A quadratic spline collocation method for the Dirichlet biharmonic problem

Abstract

A new method based on quadratic spline collocation is formulated for the solution of the Dirichlet biharmonic problem on the unit square rewritten as a coupled system of two second-order partial differential equations. This method involves the solution of an auxiliary biharmonic problem using fast Fourier transforms and the solution of a nonsymmetric Schur complement system using preconditioned BICGSTAB, at a total cost of $N^{2} \log N$ on an N × N uniform partition of the unit square. The results of numerical experiments demonstrate the optimality of the global accuracy of the method and also superconvergence results, in particular, third-order accuracy in the $L^{\infty }$ norm of the solution and its fourth-order accuracy at the partition nodes and the collocation points.

Appendix

We shall prove that the matrix P defined in (4.34) is symmetric and positive definite. It follows from (3.15) and (2.10) that

$$ Z^{T} B^{D} Z=\frac{1}{8}{\Lambda} +I_{N} \equiv {\Lambda}_{1}. $$

(A.1)

Using (2.8), we have

$$ -4 \leq \lambda_{i} < 0, \quad i = 1,\ldots,N, $$

(A.2)

which yields

$$ \frac{1}{2} \leq \frac{1}{8}\lambda_{i}+ 1, \quad i = 1,\ldots,N. $$

(A.3)

It follows from (A.1), (2.8), and (A.3) that B^D is symmetric and that the diagonal entries of Λ₁ are positive. Consequently, B^D is positive definite. Since B^D is symmetric and positive definite, it follows from (4.37) and (4.27) that − P₂₂, appearing on the right-hand side of (4.34), is symmetric and positive definite. It thus remains to show that $P_{21}P_{11}^{-1}P_{12}$, appearing on the right-hand side of (4.34), is symmetric and positive definite. To this end, we first derive a formula for $P_{21}P_{11}^{-1}P_{12}$. It follows from (4.35)–(4.37) that

$$ P_{11}= \left[ \begin{array}{cc} P_{11}^{1} & P_{11}^{2} \\ 0 & P_{11}^{1} \end{array} \right], \quad P_{12}= \left[ \begin{array}{c} P_{12}^{1} \\ P_{12}^{2} \end{array} \right], \quad P_{21}= \left[ \begin{array}{cc} P_{21}^{1} & P_{21}^{2} \end{array} \right], $$

(A.4)

where

$$ P_{11}^{1}=B^{D} \otimes A^{D} + A^{D}\otimes \left( B^{D} - \displaystyle\frac{h^{2}}{12}A^{D}\right), \quad P_{11}^{2}=-B^{D} \otimes B^{D}, $$

(A.5)

$$ P_{12}^{1}=-E \otimes B^{D}, \quad P_{12}^{2}=E \otimes A^{D} + F\otimes \left( B^{D} - \displaystyle\frac{h^{2}}{12}A^{D}\right), $$

(A.6)

$$ P_{21}^{1}=C^{D}\otimes B^{D}, \quad P_{21}^{2}=\frac{h^{2}}{12} P_{21}^{1}. $$

(A.7)

Using (A.5), (3.15), (2.10), we have

$$ (Z^{T}\otimes Z^{T})P_{11}^{1}(Z\otimes Z) =h^{-2}\left[\left( \frac{1}{8}{\Lambda} +I_{N}\right)\otimes {\Lambda} +{\Lambda} \otimes \left( \frac{1}{24}{\Lambda} +I_{N}\right)\right]\equiv {\Lambda}_{2}. $$

(A.8)

Using (A.2), we have that

$$ \frac{5}{6} \leq \frac{1}{24}\lambda_{i}+ 1, \quad i = 1,\ldots,N. $$

(A.9)

It follows from (A.8), (2.8), (A.2), (A.3), and (A.9) that P₁₁ is symmetric and that diagonal entries of Λ₂ are negative. Thus, $P_{11}^{1}$ is negative definite and hence nonsingular. With

$$ R=(P_{11}^{1})^{-1}P_{11}^{2} (P_{11}^{1})^{-1}, $$

(A.10)

we have

$$P_{11}^{-1}= \left[ \begin{array}{cc} (P_{11}^{1})^{-1} & -R \\ 0 & (P_{11}^{1})^{-1} \end{array} \right]. $$

The last equation and (A.4) yield

$$ P_{21} P_{11}^{-1} P_{12} = P_{21}^{1} (P_{11}^{1})^{-1} P_{12}^{1} - P_{21}^{1} R P_{12}^{2} + P_{21}^{2}(P_{11}^{1})^{-1} P_{12}^{2}. $$

(A.11)

