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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. $t_i=i/n$). Let
$$ S=\{s(x)\in C^1[0,1]\mid s|_{[t_i,t_{i+1}]} \text{ is a cubic polynomial, } s(0)=s(1)=0 \} $$

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

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

I did some numerical experiments to compare the condition numbers of their stiffness matrices finding that the one of the B-spline is always smaller than the one of the Hermite spline.

My problem is: how to prove this?


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

Here is one numerical example when $a=1$, $b=0$. The stiffness matrix of the 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 the 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 the Hermite basis. However, the arrangement of the basis does not affect the condition number.)

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