Question

By directly evaluating the integral in the definition, show that the Fourier transform of U_{0}(x)=\left\{\begin{array}{clr} 1+\cos (x) & : & |x|<\pi \\ 0 & z & \text { otherwise } \end{array}\right. is given by \tilde{U}_{0}(k)=\frac{2 \sin (\pi k)}{k\left(1-k^{2}\right)} The function u(z, y) satisfies the partial differential equation \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 \quad \text { in } \quad-\infty<x<\infty, \quad 0<y<\infty and the boundary conditions u \rightarrow 0 \text { as } x \rightarrow \pm \infty, \quad u=U_{0}(x) \text { on } \quad y=0, \quad u \rightarrow 0 \text { as } y \rightarrow \infty ; where Uo(z) is the function defined in part (i) above. ) Use Fourier transforms to show that \frac{\partial^{2} \tilde{u}}{\partial y^{2}}-k^{2} \tilde{u}=0 \quad \text { in } \quad 0<y<\infty, \quad-\infty<k<\infty, where ü(k, y) is the Fourier transform of u with respect to z. Write down the boundary conditions satisfied by ü(k, y) on y = 0 and as y + 0o, and hence solve for ülk, v). Use your result from (ii)(b) to obtain an integral expression for u(z, y) in the form w(x, y)=\int_{-\infty}^{\infty} K(x, y, k) d k where K(z, y, k) is a function you should determine.

Fig: 1

Fig: 2

Fig: 3

Fig: 4

Fig: 5

Fig: 6

Fig: 7

Fig: 8

Fig: 9

Fig: 10

Fig: 11

Fig: 12

Fig: 13

Fig: 14

Fig: 15

Fig: 16