Matrix-valued Bratu equation associated with the symmetric domains of type BDI and CI

We define a submanifold $\Omega(J)$ in the cone $\operatorname{Sym}^{+}_{2n+r}(\mathbb{R})$ by

$$\Omega(J)=\left\{G\in\operatorname{Sym}^{+}_{2n+r}(\mathbb{R});\;JGJ=G^{-1}\right\},$$

$$J=\begin{pmatrix}0&0&I_{n}\\ 0&I_{r}&0\\ I_{n}&0&0\end{pmatrix}.$$

Setting the set $V(J)$ by

$$V(J)=\left\{B\in\operatorname{Sym}_{2n+r}(\mathbb{R});\;JBJ=-B\right\},$$

we can consider the following map:

$$V(J)\ni B\mapsto\exp B\in\Omega(J).$$

We see that any element $B\in V(J)$ has the following form:

$$B=B(b,a,c):=\begin{pmatrix}b&a&c\\ {}^{t}\!a&0&-{}^{t}\!a\\ {}^{t}\!c&-a&-b\end{pmatrix},\quad\begin{array}[]{l}b\in\operatorname{Sym}_{n}(\mathbb{R}),\\ a\in\mathrm{M}_{n,r}(\mathbb{R}),\\ c\in\mathrm{Alt}_{n}(\mathbb{R}).\end{array}$$

For an element $B\in V(J)$, we define an $\Omega(J)$-valued function $G(s;B)$ by

$$G(s;B):=\exp(sB(b,a,c))\quad(s\in\mathbb{R}).$$

We define the following sets of matrices:

	$\displaystyle\mathcal{G}(J)=\left\{G\in\operatorname{GL}_{2n+r}(\mathbb{R});\;JGJ={}^{t}\!G^{-1}\right\}\cong\mathrm{O}(n+r,n),$
	$\displaystyle\mathcal{N}(J)=\left\{N\in\mathcal{G}(J);\;N=\begin{pmatrix}I_{n}&0&0\\ *&I_{r}&0\\ *&*&I_{n}\end{pmatrix}\right\},$
	$\displaystyle\qquad=\left\{N=\begin{pmatrix}I_{n}&0&0\\ {}^{t}\!n_{1}&I_{r}&0\\ {}^{t}\!n_{2}&-n_{1}&I_{n}\end{pmatrix},n_{2}+{}^{t}\!n_{2}=-n_{1}{}^{t}\!n_{1}\right\},$
	$\displaystyle\Omega_{0}(J)=\left\{A\in\Omega(J);\;A=\begin{pmatrix}*&0&0\\ 0&*&0\\ 0&0&*\end{pmatrix}\right\},$
	$\displaystyle\qquad=\left\{A=\begin{pmatrix}h&0&0\\ 0&I_{r}&0\\ 0&0&h^{-1}\end{pmatrix},{}^{t}\!h=h\right\}.$

We denote the variable change defined through this decomposition by

$$G\rightarrow N,A\rightarrow h,n_{1},n_{2},$$

or simply by $G\rightarrow h$.

Now we state the exponential matrix solution in [I2020] as follows.

In the rest of this section, we give an alternative proof for this proposition based on the Lagrangian calculus.

Now we derive the Euler-Lagrange equation for our Lagrangian

	$\displaystyle\mathcal{L}$	$\displaystyle=\mathcal{L}(h,n_{1},n_{2},h^{\prime},n_{1}^{\prime},n_{2}^{\prime})$
		$\displaystyle=\operatorname{tr}((h^{\prime}h^{-1})^{2})+2\operatorname{tr}(hn_{1}^{\prime}{}^{t}\!n_{1}^{\prime})+\operatorname{tr}(h{}^{t}\!mhm),$
		$\displaystyle m=n_{1}{}^{t}\!n^{\prime}_{1}+n^{\prime}_{2}.$

The result is written as follows:

