FEM, CFD, CAE, and all that.

## Tag: linear algebra

### Store a polynomial in non-symbolic language

Sometimes one wants to store a polynomial in a non-symbolic language, such as Fotran or C. A easy way to achieve this is to take advantage of the isomorphism between polynomial space $P(n)$ and space $R^{n+1}$. Namely, we can build an array with entries as the coefficients of the polynomial to be stored. Furthermore, considering these coefficients are just the coordinates of the polynomial in the space spanned by the canonical basis $1$,$x$,$x^2$,$\dots$, we can express the polynomial using any basis expansion.

### Classical iterative algorithms

For this topic, many excellent textbooks are available. Most of the modern numerical PDE solvers rely on one or another iterative algorithm taking care of the linear system given by discretization ( I know very little of direct methods). In this scenario, one has linear system

$Ax=b$

and tries to come up with some iterative scheme

$x^{(k+1)}=Tx^{(k)}+c$

with

$\lim_{k\to\infty}{x^{(k)}}=x^{\ast}$

meaning the convergence to the “real” solution, or, to be exact, the solution by direct solver if we disregard the round-off error. By this setup, one can see the solution is the fixed point of operator $T$. So the convergence of the algorithm is equivalent to two proposals: the spectral radius $\rho(A)<1$, and there exists operator norm induced by some norm of the linear space to which $x$ belongs that this norm of $T$ is less than $1$. The convergence speed is measured by average rate of convergence of $A$ (for $m$ iterations) defined by

$R(A^m)=-\log[\|A^m\|^{1/m}]=-\log[\|A^m\|]/m$

which comes from the fact that $\lim_{k\to\infty}(\|T^k\|^{1/k})=\rho(T)$. In order to put the original problem into this fixed point setup, $A$ is splitted:

$A=M-N\longrightarrow x^{(k+1)}=M^{-1}Nx^k+M^{-1}b$

If $A$ is splitted according to entries’ positions as $A=L+D+U$, one has the classical algorithms:

Jacobi: $M=D,N=-(L+U)$

Gauss-Seidel: $M=D+L,N=-U$

SOR: $x_i^{(k+1)}=\omega \bar{x}_i^{(k+1)}+(1-\omega)x_i^{(k)}$

The last one is in the component wise form in which $\bar{x}^{(k+1)}$ is the Gauss-Seidel result, and the relaxation parameter is between zero and two (even though the name “overrelaxation” comes when it is greater than one). Each of those algorithms has direct(but different in implementation) analogical block version. All these methods are stationary, i.e. in each iteration $T$ is unchanged. If this is not the case, we have unstationary methods, in which after each iteration information is re-gathered in order to determine a new $T$ for next iteration. When $A$ is a SPD matrix, a large family of unstationary methods can be formed by the solving a minimization problem(minimal residual method (MINRES)). Classical CG method I wrote about before belongs to this family. A modern member of this family is GMRES by Saad and Schultz in 1986.

### review of CG method:IV

I have shown that in method of conjuate direction, the searching directions are in some sense orthogonal to each other:

$d_i^TAd_j=0,\quad i\neq j$

Which makes more sense when the whole thing is considered under a new coordinate system, a system obtained by impose tranform of $L$, in which $L$ is the upper-triangle matrix from $A$‘s Cholesky decomposition.

$d_i^TAd_j=0\Longleftrightarrow (Ld_i)^T(Ld_j)=0\Longleftrightarrow Ld_i\perp Ld_j$

Similar conclusion includes that $d_i$ is $A$-orthogonal to $e_{i+1}$, resulting in that $d_i$ is orthogonal to $r_{i+1}$ the residual:

$d_i^Tr_{i+1}=0\Rightarrow d_i^Tr_j=0,\quad i\neq j$
This condition actually implies that one can use the residual $r_i$ processed by  Gram-Schimdt conjugation to obtain the direction $d_i$, and this is the conjugate gradient (CG) method. The advantage of CG is that, instead of finding a set of directions $d_i$ at the beginning, $r_i$ is obtained along with the iteration, because the process of Gram-Schimdt conjugation is “triagonal”, i.e., one variable is determined by its predecessors. In particular, starting from $r_0=b-Ax_0$ (the first step is the same as steepest descent method), the following searching directions $d_i$ is determined by

$d_i=r_i+\sum_{k=0}^{i-1}\alpha_{ik}d_k$

Coefficient $\alpha_{ik}$ is obtained by conjugate condition:

$d_i^TAd_j=0\Longrightarrow r_i^TAd_j+\sum_{k=0}^{i-1}\alpha_{ik}d_k^TAd_j=r_i^TAd_j+\alpha_{ij}d_j^TAd_j=0$

Then we have $\alpha_{ij}=-r_i^TAd_j/(d_j^TAd_j)$

In order to simplify this expression, remember that during this process of generating $d_i$ from $r_i$, for any $i$,

$\text{span}\{d_0,d_1,\cdots,d_i\}=\text{span}\{r_0,r_1,\cdots,r_i\}$

so $d_i^Tr_j=0\Longrightarrow r_i^Tr_j=0,\quad i\neq j$

The Gram-Schimdt process also indicates that $d_i^Tr_i=r_i^Tr_i$ (prove it). The above two identities can be used to simplify $\alpha_{ij}$, and prove that

$\alpha_{ij}=r_i^Tr_i/(l_{i-1}d_{i-1}^TAd_{i-1}),\quad j=i-1$

$\alpha_{ij}=0,\quad j

and eventually

$\alpha_{i,i-1}=r_i^Tr_i/(r_{i-1}^Tr_{i-1})$

Finally, a complete iteration, when $d_i, x_i,r_i$ is known, consists of

$l_i=r_i^Tr_i/d_i^TAd_i\longrightarrow x_{i+1}=x_i+l_id_i\longrightarrow r_{i+1}=r_i-l_iAd_i$

$\alpha_{i+1}\equiv\alpha_{i+1,i}=r_{i+1}^Tr_{i+1}/(r_i^Tr_i)\longrightarrow d_{i+1}=r_{i+1}+\alpha_{i+1}d_i$

### review of CG method:III

This Mathematica script generates the following picture indicating a typical behavior of steepest descent method: the zigzag in two perpendicular directions. In fact, in 2D this is the only case because one has only two choices, determined by starting direction, as I showed at the beginning. It is natural to ask whether one can use only two step with single change of direction to reach the minimization point. This introduces conjugate direction method.

Assume $x\in R^n$ and the iteration starts with direction $d_0$. The method of conjugate direction tries to reach a point along $d_i$ with condition that in the rest of iteration descent along $\pm d_o$ direction would never happen again, so does for any other following steps. To make sure this happen, we can require that the iterations after $d_i$ always are performed on the subspace perpendicular to $d_i$. Remembering that exact solution $x^{\ast}$ should always be in the subspace in which the coming iterations are to be performed, we come up this scheme:

$x_{i+1}=x_i+l_id_i, \quad d_i\perp (x_{i+1}-x^{\ast})=e_{i+1}=e_i+l_id_i\Longleftrightarrow d_i^T(e_i+l_id_i)=0$

If we view the above as two equations with three unknowns ($x_{i+1},l_i,x^{\ast}$), we know that it’s not going to work. However, since $A$ is SPD matrix, by Cholesky decomposition we have $A=L^TL$ in which $L$ is an upper-triagonal matrix. Then if we consider the orthogonality between $Ld_i$ and $Le_{i+1}$, instead of $d_i$ and $e_{i+1}$, we have

$Ld_i\perp Le_{i+1}\Rightarrow (Ld_i)^T Le_{i+1}=0\Rightarrow d_i^TL^TLe_{i+1}=d_i^TAe_{i+1}=d_i^TA(e_i+l_id_i)=d_i^T(Ax_i-b+Al_id_i)=d_i^T(-r_i+Al_id_i)=0$

Finally we have

$l_i=d_i^Tr_i/d_i^TAd_i$

Of course this can be obtained by using the $\frac{df(x_{i+1})}{d l_i}=0$ requirement as well, but derivation I used here gives better geometric explanation. In order to utilize the fact that we do not know $e_i$ but we know $r_i=b-Ax_i=-Ae_i$, we impose the orthogonality requirement $d_i\perp e_{i+1}$ in the space transformed by $L$. By now one can see that to consider $r_i$ as the effect of $A$ applied on $e_i$ can often serve the similar purpose of $e_i$ the error itself, that’s why sometimes we call $r_i$ a similar name: residual.

The orthogonalilty condition $Ld_i\perp Le_{i+1}\Longleftrightarrow d_i^TAe_{i+1}$ is also called $A$-orthogonality. And by $d_i$ is $A$-orthogonal to the following $e_j$ one can easily see that $d_i$ is $A$-orthogonal to the following $d_j$. Since now we have the formular for $l_i$, with a set of $A$-orthogonal $d_j, j=0,1,\cdots,n-1$ we will be done, and this set can be found by Gram-Schimdt conjugation, a $A$-orthogonal version of Gram-Schimdt orthogonazation process. With those tools in hand, it is not hard to show that this conjugate direction method, whose name comes from that the directions $d_j$ are $A$-orthogonal, or, $A$-conjugate to each other, can indeed reach $x^{\ast}$ within $n$ steps. In fact,  the goal in each interation in conjugate direction method achieved is to eliminate one dimension of the space in which the solution lies and confine the rest of search to a lower dimension hyperplane.

### review of CG method:II

Now let’s examine the convergence property of steepest descent method. For this purpose, define the error against exact solution $x^{\ast}$ at each iteration step

$e_i\equiv x_i-x^{\ast}$

By iteration scheme there is

$e_{i+1}=e_i+l_ir_i$

Since at every step matrix $A$ would be multiplied, and the iterative property of a matrix is largely determined by its eigenvalues, one’s first reaction should be look at the decomposition of the vector space under discussion into its standard forms by

$e_i=\sum_j{\xi_{(i)j}v_j}$

where $\{v_j\}$ is the orthonormal basis of eigenvector space of $A$. By this defitino, there are

$r_i=b-Ax_i=Ax^{\ast}-Ax_i=-Ae_i$

$e_{i+1}=e_i+l_ir_i=e_i+\frac{r_i^Tr_i}{r_i^TAr_i}r_i=e_i+\frac{\sum_j\xi_{(i)j}^2\lambda_j^2}{\sum_j\xi_{(i)j}^2\lambda_j^3}r_i$

In which $\lambda_j$ is eigenvalue corresponding to $v_j$. One can see if the normal/spectral condition number $\kappa(A)\equiv\lambda_{\max}(A)/\lambda_{\min}(A)=0$, i.e. all the eigenvalues of $A$ are same, the error at next step immediately drops to zero. Geometricly this simply means contours are spheres and the steepest descent direction is toward the centre. Generally, there is

$e_{i+1}=(I-l_iA)e_i\Rightarrow \|e_{i+1}\|_A^2=\|e_i\|_A^2\omega_i^2$

where the norm is energy norm induced by $A$

$\|x\|_A=\sqrt{\frac{1}{2}x^TAx}$

and $\omega_i$ (subscript emphasizes the dependence on each step) is determined by spectral condition number $\kappa(A)$ and slope $\mu_i=\xi_{(i)\max}/\xi_{(i)\min}$:

$\omega_i^2=1-\frac{(\kappa^2+\mu_i^2)^2}{(\kappa+\mu_i^2)(\kappa^3+\mu_i^2)}$

Picture below shows the dependence of $\omega$ to two variables. One can see that the slowest convergence happens when both condition number and slope are large. $A$ is referred to as ill-conditioned when $\kappa$ is large.