IVP whose forcing switches on, off, or changes abruptly. The Laplace transform handles it in one pass, which is exactly why the method exists — the classical approach requires solving on each interval and matching at the breakpoints.

The procedure

  1. Write the forcing in step form using 6.3, so the right side becomes a sum of terms $u_c(t)g(t-c)$.
  2. Transform. Each forcing term contributes $e^{-cs}G(s)$. With zero initial conditions the left side is simply $(as^2+bs+c)Y$.
  3. Solve for $Y(s)$ and factor out each exponential: $Y = e^{-c_1s}H_1(s) + e^{-c_2s}H_2(s) + cdots$
  4. Invert each $H_i$ separately by partial fractions or completing the square — ignore the exponentials while you do it.
  5. Re-attach the shifts at the very end: $e^{-cs}H(s) mapsto u_c(t)h(t-c)$.

What the answer looks like

The solution is continuous and has a continuous first derivative even though the forcing jumps; the discontinuity appears only in $y''$. Physically: kick a mass-spring system with a sudden constant force and the position and velocity stay smooth — the acceleration is what jumps.

The response to $u_c(t)$ is zero until $t = c$, then relaxes toward the new equilibrium, oscillating on the way if the system is underdamped.

Definitions and theorems

Step response
For $y' + ay = u_c(t)$ with $y(0) = 0$ and $a > 0$: $y(t) = u_c(t)cdotdfrac{1}{a}left(1 - e^{-a(t-c)} ight)$, climbing toward $1/a$.
Smoothness
If the forcing has a jump discontinuity, the solution of a second-order equation and its first derivative remain continuous; only the second derivative jumps.

Worked example

Solve $y' + 2y = u_4(t)$, $y(0) = 0$.

  1. Transform: $(s + 2)Y = dfrac{e^{-4s}}{s}$, so $Y = dfrac{e^{-4s}}{s(s+2)}$.
  2. Set the exponential aside and invert $dfrac{1}{s(s+2)}$.
  3. Partial fractions: $dfrac{1}{s(s+2)} = dfrac12left(dfrac1s - dfrac{1}{s+2} ight)$, which inverts to $h(t) = frac12left(1 - e^{-2t} ight)$.
  4. Re-attach the shift: $y(t) = u_4(t)h(t-4) = u_4(t)cdot frac12left(1 - e^{-2(t-4)} ight)$.
  5. Nothing happens until $t = 4$; after that $y$ rises smoothly from 0 toward $ frac12$.

Common mistakes

  1. Inverting with the exponential still attached. Strip $e^{-cs}$, invert the rest, then shift. Trying to partial-fraction an expression containing $e^{-cs}$ goes nowhere.
  2. Shifting the wrong function. The shift applies to $h$, the inverse of what multiplied $e^{-cs}$ — not to the original forcing.
  3. Losing a step in a multi-term forcing. A pulse gives two exponentials and therefore two shifted terms in the answer, each with its own $u_c$.
  4. Expecting the solution to jump. It does not. If your answer has a discontinuity in $y$ or $y'$, something went wrong.
  5. Evaluating before the switch. For $t < c$ every term carrying $u_c(t)$ is exactly zero — do not substitute into the formula and get a non-zero number.

Practice it

Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.

Open 6.4 →

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