How to get the desired upper bound

[evinda]

Hello! (Wave)

The backward Euler method

We consider a uniform partition such that $[0,T_f]$ and $[a,b]$

$h=\frac{b-a}{N_x+1}, \tau=\frac{T_f}{N_t}$

$x_i=a+ih, i=0,1, \dots, N_x+1$

$t_n=n \tau, n=0,1, \dots, N_t$

$u_t-u_{xx}=0 \\ u(t=0,x)=u_0(x) \\ u(t,a)=0 \forall t \\ u(t,b)=0 \forall t$

$u_t(t,x) \approx \frac{u(t,x)-u(t-\tau,x)}{\tau} \\ u_{xx}(t,x) \approx \frac{u(t,x-h)-2u(t,x)+u(t,x+h)}{h^2}$

$U_i^n \approx u(t^n, x_i)$

$U_i^{n+1}-U_i^n= \mu (U_i^{n+1}- 2 U_i^{n+1}+U_{i-1}^{n+1})$ where $\mu=\frac{\tau}{h^2}$

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

For the Backward Euler method we have:

$|T_i^{n+1}| \leq \frac{\tau}{2} M_{tt}+\frac{h^2}{12}M_{xxxx} \\ M_{tt}=||u_{tt}(t,x)||_{\infty ([a,b] \times [0,T_f])} \\ M_{xxxx}=||u_{xxxx}(t,x)||_{\infty ([a,b] \times [0,T_f])}$I have tried to show the inequality but I don't know how to continue.That's what I have tried:

We have that: $\frac{u(t_{n+1},x_i)-u(t_n,x_i)}{\tau}=\frac{u(t_{n+1},x_{i+1})-2u(t_{n+1},x_n)+u(t_{n+1},x_{i-1})}{h^2}$

$$\frac{u(t_{n+1},x_i)-u(t_n,x_i)}{\tau}=\frac{u(t_n+\tau, x_i)-u(t_n,x_i)}{\tau}=\frac{u(t_n,x_i)+ \tau u_t(t_n, x_i)+\frac{\tau^2}{2} u_{tt}(\rho_n, x_i)-u(t_n, x_i)}{\tau}=u_t(t_n, x_i)+\frac{\tau}{2} u_{tt}(\rho_n, x_i)$$Also,

$$\frac{u(t_{n+1},x_{i+1})-2u(t_{n+1},x_n)+u(t_{n+1},x_{i-1})}{h^2}=\frac{u(t_n+\tau, x_i+h)-2u(t_n+\tau,x_i)+u(t_n+\tau,x_i-h)}{h^2}=u_{xx}(t_n+\tau,x_i)+\frac{h^2}{24} (u_{xxxx}(t_n+\tau, \xi_i)+u_{xxxx}(t_n+\tau, \zeta_i)) , \xi_i \in (x_i, x_i+h), \zeta_i \in (x_i-h,x_i)$$

$T_i^{n+1}=u_t(t_n, x_i)+\frac{\tau}{2} u_{tt}(\rho_n, x_i)-u_{xx}(t_n+\tau, x_i)-\frac{h^2}{24}(u_{xxxx}(t_n+\tau, \xi_i)+u_{xxxx}(t_n+\tau, \zeta_i))$If so, then how could we continue in order to get the desired upper bound? (Thinking)

 

[jvicens]

To continue, we can use the fact that $u_{tt}(t,x)$ and $u_{xxxx}(t,x)$ are both bounded on the interval $[a,b] \times [0,T_f]$. This means that we can find some constants $M_{tt}$ and $M_{xxxx}$ such that $|u_{tt}(t,x)| \leq M_{tt}$ and $|u_{xxxx}(t,x)| \leq M_{xxxx}$ for all $t \in [0,T_f]$ and $x \in [a,b]$.

Using these bounds, we can rewrite our expression for $T_i^{n+1}$ as:

$T_i^{n+1} = u_t(t_n, x_i) + \frac{\tau}{2} u_{tt}(\rho_n, x_i) - u_{xx}(t_n+\tau, x_i) - \frac{h^2}{24} (M_{xxxx}+M_{xxxx})$

From here, we can use the triangle inequality to separate out the terms involving $u_t(t_n, x_i)$ and $u_{xx}(t_n+\tau, x_i)$:

$T_i^{n+1} \leq u_t(t_n, x_i) - u_{xx}(t_n+\tau, x_i) + \frac{\tau}{2} M_{tt} - \frac{h^2}{12} (M_{xxxx}+M_{xxxx})$

Now, we can use the Cauchy-Schwarz inequality to bound the first two terms:

$T_i^{n+1} \leq \sqrt{(u_t(t_n, x_i))^2 + (u_{xx}(t_n+\tau, x_i))^2} + \frac{\tau}{2} M_{tt} - \frac{h^2}{12} (M_{xxxx}+M_{xxxx})$

Finally, we can use the fact that $u_t(t_n, x_i)$ and $u_{xx}(t_n+\tau, x_i)$ are both bounded on the interval $[a,b] \times [0,T_f]$ to get our desired upper bound:

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