The Laplace transform of the function $f ( t )$, defined for $t \geq 0$, is denoted by $F ( s ) = \mathcal { L } [ f ( t ) ]$ and its definition is given by
$$F ( s ) = \mathcal { L } [ f ( t ) ] = \int _ { 0 } ^ { \infty } f ( t ) \exp ( - s t ) d t$$
where $s$ is a complex number. In the following, the set of all complex numbers is denoted by $\mathbb { C }$, and the set of the complex numbers with positive real parts is denoted by $\mathbb { C } ^ { + }$.
I. Consider the following function $g ( t )$ defined for $t \geq 0$ :
$$g ( t ) = \int _ { 0 } ^ { \infty } \frac { \sin ^ { 2 } ( t x ) } { x ^ { 2 } } d x$$
- Find the Laplace transform $G ( s ) = \mathcal { L } [ g ( t ) ] \left( s \in \mathbb { C } ^ { + } \right)$ of the function $g ( t )$.
- Obtain the value of the following integral using the result of Question I.1: $$\int _ { - \infty } ^ { \infty } \frac { \sin ^ { 2 } ( x ) } { x ^ { 2 } } d x$$
II. Consider the function $u ( x , t )$ that satisfies the following partial differential equation:
$$\frac { \partial u ( x , t ) } { \partial t } = \frac { \partial ^ { 2 } u ( x , t ) } { \partial x ^ { 2 } } ( 0 < x < 1 , t > 0 )$$
under the boundary conditions:
$$\left\{ \begin{array} { l }
\left. \frac { \partial u ( x , t ) } { \partial x } \right| _ { x = 0 } = 0 \quad ( t \geq 0 ) \\
u ( 1 , t ) = 1 \quad ( t \geq 0 ) \\
u ( x , 0 ) = \frac { \cosh ( x ) } { \cosh ( 1 ) } \quad ( 0 < x < 1 )
\end{array} \right.$$
- The Laplace transform of $u ( x , t )$ is denoted by $U ( x , s ) = \mathcal { L } [ u ( x , t ) ]$ ( $s \in \mathbb { C } ^ { + }$). Derive the ordinary differential equation and boundary conditions for $U ( x , s )$ with respect to the independent variable $x$. Here, the function $u ( x , t )$ can be assumed to be bounded. The following relations can also be used: $$\begin{aligned}
& \mathcal { L } \left[ \frac { \partial u ( x , t ) } { \partial x } \right] = \frac { \partial U ( x , s ) } { \partial x } \\
& \mathcal { L } \left[ \frac { \partial ^ { 2 } u ( x , t ) } { \partial x ^ { 2 } } \right] = \frac { \partial ^ { 2 } U ( x , s ) } { \partial x ^ { 2 } }
\end{aligned}$$
- Using an analytic function $Q ( s ) ( s \in \mathbb { C } )$, the function $U _ { \mathrm { c } } ( x , s )$ is defined as follows: $$U _ { c } ( x , s ) = \frac { \cosh ( x ) } { ( s - 1 ) \cosh ( 1 ) } - \frac { \cosh ( x \sqrt { s } ) } { Q ( s ) } \quad ( 0 \leq x \leq 1 )$$ When the function $U ( x , s ) = U _ { \mathrm { c } } ( x , s )$ satisfies the differential equation and the boundary conditions derived in Question II.1 for $s \in \mathbb { C } ^ { + }$, find the function $Q ( s )$.
- Using the function $Q ( s )$ derived in Question II.2, the sequence of complex numbers $\left\{ a _ { r } \right\} ( r = 1,2 , \cdots )$ is defined by arranging all of the roots of $Q ( s ) = 0 ( s \in \mathbb { C } )$ in ascending order of their absolute values. In this case, the following limits $R _ { r } ( x , t )$ are finite for $t \geq 0,0 \leq x \leq 1$, and $r \geq 1$ : $$R _ { r } ( x , t ) = \lim _ { s \rightarrow a _ { r } } \left( s - a _ { r } \right) U _ { \mathrm { c } } ( x , s ) \exp ( s t )$$ and the solution of the partial differential equation is given by $$u ( x , t ) = \sum _ { r = 1 } ^ { \infty } R _ { r } ( x , t )$$ Determine $R _ { 1 } ( x , t ) , R _ { 2 } ( x , t )$, and $R _ { r } ( x , t )$ for $r \geq 3$.