How is the Finite Difference Method Applied in Wave Approximation?

[evinda]

Hello! (Wave)

We consider the finite difference method for the approximation

$\left\{\begin{matrix}
-u''(x)+q(x)u(x)=f(x)\\
u'(a)=u'(b)=0
\end{matrix}\right.$

and let $K$ be the $(N+2) \times (N+2)$ matrix of the method. Let $v \in \mathbb{R}^{N+2}, v=\begin{pmatrix}
v_0\\
v_1\\
\dots\\
\dots\\
\dots\\
v_{N+1}
\end{pmatrix}$If $(Ku)_i \leq 0 \forall i=0,1, \dots, N-1$ then $\max_{1 \leq i \leq N} \{ v_i \} \leq \max \{ v_0, v_{N+1},0\} (\star)$Proof:

We suppose that $(\star)$ does not hold.

Then $\exists n \in \{ 1,2, \dots, N\}$ such that $v_n= \max_{1 \leq i \leq N} v_i >0$, $v_n>v_0$ and $v_n > v_{N+1}$.Now $(Ku)_n=u_{n-1}+(2+h^2 q(x_n))v_n-v_{n+1} \leq 0 \Rightarrow (2+h^2 q(x_n)) v_n \leq v_{n-1}+v_{n+1}$

but $v_n \geq v_{n-1}$ and $v_n \geq v_{n+1} \Rightarrow 2 v_n \geq v_{n-1}+v_{n+1}$

Thus, $(2+h^2 q(x_n))v_n \leq v_{n-1}+v_{n+1} \leq 2 v_n$This can only hold if $q(x_n)=0$ and $v_n=v_{n-1}=v_{n+1} \Rightarrow \dots \Rightarrow v_i=v_n \forall i$, contradiction.
Could you explain me why it holds that $(Ku)_n=u_{n-1}+(2+h^2 q(x_n))v_n-v_{n+1}$ ? (Thinking)

 

[jvicens]

Sure! Let's break down the finite difference method for this problem. We start with the second derivative of $u(x)$, which is approximated by $\frac{u_{n-1}-2u_n+u_{n+1}}{h^2}$, where $h$ is the step size. Then, we plug this into our original equation and rearrange to get:

$u_{n-1}+(2+h^2q(x_n))u_n-u_{n+1}=h^2f(x_n)$

Next, we need to account for the boundary conditions $u'(a)=u'(b)=0$. This can be done by setting $u_0=u_{N+1}=0$. Now, we can write this equation in matrix form as:

$\begin{pmatrix}
2+h^2q(x_1) & -1 & 0 & \dots & 0 \\
-1 & 2+h^2q(x_2) & -1 & \dots & 0 \\
0 & -1 & 2+h^2q(x_3) & \dots & 0 \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & -1 & 2+h^2q(x_N)
\end{pmatrix} \begin{pmatrix}
u_1 \\
u_2 \\
u_3 \\
\dots \\
u_N
\end{pmatrix} = \begin{pmatrix}
h^2f(x_1) \\
h^2f(x_2) \\
h^2f(x_3) \\
\dots \\
h^2f(x_N)
\end{pmatrix}$

So, in this matrix, the $i$th row corresponds to the $i$th equation in the original system. Now, if we multiply this matrix by the vector $v$, we get:

$Kv = \begin{pmatrix}
u_0 \\
u_1 \\
u_2 \\
\dots \\
u_N \\
u_{N+1}
\end{pmatrix} = \begin{pmatrix}
v_0 \\
v_1 \\
v_2 \\
\dots \\
v_N \\
v_{N+1}
\end{pmatrix}$

And since $u_0=u_{N+1}=0$, we can ignore the first and last equations