	$\displaystyle\text{variable }x:$	$\displaystyle\text{E-L eq. }\displaystyle\frac{d}{ds}\frac{\partial\mathcal{L}}{\partial x^{\prime}}=\frac{\partial\mathcal{L}}{\partial x}$
	$\displaystyle h:$	$\displaystyle(h^{\prime}h^{-1})^{\prime}=hn_{1}^{\prime}{}^{t}\!n_{1}^{\prime}+h{}^{t}\!mhm,$
	$\displaystyle n_{1}:$	$\displaystyle\left\{2hn_{1}^{\prime}+h{}^{t}\!mhn_{1}\right\}^{\prime}=hmhn_{1}^{\prime},$
	$\displaystyle n_{2}:$	$\displaystyle(h{}^{t}\!mh)^{\prime}=0.$

From the third equation, we get

$$h{}^{t}\!mh=\tilde{c}:\text{constant matrix},\quad\tilde{c}\in\mathrm{Alt}_{n}(\mathbb{R}).$$

Thus, the above equations become

	$\displaystyle h:$	$\displaystyle(h^{\prime}h^{-1})^{\prime}=hn_{1}^{\prime}{}^{t}\!n_{1}^{\prime}-\tilde{c}h^{-1}\tilde{c}h^{-1},$
	$\displaystyle n_{1}:$	$\displaystyle 2(hn_{1}^{\prime})^{\prime}+2(\tilde{c}n_{1})^{\prime}=0.$

From the second equation, we get

$$hn_{1}^{\prime}+{c}n_{1}=\tilde{a}:\text{constant matrix}.$$

Especially, in the case $\tilde{c}=0$, we get $hn_{1}^{\prime}=\tilde{a}$. Inserting this into the first equation, we obtain the matrix-valued Bratu equation

$\displaystyle h:$

$\displaystyle(h^{\prime}h^{-1})^{\prime}=(\tilde{a}{}^{t}\!\tilde{a})h^{-1}.$

Next we identify the constant matrices $\tilde{c},\tilde{a}$.

Defining the function $G(s)$ by the relation $G\rightarrow h,n_{1},n_{2}$, we have that

$$G^{\prime}G^{-1}=\begin{pmatrix}*&hn_{1}^{\prime}-hmhn_{1}&h{}^{t}\!mh\\ *&*&*\\ *&*&*\end{pmatrix}=\begin{pmatrix}*&\tilde{a}&\tilde{c}\\ *&*&*\\ *&*&*\end{pmatrix}.$$

On the other hand, $G(s)$ satisfies the original E-L equation

$$G^{\prime}G^{-1}=\text{(constant matrix)}.$$

This shows that $G(s)$ is given as

$$G=G(s;B(\tilde{b},\tilde{a},\tilde{c})),\quad\tilde{b}\in\operatorname{Sym}_{n}(\mathbb{R}).$$

We summarize the above calculation in the following table:

$$\begin{array}[]{ccc}\hline\cr\\ G(s;B(b,a,c))&\rightarrow&h(s),n_{1}(s),n_{2}(s)\\ \\ (G^{\prime}G^{-1})^{\prime}=0&&\begin{cases}(h^{\prime}h^{-1})^{\prime}=hn_{1}^{\prime}{}^{t}\!n_{1}^{\prime}-{c}h^{-1}{c}h^{-1}\\ hn_{1}^{\prime}+{c}n_{1}={a}\\ h{}^{t}\!(n_{1}{}^{t}\!n^{\prime}_{1}+n^{\prime}_{2})h=c\end{cases}\\ &&\\ \displaystyle\frac{1}{2}\operatorname{tr}((G^{\prime}G^{-1})^{2})&&\operatorname{tr}((h^{\prime}h^{-1})^{2})+2\operatorname{tr}(hn_{1}^{\prime}{}^{t}\!n_{1}^{\prime})+\operatorname{tr}(h{}^{t}\!mhm)\\ \\ \hline\cr\end{array}$$

The exponential matrix solution (Proposition 2.2) follows from this theorem.

