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Categories: Fiber optics, Mathematics, Nonlinear optics, Physics.

Optical soliton

In general, a soliton is a wave packet that maintains its shape as it travels over great distances. They are only explainable by nonlinear physics, but many (often unrelated) nonlinear equations give rise to solitons: the Boussinesq equations, the Korteweg-de Vries equation, the nonlinear Schrödinger (NLS) equation, and more. Here we consider waveguide optics, which is governed by the NLS equation, given in dimensionless form by:

$$\begin{aligned} i u_z + u_{tt} + r |u|^2 u = 0 \end{aligned}$$

Where $r = \pm 1$ determines the dispersion regime, and subscripts denote differentiation. We start by making the most general ansatz for the pulse envelope $u(z, t)$, namely:

$$\begin{aligned} u(z, t) = \phi(z, t) \: e^{i \theta(z, t)} \end{aligned}$$

With $\phi$ and $\theta$ both real. Note that no generality has been lost yet: we have simply split a single complex function into two real ones. The derivatives of $u$ thus become:

$$\begin{aligned} u_z &= (\phi_z + i \phi \theta_z) \: e^{i \theta} \\ u_t &= (\phi_t + i \phi \theta_t) \: e^{i \theta} \\ u_{tt} &= (\phi_{tt} + 2 i \phi_t \theta_t + i \phi \theta_{tt} - \phi \theta_t^2) \: e^{i \theta} \end{aligned}$$

Inserting $u_z$ and $u_{tt}$ into the NLS equation leads us to:

$$\begin{aligned} 0 &= i \phi_z - \phi \theta_z + \phi_{tt} + 2 i \phi_t \theta_t + i \phi \theta_{tt} - \phi \theta_t^2 + r \phi^3 \\ &= \phi_{tt} - \phi \theta_t^2 - \phi \theta_z + r \phi^3 + i (\phi \theta_{tt} + 2 \phi_t \theta_t + \phi_z) \end{aligned}$$

Since $\phi$ and $\theta$ are both real, we can split this equation into its real and imaginary parts:

$$\begin{aligned} \boxed{ \begin{aligned} 0 &= \phi_{tt} - \phi \theta_t^2 - \phi \theta_z + r \phi^3 \\ 0 &= \phi \theta_{tt} + 2 \phi_t \theta_t + \phi_z \end{aligned} } \end{aligned}$$

Still no generality has been lost so far: these coupled equation are totally equivalent to the NLS equation. But now it is time make a more specific ansatz, namely that $\phi$ and $\theta$ both have a fixed shape but move at a group velocity $v$ and phase velocity $w$, respectively:

$$\begin{aligned} \phi(z, t) &= \phi(t - v z) \\ \theta(z, t) &= \theta(t - w z) \end{aligned}$$

Meaning $\phi_z = -v \phi_t$ and $\theta_z = -w \theta_t$. Now the coupled equations are given by:

$$\begin{aligned} 0 &= \phi_{tt} - \phi \theta_t^2 + w \phi \theta_t + r \phi^3 \\ 0 &= \phi \theta_{tt} + 2 \phi_t \theta_t - v \phi_t \end{aligned}$$

We multiply the imaginary part’s equation by $\phi$ and take its indefinite integral, which can then be evaluated by recognizing the product rule of differentiation:

$$\begin{aligned} 0 &= \int \Big( \phi^2 \theta_{tt} + 2 \phi \phi_t \theta_t - v \phi \phi_t \Big) \dd{t} \\ &= \phi^2 \theta_t - \frac{v}{2} \phi^2 \end{aligned}$$

Where the integration constant has been set to zero. This implies $\theta_t = v/2$, which we insert into the real part’s equation, giving:

$$\begin{aligned} 0 &= \phi_{tt} + \frac{v}{4} (2 w - v) \phi + r \phi^3 \end{aligned}$$

Defining $B \equiv v (v - 2 w) / 4$, multiplying by $2 \phi_t$, and integrating in the same way:

$$\begin{aligned} 0 &= \int \Big( 2 \phi_t \phi_{tt} - 2 B \phi \phi_t + 2 r \phi^3 \phi_t \Big) \dd{t} \\ &= \phi_t^2 - B \phi^2 + \frac{r}{2} \phi^4 - C \end{aligned}$$

