For the proof of (a), from (
9),
\(\forall i=1,2,\ldots ,\omega\):
$$\begin{aligned} T^i_k&=\bar{G}^i+\sum ^{p}_{q=1}\sum ^{\omega }_{d=1}c^{i,d}_qT^d_{k-q}+\sum ^{p-1}_{q=1}a^i_qT^i_{k-q}+\sum ^{p-1}_{q=1}b^i_1a^i_q\left( T^i_{k-q}-T^i_{k-(q+1)}\right) \\&\quad +\sum ^{\omega }_{d=1}\left[ e_{i,d} \sum ^{p}_{q=1} m^{d}_{q} T^{d}_{k-q}\right] -\sum ^{p}_{q=1} m^{i}_{q} T^{i}_{k-q}, \end{aligned}$$
or, equivalently,
$$\begin{aligned} T^i_k&=\bar{G}^i+\sum ^{p}_{q=1}\left[ \begin{array}{cccc}c^{i,1}_q&c^{i,2}_q&\cdots&c^{i,\omega }_q\end{array}\right] \left[ \begin{array}{c}T^1_{k-q}\\ T^2_{k-q}\\ \cdots \\ T^\omega _{k-q}\end{array}\right] + \sum ^{p-1}_{q=1}a^i_qT^i_{k-q}+\sum ^{p-1}_{q=1}b^i_1a^i_q\left( T^i_{k-q}-T^i_{k-(q+1)}\right) \\&\quad +[\begin{array}{cccc}e_{i,1}&e_{i,2}&\cdots&e_{i,\omega }\end{array}] \left[ \begin{array}{c}\sum \nolimits ^{p}_{q=1} m^{1}_{q} T^{1}_{k-q}\\ \sum \nolimits ^{p}_{q=1} m^{2}_{q} T^{2}_{k-q}\\ \cdots \\ \sum \nolimits ^{p}_{q=1} m^{\omega }_{q} T^{\omega }_{k-q}\end{array}\right] -\sum ^{p}_{q=1} m^{i}_{q} T^{i}_{k-q}, \end{aligned}$$
or, equivalently,
$$\begin{aligned} T^i_k&=\bar{G}^i+\sum ^{p}_{q=1}\left[ \begin{array}{cccc}c^{i,1}_q&c^{i,2}_q&\cdots&c^{i,\omega }_q\end{array}\right] \left[ \begin{array}{c}T^1_{k-q}\\ T^2_{k-q}\\ \cdots \\ T^\omega _{k-q}\end{array}\right] \\&\quad +\sum ^{p-1}_{q=1}a^i_qT^i_{k-q}+\sum ^{p-1}_{q=1}b^i_1a^i_qT^i_{k-q}-\sum ^{p}_{q=2}b^i_1a^i_{q-1}T^i_{k-q}\\&\quad + \sum ^{p}_{q=1}\left[ \begin{array}{cccc}e_{i,1}&e_{i,2}&\cdots&e_{i,\omega }\end{array}\right] {\text {diag}}\left\{ m^i_q\right\} _{i=1,2,\ldots ,\omega } \left[ \begin{array}{c}T^{1}_{k-q}\\ T^{2}_{k-q}\\ \cdots \\ T^{\omega }_{k-q}\end{array}\right] -\sum ^{p}_{q=1} m^{i}_{q} T^{i}_{k-q}, \end{aligned}$$
or, equivalently,
$$\begin{aligned} T^i_k&=\bar{G}^i+\sum ^{p}_{q=1}\left\{ \left( \left[ \begin{array}{ccc}c^{i,1}_q&\cdots&c^{i,\omega }_q\end{array}\right] +[\begin{array}{cccc}e_{i,1}&\cdots&e_{i,\omega }\end{array}]{\text {diag}}\left\{ m^i_q\right\} _{i=1,\ldots ,\omega }\right) \left[ \begin{array}{c}T^{1}_{k-q}\\ \cdots \\ T^{\omega }_{k-q}\end{array}\right] -m^{i}_{q} T^{i}_{k-q}\right\} \\ &\quad +\sum ^{p-1}_{q=1}\left\{ a^i_qT^i_{k-q}+b^i_1a^i_qT^i_{k-q}\right\} -\sum ^{p}_{q=2}b^i_1a^i_{q-1}T^i_{k-q}, \end{aligned}$$
or, equivalently, by denoting
\(b^i_1\) as
\(b^i\)