For the set

$\displaystyle M=\left\{U=\begin{pmatrix}U_{1}\\ U_{2}\\ U_{3}\end{pmatrix}\in\mathrm{M}_{2n+r,n}(\mathbb{R});\;\operatorname{rank}U=n\right\},$

we consider the group actions

$$\operatorname{GL}(2n+r,\mathbb{R})\curvearrowright M\curvearrowleft\operatorname{GL}(n,\mathbb{R})$$

by the matrix multiplication. For the quatient

$$D:=M/\operatorname{GL}(n,\mathbb{R})=\left\{u=[U]=\begin{bmatrix}U_{1}\\ U_{2}\\ U_{3}\end{bmatrix};\;U\in M\right\},$$

we equip it with the induced topology to consider $D$ as a domain. We let

$$J=-\begin{pmatrix}0&0&I_{n}\\ 0&I_{r}&0\\ I_{n}&0&0\end{pmatrix}$$

and define a subdomain $D(J)\subset D$ by

$$M(J)=\left\{U\in M;\;{}^{t}\!UJU>0\right\},$$

$$D(J):=\left\{u=[U]\in D;\;U\in M(J)\right\}.$$

We define a subdomain $\Sigma\subset D$ (the Shilov boundary of $D(J)$) by

$$\Sigma=\left\{u=\begin{bmatrix}U_{1}\\ U_{2}\\ I_{n}\end{bmatrix}\in D;\;-{}^{t}\!U_{1}-U_{1}-{}^{t}\!U_{2}U_{2}=0\right\}.$$

We take two points $u_{0},w\in D$ as

	$\displaystyle u_{0}=[U_{0}]\in D(J),\quad w=[W]\in\Sigma,$
	$\displaystyle U_{0}=\begin{pmatrix}-I_{n}\\ 0\\ I_{n}\end{pmatrix},\quad W=\begin{pmatrix}0\\ 0\\ I_{n}\end{pmatrix}.$

For two points $u_{1},u_{2}\in D(J)$ expressed as

$$u_{j}=[U_{j}],\;U_{j}=\begin{pmatrix}U_{j,1}\\ U_{j,2}\\ U_{j,3}\end{pmatrix}\quad(j=1,2),$$

we define $\Delta(u_{1},u_{2})\in\mathrm{M}_{n}(\mathbb{R})$ by

$\displaystyle\Delta(u_{1},u_{2}):={}^{t}\!(U_{1}\Gamma_{1}^{-1})J(U_{2}\Gamma_{2}^{-1}),$

$$\Gamma_{j}={}^{t}\!WU_{j}=U_{j,3}\quad(j=1,2).$$

Consider the action $\mathcal{G}(J)\curvearrowright D(J)$ defined by

$$g\cdot u=[gU]\quad(g\in\mathcal{G}(J),u=[U]\in D(J)).$$

We also define the following subgroups:

	$\displaystyle\mathcal{P}(J)=\left\{g\in\mathcal{G}(J);\;g=\begin{pmatrix}*&*&*\\ 0&*&*\\ 0&0&*\end{pmatrix}\right\},$
	$\displaystyle\mathcal{K}(J)=\left\{g\in\mathcal{G}(J);\;{}^{t}\!g=g^{-1}\right\}\cong\mathrm{O}(n+r)\times\mathrm{O}(n),$
	$\displaystyle\mathcal{G}(J)_{u_{0}}=\left\{g\in\mathcal{G}(J);\;gu_{0}=u_{0}\right\}.$

We define a map $\mathcal{F}:\Omega(J)\xrightarrow{}D$ by

$$\mathcal{F}(G)=\begin{bmatrix}G_{3}-I_{n}\\ G_{2}\\ h\end{bmatrix},\quad G=\begin{pmatrix}h&*&*\\ G_{2}&*&*\\ G_{3}&*&*\end{pmatrix}\in\Omega(J),$$

and a map $\pi:\Omega(J)\rightarrow\operatorname{Sym}^{+}_{n}(\mathbb{R})$ by

$$\pi(G)=h,\quad G=\begin{pmatrix}h&*&*\\ G_{2}&*&*\\ G_{3}&*&*\end{pmatrix}\in\Omega(J).$$

We define an action $\mathcal{G}(J)\curvearrowright\Omega(J)$ by

$$g\cdot G={}^{t}\!g^{-1}Gg^{-1}\quad(g\in\mathcal{G}(J),G\in\Omega(J)).$$

By this lemma, we obtain the following commutative diagram: