13.4 GSL - álgebra linear

O gsl resolve problemas em álgebra linear por diversos métodos. Aqui nós vamos trabalhar apenas com a decomposição QR e sistemas tridiagonais. O pacote para álgebra linear é gsl_linalg.h.

• int gsl_linalg_QR_decomp(gsl_matrix * A, gsl_vector * tau): calcula a decomposição QR da matriz A.
• int gsl_linalg_QR_solve(const gsl_matrix * QR, const gsl_vector * tau, const gsl_vector * b, gsl_vector * x): resolve o $Ax=b$ usando a decomposição QR.

O comando

• int gsl_linalg_QR_lssolve(const gsl_matrix * QR, const gsl_vector * tau, const gsl_vector * b, gsl_vector * x, gsl_vector * residual) resolve sistema impossível (quando possui mais linhas independes que colunas) pelo critério de mínimos quadrados.

Supondo que a reta tenha equação $y=ax+b$, temos $$\begin{array}{rcll}b& =& 1& \text{(13.1)}\\ a+b& =& 0.5& \text{(13.2)}\\ 2a+b& =& -0.2& \text{(13.3)}\\ 3a+b& =& -0.7& \text{(13.4)}\end{array}$$

um sistema que pode ser resolvido pelo critério dos mínimos quadrados. Na forma matricial, temos $$\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 1\hfill \\ \hfill \hfill \end{array}\right]\left[\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0.5\hfill \\ \hfill -0.2\hfill \\ \hfill -0.7\hfill \end{array}\right].$$

Vamos resolver usando a decomposição QR.

A função

• int gsl_linalg_solve_tridiag(const gsl_vector * diag, const gsl_vector * e, const gsl_vector * f, const gsl_vector * b, gsl_vector * x) resolve o sistema tridiagonal $$\left[\begin{array}{cccccc}\hfill d_0\hfill & \hfill e_0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill f_0\hfill & \hfill d_1\hfill & \hfill e_1\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill f_1\hfill & \hfill d_2\hfill & \hfill e_2\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill & \hfill f_{n-2}\hfill & \hfill d_{n-1}\hfill & \hfill e_{n-1}\hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill f_{n-1}\hfill & \hfill d_n\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill x_3\hfill \\ \hfill ⋮\hfill \\ \hfill x_{n-1}\hfill \\ \hfill x_n\hfill \end{array}\right]=\left[\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill b_3\hfill \\ \hfill ⋮\hfill \\ \hfill b_{n-1}\hfill \\ \hfill b_n\hfill \end{array}\right].$$

O PVC do problema $$\left\{\begin{array}{c}-u_{xx}+u_x=200e^{-{\left(x-1\right)}^2},0<x<1.\hfill \\ 15u\left(0\right)+u^{\prime }\left(0\right)=500\hfill \\ 10u\left(1\right)+u^{\prime }\left(1\right)=1\hfill \end{array}\right.$$

possui a seguinte versão discreta

onde $x_i=h\left(i-1\right)$, $i=1,2,...,N$, $h=\frac{1}{N-1}$.