We next study the terms on the right-hand side in (A.11). Using (A.7), (A.6), (4.26), and (4.27), we obtain

$$ P_{21}^{1} (P_{11}^{1})^{-1} P_{12}^{1}= -\frac{1}{4h} \left( [\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes B^{D}\right) (P_{11}^{1})^{-1} \left( [\textbf{e}_{1}|\textbf{e}_{N}]\otimes B^{D}\right). $$

(A.12)

Since B^D and $P_{11}^{1}$ are symmetric, (A.12) implies that $P_{21}^{1} (P_{11}^{1})^{-1} P_{12}^{1}$ is also symmetric. From (A.7), (A.6), (4.26), (4.27), and (3.15) it follows that

$$ P_{21}^{1} =\frac{2}{h}[\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes \left( \frac{1}{8}T+I_{N}\right) =\frac{1}{4h}[\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T +\frac{2}{h}[\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}, $$

(A.13)

$$ P_{12}^{2} = \frac{1}{8h^{2}} [\textbf{e}_{1}|\textbf{e}_{N}]\otimes T+ \frac{1}{h^{2}}[\textbf{e}_{1}|\textbf{e}_{N}]\otimes \left( \!\frac{1}{24}T + I_{N}\!\right) \! = \frac{1}{6h^{2}} [\textbf{e}_{1}|\textbf{e}_{N}]\otimes T+ \frac{1}{h^{2}}[\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}. $$

(A.14)

Using the last two equations, we have

$$\begin{array}{@{}rcl@{}} P_{21}^{1} R P_{12}^{2} &=& \frac{1}{24h^{3}}([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\ &&+\frac{1}{4h^{3}} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}) \\ &&+ \frac{1}{3h^{3}} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\ &&+ \frac{2}{h^{3}} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}). \end{array} $$

(A.15)

Since B^D is symmetric, it follows from (A.5) that $P_{11}^{2}$ is symmetric. This and the symmetry of $P_{11}^{1}$ imply that R of (A.10) is symmetric since

$$R^{T}=((P_{11}^{1})^{T})^{-1}(P_{11}^{2})^{T} ((P_{11}^{1})^{T})^{-1} = (P_{11}^{1})^{-1}P_{11}^{2}(P_{11}^{1})^{-1} = R. $$

Since T of (2.7) and R are symmetric so are the matrices

$$([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T), \quad ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}), $$

appearing on the right-hand side of (A.15). It follows from (3.15) that A^DT = TA^D, B^DT = TB^D. Hence (A.5) implies that

$$ P_{11}^{2}(I_{N}\otimes T)=(I_{N}\otimes T) P_{11}^{2} $$

(A.16)

and

$$P_{11}^{1} (I_{N}\otimes T)=(I_{N}\otimes T) P_{11}^{1}. $$

The last equation yields

$$ (P_{11}^{1})^{-1}(I_{N}\otimes T)=(I_{N}\otimes T) (P_{11}^{1})^{-1}. $$

(A.17)

Using (A.10), (A.16), and (A.17), we have

$$ R(I_{N}\!\otimes T) = (P_{11}^{1})^{-1} P_{11}^{2} (P_{11}^{1})^{-1} (I_{N}\!\otimes T) = (I_{N}\!\otimes T) (P_{11}^{1})^{-1} P_{11}^{2} (P_{11}^{1})^{-1} = (I_{N}\!\otimes T) R. $$

(A.18)

For $Q=(P_{11}^{1})^{-1}$ or R, (A.17) and (A.18) yield

$$\begin{array}{@{}rcl@{}} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) Q ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}) &=& ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (I_{N} \otimes T) Q ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}) \\ &=& ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) Q (I_{N} \otimes T) ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}) \\ &=& ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) Q ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T). \end{array} $$

(A.19)

The matrices

$$([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}), \quad ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T), $$

appearing on the right-hand side of (A.15) are symmetric since the symmetry of T, R, and (A.19) with Q = R yield

$$\begin{array}{@{}rcl@{}} [([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N})]^{T} &=&([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\ &=&([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}), \end{array} $$

$$\begin{array}{@{}rcl@{}} [([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T)]^{T} &=& ([\textbf{e}_{1}|\textbf{e}_{N}]^{T}\otimes T) R ([\textbf{e}_{1}|\textbf{e}_{N}] \otimes I_{N}) \\ &=&([\textbf{e}_{1}|\textbf{e}_{N}]^{T}\otimes I_{N}) R ([\textbf{e}_{1}|\textbf{e}_{N}] \otimes T). \end{array} $$

Since all matrices on the right-hand side of (A.15) are symmetric so is $P_{21}^{1} R P_{12}^{2}$. In a similar way we show that $P_{21}^{2} (P_{11}^{1})^{-1} P_{12}^{2}$ is also symmetric. Since all matrices on the right-hand side of (A.11) are symmetric so is $P_{21}P_{11}^{-1} P_{12}$.

