Understanding Periodic Solutions in Differential Equations

A periodic solution to a differential equation is a solution $x(t)$ that repeats itself after a fixed interval of time, called the period $T$. Mathematically, this means $x(t+T) = x(t)$ for all $t$, for some $T > 0$. We need to examine each given differential equation to determine which one admits such solutions.

Analyzing Option 1: \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} + {\rm{\mu x}} = 0\)

This is a second-order linear homogeneous differential equation with constant coefficients. The standard form is $a x''(t) + b x'(t) + c x(t) = 0$. Comparing this to the given equation, we have $a=1$, $b=0$, and $c=\mu$.

To find the general solution, we look at the characteristic equation, which is obtained by replacing the derivatives with powers of a variable, say $r$:

\(r^2 + \mu = 0\)

The nature of the roots of this characteristic equation determines the form of the solution.

• If \(\mu > 0\), the roots are \(r^2 = -\mu\), so \(r = \pm \sqrt{-\mu} = \pm i\sqrt{\mu}\). These are purely imaginary roots.
• If \(\mu = 0\), the equation becomes \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} = 0\), with roots \(r=0\) (repeated).
• If \(\mu < 0\), let \(\mu = -k\) where \(k > 0\). The equation is \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} - {\rm{kx}} = 0\), with roots \(r^2 = k\), so \(r = \pm \sqrt{k}\) (real roots).

For periodic solutions to exist for a linear homogeneous equation of this type, the characteristic roots must be purely imaginary. This occurs when the coefficient of the $x'(t)$ term is zero (which it is, $b=0$) and the coefficients of $x''(t)$ and $x(t)$ have the same sign (which means $ac > 0$). In this case, $1 \cdot \mu > 0$, which requires \(\mu > 0\).

When \(\mu > 0\), the general solution corresponding to the roots \(r = \pm i\sqrt{\mu}\) is given by:

\(x(t) = C_1 \cos(\sqrt{\mu}t) + C_2 \sin(\sqrt{\mu}t)\)

This solution is a linear combination of sine and cosine functions with the same frequency \(\sqrt{\mu}\). Such functions are periodic with period \(T = \frac{2\pi}{\sqrt{\mu}}\).

Thus, the differential equation \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} + {\rm{\mu x}} = 0\) has periodic solutions when \(\mu > 0\). This equation represents simple harmonic motion, a classic example of a system with periodic behavior.

Analyzing Other Options

Let's briefly look at why the other options do not generally have periodic solutions.

• Option 2: \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} - {\rm{\mu x}} = 0\)

  The characteristic equation is \(r^2 - \mu = 0\). Assuming \(\mu > 0\), the roots are \(r = \pm \sqrt{\mu}\) (real and distinct). The general solution is \(x(t) = C_1 e^{\sqrt{\mu}t} + C_2 e^{-\sqrt{\mu}t}\). Exponential functions are not periodic (unless $C_1=C_2=0$, which is the trivial zero solution, or if roots are complex, but that needs $\mu < 0$ here). For \(\mu < 0\), it becomes the same as Option 1 with a positive coefficient, so it would have periodic solutions only if $\mu$ is meant to be a parameter that can be negative. However, in the context of asking which *one* has a periodic solution, the standard interpretation of these forms points to Option 1 with positive \(\mu\) as the archetypal periodic case.

• Option 3: \({\rm{x}}\frac{{{\rm{dx}}}}{{{\rm{dt}}}} + {\rm{\mu t}} = 0\)

  This is a first-order separable differential equation. We can rewrite it as \(x \, dx = -\mu t \, dt\). Integrating both sides gives:

  \(\int x \, dx = \int -\mu t \, dt\)

  \(\frac{1}{2}x^2 = -\frac{1}{2}\mu t^2 + C\)

  \(x^2 + \mu t^2 = 2C\)

  If \(\mu > 0\) and \(2C > 0\), this equation describes an ellipse in the \((t, x)\) phase plane. Solutions tracing such paths are periodic in the phase plane. However, when expressed as a function \(x(t)\), the solution is \(x(t) = \pm \sqrt{2C - \mu t^2}\). This solution is defined only for \(|t| \le \sqrt{2C/\mu}\), meaning it is only defined over a finite time interval, not for all \(t\). A periodic solution must be defined for all \(t\).

• Option 4: \(\frac{{{\rm{dx}}}}{{{\rm{dt}}}} + {\rm{\mu xt}} = 0\)

  This is a first-order separable differential equation. We can rewrite it as \(\frac{{{\rm{dx}}}}{{\rm{x}}} = -{\rm{\mu t \, dt}}\). Integrating both sides gives:

  \(\int \frac{{{\rm{dx}}}}{{\rm{x}}} = \int -{\rm{\mu t \, dt}}\)

  \(\ln|x| = -\frac{1}{2}\mu t^2 + C'\)

  \(|x| = e^{C'} e^{-\frac{1}{2}\mu t^2}\)

  So, \(x(t) = C e^{-\frac{1}{2}\mu t^2}\) for some constant \(C\). Assuming \(\mu > 0\), as \(t \to \pm \infty\), \(e^{-\frac{1}{2}\mu t^2} \to 0\). This solution decays to zero and is not periodic (unless $C=0$, the trivial solution).

Based on the analysis, the only differential equation among the options that standardly and generally exhibits periodic solutions for appropriate values of the parameter (\(\mu > 0\) in this case) is the one representing simple harmonic motion.

Conclusion

The differential equation \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} + {\rm{\mu x}} = 0\), which describes simple harmonic motion when \(\mu\) is a positive constant, yields solutions that are combinations of sine and cosine functions. These are the characteristic periodic solutions.