3.2 The solution by Green function

The solution has the form

$$A(z,t) = \varphi(t)e^{-iEt} \tag{26}$$

Then, there is

$$\frac{1}{2}\beta\_2\frac{\partial^2\phi}{\partial t^2} + \frac{i}{6}\beta\_3\frac{\partial^3\phi}{\partial t^3} - \frac{3k\_0}{8nA\_{\text{eff}}}\chi\_{\text{RR}}^{(3)}|\phi|^2\phi - \frac{k\_0\text{g}(\omega\_i)[1 - \text{if}(\omega\_i)]}{2nA\_{\text{eff}}}\phi\int\_{-\infty}^{+\infty}\chi\_R^{(3)}(t-\tau)|\phi(\tau)|^2d\tau = E\phi. \tag{27}$$

Let

$$
\hat{H}\_0(t) = \frac{1}{2}\beta\_2 \frac{\partial^2}{\partial t^2} + \frac{i}{6}\beta\_3 \frac{\partial^3}{\partial t^3} \tag{28}
$$

$$\hat{\mathcal{V}}(t) = \frac{-3k\_0}{8nA\_{\rm eff}} \chi\_{\rm NR}^{(3)} |\phi| - \frac{k\_0 \mathbf{g}(\alpha\_t) [1 - \dot{\mathbf{r}} f(\alpha\_t)]}{2nA\_{\rm eff}} \int\_{-\infty}^{+\infty} \chi\_{\rm R}^{(3)}(t - \tau) |\phi(\tau)|^2 d\tau \tag{29}$$

and taking the operator V t ^ ð Þ as a perturbation item, we first solve the eigen equation �∑<sup>k</sup> n¼2 i n <sup>n</sup>! β<sup>n</sup> ∂nφ <sup>∂</sup>T<sup>n</sup> ¼ Eφ.

$$\frac{1}{2}\beta\_2\frac{\partial^2\phi}{\partial T^2} + \frac{i}{6}\beta\_3\frac{\partial^3\phi}{\partial T^3} = E\phi\tag{30}$$

Assuming E ¼ 1, we get the corresponding characteristic equation:

$$-\frac{1}{2}\beta\_2 r^2 + \frac{\beta\_3}{6}r^3 = E \tag{31}$$

Its characteristic roots are r1, r2, r3. The solution can be represented as

$$
\phi = c\_1 \phi\_1 + c\_2 \phi\_2 + c\_3 \phi\_3 \tag{32}
$$

where ϕ<sup>m</sup> ¼ exp ð Þ irmt , m ¼ 1, 2, 3 and c1, c2, c<sup>3</sup> are determined by the initial pulse. The Green function of (30) is

$$(\mathbf{E} - \hat{H}\_0(t))\mathbf{G}\_0(t, t') = \delta(t - t') \tag{33}$$

By the construction method, it is

$$G\_0(t, t') = \begin{cases} a\_1 \wp\_1 + a\_2 \wp\_2 + a\_3 \wp\_3, & t > t' \\ b\_1 \wp\_1 + b\_2 \wp\_2 + b\_3 \wp\_3, & t < t' \end{cases} \tag{34}$$

At the point t ¼ t 0 , there are

$$a\_1 \phi\_1(t') + a\_2 \phi\_2(t') + a\_3 \phi\_3(t') = b\_1 \phi\_1(t') + b\_2 \phi\_2(t') + b\_3 \phi\_3(t')\tag{35}$$

$$a\_1 \phi\_1'(t') + a\_2 \phi\_2'(t') + a\_3 \phi\_3'(t') = b\_1 \phi\_1'(t') + b\_2 \phi\_2'(t') + b\_3 \phi\_3'(t')\tag{36}$$

$$a\_1 \boldsymbol{\phi\_1^\*} (t') + a\_2 \boldsymbol{\phi\_2^\*} (t') + a\_3 \boldsymbol{\phi\_3^\*} (t') - b\_1 \boldsymbol{\phi\_1^\*} (t') - b\_2 \boldsymbol{\phi\_2^\*} (t') - b\_3 \boldsymbol{\phi\_3^\*} (t') = -6 \boldsymbol{i} / \beta\_3 \tag{37}$$

Let b<sup>1</sup> ¼ b<sup>2</sup> ¼ b<sup>3</sup> ¼ 0, then

$$\mathfrak{a}\_1 = \frac{\varrho\_2 \dot{\rho}\_3 - \dot{\rho}\_2 \varrho\_3}{W(t')}, \mathfrak{a}\_2 = \frac{\varrho\_3 \dot{\rho}\_1 - \dot{\rho}\_3 \varrho\_1}{W(t')}, \mathfrak{a}\_3 = \frac{\varrho\_1 \dot{\rho}\_2 - \dot{\rho}\_1 \varrho\_2}{W(t')} \tag{38}$$

$$\mathcal{W}(t') = \begin{vmatrix} \phi\_1 & \phi\_2 & \phi\_3 \\ \phi\_1^{(1)} & \phi\_2^{(1)} & \phi\_3^{(1)} \\ \phi\_1^{(2)} & \phi\_2^{(2)} & \phi\_3^{(2)} \end{vmatrix} \tag{39}$$

u zð Þ¼ <sup>þ</sup> dz;<sup>ω</sup> exp dzD^ � � exp dzN^ � �u zð Þ ;<sup>ω</sup> (42)

exp �ik z � <sup>z</sup> � � � �<sup>0</sup>

;ω; δβ<sup>3</sup> z

<sup>6</sup> δβ<sup>3</sup> z � �<sup>0</sup> <sup>2</sup> <sup>þ</sup> is <sup>∂</sup>j j <sup>u</sup> <sup>2</sup>

