Problem 5I. A function $f ( x )$ is continuous and defined on the interval $0 \leq x \leq \pi$. If $f ( x )$ is extended to the interval $- \pi \leq x \leq \pi$ as an odd function, it can be expanded in the following Fourier sine series:
$$\begin{aligned}
& f ( x ) \doteq \sum _ { n = 1 } ^ { \infty } \left( b _ { n } \sin n x \right) \\
& b _ { n } = \frac { 2 } { \pi } \int _ { 0 } ^ { \pi } f ( x ) \sin n x \, d x \quad ( n = 1,2,3 , \cdots )
\end{aligned}$$
Here, $f ( 0 ) = f ( \pi ) = 0$.
- Find the Fourier sine series for the following function $f ( x )$: $$f ( x ) = x ( \pi - x ) \quad ( 0 \leq x \leq \pi )$$
- Derive the following equation using the result obtained in Question I.1, $$\frac { 1 } { 1 ^ { 3 } } - \frac { 1 } { 3 ^ { 3 } } + \frac { 1 } { 5 ^ { 3 } } - \frac { 1 } { 7 ^ { 3 } } + \cdots = \frac { \pi ^ { 3 } } { 32 }$$
II. A two-variable function $f ( x , y )$ is continuous and defined in the region $0 \leq x \leq \pi$ and $0 \leq y \leq \pi$. Using a similar method to Question I, $f ( x , y )$ can be expanded in the following double Fourier sine series:
$$\begin{aligned}
& f ( x , y ) = \sum _ { m = 1 } ^ { \infty } \sum _ { n = 1 } ^ { \infty } \left( B _ { m n } \sin m x \sin n y \right) \\
& B _ { m n } = \frac { 4 } { \pi ^ { 2 } } \int _ { 0 } ^ { \pi } \int _ { 0 } ^ { \pi } f ( x , y ) \sin m x \sin n y \, d x \, d y \quad ( m , n = 1,2,3 , \cdots )
\end{aligned}$$
Here, $f ( 0 , y ) = f ( \pi , y ) = f ( x , 0 ) = f ( x , \pi ) = 0$.
- Find the double Fourier sine series for the following function $f ( x , y )$: $$f ( x , y ) = x ( \pi - x ) \sin y \quad ( 0 \leq x \leq \pi , 0 \leq y \leq \pi )$$
- Function $u ( x , y , t )$ is defined in the region $0 \leq x \leq \pi , 0 \leq y \leq \pi$ and $t \geq 0$. Obtain the solution for the following partial differential equation of $u ( x , y , t )$ by the method of separation of variables: $$\frac { \partial u } { \partial t } = c ^ { 2 } \left( \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } \right)$$ where $c$ is a positive constant and the following boundary and initial conditions apply: $$\begin{aligned}
& u ( 0 , y , t ) = u ( \pi , y , t ) = u ( x , 0 , t ) = u ( x , \pi , t ) = 0 \\
& u ( x , y , 0 ) = x ( \pi - x ) \sin y
\end{aligned}$$