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.
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Try one now
Transform the initial value problem $$y'' - y' - 12y = u_{3}(t), \qquad y(0) = 0,\ y'(0) = 0.$$
Solution
- Both initial conditions are zero, so $\mathcal{L}\{y''\} = s^2Y$ and $\mathcal{L}\{y'\} = sY$ with no leftover constants.
- Left side: $\left(s^2 - s - 12\right)Y$.
- Right side: $\mathcal{L}\left\{u_{3}(t)\right\} = \dfrac{e^{-3s}}{s}$.
- Solve: $\displaystyle Y(s) = \frac{e^{-3s}}{s\left(s^2 - s - 12\right)}$.
- The $e^{-3s}$ is carried along untouched through the partial fractions, then converted back into a $u_{3}(t)$ factor with every $t$ replaced by $t - 3$ at the very end.
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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
- Write the forcing in step form using 6.3, so the right side becomes a sum of terms $u_c(t)g(t-c)$.
- Transform. Each forcing term contributes $e^{-cs}G(s)$. With zero initial conditions the left side is simply $(as^2+bs+c)Y$.
- Solve for $Y(s)$ and factor out each exponential: $Y = e^{-c_1s}H_1(s) + e^{-c_2s}H_2(s) + \cdots$
- Invert each $H_i$ separately by partial fractions or completing the square — ignore the exponentials while you do it.
- 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
Worked example
Solve $y' + 2y = u_4(t)$, $y(0) = 0$.
- Transform: $(s + 2)Y = \dfrac{e^{-4s}}{s}$, so $Y = \dfrac{e^{-4s}}{s(s+2)}$.
- Set the exponential aside and invert $\dfrac{1}{s(s+2)}$.
- Partial fractions: $\dfrac{1}{s(s+2)} = \dfrac12\left(\dfrac1s - \dfrac{1}{s+2}\right)$, which inverts to $h(t) = \tfrac12\left(1 - e^{-2t}\right)$.
- Re-attach the shift: $y(t) = u_4(t)h(t-4) = u_4(t)\cdot\tfrac12\left(1 - e^{-2(t-4)}\right)$.
- Nothing happens until $t = 4$; after that $y$ rises smoothly from 0 toward $\tfrac12$.
Common mistakes
- 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.
- Shifting the wrong function. The shift applies to $h$, the inverse of what multiplied $e^{-cs}$ — not to the original forcing.
- 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 $u_c$.
- Expecting the solution to jump. It does not. If your answer has a discontinuity in $y$ or $y'$, something went wrong.
- 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.