We next show that $P_{21}P_{11}^{-1} P_{12}$ is positive definite. Using (A.12), (3.15), and (A.19) with $Q=(P_{11}^{1})^{-1}$, we have that

$$\begin{array}{@{}rcl@{}} P_{21}^{1} (P_{11}^{1})^{-1} P_{12}^{1} &=& -\frac{1}{4h} \left( [\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes \left( \frac{1}{8}T+I_{N}\right)\right)\\ &&\times(P_{11}^{1})^{-1} \left( [\textbf{e}_{1}|\textbf{e}_{N}]\otimes \left( \frac{1}{8}T+I_{N}\right)\right)\\ &=& -\frac{1}{256h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T)\\ &&-\frac{1}{32h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N})\\ &&-\frac{1}{32h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\ &&-\frac{1}{4h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N})\\ &=& -\frac{1}{256h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T)\\ &&-\frac{1}{16h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\&&- \frac{1}{4h}([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}). \end{array} $$

Using (A.7), (4.26), (3.15), (A.14), and (A.19) with Q = (P₁₁)^− 1, we have

$$\begin{array}{@{}rcl@{}} P_{21}^{2} (P_{11}^{1})^{-1} P_{12}^{2} &=&\frac{h}{6} \left( [\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes \left( \frac{1}{8}T+I_{N}\right)\right)\\ &&\times(P_{11}^{1})^{-1} \left( \frac{1}{6h^{2}} [\textbf{e}_{1}|\textbf{e}_{N}]\otimes T+ \frac{1}{h^{2}}[\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}\right)\\ &=& \frac{1}{288h}([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\&&+ \frac{1}{48h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N})\\ &&+ \frac{1}{36h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\&&+ \frac{1}{6h}([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N})\\ &=& \frac{1}{288h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\&&+\frac{7}{144h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T)\\ &&+ \frac{1}{6h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}). \end{array} $$

The last two equations yield

$$\begin{array}{@{}rcl@{}} P_{21}^{1} (P_{11}^{1})^{-1} P_{12}^{1}+P_{21}^{2} (P_{11}^{1})^{-1} P_{12}^{2} \!&=&\! -\frac{1}{2304h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\ &&\!-\frac{1}{72h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\ &&\!- \frac{1}{12h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}) \\ \!&=&\! -\frac{1}{72h} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (P_{11}^{1})^{-1} \\ &&\!\times(I_{N}\otimes (T + 6 I_{N}))([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}) \\ &&\!- \frac{1}{2304h}([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes T) (P_{11}^{1})^{-1} \\ &&\times ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T). \end{array} $$

(A.20)

Using (2.10) and (A.8), we have

$$\begin{array}{@{}rcl@{}} &&-(Z^{T}\otimes Z^{T}) \left[(P_{11}^{1})^{-1} (I_{N}\otimes \left( T + 6 I_{N}\right)\right] (Z\otimes Z)\\ &=&-[(Z^{T}\otimes Z^{T}) P_{11}^{1} (Z\otimes Z)]^{-1} (Z^{T}\otimes Z^{T})\\ &&\times\left[ (I_{N}\otimes (T + 6 I_{N}) \right] (Z\otimes Z) \\ &=&-{\Lambda}_{2}^{-1} \left[I_{N}\otimes ({\Lambda} + 6I_{N})\right]. \end{array} $$

(A.21)

It follows from (2.8) and (A.2) that the diagonal entries of Λ + 6I_N are positive. Recall that the diagonal entries of Λ₂ are negative. Thus it follows from (A.21) that $-(P_{11}^{1})^{-1} \left [ (I_{N}\otimes (T + 6I_{N}) \right ]$ is symmetric and its eigenvalues are positive. Consequently $-(P_{11}^{1})^{-1} \left [ (I_{N}\otimes (T + 6I_{N}) \right ]$ is positive definite. Since $P_{11}^{1}$ is negative definite, $-(P_{11}^{1})^{-1}$ is positive definite. It follows from (2.10) and (2.8) that the eigenvalues of T are nonzero which makes T nonsingular. Hence, for nonzero v in R^2N, w = ([e₁|e_N] ⊗ T)v, z = ([e₁|e_N] ⊗ I_N)v are nonzero. Consequently, the positive definiteness of $-(P_{11}^{1})^{-1} \left [ (I_{N}\otimes (T + 6I_{N}) \right ]$ and $-(P_{11}^{1})^{-1}$ yield

$$\begin{array}{@{}rcl@{}} &&-\textbf{v}^{T} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T}\otimes T) (P_{11}^{1})^{-1} \left[ (I_{N}\otimes (T + 6I_{N}) \right] ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \textbf{v}\\ &=& -\textbf{w}^{T} (P_{11}^{1})^{-1}\left[ (I_{N}\otimes (T + 6I_{N})\right] \textbf{w}>0,\\ &&-\textbf{v}^{T} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T}\otimes T) (P_{11}^{1})^{-1} ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \textbf{v}\\ &=& -\textbf{z}^{T} (P_{11}^{1})^{-1} \textbf{z}>0. \end{array} $$