$$\begin{aligned}T^i_k&=\bar{G}^i+\left( \left[ \begin{array}{ccc}c^{i,1}_1&\cdots&c^{i,\omega }_1\end{array}\right] +[\begin{array}{cccc}e_{i,1}&\cdots&e_{i,\omega }\end{array}]{\text {diag}}\left\{ m^i_1\right\} _{i=1,\ldots ,\omega }\right) \left[ \begin{array}{c}T^{1}_{k-1}\\ \cdots \\ T^{\omega }_{k-1}\end{array}\right] \\&\quad +\left( a^i_1+b^i_1a^i_1-m^{i}_{1}\right) T^i_{k-1}+\sum ^{p-1}_{q=2}\left\{ \left( \left[ \begin{array}{ccc}c^{i,1}_q&\cdots&c^{i,\omega }_q\end{array}\right] \right. \right. \\&\quad \left. +\,[\begin{array}{cccc}e_{i,1}&\cdots&e_{i,\omega }\end{array}]{\text {diag}}\left\{ m^i_q\right\} _{i=1,\ldots ,\omega }\right) \left[ \begin{array}{c}T^{1}_{k-q}\\ \cdots \\ T^{\omega }_{k-q}\end{array}\right] \\&\quad + \left. \left( a^i_q+b^i\left( a^i_q-a^i_{q-1}\right) -m^{i}_{q}\right) T^i_{k-q}\right\} +\left( \left[ \begin{array}{ccc}c^{i,1}_p&\cdots&c^{i,\omega }_p\end{array}\right] \right. \\&\quad \left. +\, [\begin{array}{cccc}e_{i,1}&\cdots&e_{i,\omega }\end{array}]{\text {diag}}\left\{ m^i_p\right\} _{i=1,\ldots ,\omega }\right) \left[ \begin{array}{c}T^{1}_{k-p}\\ \cdots \\ T^{\omega }_{k-p}\end{array}\right] +\left( b^i_1a^i_p-m^{i}_{p}\right) T^i_{k-p}, \end{aligned}$$
or, equivalently, in matrix form,
$$\begin{aligned}T_k&=\bar{G}+\left( C^{(1)}+EM^{(1)}+A^{(1)}+BA^{(1)}-M^{(1)}\right) T^i_{k-1}\\&\quad + \sum ^{p-1}_{q=2}\left\{ \left( C^{(q)}+EM^{(q)}+A^{(q)}+B\left( A^{(q)}-A^{(q-1)}\right) -M^{(q)}\right) T_{k-q}\right\} \\&\quad + \left( C^{(p)}+EM^{(p)}+BA^{(p)}-M^{(p)}\right) T^i_{k-p}, \end{aligned}$$
which leads to (
10). Let
\(T^*\) be the equilibrium of (
10). Then, for
\({\text {det}}[\sum _{q=1}^{p-1}\{C^{(q)}+A^{(q)}+(E-I_\omega )M^{(q)}\}+C^{(p)}+(E-I_\omega )M^{(p)}+BA^{(1)}]\ne 0\) we have
$$\begin{aligned} T^*&=\bar{G}+\left\{ C^{(1)}+A^{(1)}+BA^{(1)}+(E-I_\omega )M^{(1)}\right. \\&\quad + \sum _{q=2}^{p-1}\left[ C^{(q)}+A^{(q)}+B\left( A^{(q)}-A^{(q-1)}\right) + (E-I_\omega )M^{(q)}\right] \\&\quad \left. +\, C^{(p)}-BA^{(p-1)}+(E-I_\omega )M^{(p)}\right\} T^* \end{aligned}$$
or, equivalently,
$$T^*=\left[ \sum _{q=1}^{p-1}\left\{ C^{(q)}+A^{(q)}+(E-I_\omega )M^{(q)}\right\} +C^{(p)}+(E-I_\omega )M^{(p)}+BA^{(1)}\right] ^{-1}\bar{G}.$$
We are interested in the asymptotic stability of the equilibrium, i.e. when
\(lim_{k\longrightarrow \infty }T_k=T^*\). By adopting the notation
$$\begin{array}{c} T_{k-p}=Y^1_k,\\ T_{k-p+1}=Y^2_k,\\ \vdots \\ T_{k-2}=Y^{p-1}_k\\ T_{k-1}=Y^p_k \end{array}$$
(14)
and
