Solve Periodic Problem: Find Function $u \in C^2(\mathbb{R})$

[evinda]

Hello! (Wave)

Periodic problem

We are looking for a periodic function $u \in C^2(\mathbb{R})$ with period $(b-a)$

$$-u''+qu=f \text{ where } q,f \text{ periodic functions with period } (b-a) \\ u(a)=u(b) \\ u(x)=u(x+(b-a))$$

$x_i=a+ih \\ h=\frac{b-a}{N+1}$

$\mathbb{R}_{\text{per}}^{N+1}=\{ U=(u_i)_{i \in \mathbb{Z}}: u_i \in \mathbb{R} \text{ and } u_{i+N+1}=u_i, i \in \mathbb{Z}\}$

$-\frac{u_{i-1}-2u_i+u_{i+1}}{h^2}+q(x_i) u_i =f(x_i), i=0,1, \dots, N (\star)$

$u_{-1}=u_N \\ u_{N+1}=u_0$

$U=\begin{bmatrix}
u_0\\
u_1\\
\dots\\
\dots\\
u_N
\end{bmatrix}$

$i=0 \overset{\star}{\Rightarrow} -\frac{u_N-2u_0+u_1}{h^2}+q(x_0) u_0=f(x_0)$

$\dots$

$i=1 \overset{\star}{\Rightarrow} -\frac{u_{N-1}-2u_N+u_0}{h^2}+q(x_N) u_N=f(x_N)$
Could you explain to me why we want that $u_{i+N+1}=u_i, i \in \mathbb{Z}$ ?
Do we suppose that $b=N+1$, $a=0$ ? (Thinking)

 

[I like Serena]

Could you explain to me why we want that $u_{i+N+1}=u_i, i \in \mathbb{Z}$ ?
Do we suppose that $b=N+1$, $a=0$ ? (Thinking)

Hey evinda! (Smile)

That's because:
$$u_{i+N+1} = u(x_{i+N+1}) = u(a+(i+N+1)h) = u((a+ih) + (N+1)h) = u(x_i+(b-a)) = u(x_i) = u_i$$
(Mmm)

And no, we wouldn't suppose that $b=N+1$, $a=0$. (Shake)

 

[evinda]

That's because:
$$u_{i+N+1} = u(x_{i+N+1}) = u(a+(i+N+1)h) = u((a+ih) + (N+1)h) = u(x_i+(b-a)) = u(x_i) = u_i$$
(Mmm)

And no, we wouldn't suppose that $b=N+1$, $a=0$. (Shake)

I understand... Thanks a lot! (Smile)