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

Question

Quizplus · Accepted Answer

The correct solution involves a sine series for $u(x, t)$ because it typically represents solutions to problems with certain boundary conditions (like Dirichlet conditions), and the coefficient $c_n$ is calculated using an orthogonal projection of the initial condition $f(x)$ onto the basis functions, which in this case are sine functions. The factor of $2/L$ is necessary for normalization in the context of Fourier series on the interval $[0, L]$.

In the Previous Three Problems, the Solution of the Original u(x,t)=∑n=1∞cnsin⁡(nπx/L)e−(nπ/L)2tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t }u(x,t)=∑n=1∞​cn​sin(nπx/L)e−(nπ/L)2t

In the Previous Three Problems, the Solution of the Original $u ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t }$