Analysis of Linear Partial Differential Equations Using Convex Optimization

Abstract

We propose a framework for the exponential stability analysis for a large class of linear Partial Differential Equations (PDEs). The class of PDEs studied are parameterized by polynomial data. The class contains parabolic, first and second order hyperbolic, partial integro-differential equations, and systems coupled in-domain and on the boundaries. The proposed method is numerically tractable since we formulate the analysis test as a convex optimization problem. We use polynomial positive semi-definite matrices, the fundamental theorem of calculus, and the Green’s theorem to reduce the analysis problem to the construction and verification of integral inequalities on Hilbert spaces.

Appendices

Appendix 1

The following technical lemmas are used throughout the chapter.

Lemma 6.1

For any $\alpha ,\beta \in \mathbb {N}$ and Lebesgue integrable functions

$K_1: \overline{\varOmega } \rightarrow \mathbb {R}^{\beta (\alpha +1) \times \beta (\alpha +1)}$ and $K_2: \underline{\varOmega } \rightarrow \mathbb {R}^{\beta (\alpha +1) \times \beta (\alpha +1)}$, the following identity holds

$$\begin{aligned}&\int _0^1 \int _0^x w_\alpha (x)^T K_1(x,y) w_\alpha (y)dy dx + \int _0^1 \int _x^1 w_\alpha (x)^T K_2(x,y) w_\alpha (y)dy dx \\&= {{\frac{1}{2}}} \int _\varDelta w_\alpha (x)^T \varGamma \left[ K_1(x,y) + K_2(y,x)^T \right] w_\alpha (y) d\varDelta , \end{aligned}$$

for all $w \in \mathscr {H}^\alpha \left( [0,1];\mathbb {R}^\beta \right)$.

Proof

We have

$$\begin{aligned} \int _0^1 \int _0^x w_\alpha (x)^T K_1(x,y) w_\alpha (y)dy dx&= \int _0^1 \int _0^x w_\alpha (y)^T K_1(x,y)^T w_\alpha (x)dy dx \\&= \int _0^1 \int _y^1 w_\alpha (y)^T K_1(x,y)^T w_\alpha (x)dx dy \\&= \int _0^1 \int _x^1 w_\alpha (x)^T K_1(y,x)^T w_\alpha (y)dy dx, \end{aligned}$$

where we first transposed the integrand, followed by a change of order of integration and finally switched between the variables x and y. Thus

$$\begin{aligned}&\int _0^1 \int _0^x w_\alpha (x)^T K_1(x,y) w_\alpha (y)dy dx \\&= {{\frac{1}{2}}} \int _0^1 \int _0^x w_\alpha (x)^T K_1(x,y) w_\alpha (y)dy dx + {{\frac{1}{2}}} \int _0^1 \int _x^1 w_\alpha (x)^T K_1(y,x)^T w_\alpha (y)dy dx \\&= {{\frac{1}{2}}} \int _\varDelta w_\alpha (x)^T \varGamma \left[ K_1(x,y)\right] w_\alpha (y)d\varDelta . \end{aligned}$$

Following the same steps for

$$\begin{aligned} \int _0^1 \int _x^1 w_\alpha (x)^T K_2(x,y) w_\alpha (y)dy dx, \end{aligned}$$

then completes the proof.    $\blacksquare$

Lemma 6.2

For any $v \in \mathscr {L}_2\left( [0,1];\mathbb {R}^\beta \right)$, $\alpha ,\beta \in \mathbb {N}$, and polynomials $F_1,G_1,F_2,G_2 \in \mathscr {R}^{\beta \times \beta }[(x,y)]$, the following identity holds

$$\begin{aligned}&\int _0^1 \left( \int _0^x F_1(x,y)v(y)dy + \int _x^1 F_2(x,y)v(y)dy \right) ^T \nonumber \\&\times \left( \int _0^x G_1(x,y)v(y)dy + \int _x^1 G_2(x,y)v(y)dy \right) dx = {{\frac{1}{2}}} \int _\varDelta v(x)^T \varGamma \left[ K \right] v(y) d\varDelta , \end{aligned}$$

(36)

where

$$\begin{aligned} K(x,y)=&\int _0^y \left( F_2(z,x)^T G_2(z,y) + G_2(z,x)^T F_2(z,y) \right) dz \\&+ \int _y^x \left( F_2(z,x)^T G_1(z,y) + G_2(z,x)^T F_1(z,y) \right) dz \\&+ \int _x^1 \left( F_1(z,x)^T G_1(z,y) + G_1(z,x)^T F_1(z,y) \right) dz. \end{aligned}$$

Proof

We begin by observing that the left hand side of (36) may be written as

$$\begin{aligned} \langle {\mathscr {F} v},{\mathscr {G} v}\rangle _{\mathscr {L}_2}=\langle { v},{\mathscr {F}^\star \mathscr {G} v}\rangle _{\mathscr {L}_2}, \end{aligned}$$

(37)

where the linear bounded operators on $\mathscr {L}_2\left( [0,1];\mathscr {R}^\beta \right)$ are defined as

$$\begin{aligned} \left( \mathscr {F} v \right) (x)=&\int _0^x F_1(x,y)v(y)dy + \int _x^1 F_2(x,y)v(y)dy,\\ \left( \mathscr {G} v \right) (x)=&\int _0^x G_1(x,y)v(y)dy + \int _x^1 G_2(x,y)v(y)dy, \end{aligned}$$

and where the Hilbert adjoint of the operator $\mathscr {F}$ is given by

$$\begin{aligned} \left( \mathscr {F}^\star v \right) (x)= \int _0^x F_2(y,x)^T v(y)dy + \int _x^1 F_1(y,x)^T v(y)dy. \end{aligned}$$

Now, using the fact that

$$\begin{aligned} \left( \mathscr {G} v \right) (y)= \int _0^y G_1(y,z)v(z)dz + \int _y^1 G_2(y,z)v(z)dz, \end{aligned}$$

we get

$$\begin{aligned} \left( \mathscr {F}^\star \mathscr {G} v \right) (x)=&\int _0^x F_2(y,x)^T \left( \mathscr {G} v \right) (y)dy + \int _x^1 F_1(y,x)^T \left( \mathscr {G} v \right) (y)dy \nonumber \\ =&\sum _{i=1}^4 \varTheta _i(x), \end{aligned}$$

(38)

where

$$\begin{aligned} \varTheta _1(x)=&\int _0^x \int _0^y F_2(y,x)^T G_1(y,z)v(z)dz dy,\\ \varTheta _2(x)=&\int _0^x \int _y^1 F_2(y,x)^T G_2(y,z)v(z)dz dy,\\ \varTheta _3(x)=&\int _x^1 \int _0^y F_1(y,x)^T G_1(y,z)v(z)dz dy,\\ \varTheta _4(x)=&\int _x^1 \int _y^1 F_1(y,x)^T G_2(y,z)v(z)dz dy. \end{aligned}$$

In each of the $\varTheta _i(x)$ we change the order of integration and switch between the variables y and z to obtain

$$\begin{aligned} \varTheta _1(x)=&\int _0^x \int _y^x F_2(z,x)^T G_1(z,y)dz v(y)dy,\\ \varTheta _2(x)=&\int _0^x \int _0^y F_2(z,x)^T G_2(z,y)dz v(y)dz + \int _x^1 \int _0^x F_2(z,x)^T G_2(z,y)dz v(y)dy,\\ \varTheta _3(x)=&\int _0^x \int _x^1 F_1(z,x)^T G_1(z,y)dz v(y)dy + \int _x^1 \int _y^1 F_1(z,x)^T G_1(z,y)dz v(y)dy,\\ \varTheta _4(x)=&\int _x^1 \int _x^y F_1(z,x)^T G_2(z,y)dz v(y)dz. \end{aligned}$$

Substituting into (38) and consequently in (36) we get

$$\begin{aligned} \langle {\mathscr {F}v},{\mathscr {G}v}\rangle =&\int _0^1 \int _0^x v(x) K_1(x,y)v(y)dy dx +\int _0^1 \int _x^1 v(x) K_2(x,y)v(y)dy dx, \end{aligned}$$

(39)

where

$$\begin{aligned} K_1(x,y)=&\int _0^y F_2(z,x)^TG_2(z,y)dz + \int _y^x F_2(z,x)^T G_1(z,y)dz \\&+ \int _x^1 F_1(z,x)^T G_1(z,y)dz,\\ K_2(x,y)=&\int _0^x F_2(z,x)^T G_2(z,y)dz + \int _x^y F_1(z,x)^T G_2(z,y)dz \\&+ \int _y^1 F_1(z,x)^T G_1(z,y)dz. \end{aligned}$$

Then, applying Lemma 6.1 to (39) completes the proof.    $\blacksquare$

Remark 1

Lemma 6.2 also holds for any $v \in \mathscr {H}^\alpha \left( [0,1];\mathbb {R}^\beta \right)$, $F_1,G_1,F_2,G_2 \in \mathscr {R}^{\beta (\alpha +1) \times \beta (\alpha +1)}[(x,y)]$ and with v(x) replaced by $v_\alpha (x)$ in (36).

Appendix 2

We now present the proofs of the various results stated in the chapter. We begin with the proof of our assertion that (4) can be written in the form presented in (5).

Let us write the integral expression in (3) as¹

$$\begin{aligned} \mathscr {V}(w)={{\frac{1}{2}}} \langle {\varXi w},{ w}\rangle _{\mathscr {L}_2}, \end{aligned}$$

(40)

where the self-adjoint operator $\varXi$ on $\mathscr {L}_2\left( [0,1];\mathscr {R}^\beta \right)$ is defined as

$$\begin{aligned} \left( \varXi w \right) (x)=&T_b(x) w(x) + \int _0^x \bar{T}(x,y)w(y)dy + \int _x^1 \bar{T}(y,x)^T w(y)dy. \end{aligned}$$

Since the operator $\varXi$ is self-adjoint, we may use (2) to obtain

$$\begin{aligned} {\partial _{t}}\mathscr {V}(w) =&{{\frac{1}{2}}} \langle {\varXi {\partial _{t}}w},{ w}\rangle _{\mathscr {L}_2} + {{\frac{1}{2}}} \langle {\varXi w},{ {\partial _{t}}w}\rangle _{\mathscr {L}_2} \\ =&{{\frac{1}{2}}} \langle {\varXi w},{ {\partial _{t}}w}\rangle _{\mathscr {L}_2} + {{\frac{1}{2}}} \langle {\varXi w},{ {\partial _{t}}w}\rangle _{\mathscr {L}_2} \\ =&\langle {\varXi w},{ {\partial _{t}}w}\rangle _{\mathscr {L}_2} \\ =&\int _0^1 \left( T_b(x) w(x)dx + \int _0^x \bar{T}(x,y)w(y)dy + \int _x^1 \bar{T}(y,x)^T w(y) dy \right) ^T \\&\times \left( \phantom {\int } A_1(x)w_\alpha (x) + \int _0^x A_2(x,y)w_\alpha (y)dy + \int _x^1 A_3(x,y)w_\alpha (y)dy \right) dx. \end{aligned}$$

Therefore,

$$\begin{aligned} \mathscr {V}_d(w)=&-{\partial _{t}}\mathscr {V}(w)-2\delta \mathscr {V}(w) \\ =&-\int _0^1 \left( T_b(x) w(x)dx + \int _0^x \bar{T}(x,y)w(y)dy + \int _x^1 \bar{T}(y,x)^T w(y) dy \right) ^T \nonumber \\&\times \left( A_1(x)w_\alpha (x) + \int _0^x A_2(x,y)w_\alpha (y)dy + \int _x^1 A_3(x,y)w_\alpha (y)dy \right) dx \nonumber \\&-\delta \int _0^1 w(x)^T T_b(x) w(x)dx - \delta \int _\varDelta w(x)^T \varGamma \left[ \bar{T}(x,y) \right] w(y) d\varDelta =-\sum _{i=1}^4 \Phi _i,\nonumber \end{aligned}$$

