In the previous three problems, the solution of the original problem is
A) $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L )$ , where $c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / \sinh ( - n \pi H / L )$
B) $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L )$ , where $c _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )$
C) $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \cosh ( n \pi ( y - H ) / L )$ , where $c _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )$
D) $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \cosh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right)$ , where $c _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )$
E) $u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) \sinh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right)$ , where $c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / \sinh ( - n \pi H / L )$

Question

Quizplus · Accepted Answer

The correct solution involves a sine series expansion for $u(x,t)$ and a coefficient $c_n$ that is determined by an integral involving the sine function, not the cosine function, and includes a normalization factor of $2/L$. The hyperbolic sine function ($\sinh$) correctly describes the behavior in the $y$ direction for problems involving certain boundary conditions and physical setups, such as heat conduction or wave equations in a bounded domain. The negative sign in the denominator of the coefficient formula is necessary due to the properties of the hyperbolic sine function and the boundary conditions at $y = H$.

In the Previous Three Problems, the Solution of the Original u(x,t)=∑n−1∞cnsin⁡(nπx/L)sinh⁡(nπ(y−H)/L)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L )u(x,t)=∑n−1∞​cn​sin(nπx/L)sinh(nπ(y−H)/L)

In the Previous Three Problems, the Solution of the Original $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L )$