Where $C$ is an integration constant. Rearranging this yields a powerful equation, which can be interpreted as a “pseudoparticle” with kinetic energy $\phi_t^2$ moving in a potential $-P(\phi)$:

$$\begin{aligned} \boxed{ \phi_t^2 = P(\phi) \equiv -\frac{r}{2} \phi^4 + B \phi^2 + C } \end{aligned}$$

We further restrict the set of acceptable solutions by demanding that $\phi(t)$ is localized, meaning $\phi \to \phi_\infty$ when $t \to \pm \infty$, for a finite constant $\phi_\infty$. This implies $\phi_t \to 0$ and $\phi_{tt} \to 0$: the former clearly requires $P(\phi_\infty) = 0$. Regarding the latter, we differentiate the pseudoparticle equation with respect to $t$, which tells us for $t \to \pm \infty$:

$$\begin{aligned} 0 = \phi_{tt} &= \frac{1}{2} P'(\phi_\infty) = (B - r \phi_\infty^2) \phi_\infty \end{aligned}$$

Here we have two options: the “bright” case $\phi_\infty = 0$, and the “dark” case $\phi_\infty^2 = r B$. Before we investigate those further, let us finish finding $\theta$: we know that $\theta_t = v/2$, so:

$$\begin{aligned} \theta(t - w z) = \int \theta_t \dd{(t - w v)} = \frac{v}{2} (t - w v) \end{aligned}$$

Where we can ignore the integration constant because the NLS equation has Gauge symmetry, i.e. it is invariant under a transformation of the form $u \to u e^{i a}$ with constant $a$. Finally, we rewrite this result to eliminate $w$ in favor of $B$:

$$\begin{aligned} \theta(z, t) = \frac{v}{2} t - \bigg( \frac{v^2}{4} - B \bigg) z \end{aligned}$$

Bright solitons

First we consider the “bright” option $\phi_\infty = 0$, where our requirement that $P(\theta_\infty) = 0$ clearly means that we must set $C = 0$. We are therefore left with:

$$\begin{aligned} \phi_t^2 = P(\phi) = -\frac{r}{2} \phi^4 + B \phi^2 \end{aligned}$$

We must consider $r = 1$ and $r = -1$, and the sign of $B$; the possible forms of $P(\phi)$ are shown in the sketch below. Because $\phi_t$ is real by definition, valid solutions can only exist in the shaded regions where $P(\phi) \ge 0$:

However, in order to have stable solutions where $\phi$ does not grow uncontrolably, we must restrict ourselves to shaded regions with a finite area. Otherwise, if they are infinite (as for $r = -1$), then a positive feedback loop arises: $\phi_t^2$ grows, so $|\phi|$ increases, then according to the sketch $\phi_t^2$ grows even more, etc. While mathematically correct, that would be physically unacceptable, so the only valid case here is $r = 1$ with $B > 0$.

Armed with this knowledge, we are now ready to integrate the pseudoparticle integration. First, we rewrite it as follows, defining $x \equiv t - vz$:

$$\begin{aligned} \phi_t = \pdv{\phi}{x} = \pm \sqrt{P(\phi)} = \pm \phi \sqrt{B - \phi^2 / 2} \end{aligned}$$

This can be rearranged such that the differential elements $\dd{x}$ and $\dd{\phi}$ are on opposite sides, which can then each be wrapped in an integral, like so:

$$\begin{aligned} \dd{x} = \pm \frac{\sqrt{2}}{\phi \sqrt{2 B - \phi^2}} \dd{\phi} \qquad\implies\qquad \int_{x_0}^{x} \dd{\xi} = \pm \sqrt{2} \int_{\phi_0}^{\phi} \frac{1}{\psi \sqrt{2 B - \psi^2}} \dd{\psi} \end{aligned}$$

Note that these are indefinite integrals, which have been written as definite integrals by placing the constants $x_0$ and $\phi_0$ and target variables $x$ and $\phi$ in the limits.