<sup>∂</sup><sup>z</sup> � <sup>D</sup>^ � <sup>N</sup>^ and

<sup>3</sup> <sup>þ</sup> δβ3, u zð Þ¼ ;<sup>ω</sup> <sup>u</sup><sup>0</sup>ð Þþ <sup>z</sup>;<sup>ω</sup> <sup>δ</sup>u zð Þ ;<sup>ω</sup> , then

0 ;ω

;ω � � and u<sup>0</sup> z

0 ;ω; β<sup>0</sup> 3 � � is

<sup>δ</sup>u zð Þ ;<sup>ω</sup> <sup>=</sup>∂ω<sup>2</sup> � �>0, so

<sup>0</sup> � �; <sup>u</sup><sup>0</sup> <sup>z</sup>

∂u<sup>0</sup> ∂ω

� � (45)

� � � � dz<sup>0</sup>

ω<sup>3</sup>u<sup>0</sup> z 0

<sup>ω</sup> � <sup>1</sup> u0

� �d<sup>ω</sup>

n o h i , and <sup>Γ</sup>

ik � <sup>D</sup>^ � <sup>N</sup>^ dk (43)

<sup>∂</sup><sup>t</sup> <sup>þ</sup> is uj j<sup>2</sup> <sup>∂</sup>

∂t

(44)

<sup>6</sup>ω<sup>3</sup>β3, <sup>N</sup>^ <sup>¼</sup> <sup>Γ</sup> <sup>i</sup><sup>γ</sup> exp ð Þ �2α<sup>z</sup> j j <sup>u</sup>

; we obtain the Green function

ðþ<sup>∞</sup> �∞

2π

The minimum value of <sup>δ</sup>u zð Þ ;<sup>ω</sup> satisfies <sup>∂</sup>δu zð Þ ;<sup>ω</sup> <sup>=</sup>∂<sup>ω</sup> <sup>¼</sup> <sup>0</sup>, R <sup>∂</sup><sup>2</sup>

�∞

� 1 G ∂G <sup>∂</sup><sup>ω</sup> � <sup>3</sup>

where <sup>D</sup>^ <sup>¼</sup> <sup>i</sup>

;<sup>ω</sup> � � <sup>¼</sup> <sup>δ</sup> <sup>z</sup> � <sup>z</sup> � �<sup>0</sup>

The Green algorithm for solving NLSE.

Nonlinear Schrödinger Equation

DOI: http://dx.doi.org/10.5772/intechopen.81093

LG z ^ ; z 0

Figure 3.

there is

23

where Z z<sup>0</sup>

determined by (42).

<sup>2</sup>ω<sup>2</sup>β<sup>2</sup> � <sup>i</sup>

Constructing the iteration <sup>β</sup><sup>3</sup> <sup>¼</sup> <sup>β</sup><sup>0</sup>

δu zð Þ¼ ;ω

;ω; δβ<sup>3</sup> z � �<sup>0</sup> ; u<sup>0</sup> z 0 ;<sup>ω</sup> � � � � ¼ � <sup>i</sup>

represents the Fourier transform [32]. Let <sup>L</sup>^ <sup>¼</sup> <sup>∂</sup>

G z; z 0 ;ω � � <sup>¼</sup> <sup>1</sup>

> ð G z; z 0 ;ω � �Z z<sup>0</sup>

δβ<sup>3</sup> <sup>¼</sup> exp <sup>ð</sup>þ<sup>∞</sup>

Finally, the solution of (27) can be written with the eigen function and Green function:

$$\begin{split} \boldsymbol{\rho}(t) &= \boldsymbol{\phi}(t) + \int \mathbf{G}\_{0}(t, \mathbf{t}') \mathbf{V}(t') \boldsymbol{\rho}(t') dt' \\ &= \boldsymbol{\phi}(t) + \int \mathbf{G}\_{0}(t, \mathbf{t'}, \mathbf{E}) \mathbf{V}(t') \boldsymbol{\phi}(t') dt' + \int \boldsymbol{d}t' \mathbf{G}\_{0}(t, \mathbf{t'}, \mathbf{E}) \mathbf{V}(t') \int \mathbf{G}\_{0}(t', \mathbf{t'}, \mathbf{E}) \mathbf{V}(t') \boldsymbol{\rho}(t') dt' \\ &= \boldsymbol{\phi}(t) + \int \mathbf{G}\_{0}(t, \mathbf{t'}, \mathbf{E}) \mathbf{V}(t') \boldsymbol{\phi}(t') dt' + \int \boldsymbol{d}t' \mathbf{G}\_{0}(t, \mathbf{t'}, \mathbf{E}) \mathbf{V}(t') \int \mathbf{G}\_{0}(t', \mathbf{t'}, \mathbf{E}) \mathbf{V}(t') \boldsymbol{\phi}(t') dt' + \cdots \\ &+ \underbrace{\int \mathbf{d}t' \mathbf{G}\_{0}(t, \mathbf{t'}) \mathbf{V}(t') \left[ \int \mathbf{G}\_{0}(t', \mathbf{t'}) \mathbf{V}(t') dt' \cdots \right] \mathbf{G}\_{0}(t', \mathbf{t'}^{l+1}) \boldsymbol{V}(t'^{l+1}) \boldsymbol{\rho}(t'^{l+1}) dt'^{l+1}}\_{\tag{40}} \tag{41}$$

The accuracy can be estimated by the last item of (40). The algorithm is plotted in Figure 3.