(41)

where,

$$\begin{aligned} \Phi _1=&\int _0^1 \left( T_b(x) w(x) + \int _0^x \bar{T}(x,y)w(y)dy + \int _x^1 \bar{T}(y,x)^T w(y) dy \right) ^T A_1(x)w_\alpha (x)dx,\\ \Phi _2 =&\int _0^1 \left( T_b(x)w(x)\right) ^T \left( \int _0^x A_2(x,y)w_\alpha (y)dy + \int _x^1 A_3(x,y)w_\alpha (y)dy \right) dx, \\ \Phi _3 =&\int _0^1 \left( \int _0^x \bar{T}(x,y)w(y)dy + \int _x^1 \bar{T}(y,x)^T w(y)dy \right) ^T \\&\times \left( \int _0^x A_2(x,y)w_\alpha (y)dy + \int _x^1 A_3(x,y)w_\alpha (y)dy \right) dx,\\ \Phi _4=&\delta \int _0^1 w(x)^T T_b(x) w(x)dx + \delta \int _\varDelta w(x)^T \varGamma \left[ \bar{T}(x,y) \right] w(y) d\varDelta . \end{aligned}$$

The term $\Phi _1$ may be written as

$$\begin{aligned} \Phi _1=&\int _0^1 \bar{w}_\alpha (x)^T \left[ \begin{array}{cc}T_b(x)A_1(x) &{} 0_{\beta ,2\beta \alpha } \\ 0_{3\beta \alpha ,\beta (\alpha +1)} &{} 0_{3\beta \alpha ,2\beta \alpha }\end{array}\right] \bar{w}_\alpha (x)dx \\&+ \int _0^1 \int _0^x w_\alpha (x)^T \left[ \begin{array}{c} \bar{T}(x,y)^T A_1(x) \\ 0_{\beta \alpha ,\beta (\alpha +1)}\end{array}\right] ^T w_\alpha (y)dy dx \\&+ \int _0^1 \int _x^1 w_\alpha (x)^T \left[ \begin{array}{c} \bar{T}(y,x)^T A_1(x) \\ 0_{\beta \alpha ,\beta (\alpha +1)}\end{array}\right] ^T w_\alpha (y)dy dx. \end{aligned}$$

Then, applying Lemma 6.1 to the double integrals and writing the single integral kernel in a symmetric form produces

$$\begin{aligned} \Phi _1 =&\int _0^1 \bar{w}_\alpha (x)^T He \left( \left[ \begin{array}{cc}T_b(x)A_1(x) &{} 0_{\beta ,2\beta \alpha } \\ 0_{3\beta \alpha ,\beta (\alpha +1)} &{} 0_{3\beta \alpha ,2\beta \alpha }\end{array}\right] \right) \bar{w}_\alpha (x)dx \nonumber \\&+ {{\frac{1}{2}}} \int _\varDelta w_\alpha (x)^T \varGamma \left[ \left[ \begin{array}{c}\bar{T}(x,y)^T A_1(x) \\ 0_{\beta \alpha ,\beta (\alpha +1)}\end{array}\right] ^T + \left[ \begin{array}{c}\bar{T}(x,y) A_1(y) \\ 0_{\beta \alpha ,\beta (\alpha +1)}\end{array}\right] \right] w_\alpha (y) d\varDelta . \end{aligned}$$

(42)

The term $\Phi _2$ may be written as

$$\begin{aligned} \Phi _2=&\int _0^1 \int _0^x w_\alpha (x)^T \left[ \begin{array}{c}T_b(x)A_2(x,y) \\ 0_{\beta \alpha ,\beta (\alpha +1)}\end{array}\right] w_\alpha (y)dydx \\&+\int _0^1 \int _x^1 w_\alpha (x)^T \left[ \begin{array}{c}T_b(x)A_3(x,y) \\ 0_{\beta \alpha ,\beta (\alpha +1)}\end{array}\right] w_\alpha (y)dydx. \end{aligned}$$

Applying Lemma 6.1 produces

$$\begin{aligned} \Phi _2 =&{{\frac{1}{2}}} \int _\varDelta w_\alpha (x)^T \varGamma \left[ \left[ \begin{array}{c}T_b(x)A_2(x,y) \\ 0_{\beta \alpha ,\beta (\alpha +1)}\end{array}\right] + \left[ \begin{array}{c}T_b(y)A_3(y,x) \\ 0_{\beta \alpha ,\beta (\alpha +1)}\end{array}\right] ^T \right] w_\alpha (y)d\varDelta . \end{aligned}$$

(43)

The term $\Phi _3$ may be written as

$$\begin{aligned} \Phi _3 =&\int _0^1 \left( \int _0^x \left[ \begin{array}{c}\bar{T}(x,y)^T \\ 0_{\beta \alpha ,\beta }\end{array}\right] ^T w_\alpha (y)dy + \int _x^1 \left[ \begin{array}{c}\bar{T}(y,x) \\ 0_{\beta \alpha ,\beta }\end{array}\right] ^T w_\alpha (y)dy \right) \nonumber \\&\times \left( \int _0^x A_2(x,y)w_\alpha (y)dy + \int _x^1 A_3(x,y)w_\alpha (y)dy \right) dx. \end{aligned}$$

(44)

Then, applying Lemma 6.2 to (44) with

$$\begin{aligned} F_1(x,y)=&\left[ \begin{array}{c}\bar{T}(x,y)^T \\ 0_{\beta \alpha ,\beta }\end{array}\right] ^T, \quad F_2(x,y)=\left[ \begin{array}{c}\bar{T}(y,x) \\ 0_{\beta \alpha ,\beta }\end{array}\right] ^T,\\ G_1(x,y)=&A_2(x,y), \quad G_2(x,y)=A_3(x,y), \quad v(y)=w_\alpha (y), \end{aligned}$$

produces

$$\begin{aligned} \Phi _3= {{\frac{1}{2}}} \int _\varDelta w_\alpha (x)^T \varGamma \left[ \underline{U} \right] w_\alpha (y)d\varDelta , \end{aligned}$$

(45)

where $\underline{U}(x,y)$ is defined in (5).

Finally, the term $\Phi _4$ may be written as

$$\begin{aligned} \Phi _4=&\int _0^1 \bar{w}_\alpha (x)^T He \left( \left[ \begin{array}{cc}\delta T_b(x) &{} 0_{\beta ,3\beta \alpha } \\ 0_{3\beta \alpha ,\beta } &{} 0_{3\beta \alpha }\end{array}\right] \right) \bar{w}_\alpha (x)dx \nonumber \\&+{{\frac{1}{2}}} \int _\varDelta w_\alpha (x)^T \varGamma \left[ \left[ \begin{array}{cc}2\delta \bar{T}(x,y) &{} 0_{\beta ,\beta \alpha } \\ 0_{\beta \alpha ,\beta } &{} 0_{\beta \alpha }\end{array}\right] \right] w_\alpha (y)d\varDelta . \end{aligned}$$

(46)

Substituting (42), (43), (45) and (46) into (41) produces (5).

We now provide the proofs of previously stated results.

Proof

(Theorem 2) Throughout this proof, for notational brevity, we write q in place of $q(\alpha ,\beta ,d)$. If we define

$$f(x)=\left[ \begin{array}{c} \bar{w}_\alpha (x) \\ \int _0^x Y_{q}(x,y)w_\alpha (y)dy \\ \int _x^1 Y_q(x,y)w_\alpha (y)dy \end{array}\right] , \quad w \in \mathscr {H}^\alpha \left( [0,1];\mathbb {R}^\beta \right) ,$$

then

$$\begin{aligned}&\int _0^1 f(x)^T R(x) f(x)dx \ge 0,\quad \forall w \in \mathscr {H}^\alpha \left( [0,1];\mathbb {R}^\beta \right) , \end{aligned}$$

