Using Time-averaging Operators

Another way to approximate (3.1) involves the use of temporal averaging operators, as introduced in Section 2.2. For example,

$$\delta_{tt}u = -\omega_{0}^2\mu_{t\cdot}u$$ (3.42)

is also a second order approximation to (3.1), where it is to be recalled that the averaging operator $\mu _{t\cdot }$ approximates the identity operation. Second order accuracy is easy enough to determine by inspection of the operator $\delta _{tt}$, which is a second order accurate approximation to $\frac{d^{2}}{dt^2}$.

This scheme, when the action of the operators $\delta _{tt}$ and $\mu _{t\cdot }$ is expanded out, leads to

$$\frac{1}{k^2}\left(u^{n+1}-2u^{n}+u^{n-1}\right) = -\frac{\omega_{0}^2}{2}\left(u^{n+1}+u^{n-1}\right)$$ (3.43)

which again involves values of the time series at three levels $n+1$, $n$, and $n-1$, and which can again be solved for $u^{n+1}$, as

$$u^{n+1} = \frac{2}{1+\omega_{0}^2 k^2/2}u^n-u^{n+1}$$ (3.44)

The characteristic polynomial is now

$$z-\frac{2}{1+\omega_{0}^2 k^2/2} + z^{-1} = 0$$ (3.45)

which has roots

$$z_{\pm} = \frac{1\pm\sqrt{1-4\left(1+\omega_{0}^2 k^2/2\right)^2}}{1+\omega_{0}^2 k^2/2}$$ (3.46)

The solution will again evolve according to (3.24) when $z_{+}$ and $z_{-}$ are distinct.

In this case, however, it is not difficult to show that the roots $z_{\pm}$ will be complex conjugates of unit magnitude for any choice of $k$. Thus $z_{\pm} = e^{\pm j\omega k}$, where the $\omega$ is given by

$$\omega = \frac{1}{k}\cos^{-1}\left(\frac{1}{1+\omega_{0}^2 k^2/2}\right)$$ (3.47)

Again, $\omega\neq \omega_{0}$, but, in contrast to the case of scheme (3.18), one now has $\omega<\omega_{0}$. One now has a well-behaved sinusoidal solution of the form of (3.27) for any value of $k$; there is no stability condition of the form of (3.25).

The energetic analysis mirrors this behaviour. Multiplying (3.42) by $\delta_{t\cdot}u$ gives

$$(\delta_{t\cdot}u)\delta_{tt}u+\omega_{0}^2(\delta_{t\cdot}u)\mu_... ...c{1}{2}\left(\delta_{t-}u\right)^2+\frac{\omega_{0}^2}{2}\mu_{t-}u^2\right) = 0$$ (3.48)

or

$$\delta_{t+}{\mathfrak{h}} = 0\qquad {\rm with}\qquad {\mathfrak{h}} = {\mathfrak{t}}+{\mathfrak{v}}$$ (3.49)

and

$${\mathfrak{t}} = \frac{1}{2}(\delta_{t-}u)^2\qquad {\mathfrak{v}} = \frac{\omega_{0}^2}{2}\mu_{t-}u^2$$ (3.50)

Now, ${\mathfrak{t}}$ and ${\mathfrak{v}}$, and as a result ${\mathfrak{h}}$ are non-negative for any choice of $k$. A bound on solution size follows immediately. See Problem 3.2.