= Long-wave transport along a depth step
{title2=$c_T=fdR_D/H_0$}
The long <step-trapped topographic Rossby wave> has equal <phase velocity> and <group velocity>, $c_T=fdR_D/H_0$. For an outer surface profile $\eta=\eta_0\operatorname{sgn}(x)-A(x,t)e^{-|y|/R_D}$, cross-step matching gives $A_t+c_TA_x=2c_T\eta_0\delta(x)$. With $A(x,0)=0$, its weak solution is $A=\eta_0[\operatorname{sgn}(x)-\operatorname{sgn}(x-c_Tt)]$. This describes the slowly varying outer response; the fast adjustment region at the origin is unresolved.
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