Isogeometric Galerkin discretization — Examples

The degree-one Bernstein functions are $1-\xi ,\xi$ and $1-\eta ,\eta$. With all weights one, [F2] gives $W=1$, so the tensor rational basis is the ordinary bilinear basis. Its first component is $\xi$ and its second component is $(1-\xi )\eta +\xi (-\eta )=\eta (1-2\xi )$.

Differentiation gives $DF=\left(\begin{array}{cc}1& 0\\ -2\eta & 1-2\xi \end{array}\right),\det DF=1-2\xi \mathrm{.}$ Thus the Jacobian is zero on $\xi =1\mathrm{/}2$, positive for $\xi <1\mathrm{/}2$, and negative for $\xi >1\mathrm{/}2$. Moreover, $F(1\mathrm{/}2,0)=F(1\mathrm{/}2,1)=(1\mathrm{/}2,0)$ with distinct parameter points. These violate two separate admissibility requirements in [F1] despite all weights being positive.

Put $a=\sqrt{2}$ and $x_0(t)=(1-t{)}^2+at(1-t),y_0(t)=t^2+at(1-t),D(t)=(1-t{)}^2+at(1-t)+t^2\mathrm{.}$ The weighted middle Bernstein coefficient is $2(1\mathrm{/}\sqrt{2})t(1-t)=at(1-t)$, so the angular map is $(X,Y)=(x_0\mathrm{/}D,y_0\mathrm{/}D)$. Direct expansion gives $x_0^2+y_0^2-D^2=-2t^2(1-t{)}^2+a^2{t}^2(1-t{)}^2=0,$ so $X^2+Y^2=1$; also $X,Y\ge 0$ and $D>0$. The tensor-product weights and controls therefore give $F(s,t)=(\rho (s)X(t),\rho (s)Y(t))$.

Expanding the numerator derivatives gives $x_0{y}_0^{\mathrm{\prime }}-y_0{x}_0^{\mathrm{\prime }}=a+2(1-a)t(1-t)=aD(t),$ because $D(t)=1+(a-2)t(1-t)$ and $a(a-2)=2(1-a)$. Consequently $X{Y}^{\mathrm{\prime }}-Y{X}^{\mathrm{\prime }}=a\mathrm{/}D(t)>0$. Since $X^2+Y^2=1$, the angular coordinate $\theta (t)=\arctan (Y\mathrm{/}X)$ has derivative ${\theta }^{\mathrm{\prime }}=X{Y}^{\mathrm{\prime }}-Y{X}^{\mathrm{\prime }}=a\mathrm{/}D(t)>0$ on the interior and extends monotonically from $0$ to $\pi \mathrm{/}2$. Thus $t\mapsto (X,Y)$ traverses the first-quadrant unit-circle arc exactly once. Since $\rho$ increases bijectively from $r_0$ to $1$, $F$ is injective and its image is exactly the stated quarter annulus.

The Jacobian is $\det DF(s,t)={\rho }^{\mathrm{\prime }}(s)\rho (s)(X{Y}^{\mathrm{\prime }}-Y{X}^{\mathrm{\prime }})=(1-r_0)\rho (s)\frac{\sqrt{2}}{D(t)}>0.$ On the closed square, $\rho \ge r_0>0$ and $D$ is continuous and positive, so this determinant has positive minimum and finite maximum; $DF$ is continuous and bounded, and the inverse matrix is bounded by its cofactor formula divided by the positive determinant. The physical inverse first recovers $\rho =\sqrt{x^2+y^2}$ and the angle $\theta =\mathrm{atan2}(y,x)$. This angle is Lipschitz on the annular sector because $\rho \ge r_0$, and the inverse parameter $t(\theta )$ is Lipschitz because ${\theta }^{\mathrm{\prime }}(t)=\sqrt{2}\mathrm{/}D(t)$ has a positive minimum. Together with the linear inverse for $s$, this makes $F^{-1}$ Lipschitz; bounded $DF$ makes $F$ Lipschitz. The quarter annulus has a finite piecewise smooth Lipschitz boundary, so all admissibility hypotheses in [F1] hold. Applying [F3] gives the announced pullback norm equivalence.

The two constants are square integrable on their bounded rectangles, hence $v\in L^2(D)$. Choose $\eta \in C_c^{\mathrm{\infty }}((0,1))$ with ${\int }_0^1\eta (y)dy\mathrm{\ne }0$, and choose $\psi \in C_c^{\mathrm{\infty }}((-1,1))$ with $\psi (0)=1$. For $0<\epsilon <1$, set ${\varphi }_{\epsilon }(x,y)=\psi ((x-1)\mathrm{/}\epsilon )\eta (y)\in C_c^{\mathrm{\infty }}(D)$.

Since $v$ vanishes for $x<1$ and equals one for $x>1$, a change of variable $t=(x-1)\mathrm{/}\epsilon$ gives ${\int }_{D}v{\mathrm{\partial }}_x{\varphi }_{\epsilon }dxdy=\left({\int }_0^1\eta \right){\int }_1^2{\epsilon }^{-1}{\psi }^{\mathrm{\prime }}((x-1)\mathrm{/}\epsilon )dx=-\psi (0){\int }_0^1\eta ,$ which is a fixed nonzero number. If an $L^2$ weak derivative $g={\mathrm{\partial }}_{x}v$ existed, [F2] and Cauchy--Schwarz would make the absolute value at most $\mathrm{\parallel }g{\mathrm{\parallel }}_2\mathrm{\parallel }{\varphi }_{\epsilon }{\mathrm{\parallel }}_2$. But $\mathrm{\parallel }{\varphi }_{\epsilon }{\mathrm{\parallel }}_2=\sqrt{\epsilon }\mathrm{\parallel }\psi {\mathrm{\parallel }}_2\mathrm{\parallel }\eta {\mathrm{\parallel }}_2\to 0$, contradiction. Therefore $v\mathrm{\notin }{H}^1(D)$. Its two constant traces across $x=1$ are 0 and 1, violating [F1] even though the geometry is matching; the sufficient condition of [F3] does not apply.

The degree-two Bernstein functions are $B_0^2(t)=(1-t{)}^2$, $B_1^2(t)=2t(1-t)$, and $B_2^2(t)=t^2$. Thus ${\sum }_{i=0}^2{B}_i^2(t)=1$ and ${\sum }_{i=0}^2(i\mathrm{/}2)B_i^2(t)=t$. For $P_{ij}=(i\mathrm{/}2,j\mathrm{/}2)$, unit weights make the tensor denominator one; summing the first and second control coordinates yields $\xi$ and $\eta$. Hence the elevated representation is still exactly the identity map, with positive weights.

The midpoint $1\mathrm{/}2$ is new and lies strictly inside the single nonempty span of the clamped degree-two Bernstein knot vector. Apply [F3] to each fixed-coordinate control-net slice first in $\xi$, then in $\eta$. Each insertion preserves the entire homogeneous geometry and positive weights, so the final patch is exactly $F(\xi ,\eta )=(\xi ,\eta )$. The resulting degree-two univariate knot vectors have a simple interior knot; [F2] gives $C^1$ continuity in both coordinate directions. This is the stated k-refinement, and no general p/k geometry-preservation claim is needed.