Let
$$
l=\frac\pi L,
\qquad
\Theta(z)=1-\mathcal H(z-H),
$$
where $\mathcal H$ is the <Heaviside step function>. Then
$$
\nabla\mathbin\cdot\overline{\mathbf F}
=-F_0\delta(z-H)\sin^2(ly),
$$
so the <Eliassen equation for residual circulation> is
$$
f_0^2\overline\chi_{a,zz}^*
+N^2\overline\chi_{a,yy}^*
=f_0F_0\delta'(z-H)\sin^2(ly).
$$
Take no normal residual flow at $y=0,L$ and $z=0$, decay as $z\to\infty$, and choose the streamfunction constant on the connected rigid boundary to be zero. Thus
$$
\overline\chi_a^*=0
\quad\text{on }y=0,L\text{ and }z=0,
\qquad
\overline\chi_a^*\to0
\quad(z\to\infty).
$$

The required <Fourier series> in sine modes is
$$
\sin^2\frac{\pi y}{L}
=\sum_{\substack{n\geq1\\n\text{ odd}}}
b_n\sin\frac{n\pi y}{L},
\qquad
\boxed{
b_n=\frac8{\pi n(4-n^2)}}.
$$
Define
$$
\lambda_n=\frac{Nn\pi}{|f_0|L},
\qquad
C_n=\frac{F_0b_n}{f_0}.
$$
For each mode, the vertical equation is
$$
\chi_n''-\lambda_n^2\chi_n
=C_n\delta'(z-H).
$$
The <Dirac delta function> requires
$$
[\chi_n]_{H^-}^{H^+}=C_n,
\qquad
[\chi_n']_{H^-}^{H^+}=0.
$$
The solution satisfying the boundary conditions is therefore
$$
\boxed{
\overline\chi_a^*(y,z)
=\frac{F_0}{f_0}
\sum_{\substack{n\geq1\\n\text{ odd}}}
b_nS_n(z)\sin\frac{n\pi y}{L}},
$$
where
$$
\boxed{
S_n(z)=
\begin{cases}
-e^{-\lambda_nH}\sinh(\lambda_nz),&0<z<H,\\
\cosh(\lambda_nH)e^{-\lambda_nz},&z>H.
\end{cases}}
$$

The momentum equation gives
$$
\overline u_t
=f_0\overline\chi_{a,z}^*
-F_0\delta(z-H)\sin^2\frac{\pi y}{L}.
$$
The jump of $\overline\chi_a^*$ supplies an equal positive delta function in $f_0\overline\chi_{a,z}^*$, so the singular terms cancel. The regular acceleration is
$$
\boxed{
\overline u_t
=-F_0
\sum_{\substack{n\geq1\\n\text{ odd}}}
b_n\lambda_nR_n(z)
\sin\frac{n\pi y}{L}},
$$
where
$$
R_n(z)=
\begin{cases}
e^{-\lambda_nH}\cosh(\lambda_nz),&0<z<H,\\
\cosh(\lambda_nH)e^{-\lambda_nz},&z>H.
\end{cases}
$$
Finally, the transformed density equation gives
$$
\boxed{
\overline\rho_t
=\frac{d\rho_s}{dz}
\overline\chi_{a,y}^*
=\frac{d\rho_s}{dz}
\frac{F_0}{f_0}
\sum_{\substack{n\geq1\\n\text{ odd}}}
b_n\frac{n\pi}{L}S_n(z)
\cos\frac{n\pi y}{L}}.
$$
These exponentially decaying modes are the balanced mean response to <wave-activity deposition> at $z=H$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 333 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-333/4/d/solution