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Apply the Fourier Trans<strong>for</strong>m to problem (C.1):<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

(−ik) 2 ( π<br />

l)2<br />

ˆφ+ ˆφζζ = 0, on−h 2 /L≤ζ≤ 0,<br />

ˆ<br />

φ ζ (k, 0)=ĝ(k),<br />

ˆ<br />

φ ζ (k,−h 2 /L)=0.<br />

(C.3)<br />

The solution <strong>for</strong> problem (C.3) is<br />

φ(k,ζ)=<br />

ˆ ĝ(k) cosh ( kπ<br />

l<br />

πk/l<br />

There<strong>for</strong>e, <strong>for</strong> C 1 initial Neumann data g,<br />

φ(ξ,ζ)= 1<br />

2π<br />

= 1<br />

2π<br />

∞∑<br />

k=−∞<br />

∞∑<br />

k=−∞<br />

k0<br />

sinh ( kπ<br />

l<br />

ˆ<br />

φ(k,ζ)e ikξ ,<br />

ĝ(k) cosh ( kπ<br />

l<br />

πk/l<br />

( ))<br />

ζ+<br />

h 2<br />

L<br />

) , k0.<br />

h 2<br />

L<br />

sinh ( kπ<br />

l<br />

( ))<br />

ζ+<br />

h 2<br />

L<br />

) e ikξ +<br />

h 2<br />

L<br />

ˆ<br />

φ(0)<br />

2π ,<br />

the convergence is uni<strong>for</strong>m in 0≤ξ≤2π and also in−h 2 /L≤ζ≤ 0 since<br />

∣<br />

cosh ( kπ<br />

l<br />

<strong>for</strong> all integer k0.<br />

sinh ( kπ<br />

l<br />

( ))<br />

ζ+<br />

h 2<br />

∣ ∣∣∣∣∣∣∣<br />

L<br />

)<br />

∣ ≤ cosh ( kπ<br />

l<br />

sinh ( kπ<br />

We wantφ ξ ; returning to the original variables<br />

φ(ξ,ζ)= 1<br />

2π<br />

h 2<br />

L<br />

∞∑<br />

k=−∞<br />

k0<br />

ĝ(k) cosh ( kπ<br />

l<br />

πk/l<br />

l<br />

h 2<br />

L<br />

h 2<br />

L<br />

sinh ( kπ<br />

l<br />

)<br />

∣ ∣∣∣∣∣<br />

)<br />

π coth( ∣ = l<br />

( ))<br />

ζ+<br />

h 2<br />

L<br />

) e ikξπ/l +<br />

h 2<br />

L<br />

)∣<br />

h ∣∣∣∣∣<br />

2<br />

, (C.4)<br />

L<br />

ˆ<br />

φ(0)<br />

2π .<br />

95

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