7. Integrating the Lax pair system

It is clear that the first-order equation in (67) for QðsÞ is separable and can be integrated. The integral depends on whether K is zero or nonzero:

$$Q(\mathbf{s}) = \frac{1}{\varepsilon \mathbf{s} + \boldsymbol{\gamma}}, \quad \mathbf{K} = 0; \qquad \log\left(\frac{2(\mathbf{K} + \sqrt{\mathbf{K}}\sqrt{\mathbf{Q}^2 + \mathbf{K}})}{Q}\right) = \varepsilon\sqrt{\mathbf{K}}\mathbf{s} + \boldsymbol{\gamma}, \quad \mathbf{K} \neq \mathbf{0}.\tag{72}$$

Here ε ¼ �1 and γ is the last constant of integration. Taking specific choices for the constants, for example, <sup>e</sup><sup>γ</sup> <sup>¼</sup> <sup>2</sup> ffiffiffi K <sup>p</sup> when <sup>K</sup>≠0 and <sup>a</sup> <sup>¼</sup> ffiffiffi K <sup>p</sup> , the set of solutions (72) for <sup>Q</sup>ðs<sup>Þ</sup> can be summarized below.

$$\begin{array}{ccccc} \text{Dom(s)} & Q(s) & \text{Dom(s)} & Q(s) \\ \hline s > 0 & \frac{1}{s} & s < 0 & -\frac{1}{s} \\ 0 < s < \frac{\pi}{a} & \frac{a}{\sin(as)} & -\frac{\pi}{a} < s < 0 & -\frac{a}{\sin(as)} \\ s > 0 & \frac{a}{\sinh(as)} & s < 0 & -\frac{a}{\sinh(as)} \end{array} \tag{73}$$

It is presumed that other choices of the constants can be geometrically eliminated in favor of Eq. (73). The solutions (73) are then substituted back into linear system (67). The first equation in (67) implies that either

$$
\psi \equiv 0, \mod \frac{\pi}{2}; \qquad \frac{2\psi\_s}{\sin \left(2\psi\right)} = -Q. \tag{74}
$$

Substitute ψ≡0 into the second equation in (67). It implies that ðlogQÞ<sup>s</sup> ¼ Q and ψ ¼ π=2 gives ðlogQÞ<sup>s</sup> ¼ −Q. In both cases QðsÞ is a solution which already appears in Eq. (73).

For the second case in Eq. (74), the equation can be put in the form

$$(\log|\tan\left(\psi\right)|)\_s = -Q.$$

Integrating we have for some function yðtÞ to be determined,

$$\tan\left(\psi\right) = e^{-\int Q(s)ds} \cdot \boldsymbol{y}(t). \tag{75}$$

Therefore, tan ðψÞ can be obtained by substituting for QðsÞ for each of the three cases in Eq. (73). The upper sign holds for s > 0 and the lower sign holds if s < 0.

$$\begin{aligned} \text{s.t. } \qquad Q(s) &= \pm s^{-1}, \ \left[ Q(s)ds = \log|s|^{\mp} \text{ and} \right. \\\\ \tan \left( \psi \right) &= s^{\mp} \cdot y(t). \end{aligned} \tag{76}$$

$$\begin{aligned} \text{iii.} \qquad Q(\mathbf{s}) &= \pm \frac{a}{\sin(\mathbf{s})'} - \int Q(\mathbf{s}) d\mathbf{s} = \log|\csc(\mathbf{as}) - \cot(\mathbf{as})|^\mp \text{ and} \\\\ \tan(\psi) &= \left(\tan\left(\frac{as}{2}\right)\right)^\mp \cdot y(t). \end{aligned} \tag{77}$$

$$\text{iii.} \qquad Q(\mathbf{s}) = \pm \frac{a}{\sinh(a\mathbf{s})^{\prime}} \cdot \left[ Q(\mathbf{s}) d\mathbf{s} = \mp \text{arctanh}(e^{a\mathbf{s}}) \text{, and } \mathbf{s} \right]$$

An Intrinsic Characterization of Bonnet Surfaces Based on a Closed Differential Ideal http://dx.doi.org/10.5772/67008 39

$$\tan\left(\psi\right) = \left(\tanh\left(\frac{a\mathbf{s}}{2}\right)\right)^{\mp} \cdot y(t). \tag{78}$$

In case ðiiÞ, if s > 0 and yðt޼�1 then ψ ¼ �<sup>1</sup> <sup>2</sup>ðas þ πÞ, modπ, and if s < 0 and yðt޼�1, then <sup>ψ</sup> ¼ � <sup>1</sup> <sup>2</sup> as, modπ.

It remains to integrate the second equation of the Lax pair (67) using solutions for both QðsÞ and tan ðψÞ. The first case ðiÞ is not hard and will be shown explicitly here. The others can be done, and more complicated cases are considered in the Appendix.

<sup>ð</sup>i<sup>Þ</sup> Consider <sup>Q</sup>ðsÞ ¼ <sup>s</sup><sup>−</sup><sup>1</sup> and tan <sup>ð</sup>ψÞ ¼ <sup>s</sup><sup>−</sup><sup>1</sup> � <sup>y</sup>ðtÞ. The second equation in (67) simplifies considerably to yt ¼ −1, therefore,

$$y(t) = -(t+\sigma), \qquad \tan\left(\psi\right) = -\frac{(t+\sigma)}{s}.\tag{79}$$

For <sup>Q</sup>ðsÞ ¼ <sup>−</sup>s<sup>−</sup><sup>1</sup> and tan <sup>ð</sup>ψÞ ¼ <sup>s</sup> � <sup>y</sup>ðtÞ, the second equation of (67) becomes yt <sup>¼</sup> <sup>−</sup>y2, therefore,

$$y(t) = \frac{1}{t + \sigma}, \qquad \tan\left(\psi\right) = \frac{s}{t + \sigma}.\tag{80}$$
