Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Background

Consider the one dimensional second order elliptic PDE, $$ \left\{\!\! \begin{aligned} & -(a(x)u'(x))'+b(x)u(x)=f(x)\qquad x\in[0,1]\\ & u(0)=u(1)=0 \end{aligned} \right. $$ where $a(x)\ge a_0>0$ and $b(x)\ge0$.
Its weak form is $$ \alpha(u,v)=(f,v),\quad\forall v\in H_0^1([0,1]) $$ where $$ \begin{align*} \alpha(u,v) & =\int_0^1 a(x)u'(x)v'(x)\,dx+\int_0^1b(x)u(x)v(x)\,dx\\ (f,v) & =\int_0^1 f(x)v(x)\,dx\qquad\forall u,v\in H_0^1([0,1]) \end{align*} $$

Then we choose the finite element space. Suppose $$ 0=t_0<t_1<t_2<\cdots<t_n=1 $$ are equally spaced knots. (i.e. $t_i=i/n$)
$$ S=\{s(x)\in C^1[0,1]\mid s|_{[t_i,t_{i+1}]} \text{ is a cubic polynomial, } s(0)=s(1)=0 \} $$

We will use two different basis for this space (whose dimension is $2n$).

First, we use B-spline with double knots. Say B-spline defined on knots $$(0,0,0,0,\frac{1}{n},\frac{1}{n},\frac{2}{n},\frac{2}{n},\ldots,1,1,1,1)$$

Then, we use Hermite finite element, which is defined as $$ f_i(x)= \begin{cases} \frac{(x-t_{i-1})^2}{(t_i-t_{i-1})^3}[2(t_i-x)+(t_i-t_{i-1})] & t_{i-1}\le x\le t_i\\ \frac{(t_{i+1}-x)^2}{(t_{i+1}-t_i)^3}[2(t_{i+1}-x)-(t_{i+1}-t_i)] & t_{i}\le x\le t_{i+1}\\ 0 & \text{otherwise} \end{cases} $$ $$ g_i(x)= \begin{cases} \frac{(x-t_{i-1})^2(x-t_i)}{(t_i-t_{i-1})^2} & t_{i-1}\le x\le t_i\\ \frac{(x-t_{i+1})^2(x-t_i)}{(t_{i+1}-t_i)^2}& t_{i}\le x\le t_{i+1}\\ 0 & \text{otherwise} \end{cases} $$

Then I did some numeric experiment to compare the condition numbers of their stiffness matrix finding that the one of B-spline is always smaller than the one of Hermite spline.


My problem is: how to prove this?

EDIT:

Or if the condition number is difficult to compare, is there other way to show that B-spline is numerically better than Hermite spline?

Here is one numerical example when $a=1$, $b=0$. The stiffness matrix of B-spline finite element is: $$ \left( \begin{array}{cccccccccc} 6 & 0 & -\frac{3}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 0 & -\frac{15}{8} & -\frac{3}{8} & 0 & 0 & 0 & 0 & 0 \\ -\frac{3}{2} & 0 & 6 & -\frac{15}{8} & -\frac{15}{8} & 0 & 0 & 0 & 0 & 0 \\ 0 & -\frac{15}{8} & -\frac{15}{8} & 6 & 0 & -\frac{15}{8} & -\frac{3}{8} & 0 & 0 & 0 \\ 0 & -\frac{3}{8} & -\frac{15}{8} & 0 & 6 & -\frac{15}{8} & -\frac{15}{8} & 0 & 0 & 0 \\ 0 & 0 & 0 & -\frac{15}{8} & -\frac{15}{8} & 6 & 0 & -\frac{15}{8} & -\frac{3}{8} & 0 \\ 0 & 0 & 0 & -\frac{3}{8} & -\frac{15}{8} & 0 & 6 & -\frac{15}{8} & -\frac{15}{8} & 0 \\ 0 & 0 & 0 & 0 & 0 & -\frac{15}{8} & -\frac{15}{8} & 6 & 0 & -\frac{3}{2} \\ 0 & 0 & 0 & 0 & 0 & -\frac{3}{8} & -\frac{15}{8} & 0 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -\frac{3}{2} & 0 & 6 \end{array} \right) $$ whose condition number (infinity norm) is about 15.6138.

And the stiffness matrix of Hermite finite element is: $$ \left( \begin{array}{cccccccccc} 12 & -6 & 0 & 0 & -\frac{1}{10} & 0 & \frac{1}{10} & 0 & 0 & 0 \\ -6 & 12 & -6 & 0 & 0 & -\frac{1}{10} & 0 & \frac{1}{10} & 0 & 0 \\ 0 & -6 & 12 & -6 & 0 & 0 & -\frac{1}{10} & 0 & \frac{1}{10} & 0 \\ 0 & 0 & -6 & 12 & 0 & 0 & 0 & -\frac{1}{10} & 0 & \frac{1}{10} \\ -\frac{1}{10} & 0 & 0 & 0 & \frac{2}{75} & -\frac{1}{150} & 0 & 0 & 0 & 0 \\ 0 & -\frac{1}{10} & 0 & 0 & -\frac{1}{150} & \frac{4}{75} & -\frac{1}{150} & 0 & 0 & 0 \\ \frac{1}{10} & 0 & -\frac{1}{10} & 0 & 0 & -\frac{1}{150} & \frac{4}{75} & -\frac{1}{150} & 0 & 0 \\ 0 & \frac{1}{10} & 0 & -\frac{1}{10} & 0 & 0 & -\frac{1}{150} & \frac{4}{75} & -\frac{1}{150} & 0 \\ 0 & 0 & \frac{1}{10} & 0 & 0 & 0 & 0 & -\frac{1}{150} & \frac{4}{75} & -\frac{1}{150} \\ 0 & 0 & 0 & \frac{1}{10} & 0 & 0 & 0 & 0 & -\frac{1}{150} & \frac{2}{75} \end{array} \right) $$ whose condition number is about 1270.08.

(Well, the band width of the latter may be relatively large due to the arrangement of Hermite basis. However, the arrangement of basis does not affect the condition number. )

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.