8. A third-order equation for H and fundamental forms

Since <sup>ξ</sup><sup>12</sup> ¼ ðlogeðsÞÞ′ dt, using Eq. (60) ω<sup>12</sup> can be written as

$$d\omega\_{12} = \xi\_{12} \text{--} d\psi = \left(\log e(\mathbf{s})\right)' dt \text{--} d\psi. \tag{81}$$

Using Eqs. (14) and (64) for α1, it follows that

Here ε ¼ �1 and γ is the last constant of integration. Taking specific choices for the constants,

DomðsÞ QðsÞ DomðsÞ QðsÞ

π a

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

Substitute ψ≡0 into the second equation in (67). It implies that ðlogQÞ<sup>s</sup> ¼ Q and ψ ¼ π=2 gives

ðlogj tan ðψÞjÞ<sup>s</sup> ¼ −Q:

− ð

Therefore, tan ðψÞ can be obtained by substituting for QðsÞ for each of the three cases in

2 ;

ðlogQÞ<sup>s</sup> ¼ −Q. In both cases QðsÞ is a solution which already appears in Eq. (73).

tan ðψÞ ¼ e

<sup>∓</sup> and

tan ðψÞ ¼ s

<sup>Q</sup>ðsÞds <sup>¼</sup> logjcscðasÞ<sup>−</sup> cotðasÞj<sup>∓</sup> and

as 2 � � � � <sup>∓</sup>

tan ðψÞ ¼ tan

<sup>Q</sup>ðsÞds <sup>¼</sup> <sup>∓</sup>arctanhðeasÞ, and

Eq. (73). The upper sign holds for s > 0 and the lower sign holds if s < 0.

QðsÞds ¼ logjsj

sinhðas<sup>Þ</sup> <sup>s</sup> <sup>&</sup>lt; <sup>0</sup> <sup>−</sup> <sup>a</sup>

2ψ<sup>s</sup> sin ð2ψÞ

s < 0 −

<sup>&</sup>lt; <sup>s</sup> <sup>&</sup>lt; <sup>0</sup> <sup>−</sup> <sup>a</sup>

<sup>p</sup> , the set of solutions (72) for <sup>Q</sup>ðs<sup>Þ</sup> can be summa-

1 s

sin ðasÞ

(73)

sinhðasÞ

¼ −Q: (74)

<sup>Q</sup>ðsÞds � <sup>y</sup>ðtÞ: (75)

<sup>∓</sup> � <sup>y</sup>ðtÞ: (76)

� yðtÞ: (77)

K

for example, <sup>e</sup><sup>γ</sup> <sup>¼</sup> <sup>2</sup> ffiffiffi

38 Manifolds - Current Research Areas

in (67) implies that either

i. <sup>Q</sup>ðs޼�s<sup>−</sup>1, <sup>−</sup>

ii. <sup>Q</sup>ðs޼� <sup>a</sup>

iii. <sup>Q</sup>ðs޼� <sup>a</sup>

rized below.

K

<sup>p</sup> when <sup>K</sup>≠0 and <sup>a</sup> <sup>¼</sup> ffiffiffi

<sup>s</sup> <sup>&</sup>gt; <sup>0</sup> <sup>1</sup>

π a

<sup>s</sup> <sup>&</sup>gt; <sup>0</sup> <sup>a</sup>

<sup>ψ</sup>≡0; mod <sup>π</sup>

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

Integrating we have for some function yðtÞ to be determined,

ð

sin ðasÞ , − ð

sinhðasÞ , − ð

0 < s <

s

a sin <sup>ð</sup>as<sup>Þ</sup> <sup>−</sup>

$$d\log(f) = Q(\cos(2\psi)\,\mathrm{d}s - \sin(2\psi)\,\mathrm{d}t) - 2 \ast (\psi\_t\,\mathrm{d}t + \psi\_s\,\mathrm{d}s) + 2 \ast (\log(e(s)))^\cdot\,\mathrm{d}t.$$

when ω<sup>i</sup> are put in the s;t coordinates, using �ω<sup>1</sup> ¼ ω2, it can be stated that �ds ¼ dt and �dt ¼ −ds. Consequently, dlogðJÞ simplifies to

$$d\log(I) = \left(Q\cos\left(2\psi\right) + 2\psi\_i - 2\left(\log(e(s))\right)'\right)ds + \left(-Q\sin\left(2\psi\right) - 2\psi\_s\right)dt.\tag{82}$$

