91Ó°ÊÓ

Consider \(f(x)=e^{-\alpha x}, \alpha>0(x \geqslant 0)\). (a) Derive the Fourier sine transform of \(f(x)\). (b) Derive the Fourier cosine transform of \(f(x)\).

Show that the inverse Fourier transform is a linear operator; that is, show that: (a) \(\mathscr{F}^{-1}\left[c_{1} F(\omega)+c_{2} G(\omega)\right]=c_{1} f(x)+c_{2} g(x)\) (b) \(\mathscr{F}^{-1}[F(\omega) G(\omega)] \neq f(x) g(x)\)

Solve $$ \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 $$ for \(x>0, y>0\) subject to: (a) \(u(0, y)=0\) and \(\frac{\partial u}{\partial y}(x, 0)=f(x)\) (b) \(u(0, y)=0\) and \(u(x, 0)=f(x)\)

There is an interesting convolution-type theorem for Fourier sine transforms. Suppose that we want \(h(x)\), but know its sine transform \(H(\omega)\) to be a product $$ H(\omega)=\bar{S}(\omega) \bar{C}(\omega), $$ where \(\bar{S}(\omega)\) is the sine transform of \(s(x)\) and \(\bar{C}(\omega)\) is the cosine transform of \(c(x)\). Assuming that \(c(x)\) is even and \(s(x)\) is odd, show that $$ h(x)=\frac{1}{\pi} \int_{0}^{\infty} s(\bar{x})[c(x-\bar{x})-c(x+\bar{x})] d \bar{x}=\frac{1}{\pi} \int_{0}^{\infty} c(\bar{x})[s(x+\bar{x})-s(\bar{x}-x)] d \bar{x} . $$

Solve $$ \frac{\partial u}{\partial t}=k_{1} \frac{\partial^{2} u}{\partial x^{2}}+k_{2} \frac{\partial^{2} u}{\partial y^{2}} $$ subject to the initial condition $$ u(x, y, 0)=f(x, y) $$

\- Consider $$ \frac{\partial u}{\partial t}=k\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right) \quad \begin{aligned} &00 \end{aligned} $$ subject to the initial condition $$ u(x, y, 0)=f(x, y) . $$ Solve with the following boundary conditions: *(a) \(u(0, y, t)=0, \quad u(L, y, t)=0, \quad u(x, 0, t)=0\) (b) \(u(0, y, t)=0, \quad u(L, y, t)=0, \quad \frac{\partial u}{\partial y}(x, 0, t)=0\) (c) \(\frac{\partial u}{\partial x}(0, y, t)=0, \quad \frac{\partial u}{\partial x}(L, y, t)=0, \quad \frac{\partial u}{\partial y}(x, 0, t)=0\)