In order to integrate by substitution, we define the new variable $f \equiv \psi / \sqrt{2 B}$ and update the limits accordingly to $F \equiv \phi / \sqrt{2 B}$ and $F_0 \equiv \phi_0 / \sqrt{2 B}$:

$$\begin{aligned} x - x_0 &= \pm \sqrt{2} \int_{F_0}^{F} \frac{\sqrt{2 B}}{f \sqrt{2 B} \sqrt{2 B - 2 B f^2}} \dd{f} \\ &= \pm \frac{1}{\sqrt{B}} \int_{F_0}^{F} \frac{1}{f \sqrt{1 - f^2}} \dd{f} \end{aligned}$$

We look up this integrand, and discover that it is in fact the derivative of the inverse $\sech^{-1}$ of the hyperbolic secant function, so we arrive at:

$$\begin{aligned} x - x_0 &= \pm \frac{1}{\sqrt{B}} \int_{F_0}^{F} \dv{}{f} \Big( \sech^{-1}(f) \Big) \dd{f} \\ &= \pm \frac{1}{\sqrt{B}} \sech^{-1}(F) \mp \frac{1}{\sqrt{B}} \sech^{-1}(F_0) \end{aligned}$$

Rearranging and combining the integration constants $x_0$ and $F_0$ into a single $t_0$, we get:

$$\begin{aligned} \sech^{-1}(F) = \pm \sqrt{B} (x - t_0) \qquad\qquad t_0 \equiv x_0 \mp \frac{1}{\sqrt{B}} \sech^{-1}(F_0) \end{aligned}$$

Then, wrapping everything in $\sech$ (which is an even function, so we can discard the $\pm$) and using $F \equiv \phi / \sqrt{2 B}$, we finally arrive at the desired solution for $\phi$:

$$\begin{aligned} \phi(x) = \sqrt{2 B} \sech\!\Big( \sqrt{B} (x - t_0) \Big) \end{aligned}$$

Combining this result with our earlier solution for $\theta$, we find that the full so-called bright soliton $u$ is as follows, controlled by two real parameters $B > 0$ and $v$:

$$\begin{aligned} \boxed{ u(z, t) = \sqrt{2 B} \sech\!\bigg( \sqrt{B} (t - v z - t_0) \bigg) \exp\!\bigg( i \frac{v}{2} t - i \Big( \frac{v^2}{4} - B \Big) z \bigg) } \end{aligned}$$

It is always possible to transform the NLS equation into a new moving coordinate system such that $v = 0$, yielding a stationary soliton given by:

$$\begin{aligned} \boxed{ u(z, t) = \sqrt{2 B} \sech\!\Big( \sqrt{B} (t - t_0) \Big) \exp(i B z) } \end{aligned}$$

You may be wondering how we can set $v = 0$ without affecting $B$; a more correct way of saying it would be that we take the limits $v \to 0$ and $w \to -\infty$.

That was for the dimensionless form of the NLS equation; let us specialize this to its usual form in fiber optics. We thus make a transformation $u \to U/U_c$, $t \to T/T_c$ and $z \to Z/Z_c$:

$$\begin{aligned} \frac{U(Z, T)}{U_c} &= \sqrt{2 B} \sech\!\bigg( \sqrt{B} \: \frac{T - T_0}{T_c} \bigg) \exp\!\bigg( i B \frac{Z}{Z_c} \bigg) \end{aligned}$$

Where $U_c$, $T_c$ and $Z_c$ are scale constants determined during non-dimensionalization to obey the relations below. We only have two relations, so we can choose one value freely, say, $U_c$:

$$\begin{aligned} Z_c = \frac{1}{\gamma_0 U_c^2} \qquad\qquad T_c = \sqrt{\frac{- \beta_2}{2 \gamma_0 U_c^2}} \end{aligned}$$

Note that $r = 1$ implies $\beta_2 < 0$ assuming $\gamma_0 > 0$. In other words, bright solitons only exist in the anomalous dispersion regime of an optical fiber. Inserting these relations into the expression and defining the peak power $P_0 \equiv 2 B U_c^2$ yields:

$$\begin{aligned} U(Z, T) &= \sqrt{P_0} \sech\!\Bigg( \sqrt{\frac{\gamma_0 P_0}{- \beta_2}} (T - T_0) \Bigg) \exp\!\bigg( i \frac{\gamma_0 P_0}{2} Z \bigg) \end{aligned}$$