First-order system (67) permits this to be written using <sup>e</sup>ðsÞ ¼ ffiffiffiffiffiffiffiffiffi <sup>E</sup>ðs<sup>Þ</sup> <sup>p</sup> as

$$\left(\log(\!/\!)\right)^{\circ} + \left(\log(E)\right)^{\circ} = \left(\log(Q)\right)^{\circ}.\tag{83}$$

Hence, there exists a constant τ independent of s such that E � J ¼ τQ or

$$E = \tau \frac{Q}{I} = \tau \frac{Q^2}{H} \,. \tag{84}$$

This result (84) for E is substituted into the Gauss equation −((log(E))ss+(log(E))tt)=2E(H<sup>2</sup> −J 2 ) giving

$$\left(\log(E)\right)^{\*} = 2\left(\log(Q)\right)^{\*} \text{--} \left(\log(H\_{s})\right)^{\*} = 2Q^{2} \text{--} \left(\frac{H}{H}^{\*}\right)^{\*}.\tag{85}$$

Therefore, the Gauss equation transforms into a third-order differential equation in the s variable,

$$\left(\frac{H}{H}^\*\right)^\prime + 2\tau H = 2Q^2 \left(1 + \tau \frac{H^2}{H}\right). \tag{86}$$

Thus, a characterization of Bonnet surfaces is reached by means of the solutions to these equations. This equation determines the function HðsÞ and after that the functions JðsÞ and EðsÞ. Therefore, Bonnet surfaces have as first fundamental form the expression

$$I = E(s)(ds^2 + dt^2), \qquad E(s) = \tau \frac{Q^2(s)}{H'(s)}.\tag{87}$$

Since ψ is the angle from the principal axis e<sup>1</sup> to the s curve with t equals constant, the second fundamental form is given by

$$
\Pi = \mathcal{L} \,\mathrm{d}\mathbf{s}^2 + 2\mathcal{M} \,\mathrm{d}\mathbf{s} \,\mathrm{d}t + \mathcal{N} \,\mathrm{d}t^2. \tag{88}
$$

where the coefficients L; M; N are given by

$$L = E(H + J \cos \left(2\psi\right)) = EH + \tau Q \cos \left(2\psi\right),$$

$$M = -Ej \sin \left(2\psi\right) = -\tau Q \sin \left(2\psi\right),\tag{89}$$

$$N = E(H \neg j \cos \left(2\psi\right)).$$

#### Appendix

It is worth seeing how the second equation in (67) can be integrated for cases (ii) and (iii). Only the case s > 0 will be done with QðsÞ taken from Eq. (73).

ðaÞ Differentiating tan ðψÞ given in Eq. (77), we obtain that

$$
\psi\_t = \frac{\tan\left(\frac{as}{2}\right)}{\tan^2\left(\frac{as}{2}\right) + y^2} y\_t(t).
$$

The following identities are required to simplify the result,

An Intrinsic Characterization of Bonnet Surfaces Based on a Closed Differential Ideal http://dx.doi.org/10.5772/67008 41

$$\tan\left(as\right) = \frac{2\tan\left(\frac{as}{2}\right)}{1-\tan^2\left(\frac{as}{2}\right)}, \qquad \cos\left(2\psi\right) = \frac{\tan^2\left(\frac{as}{2}\right) - y^2}{\tan^2\left(\frac{as}{2}\right) + y^2}$$

Substituting ψ<sup>t</sup> into Eq. (67), we obtain

$$\frac{2\tan\left(\frac{as}{2}\right)}{\tan^2\left(\frac{as}{2}\right) + y^2}y\_t = -a\cot\left(as\right) - \frac{a}{\sin\left(as\right)}\frac{\tan^2\left(\frac{as}{2}\right) - y^2}{\tan^2\left(\frac{as}{2}\right) + y^2}.$$

Simplifying this, we get

E ¼ τ Q <sup>J</sup> <sup>¼</sup> <sup>τ</sup>

<sup>ð</sup>logðEÞÞ″ <sup>¼</sup> <sup>2</sup>ðlogðQÞÞ″

H″ H′ ′

EðsÞ. Therefore, Bonnet surfaces have as first fundamental form the expression

<sup>I</sup> <sup>¼</sup> <sup>E</sup>ðsÞðds<sup>2</sup> <sup>þ</sup> dt<sup>2</sup>

giving

40 Manifolds - Current Research Areas

variable,

Appendix

fundamental form is given by

where the coefficients L; M; N are given by

the case s > 0 will be done with QðsÞ taken from Eq. (73). ðaÞ Differentiating tan ðψÞ given in Eq. (77), we obtain that

The following identities are required to simplify the result,

This result (84) for E is substituted into the Gauss equation −((log(E))ss+(log(E))tt)=2E(H<sup>2</sup>

