A Damped Spiral That Rings Down to One, Not to Zero

The characteristic roots of u'' + u' + u are the primitive cube roots of unity, so the homogeneous part decays with envelope e to the minus x over two and oscillates with period 4 pi over root 3, which is 7.2552. Substituting that homogeneous solution back into the equation leaves a residual of exactly minus one, and that residual is the whole distance between the common wrong answer and the right one. The constant u = 1 solves the equation by itself, so every solution settles on 1, and the constant trial works only because the coefficient on u is not zero.

Solve $u'' + u' + u = 1$ . No initial conditions are supplied, which is a hint about what is being asked: not a particular curve, but where every curve ends up.

The answer is that every solution settles on the line $u = 1$ , and most of them ring their way down to it. The most common wrong answer is zero, and it is wrong for a reason that takes one line to state and is easy to walk past.

The oscillation, which everyone gets right

Set the right-hand side aside for a moment and look for solutions of the form $e^{\lambda x}$ . Substituting gives the characteristic equation:

\lambda^2 + \lambda + 1 = 0, \qquad \lambda = -\tfrac{1}{2} \pm i\,\tfrac{\sqrt{3}}{2}

(1)

The discriminant is $1 - 4 = -3$ , so the roots are a conjugate pair. Their real part is negative, which means decay, and their imaginary part is non-zero, which means oscillation. Combining the two complex exponentials into real functions gives the homogeneous solution:

u_h(x) = e^{-x/2}\left[a\cos\frac{\sqrt{3}\,x}{2} + b\sin\frac{\sqrt{3}\,x}{2}\right]

(2)

Those two roots happen to be the primitive cube roots of unity. They sit on the unit circle at $\pm 120$ degrees, so their modulus is exactly 1 while their real part is $-1/2$ , and it is the real part alone that governs how fast the amplitude falls.

Fig. 1 — The two characteristic roots. On the unit circle, at plus and minus 120 degrees, with a real part of −1/2 that fixes the decay rate and an imaginary part that fixes the frequency.

Why the answer is not zero

Equation (2) does decay to zero, and stopping there is the mistake. The test is mechanical: substitute the candidate back into the equation you were actually given. Every derivative of $u_h$ is another expression of the same form, and the three terms cancel exactly, so what is left over is the constant on the right:

u_h'' + u_h' + u_h - 1 = -1 \qquad \text{for every } a, b

(3)

A residual of $-1$ rather than 0 says the homogeneous solution solves a different problem, the one with zero on the right. The forcing was dropped somewhere between reading the question and writing the characteristic equation, and nothing downstream notices, because every step after that point is correct.

One constant does all the work

The equation is linear, so the general solution is any single solution plus the whole homogeneous family. Finding a single solution here takes no machinery at all. Try a constant: both derivatives vanish, so $u'' + u' + u$ collapses to $u$ , and $u = 1$ satisfies the equation on its own. Adding it to (2):

u(x) = e^{-x/2}\left[a\cos\frac{\sqrt{3}\,x}{2} + b\sin\frac{\sqrt{3}\,x}{2}\right] + 1

(4)

Substituting (4) into $u'' + u' + u - 1$ returns exactly 0 for every $a$ and $b$ , so the constant has done the whole repair. The two free constants are fixed by two initial conditions, $a = u(0) - 1$ and $b = \tfrac{2}{\sqrt{3}}\!\left(u'(0) + \tfrac{a}{2}\right)$ , and the problem withholds both on purpose. Neither one changes where the curve ends up.

General solution of a linear equation

For a linear equation $L[u] = r$ with $L$ a linear operator, if $u_p$ satisfies $L[u_p] = r$ and $u_h$ ranges over the solutions of $L[u_h] = 0$ , then $u_p + u_h$ is exactly the solution set. Any two solutions differ by a homogeneous one, so nothing is missed and nothing is counted twice. The forcing term selects $u_p$ ; the order of the equation fixes how many free constants live in $u_h$ .

What the curve looks like

Three numbers describe the picture, and all three come from (1). The envelope is $e^{-x/2}$ , which halves every $2\ln 2 = 1.3863$ in $x$ . The angular frequency is $\sqrt{3}/2$ , so the period is $4\pi/\sqrt{3} = 7.2552$ . And the resting value is 1, so the curve crosses the line $u = 1$ once every half period, at spacings of 3.6276, infinitely often.

Combining the first two gives the amount by which each swing is smaller than the one before it: a full period costs a factor of $e^{-\text{period}/2} = e^{-3.6276} = 0.02658$ . So each peak is under 3 percent of its predecessor, and after two full cycles the wobble is invisible on any plot. That is why a numerical integration reaches 1 to eight decimal places long before the interval ends.

Fig. 2 — The solution from u(0) = 3 and u'(0) = 0. The envelope collapses onto u = 1, not onto 0, and the marked crossings are 3.63 apart, half a period.

A fourth-order Runge-Kutta integration at a step of $10^{-4}$ over 60 units, started from five different pairs of initial conditions, ends at 1 every time to within $10^{-8}$ , while the same integrator run on the equation with zero on the right ends at 0. Measured on the trajectory rather than read off the formula, the peak spacing comes out at the period and the peak heights shrink by that factor of 0.02658 per cycle.

Where the trick stops working

Guessing a constant is not a general method, and it is worth knowing precisely when it fails. For $u'' + p\,u' + q\,u = r$ with $q \ne 0$ , the constant $u = r/q$ always works, because the derivative terms disappear. Here $q = 1$ and $r = 1$ , which is why the particular solution is the unadorned 1.

Set $q = 0$ and the method dies. In $u'' + u' = 1$ no constant can produce a non-zero right-hand side, and the particular solution is $u = x$ instead. The general diagnosis is resonance: the constant trial works when 0 is not a root of the characteristic polynomial, and it has to be multiplied by $x$ once for each time 0 is a root. Since $\lambda^2 + \lambda + 1$ has value 1 at $\lambda = 0$ , we are clear.

Two smaller caveats. The word every in every solution rings down is not quite right: taking $a = b = 0$ gives the constant solution $u \equiv 1$ , which settles without ever ringing. It still settles at 1, so the claim about the resting point survives the degenerate case even though the picture does not. And decay requires the real part of both roots to be negative, which for a real quadratic means $p > 0$ and $q > 0$ . Change the sign of the damping to $u'' - u' + u = 1$ and the same 1 is still a solution, but now the oscillation grows around it and nothing settles anywhere.

Sources and further reading

The structure used in (4) — Linear differential equation
The step that produced (1) — Characteristic equation
The physical reading of the homogeneous part — Damped harmonic oscillator
The general recipe behind the constant guess, including the resonance correction — Method of undetermined coefficients
Where the two roots also turn up — Root of unity

Both candidate answers were substituted symbolically: (4) leaves a residual of exactly 0 for all $a$ and $b$ , and (2) leaves exactly $-1$ . The closed form was then checked against an independent numerical integration rather than against itself, with the constants read off the initial conditions and the two curves agreeing to $10^{-8}$ over twenty units.