Substitute

W = f (z) e^{i (k x - ω t)}

into the transformed equation. It yields

f^{''} + (\frac{αg k ^{2}}{ω ^{2}} - k^{2} - \frac{α ^{2}}{4}) f = 0, f (0) = f (h) = 0.

f = A sin (nπ z / h)

n \geq 1

. Hence, with

N = αg

and

b_{n}^{2} = (nπ / h)^{2} + α^{2} /4

ω^{2} = \frac{N ^{2} k ^{2}}{k ^{2} + b _{n}^{2}} = \frac{4 αg h ^{2} k ^{2}}{4 n ^{2} π ^{2} + h ^{2} ( α ^{2} + 4 k ^{2} )} .

The physical vertical velocity includes the factor

e^{α z /2} sin (nπ z / h)

. The zero horizontal wavenumber is the zero-frequency limiting case, not a propagating oscillation.

ii

Solution

Choose the rightgoing signed dispersion relation

ω (k) = N k / k^{2} + b_{n}^{2}

c_{p} = \frac{ω}{k} = \frac{N}{k ^{2} + b _{n}^{2}}, c_{g} = \frac{d ω}{d k} = \frac{N b _{n}^{2}}{( k ^{2} + b _{n}^{2} ) ^{3/2}} .

The leftgoing branch has their negatives. For a convention using positive frequency at both signs of

k

, multiply each displayed speed by

sgn k

. For every nonzero

k

0 < c_{g} < c_{p}

c_{g} \leq N / b_{n}

, approached as

k \to 0

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