The last two equations show, on using (A.20), that $P_{21}^{1} (P_{11}^{1})^{-1} P_{12}^{1}+P_{21}^{2} (P_{11}^{1})^{-1} P_{12}^{2}$ is positive definite. Using (A.15), (A.19) with Q = R, and (A.18), we have

$$\begin{array}{@{}rcl@{}} -P_{21}^{1} R P_{12}^{2} &=& - \frac{1}{24h^{3}} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) (I_{N} \otimes T) R (I_{N}\otimes T) ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}) \\ &&-\frac{7}{12h^{3}} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes T) \\ &&- \frac{2}{h^{3}}([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) R ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}) \\ &=& -\frac{1}{24h^{3}} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) \left[ R (I_{N} \otimes T) (I_{N}\otimes T) \right.\\ &&\left.+ 14 R I_{N}\otimes T + 48R \right] ([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}) \\ &=& -\frac{1}{24h^{3}} ([\textbf{e}_{1}|\textbf{e}_{N}]^{T} \otimes I_{N}) R \left[ I_{N} \otimes \left( T^{2}+ 14 T + 48 I_{N} \right) \right]\\ &&\times([\textbf{e}_{1}|\textbf{e}_{N}]\otimes I_{N}). \end{array} $$

(A.22)

Using (A.10), (A.5), (2.10), (A.8), and (A.1), we have

$$\begin{array}{@{}rcl@{}} &&-(Z^{T}\otimes Z^{T}) R \left[ I_{N}\otimes (T^{2}+ 14 T + 48I_{N}) \right] (Z\otimes Z) \\ &=& [(Z^{T}\otimes Z^{T}) P_{11}^{1} (Z\otimes Z)]^{-1} (Z^{T}\otimes Z^{T}) (B^{D}\otimes B^{D}) (Z\otimes Z) \\ &&((Z^{T}\otimes Z^{T}) P_{11}^{1} (Z\otimes Z))^{-1} \\ &&(Z^{T}\otimes Z^{T}) \left[ I_{N}\otimes (T^{2}+ 14 T + 48I_{N}) \right] (Z\otimes Z) \\ &=& {\Lambda}_{2}^{-2} ({\Lambda}_{1}\otimes {\Lambda}_{1}) \left[ I_{N}\otimes ({\Lambda}^{2}+ 14 {\Lambda} + 48 I_{N}) \right]. \end{array} $$

(A.23)

Since the diagonal entries of Λ₁ are positive and the diagonal entries of Λ₂ are negative, the diagonal entries of ${\Lambda }_{2}^{-2} ({\Lambda }_{1} \otimes {\Lambda }_{1})$ are positive. The diagonal entries of Λ² + 14Λ + 48I_N are positive since

$$x^{2}+ 14 x + 48=(x + 6)(x + 8)> 0, \quad -4\leq x<0. $$

It follows from (A.23) that − R [(I_N ⊗ (T² + 14T + 48I_N)] is symmetric and that its eigenvalues are positive. Consequently, the matrix − R [(I_N ⊗ (T² + 14T + 48I_N)] is positive definite. For nonzero v in R^2N, w = ([e₁|e_N] ⊗ I_N)v is nonzero, and hence, (A.22) and the positive definiteness of − R [(I_N ⊗ (T² + 14T + 48I_N)] yield

$$-\textbf{v}^{T} P_{21}^{1} R P_{12}^{2} \textbf{v}=-\frac{1}{24h^{3}} \textbf{w}^{T} R \left[ (I_{N}\otimes (T^{2}+ 14 T + 48I_{N})\right] \textbf{w}>0. $$

This shows that $-P_{21}^{1} R P_{12}^{2}$ is positive definite. Since $P_{21}^{1} (P_{11}^{1})^{-1} P_{12}^{1}+P_{21}^{2} (P_{11}^{1})^{-1} P_{12}^{2}$ and $-P_{21}^{1} R P_{12}^{2}$ are positive definite, it follows from (A.11) that $P_{21}P_{11}^{-1}P_{12}$ is also positive definite. This completes the proof that P of (4.34) is symmetric and positive definite.