Metoda numerica

Metoda numerica propriu-zisa consta in reiterarea urmatorilor pasi:

1. calculam solutia numerica si derivata sa in vecinatatea lui $x_0^{(k)}$ (in acest paragraf, indicele $\omega$ nu este esential):
   $$u\left( x_0^{(k)}+\delta _k\right) =\sum\limits_{n=0}^{N_k}c_n^{(k)}(\delta _k)^n=c_0^{(k+1)}$$

   $$u^{\prime }\left( x_0^{(k)}+\delta _k\right) =\sum\limits_{n=1}^{... ... _k)^{n-1}=\sum\limits_{n=0}^{N_k-1}(n+1)c_{n+1}^{(k)}(\delta _k)^n=c_1^{(k+1)}$$ (5.3.19)

   
   

2. calculam (5.2.15) si (5.2.16) in punctele curente $x_0^{(k)}$, utilizand $u(x_0^{(k)})$ si $u^{\prime }(x_0^{(k)})$.

3. utilizand (5.2.9) si (5.3.19) calculam solutia in $x_0^{(k+1)}=x_0^{(k)}+\delta _k$, si derivata sa.

4. calculam in $x_0^{(k+1)}$ marimile (5.2.15) si (5.2.16), valoarea lui (5.2.17), si apoi verificam conditia (5.2.18) pentru o valoare fixata a lui $\varepsilon$, ajustand valorile pentru $N_k$ si $\delta _k$ daca e necesar

   si iteram pana la parcurgerea intregului interval de studiu $(x_0;x_F)$.

Eroarea de trunchiere se poate estima usor in cazul unui pas constant $\delta$ si al unei trunchieri uniforme (dupa $N$ termeni). Avem ( $\xi _k\in (0,\delta )$):

$$c_0^{(k+1)}=\sum\limits_{n=0}^Nc_n^{(k)}(\delta _k)^n+\frac 1{(N+1)!}u^{\prime (N+1)}\left( x_0^{(k)}+\xi _k\right) \delta ^{N+1}$$

$$c_1^{(k+1)}=\sum\limits_{n=0}^{N-1}(n+1)c_{n+1}^{(k)}(\delta _k)^n+\frac 1{N!}u^{\prime (N+1)}\left( x_0^{(k)}+\xi _k\right) \delta ^N$$

care se poate scrie, utilizand relatiile de recurenta (5.2.9) ca:

$$c_0^{(k+1)}=f_kc_0^{(k)}+g_kc_1^{(k)}+\frac 1{(N+1)!}u^{\prime (N+1)}\left( x_0^{(k)}+\xi _k\right) \delta ^{N+1}$$

$$c_1^{(k+1)}=m_kc_0^{(k)}+n_kc_1^{(k)}+\frac 1{N!}u^{\prime (N+1)}\left( x_0^{(k)}+\xi _k\right) \delta ^N$$

unde marimile $f_k$, $g_k$, $m_k$ si $n_k$ pot fi estimate direct utilizand metoda iterativa de mai sus. Cu notatiile:

$$C_k=\left( \begin{array}{l} c_0^{(k)} \\ c_1^{(k)} \end{array}\right)$$

$$\Delta =\left( \begin{array}{l} \frac \delta {N+1} \\ 1 \end{array}\right)$$

$$\alpha _k=u^{\prime (N+1)}\left( x_0^{(k)}+\xi _k\right)$$

$$\hat{A}_k=\left( \begin{array}{ll} f_k & g_k \\ m_k & n_k \end{array}\right)$$

se obtine relatia de recurenta:

$$C_{k+1}=\hat{A}_kC_k+\frac 1{N!}\alpha _k\delta ^N\Delta$$

a carei solutie este:

$$C_{k+1}=\hat{A}_k\cdot \cdot \cdot \hat{A}_0C_0+\frac 1{N!}\delta... ...k-i}\hat{A}_k\cdot \cdot \cdot \hat{A} _{k-i+1}+\alpha _k\hat{I}\right) \Delta$$

Desigur, primul termen corespunde solutiei numerice trunchiate dupa $N$ termeni, in timp ce al doilea este eroarea de trunchiere. Pentru o estimare calitativa putem lua valori egale pentru $\alpha _k(=\alpha )$:

$$E_t=\frac 1{N!}\delta ^N\alpha \left( \sum\limits_{i=0}^{k-1}\prod\limits_{j=0}^i\hat{A}_{k-j}+\hat{I}\right) \Delta$$