We assume (in this question only) that $c ( x ) = 0$ and $f ( x ) = 1$ for all $x \in [ 0,1 ]$. Let $n \in \mathbb{N}^*$, $h = \frac{1}{n+1}$, and $x_i = ih$. We denote by $u$ the exact solution of problem (1): $$\left\{ \begin{array} { l }
- u ^ { \prime \prime } ( x ) + c ( x ) u ( x ) = f ( x ) , x \in [ 0,1 ] \\
u ( 0 ) = u ( 1 ) = 0
\end{array} \right.$$ and $(u_i)_{0 \leq i \leq n+1}$ the unique family satisfying $$\left\{ \begin{array} { l }
- \frac { 1 } { h ^ { 2 } } \left( u _ { i + 1 } + u _ { i - 1 } - 2 u _ { i } \right) + c \left( x _ { i } \right) u _ { i } = f \left( x _ { i } \right) , \text { for } 1 \leq i \leq n \\
u _ { 0 } = u _ { n + 1 } = 0
\end{array} \right.$$ Show that for all $i \in \{ 0 , \ldots , n + 1 \}$, we have $$u _ { i } = u \left( x _ { i } \right) = \frac { 1 } { 2 } x _ { i } \left( 1 - x _ { i } \right)$$
We assume (in this question only) that $c ( x ) = 0$ and $f ( x ) = 1$ for all $x \in [ 0,1 ]$. Let $n \in \mathbb{N}^*$, $h = \frac{1}{n+1}$, and $x_i = ih$. We denote by $u$ the exact solution of problem (1):
$$\left\{ \begin{array} { l }
- u ^ { \prime \prime } ( x ) + c ( x ) u ( x ) = f ( x ) , x \in [ 0,1 ] \\
u ( 0 ) = u ( 1 ) = 0
\end{array} \right.$$
and $(u_i)_{0 \leq i \leq n+1}$ the unique family satisfying
$$\left\{ \begin{array} { l }
- \frac { 1 } { h ^ { 2 } } \left( u _ { i + 1 } + u _ { i - 1 } - 2 u _ { i } \right) + c \left( x _ { i } \right) u _ { i } = f \left( x _ { i } \right) , \text { for } 1 \leq i \leq n \\
u _ { 0 } = u _ { n + 1 } = 0
\end{array} \right.$$
Show that for all $i \in \{ 0 , \ldots , n + 1 \}$, we have
$$u _ { i } = u \left( x _ { i } \right) = \frac { 1 } { 2 } x _ { i } \left( 1 - x _ { i } \right)$$