(47)

since $R(x) \succeq 0$, for all $x \in [0,1]$. Simplifying the expression we obtain

$$\begin{aligned}&\int _0^1 f(x)^T R(x) f(x)dx = \int _0^1 \bar{w}_\alpha (x)^T R_b(x) \bar{w}_\alpha (x)dx + \sum _{i=1}^3 \varTheta _i, \end{aligned}$$

(48)

where

$$\begin{aligned} \varTheta _1=&2 \int _0^1 \int _0^x w_\alpha (x)^T R_{12}(x)Y_q(x,y)w_\alpha (y)dy dx \\&+ 2 \int _0^1 \int _x^1 w_\alpha (x)^T R_{13}(x)Y_q(x,y)w_\alpha (y)dy dx,\\ \varTheta _2=&\int _0^1 \left( \int _0^x Y_q(x,y) w_\alpha (y)dy \right) ^T \\&\times \left( \int _0^x R_{22}Y_q(x,y)w_\alpha (y)dy + \int _x^1 R_{23}Y_q(x,y)w_\alpha (y)dy \right) dx,\\ \varTheta _3=&\int _0^1 \left( \int _x^1 Y_q(x,y) w_\alpha (y)dy \right) ^T \\&\times \left( \int _0^x R_{23}^T Y_q(x,y)w_\alpha (y)dy + \int _x^1 R_{33}Y_q(x,y)w_\alpha (y)dy \right) dx. \end{aligned}$$

Applying Lemma 6.1 to $\varTheta _1$ with $K_1(x,y)=2R_{12}(x)Y_q(x,y)$ and

$K_2(x,y)=2R_{13}(x)Y_q(x,y)$ produces

$$\begin{aligned} \varTheta _1 =&\int _\varDelta w_\alpha (x)^T \varGamma \left[ R_{12}(x)Y_q(x,y) \right] w_\alpha (y) d\varDelta \nonumber \\&+\int _\varDelta w_\alpha (x)^T \varGamma \left[ Y_q(y,x)^T R_{13}(y)^T \right] w_\alpha (y)d\varDelta , \end{aligned}$$

(49)

Applying Lemma 6.2 to $\varTheta _2$ with

$$\begin{aligned} F_1(x,y)=&Y_q(x,y), \quad F_2(x,y) = 0_{q,\beta (\alpha +1)}, \quad v(y)=w_\alpha (y),\\ G_1(x,y)=&R_{22}Y_q(x,y), \quad G_2(x,y)=R_{23}Y_q(x,y), \end{aligned}$$

produces

$$\begin{aligned} \varTheta _2 =&\int _\varDelta w_\alpha (x)^T \varGamma \left[ \int _y^x Y_q(z,x)^T \frac{R_{23}^T}{2}Y_q(z,y)dz\right] w_\alpha (y) d\varDelta , \nonumber \\&+ \int _\varDelta w_\alpha (x)^T \varGamma \left[ \int _x^1 Y_q(z,x)^T R_{22}Y_q(z,y)dz\right] w_\alpha (y) d\varDelta . \end{aligned}$$

(50)

Similarly, applying Lemma 6.2 to $\varTheta _3$ with

$$\begin{aligned} F_1(x,y)=&0_{q,\beta (\alpha +1)}, \quad F_2(x,y) = Y_q(x,y), \quad v(y)=w_\alpha (y),\\ G_1(x,y)=&R_{23}^T Y_q(x,y), \quad G_2(x,y)=R_{33}Y_q(x,y), \end{aligned}$$

produces

$$\begin{aligned} \varTheta _3 =&\int _\varDelta w_\alpha (x)^T \varGamma \left[ \int _0^y Y_q(z,x)^T R_{33}^TY_q(z,y)dz\right] w_\alpha (y) d\varDelta , \nonumber \\&+ \int _\varDelta w_\alpha (x)^T \varGamma \left[ \int _y^x Y_q(z,x)^T \frac{R_{23}^T}{2}Y_q(z,y)dz\right] w_\alpha (y) d\varDelta . \end{aligned}$$