In practice, most authors write this as follows, where $T_\mathrm{w}$ determines the width of the pulse:

$$\begin{aligned} \boxed{ U(Z, T) = \sqrt{P_0} \sech\!\bigg( \frac{T - T_0}{T_\mathrm{w}} \bigg) \exp\!\bigg( i \frac{\gamma_0 P_0}{2} Z \bigg) } \end{aligned}$$

Clearly, for this to be a valid solution of the NLS equation, $T_\mathrm{w}$ must be subject to a constraint involving the so-called soliton number $N_\mathrm{sol}$:

$$\begin{aligned} \boxed{ N_\mathrm{sol}^2 \equiv \frac{L_D}{L_N} = \frac{\gamma_0 P_0 T_\mathrm{w}^2}{|\beta_2|} = 1 } \end{aligned}$$

Where $L_D \equiv T_0 / |\beta_2|$ is the linear length scale of dispersive broadening, and $L_N \equiv 1 / (\gamma_0 P_0)$ is the nonlinear length scale of self-phase modulation. A first-order soliton has $N_\mathrm{sol} = 1$ and simply maintains its shape, whereas higher-order solitons have complicated periodic dynamics.

Dark solitons

The other option to satisfy $P'(\phi_\infty) = 0$ is $\phi_\infty^2 = r B$, which implies $r B > 0$ such that $\phi_\infty$ is real. With this in mind, we again sketch all remaining candidates for $P(\phi)$:

At a glance, there are plenty of solutions here, even stable ones! However, as explained earlier, our localization requirement means that we need $P(\phi_\infty) = 0$ and $P'(\phi_\infty) = 0$. The latter is only satisfied by the solid curve above, so we must limit ourselves to $r = -1$ and $B < 0$, with $C = C_0$ for some positive $C_0$. The next step is to find $C_0$.

We notice that the target curve has two double roots at $\pm \phi_\infty$, so we can rewrite:

$$\begin{aligned} P(\phi) &= \frac{1}{2} \Big( \phi^4 + 2 B \phi^2 + 2 C \Big) \\ &= \frac{1}{2} \Big( \phi^4 + 2 B \phi^2 + B^2 - B^2 + 2 C \Big) \\ &= \frac{1}{2} \big( \phi^2 + B \big)^2 - \frac{1}{2} \big( B^2 - 2 C \big) \end{aligned}$$

Here we see that $P(\phi_\infty)$ can only have a double root when $C = C_0 = B^2 / 2$, in which case the root is clearly $\phi_\infty = \pm \sqrt{-B}$. We are therefore left with:

$$\begin{aligned} \phi_t^2 = P(\phi) = \frac{1}{2} \big( \phi^2 + B \big)^2 \end{aligned}$$

Now we are ready to integrate this equation. Taking the square root with $x \equiv t - v z$:

$$\begin{aligned} \phi_t = \pdv{\phi}{x} = \pm \sqrt{P(\phi)} = \pm \frac{1}{\sqrt{2}} (\phi^2 + B) \end{aligned}$$

We put the differential elements $\dd{\phi}$ and $\dd{x}$ on opposite sides and take the integrals:

$$\begin{aligned} \dd{x} = \pm \frac{\sqrt{2}}{\phi^2 + B} \dd{\phi} \qquad\implies\qquad \int_{x_0}^{x} \dd{\xi} = \pm \sqrt{2} \int_{\phi_0}^{\phi} \frac{1}{\psi^2 + B} \dd{\psi} \end{aligned}$$

Then we define $f \equiv \psi / \sqrt{-B}$, and update the limits to $F = \phi / \sqrt{-B}$ and $F_0 = \phi_0 / \sqrt{-B}$, in order to integrate by substitution:

$$\begin{aligned} x - x_0 &= \pm \sqrt{2} \int_{F_0}^{F} \frac{\sqrt{-B}}{- B f^2 + B} \dd{f} \\ &= \pm \sqrt{-\frac{2}{B}} \int_{F_0}^{F} \frac{1}{1 - f^2} \dd{f} \end{aligned}$$

