Computation of breaking source term

The surf breaking dissipation of Battjes and Janssen (1978) reads

  $$D_{\rm tot} = -\alpha_{\rm BJ} Q_{b} \tilde{\sigma} \frac{H^2_{\rm max}}{8\pi}$$ (3.136)

The surf breaking source term for each spectral bin $i$ is

  $$S_{i} = \frac{D_{\rm tot}}{E_{\rm tot}} E_i = {\tilde D}\, E_i$$ (3.137)

with the normalized total dissipation

  $${\tilde D} = -\frac{\alpha_{\rm BJ} \tilde{\sigma} Q_{b}}{\pi {\cal B}} < 0$$ (3.138)

and

  $${\cal B} = \frac{8E_{\rm tot}}{H^2_{\rm max}} = \left( \frac{H_{\rm rms}}{\gamma d} \right)^2$$ (3.139)

Since, the source term is strongly nonlinear in $E$ (since ${\tilde D}$ depends on $E$ through ${\cal B}$), we apply the Newton linearisation to approximate the source term at iteration level $n+1$, as follows

  $$S^{n+1}_{i} \approx {\tilde D} E_i^{n} + \left( \frac{\partial S}{\partial E} \right)_i^n (E_i^{n+1} - E_i^n)$$ (3.140)

In SWAN, this approximation has been slightly adapted for reasons of numerical stability; the first term in the right hand side, ${\tilde D} E_i^{n}$, is replaced by ${\tilde D} E_i^{n+1}$. This preserves positivity of energy density $E$, if the following inequality holds

  $$\frac{\partial S}{\partial E} < 0$$ (3.141)

We derive an expression for this derivative as follows. From (3.137), we have

  $$\frac{\partial S}{\partial E} \vert _i = \frac{\partial {\tilde D}}{\partial E}\vert _i E_i + {\tilde D}$$ (3.142)

The normalized dissipation ${\tilde D}$ is a function of ${\cal B}$ which is proportional to $E$, so

  $$\frac{\partial S}{\partial E} \vert _i = \frac{\partial {\tilde D}}{\partial {\cal B}}\vert _i {\cal B}_i + {\tilde D}$$ (3.143)

Since, $Q_b$ is a function of ${\cal B}$, we get (using the quotient rule)

  $$\frac{\partial S}{\partial E} \vert _i = -\frac{\alpha_{\rm BJ} \tilde{\sigma}}{\pi}\, \frac{\partial Q_b}{\partial {\cal B}}$$ (3.144)

Since,

  $$1- Q_b + {\cal B} \ln Q_b = 0$$ (3.145)

the derivative of $Q_b$ is found by differentiating this with respect to ${\cal B}$:

  $$-{Q'}_{b} + \ln Q_b + \frac{\cal B}{Q_b} {Q'}_{b} = 0$$ (3.146)

Hence,

  $${Q'}_{b} = \frac{\ln Q_b}{1 - {\cal B}/Q_b} = \frac{Q_b}{\cal B} \frac{Q_b - 1}{Q_b - {\cal B}}$$ (3.147)

using Eq. (3.145). Now, ${Q'}_{b} > 0$, because $0 < Q_b < 1$ and ${\cal B} > Q_b$. Substitution in (3.144) gives

  $$\frac{\partial S}{\partial E} \vert _i = {\tilde D} \, \frac{Q_b - 1}{Q_b - {\cal B}} \vert _i < 0$$ (3.148)

Finally, the approximation of the source term reads

  $$S^{n+1}_{i} = {\tilde D} \left( 1 + \frac{Q_b - 1}{Q_b - {\cal B}... ...)_i^n E_i^{n+1} - {\tilde D} \, \frac{Q_b - 1}{Q_b - {\cal B}} \vert^n_i E_i^n$$ (3.149)