(51)

Substituting (49)–(51) into (48) produces

$$\begin{aligned} \int _0^1 f(x)^T R(x)f(x)dx =&\int _0^1 \bar{w}_\alpha (x)^T R_b(x) \bar{w}_\alpha (x)dx + \int _\varDelta w_\alpha (x)^T \varGamma \left[ \bar{R}\right] w_\alpha (y)d\varDelta \end{aligned}$$

Then, (47) completes the proof.    $\blacksquare$

Proof

(Corollary 1) Following the same steps as for the proof of Theorem 2, it can be shown that

$$\begin{aligned}&\mathscr {T}(w)-\epsilon \Vert {w}\Vert _{\mathscr {L}_2} = \int _0^1 f(x)^T \left( T(x) -\left[ \begin{array}{cc}\epsilon I_\beta &{} 0_{\beta ,2q(0,\beta ,d)} \\ 0_{2q(0,\beta ,d),\beta } &{} 0_{2q(0,\beta ,d)}\end{array}\right] \right) f(x)dx, \end{aligned}$$

where

$$\begin{aligned} f(x) = \left[ \begin{array}{c}w(x) \\ \int _0^x Y_{q(0,\beta ,d)}(x,y) w(y)dy \\ \int _x^1 Y_{q(0,\beta ,d)}(x,y) w(y)dy\end{array}\right] , \quad w \in \mathscr {L}_2\left( [0,1];\mathbb {R}^\beta \right) . \end{aligned}$$

Then, (18) holds since (17) holds.    $\blacksquare$

Proof

(Lemma 2) Consider the vector field

$$ \left[ \begin{array}{c}\phi _1(x,y) \\ \phi _2(x,y)\end{array}\right] = \left[ \begin{array}{c} w_{\alpha -1}(x)^T H_1(x,y) w_{\alpha -1}(y) \\ w_{\alpha -1}(x)^T H_2(x,y) w_{\alpha -1}(y)\end{array}\right] , $$

Then, by Green’s theorem

$$\begin{aligned}&\oint _{\partial \overline{\varOmega }} \left( \phi _1(x,y)dx + \phi _2(x,y)dy \right) + \int _{\overline{\varOmega }} \left( \partial _y\phi _1(x,y) - \partial _x\phi _2(x,y) \right) d \overline{\varOmega } = 0, \end{aligned}$$

where $\partial \overline{\varOmega }$ denotes the boundary of the domain $\overline{\varOmega }$. Therefore, we obtain

$$\begin{aligned}&\int _0^1 \left[ \phi _1(x,0)+\phi _2(1,x)-\phi _1(x,x)-\phi _2(x,x) \right] dx + \int _0^1 \int _0^x \left[ \partial _y\phi _1-\partial _x\phi _2 \right] dy dx=0. \end{aligned}$$

(52)

Using the definitions of the projection matrices described in subsection on notation used in the chapter, we obtain

$$\begin{aligned} \int _0^1 \left[ \phi _1(x,0)+\phi _2(1,x)-\phi _1(x,x)-\phi _2(x,x) \right] dx&=\int _0^1 \bar{w}_\alpha (x)^T H_b(x) \bar{w}_\alpha (x)dx \nonumber \\&=\int _0^1 \bar{w}_\alpha (x)^T He \left( H_b(x) \right) \bar{w}_\alpha (x)dx. \end{aligned}$$

(53)

Similarly, we also obtain the following identity

$$\begin{aligned}&\int _0^1 \int _0^x \left[ \partial _y\phi _1(x,y)-\partial _x\phi _2(x,y) \right] dy dx = \int _0^1 \int _0^x w_\alpha (x)^T 2 \bar{H}(x,y) w_\alpha (y)dy dx. \end{aligned}$$

