Question: A slowly moving viscous fluid occupies the right half-plane \(x>0,-\infty \[\nabla^{4} \psi=0\]

where \(\psi(x, y)\) is called the stream function from which the velocity components in the \(x\) and \(y\) directions are given by

and where \(\nabla^{4}\) is the biharmonic operator defined by \(\nabla^{4} \equiv\left[\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right]^{2}\).

Suppose that the boundary \(x=0\) is solid everywhere except the gap \(|y|

Hence the problem to solve is the biharmonic equation subject to

and where

Note that \(f(y)=0\) if \(|y| \geq c\)

1. Take the Fourier transform of the biharmonic equation with respect to \(y\), such that \(S(x, k)=\mathcal{F}\{\psi(x, y)\}\), and show that the transform of the stream function satisfies
   \[\left(\frac{d^{2}}{d x^{2}}-k^{2}\right)^{2} S(x, k)=0\]
   and show that this has a solution, which tends to zero as \(x \rightarrow \infty\), of the form
   \[S(x, k)=-i \frac{F(k)}{k}(1+|k| x) e^{-|k| x}\]
   where \(F(k)=\mathcal{F}\{f(x)\}\)
2. Find the function \(g(x, y)\) such that
   \[g(x, y)=\mathcal{F}^{-1}\left\{\frac{-i}{k}(1+|k| x) e^{-|k| x}\right\}\]
   and hence use the convolution theorem to write down the solution for \(\psi(x, y)\). You may use without proof that \(\mathcal{F}\left\{\frac{e^{-|x|}}{x}\right\}=-i \sqrt{\frac{2}{\pi}} \tan ^{-1}(k)\).
3. Find the stream function when fluid is introduced through the gap at constant velocity \(v_{0}\), i.e. when \(f(y)=-v_{0}\) for \(|y|

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