Therefore, the Gauss equation transforms into a third-order differential equation in the s

<sup>þ</sup> <sup>2</sup>τ<sup>H</sup> <sup>¼</sup> <sup>2</sup>Q<sup>2</sup> <sup>1</sup> <sup>þ</sup> <sup>τ</sup>

Thus, a characterization of Bonnet surfaces is reached by means of the solutions to these equations. This equation determines the function HðsÞ and after that the functions JðsÞ and

Since ψ is the angle from the principal axis e<sup>1</sup> to the s curve with t equals constant, the second

II <sup>¼</sup> L ds<sup>2</sup> <sup>þ</sup> <sup>2</sup>M ds dt <sup>þ</sup> N dt<sup>2</sup>

L ¼ EðH þ J cos ð2ψÞÞ ¼ EH þ τQ cos ð2ψÞ; M ¼ −EJ sin ð2ψÞ ¼ −τQ sin ð2ψÞ; N ¼ EðH−J cos ð2ψÞÞ:

It is worth seeing how the second equation in (67) can be integrated for cases (ii) and (iii). Only

2 Þ

<sup>2</sup> Þ þ <sup>y</sup><sup>2</sup> yt

ðtÞ:

<sup>ψ</sup><sup>t</sup> <sup>¼</sup> tan <sup>ð</sup>as

tan <sup>2</sup>ðas

Þ; EðsÞ ¼ τ

Q2

<sup>−</sup>ðlogðHsÞÞ″ <sup>¼</sup> <sup>2</sup>Q<sup>2</sup>

H2 H′ 

> Q2 ðsÞ H′ ðsÞ

<sup>H</sup>′ : (84)

<sup>−</sup> <sup>H</sup>″ H′ ′ −J 2 )

(89)

: (85)

: (86)

: (87)

: (88)

$$\frac{4}{a}y\_t = -\frac{1}{2}\left(1 - \tan^2\left(\frac{as}{2}\right)\right) - \frac{1}{2}\left(\cot^2\left(\frac{as}{2}\right) - 1\right)y^2 - \sec^2\left(\frac{as}{2}\right) + \csc^2\left(\frac{as}{2}\right)y^2 - \dots$$

This simplifies to the elementary equation,

$$y\_t = \frac{a}{2}(y^2 - 1), \qquad y(t) = -\tanh\left(\frac{at}{2} + \eta\right).$$

Here η is an integration constant. To summarize then,

$$
\tan\left(\psi\right) = \tanh\left(\frac{at}{2} + \eta\right) \cdot \tan\left(\frac{as + \pi}{2}\right).
$$

ðbÞ Consider now s > 0 and take QðsÞ from the last line of Eq. (73). Differentiating tan ðψÞ from (78), we get

$$
\psi\_t = \frac{\coth\left(\frac{as}{2}\right)}{1 + \coth^2\left(\frac{as}{2}\right)y^2} y\_t(t).
$$

In this case, the following identities are needed,

$$\tanh(as) = \frac{2\tanh(\frac{\omega}{2})}{1 + \tanh^2(\frac{\omega}{2})}, \qquad \cos\left(2\psi\right) = \frac{1 - \coth^2(\frac{\omega}{2})y^2}{1 + \coth^2(\frac{\omega}{2})y^2}.$$

Therefore, Eq. (67) becomes

$$2\frac{\coth(\frac{\alpha}{2})}{1+\coth^2(\frac{\alpha}{2})y^2}y\_t = -\text{acoth}\left(as\right) - \frac{a}{\sinh(as)}\frac{\tanh^2(\frac{\alpha}{2}) - y^2}{\tanh^2(\frac{\alpha}{2}) + y^2}.$$

This reduces to

$$-\frac{4}{a}y\_t = \left(1 + \tanh^2\left(\frac{as}{2}\right) + \mathrm{sech}^2\left(\frac{as}{2}\right)\right) + \left(\coth^2\left(\frac{as}{2}\right) + 1 - \mathrm{csch}^2\left(\frac{as}{2}\right)\right)y^2.$$

Simplifying and integrating, it has been found that

$$y\_t = -\frac{a}{2}(1+y^2), \qquad y(t) = -\tan\left(\frac{at}{2} + \eta\right).$$

To summarize then, it has been shown that,

$$\tan\left(\psi\right) = \cot\left(\frac{at}{2} + \eta\right) \cdot \coth\left(\frac{as}{2}\right).$$

These results apply to the case s > 0 and similar results can be found for the case s < 0 as well.

MSCs: 53A05, 58A10, 53B05
