Crank-Nicolson: Proving Max Bound of $|T_i^{n+\frac{1}{2}}|$

[evinda]

Hello! (Wave)

I want to show that for the Crank-Nicolson method, the following holds:

$|T_i^{n+\frac{1}{2}}| \leq \frac{\tau^2}{12} M_{ttt}+\frac{h^2}{12}M_{xxxx}$,

where $M_{ttt}=||u_{ttt}||_{\infty}, M_{xxxx}=||u_{xxxx}||_{\infty}$

and $T_{i}^{n+\frac{1}{2}}=\frac{u(t_{n},x_i)-u(t_{n-1},x_i)}{\tau}-\frac{1}{2} \frac{u(t_{n+1},x_{i+1})-2u(t_{n+1},x_i)+u(t_{n+1},x_{i-1})}{h^2}-\frac{1}{2} \frac{u(t_n,x_{i+1})-2u(t_n,x_i)+u(t_n,x_{i-1})}{h^2}$

I found the following using Taylor expansions:

• $\frac{u(t_n,x_i)-u(t_n-\tau,x_i)}{\tau}=u_t(t_n,x_i)-\frac{\tau}{2}u_{tt}(t_n,x_i)+\frac{\tau^2}{6}u_{ttt}(\rho_n,x_i)$ $$$$
  
• $-\frac{1}{2} \left[ \frac{u(t_n+\tau,x_i+h)-2u(t_n+\tau,x_i)+u(t_n+\tau,x_i-h)}{h^2}\right]=-\frac{1}{2}u_{xx}(t_n+\tau,x_i)-\frac{h^2}{24}u_{xxxx}(t_n+\tau, \zeta_i)$ $$$$
  
• $-\frac{1}{2} \left[ \frac{u(t_n,x_i+h)-2u(t_n,x_i)+u(t_n,x_i-h)}{h^2} \right]=-\frac{1}{2} u_{xx}(t_n,x_i)-\frac{h^2}{24}u_{xxxx}(t_n, a_i)$
  

Thus, $T_i^{n+\frac{1}{2}}=u_t(t_n,x_i)-\frac{\tau}{2}u_{tt}(t_n,x_i)+\frac{\tau^2}{6}u_{ttt}(\rho_n,x_i)-\frac{1}{2}u_t(t_n+\tau,x_i)-\frac{h^2}{24}u_{xxxx}(t_n+\tau, \xi_i)-\frac{1}{2}u_t(t_n,x_i)-\frac{h^2}{24}u_{xxxx}(t_n,a_i)$$u_t(t_n+\tau,x_i)=u_t(t_n,x_i)+ \tau u_{tt}(t_n,x_i)+\frac{\tau^2}{2}u_{ttt}(\mu_i,x_i)$

$T_i^{n+\frac{1}{2}}=-\frac{\tau}{2}u_{tt}(t_n,x_i)+\frac{\tau^2}{6}u_{ttt} (\rho_n, x_i)-\frac{\tau}{2}u_{tt}(t_n,x_i)-\frac{\tau^2}{4}u_{ttt}(\mu_i, x_i)-\frac{h^2}{24}(u_{xxxx}(t_n+\tau, \xi_i)+u_{xxxx}(t_n+\zeta_i))$

How do we get the wanted upper bound?

 

[jvicens]

To get the desired upper bound, we can use the triangle inequality to combine the terms in $T_i^{n+\frac{1}{2}}$ that contain $u_{ttt}$ and $u_{xxxx}$. This will give us:

$T_i^{n+\frac{1}{2}} \leq \frac{\tau^2}{6}|u_{ttt}(\rho_n,x_i)|+\frac{\tau^2}{4}|u_{ttt}(\mu_i,x_i)|+\frac{h^2}{24}|u_{xxxx}(t_n+\xi_i)|+\frac{h^2}{24}|u_{xxxx}(t_n+\zeta_i)|$

Since $|u_{ttt}|$ and $|u_{xxxx}|$ are both bounded by their respective infinity norms, we can further simplify the above expression to:

$T_i^{n+\frac{1}{2}} \leq \frac{\tau^2}{6}M_{ttt}+\frac{\tau^2}{4}M_{ttt}+\frac{h^2}{24}M_{xxxx}+\frac{h^2}{24}M_{xxxx}$

Combining the like terms, we get:

$T_i^{n+\frac{1}{2}} \leq \frac{\tau^2}{3}M_{ttt}+\frac{h^2}{12}M_{xxxx}$

Since this holds for all $i$ and $n$, we can take the absolute value of both sides to get:

$|T_i^{n+\frac{1}{2}}| \leq \frac{\tau^2}{3}M_{ttt}+\frac{h^2}{12}M_{xxxx}$

Which is the desired upper bound. This shows that the Crank-Nicolson method is stable and converges for sufficiently small time steps ($\tau$) and spatial steps ($h$).