Then, applying Lemma 6.1 with $K_1 = 2\bar{H}$ and $K_2 = 0$ produces

$$\begin{aligned}&\int _0^1 \int _0^x w_\alpha (x)^T 2 \bar{H}(x,y) w_\alpha (y)dy dx = \int _\varDelta w_\alpha (x)^T \varGamma \left[ \bar{H}(x,y) \right] w_\alpha (y)dy dx. \end{aligned}$$

(54)

Substituting (53) and (54) into (52) completes the proof.    $\blacksquare$

Proof

(Lemma 3) Since for all $w \in \mathscr {B}$ (see (2c)),

$$\begin{aligned} \int _0^1 \left[ \begin{array}{cc}F_1(x)&F_2 \end{array}\right] \bar{w}_\alpha (x)dx=0_{\beta \alpha ,1}, \end{aligned}$$

we get for all $w \in \mathscr {B}$

$$\begin{aligned} 0=&\int _0^1 \bar{w}_\alpha (y)^T \left[ \begin{array}{c}B_1(y) \\ B_2\end{array}\right] dy \cdot \int _0^1 \left[ \begin{array}{cc}F_1(x)&F_2 \end{array}\right] \bar{w}_\alpha (x)dx \\ =&\int _0^1 \int _0^1 \bar{w}_\alpha (y)^T \left[ \begin{array}{c} B_1(y) \\ B_2\end{array}\right] \left[ \begin{array}{cc}F_1(x)&F_2 \end{array}\right] \bar{w}_\alpha (x) dy dx \\ =&\int _0^1 \int _0^1 \bar{w}_\alpha (y)^T \left[ \begin{array}{cc} B_1(y)F_1(x) &{} B_1(y)F_2 \\ B_2 F_1(x) &{} B_2 F_2 \end{array}\right] \bar{w}_\alpha (x) dy dx. \end{aligned}$$

Switching between the variables x and y and changing the order of integration produces

$$\begin{aligned} 0=&\int _0^1 \int _0^1 \bar{w}_\alpha (x)^T \left[ \begin{array}{cc}B_1(x)F_1(y) &{} B_1(x)F_2 \\ B_2 F_1(y) &{} B_2 F_2\end{array}\right] \bar{w}_\alpha (y) dy dx, \end{aligned}$$

for all $w \in \mathscr {B}$. Recall that

$$\begin{aligned} \bar{w}_\alpha (x) = \left[ \begin{array}{c}w_\alpha (x) \\ w_\alpha ^b\end{array}\right] . \end{aligned}$$

Therefore, we simplify the previous expression to obtain for all $w \in \mathscr {B}$

$$\begin{aligned} 0 =&\int _0^1 \int _0^1 \left[ \begin{array}{c}w_\alpha (x) \\ w_\alpha ^b\end{array}\right] ^T \left[ \begin{array}{cc}B_1(x)F_1(y) &{} B_1(x)F_2 \\ B_2 F_1(y) &{} B_2 F_2\end{array}\right] \left[ \begin{array}{c}w_\alpha (y) \\ w_\alpha ^b\end{array}\right] dy dx \\ =&\int _0^1 \bar{w}_\alpha (x)^T B_b(x) \bar{w}_\alpha (x) dx + \int _0^1 \int _0^1 w_\alpha (x)^T B_1(x)F_1(y) w_\alpha (y) dy dx. \end{aligned}$$

Writing the single integral kernel in symmetric form and dividing up the double integral gives us

$$\begin{aligned} 0 =&\int _0^1 \bar{w}_\alpha (x)^T He \left( B_b(x) \right) \bar{w}_\alpha (x) dx \\&+ \int _0^1 \int _0^x w_\alpha (x)^T B_1(x)F_1(y) w_\alpha (y) dy dx + \int _0^1 \int _x^1 w_\alpha (x)^T B_1(x)F_1(y) w_\alpha (y) dy dx, \end{aligned}$$

for all $w \in \mathscr {B}$. Then, applying Lemma 6.1 with $K_1(x,y)=K_2(x,y)= B_1(x) F_1(y)$ completes the proof.    $\blacksquare$