## Abstract

The theoretical characteristics of a 1-D model of a steady-state laser with uniform saturable gain and distributed losses are calculated by iterative solution, using integration of a series representation of the two-way intensity distribution over the length of the laser. Results for this model and for the model with point losses at the end mirrors show that the output can be the same for both models over a range of the point loss factor. This factor becomes unique if the losses are also set equal, but the internal intensity distributions are similar only when the loss is low.

© 1987 Optical Society of America

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### Equations (23)

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(1)
$$g(z)={g}_{0}/(1+{\beta}_{+}+{\beta}_{-}).$$
(2)
$$\frac{1}{{\beta}_{+}}\frac{d{\beta}_{+}}{dz}=-\frac{1}{{\beta}_{-}}\frac{d{\beta}_{-}}{dz}=g(z)-\alpha .$$
(3)
$${\beta}_{+}{\beta}_{-}=C={\beta}_{0}^{2}={\beta}_{1}{\beta}_{4}={\beta}_{2}{\beta}_{3}.$$
(4)
$${\beta}_{3}={r}_{2}{\beta}_{2},\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}{\beta}_{1}={r}_{1}{\beta}_{4}.$$
(5)
$${r}_{1}=1-{a}_{1}-{t}_{1},\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}{r}_{2}=1-{a}_{2}-{t}_{2}.$$
(6)
$$\left(1+\frac{1}{{\beta}_{+}}+\frac{{\beta}_{0}^{2}}{{\beta}_{+}^{2}}\right)\frac{d{\beta}_{+}}{dz}={g}_{0}-\alpha (1+{\beta}_{+}+{\beta}_{-}).$$
(7)
$${\beta}_{0}\left(\frac{1-{r}_{2}}{\surd {r}_{2}}+\frac{1-{r}_{1}}{\surd {r}_{1}}\right)-\xbd\hspace{0.17em}\text{ln}({r}_{1}{r}_{2})={g}_{0}L-\alpha L-\alpha {\int}_{0}^{L}({\beta}_{+}+{\beta}_{-})dz.$$
(8)
$$\frac{1}{{\beta}_{+}}\frac{d{\beta}_{+}}{d\zeta}=\frac{{g}_{0}L}{1+{\beta}_{+}+{\beta}_{0}^{2}/{\beta}_{+}}-\alpha L,$$
(9)
$$-\frac{1}{{\beta}_{-}}\frac{d{\beta}_{-}}{d\zeta}=\frac{{g}_{0}L}{1+{\beta}_{-}+{\beta}_{0}^{2}/{\beta}_{-}}-\alpha L.$$
(10)
$${\beta}_{-}(-\zeta )={\beta}_{+}(\zeta )$$
(11)
$$(1+{\beta}_{+}+{\beta}_{-})d{\beta}_{+}/d\zeta =[{g}_{0}L-\alpha L(1+{\beta}_{+}+{\beta}_{-})]{\beta}_{+}.$$
(12)
$${\beta}_{+}(\zeta )=\sum _{n=0}^{\infty}{I}_{n}{\zeta}^{n}/n!,$$
(13)
$$\begin{array}{l}{I}_{1}=(s-\alpha L){\beta}_{0},\\ {I}_{2}={(s-\alpha L)}^{2}{\beta}_{0},\\ {I}_{3}={(s-\alpha L)}^{2}\left[\frac{s}{1+2{\beta}_{0}}-\alpha L\right]\hspace{0.17em}{\beta}_{0},\\ {I}_{4}={(s-\alpha L)}^{3}\left[s\hspace{0.17em}\left(\frac{1-6{\beta}_{0}}{1+2{\beta}_{0}}\right)-\alpha L\right]\hspace{0.17em}{\beta}_{0},\end{array}$$
(14)
$$s={g}_{0}L/(1+2{\beta}_{0}).$$
(15)
$$\sigma =1+\frac{{(s-\alpha L)}^{2}}{{2}^{2}\xb73!}+\frac{{(s-\alpha L)}^{3}}{{2}^{4}\xb75!}\left[s\left(\frac{1-6{\beta}_{0}}{1+2{\beta}_{0}}\right)-\alpha L\right]+\dots .$$
(16)
$${\beta}_{0}=\frac{{r}^{1/2}({g}_{0}L+\text{ln}r-\alpha L)}{2(1-r-\alpha L\sigma {r}^{1/2})}.$$
(17)
$${\beta}_{2}t=(t/2)[{g}_{0}L+\text{ln}(1-a-t)]/(t+a),$$
(18)
$${\beta}_{2}t=(t/2)[{g}_{0}L-\alpha L+\text{ln}(1-t)]/[t+\alpha L{(1-t)}^{1/2}\sigma ].$$
(19)
$$\frac{{\beta}_{+}^{2}+{\beta}_{0}^{2}}{{\beta}_{+}}+\text{ln}\frac{{\beta}_{+}}{{\beta}_{0}}=\zeta [{g}_{0}L-\alpha L-2\alpha L{\beta}_{0}\sigma (\zeta )],$$
(20)
$$\sigma (\zeta )=1+{(s-\alpha L)}^{2}\frac{{\zeta}^{2}}{3!}+{(s-\alpha L)}^{3}\left[s\left(\frac{1-6{\beta}_{0}}{1+2{\beta}_{0}}\right)-\alpha L\right]\frac{{\zeta}^{4}}{5!}+\dots .$$
(21)
$$I(\zeta )={\beta}_{0}\hspace{0.17em}\text{exp}[(s-\alpha L)\zeta ]$$
(22)
$$s-\alpha L\simeq \text{ln}(1/r).$$
(23)
$${\beta}_{0}\simeq \frac{1}{2}\left[\frac{{g}_{0}L}{\alpha L+\text{ln}(1/r)}-1\right].$$