The integrand can be looked up: it turns out be the derivative of $\tanh^{-1}$, the inverse hyperbolic tangent function, so we arrive at:

$$\begin{aligned} x - x_0 &= \pm \sqrt{-\frac{2}{B}} \int_{F_0}^{F} \dv{}{f} \Big( \tanh^{-1}(f) \Big) \dd{f} \\ &= \pm \sqrt{-\frac{2}{B}} \tanh^{-1}(F) \mp \sqrt{-\frac{2}{B}} \tanh^{-1}(F_0) \end{aligned}$$

Rearranging, and combining the integration constants $x_0$ and $F_0$ into a single $t_0$, yields:

$$\begin{aligned} \tanh^{-1}(F) &= \pm \sqrt{-\frac{B}{2}} (x - t_0) \qquad\qquad t_0 \equiv x_0 \mp \sqrt{-\frac{2}{B}} \tanh^{-1}(F_0) \end{aligned}$$

Next, we take the $\tanh$ of both sides. It is an odd function, so the $\pm$ can be moved outside, where it can be ignored entirely thanks to the NLS equation’s Gauge symmetry. Using $F = \phi / \sqrt{-B}$:

$$\begin{aligned} \phi(x) &= \sqrt{-B} \tanh\!\Bigg( \sqrt{-\frac{B}{2}} (x - t_0) \Bigg) \end{aligned}$$

Combining this with our expression for $\theta$, we arrive at the full dark soliton solution for $u$:

$$\begin{aligned} \boxed{ u(z, t) = \sqrt{-B} \tanh\!\Bigg( \sqrt{-\frac{B}{2}} (t - v z - t_0) \Bigg) \exp\!\bigg( i \frac{v}{2} t - i \Big( \frac{v^2}{4} - B \Big) z \bigg) } \end{aligned}$$

There are two free parameters here: $B < 0$ and $v$. Once again, we can always transform to a moving coordinate system such that $v = 0$, resulting in a stationary soliton:

$$\begin{aligned} \boxed{ u(z, t) = \sqrt{-B} \tanh\!\Bigg( \sqrt{-\frac{B}{2}} (t - t_0) \Bigg) \exp(i B z) } \end{aligned}$$

Like we did for the bright solitons, let us specialize this result to fiber optics. Making a similar transformation $u \to U/U_c$, $t \to T/T_c$ and $z \to Z/Z_c$ yields:

$$\begin{aligned} \frac{U(Z, T)}{U_c} = \sqrt{-B} \tanh\!\Bigg( \sqrt{-\frac{B}{2}} \frac{T - T_0}{T_c} \Bigg) \exp\!\bigg( i B \frac{Z}{Z_c} \bigg) \end{aligned}$$

Where we again choose $U_c$ manually, and then find $T_c$ and $Z_c$ using these relations (note the opposite signs because $r = -1$ in this case):

$$\begin{aligned} Z_c = \frac{-1}{\gamma_0 U_c^2} \qquad\qquad T_c = \sqrt{\frac{\beta_2}{2 \gamma_0 U_c^2}} \end{aligned}$$

Recall that $r = -1$ implies $\beta_2 > 0$ assuming $\gamma_0 > 0$, meaning dark solitons can only exist in the normal dispersion regime. Inserting this into the expression and defining the background power $P_0 \equiv -B U_c^2$ such that $|U|^2 \to P_0$ for $t \to \pm \infty$, we arrive at:

$$\begin{aligned} U(Z, T) = \sqrt{P_0} \tanh\!\Bigg( \sqrt{\frac{\gamma_0 P_0}{\beta_2}} (T - T_0) \Bigg) \exp(i \gamma_0 P_0 Z) \end{aligned}$$

Which, as for bright solitons, can be rewritten with a pulse width $T_\mathrm{w}$ satisfying $N_\mathrm{sol} = 1$:

$$\begin{aligned} \boxed{ U(Z, T) = \sqrt{P_0} \tanh\!\bigg( \frac{T - T_0}{T_\mathrm{w}} \bigg) \exp(i \gamma_0 P_0 Z) } \end{aligned}$$

References

1. A. Scott, Nonlinear science: emergence and dynamics of coherent structures, 2nd edition, Oxford.
2. O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.

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