$$\begin{aligned} \begin{array}{cc} &T_{k-p+1}=Y^1_{k+1}=Y^2_k,\\ &T_{k-p+2}=Y^2_{k+1}=Y^3_k,\\ &\vdots \\ &T_{k-1}=Y^{p-1}_{k+1}=Y^p_k\\ \end{array}\end{aligned}$$
$$\begin{aligned} T_{k}&=Y^p_{k+1}=\bar{G}+\left( C^{(1)}+A^{(1)}+BA^{(1)}+(E-I_\omega )M^{(1)}\right) Y^p_k\\ &\quad +\sum \limits _{q=2}^{p-1}\left\{ (C^{(q)}+A^{(q)}+B\left( A^{(q)}-A^{(q-1)}\right) +(E-I_\omega )M^{(q)})Y^q_k\right\} \\ &\quad +\left( C^{(p)}-BA^{(p-1)}+(E-I_\omega )M^{(p)}\right) Y^1_k. \end{aligned}$$
Then, the matrix difference equation (
10) takes the form
$$Y_{k+1}=FY_k+\left[ \begin{array}{c}0_{\omega ,1}\\ \vdots \\ 0_{\omega ,1}\\ G\end{array}\right] .$$
The stability of the above system depends on the eigenvalues of
F. Let
\(\lambda\) be an eigenvalue of
F. Then, the equilibrium of (
10) is asymptotic stable if and only if
$$\begin{array}{cc}\left| \lambda \right| <1,&\lambda \in {\mathbb {C}}.\end{array}$$
In addition, if
\(U=\left[ \begin{array}{c}U_1\\ U_2\\ \vdots \\ U_p\end{array}\right]\) is an eigenvector of the eigenvalue
\(\lambda\), then
$$\begin{aligned} \left[ \begin{array}{cccccc} 0_{\omega ,\omega }&\quad I_\omega &\quad 0_{\omega ,\omega }&\quad \cdots &\quad 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }\\ 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }&\quad I_\omega &\quad \cdots &\quad 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }\\ \vdots &\quad \vdots &\quad \vdots &\quad \ddots &\quad \vdots &\quad \vdots \\ 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }&\quad \cdots &\quad 0_{\omega ,\omega }&\quad I_\omega \\ v_1&\quad v_2&\quad v_3&\quad \cdots &\quad v_{p-1}&\quad v_p\end{array}\right] U=\lambda U, \end{aligned}$$
or, equivalently,
$$\begin{array}{c} U_2=\lambda U_1,\\ U_3=\lambda U_2,\\ \vdots \\ U_p=\lambda U_{p-1},\\ \sum \nolimits _{i=1}^pv_iU_i=\lambda U_p, \end{array}$$
or, equivalently,
$$\begin{array}{c} U_2=\lambda U_1,\\ U_3=\lambda ^2 U_1,\\ \vdots \\ U_p=\lambda ^{p-1} U_1,\\ v_1U_1+v_2U_2+\cdots +v_pU_p=\lambda U_p. \end{array}$$
By replacing the first
\(p-1\) equations into the
pth equation of the above expression, we get
$$v_1U_1+\lambda v_2 U_1+\cdots +\lambda ^{p-1}v_pU_1=\lambda ^p U_1,$$
or, equivalently,
$$\left[ v_1+\lambda v_2+\cdots +\lambda ^{p-1}v_p-\lambda ^pI_\omega \right] U_1=0.$$
Hence, the eigenvalues
\(\lambda\) are the roots of the polynomial
\(\phi (\lambda )={\hbox{det}}[v_1+\lambda v_2+\cdots +\lambda ^{p-1}v_p-\lambda ^pI_\omega ]\), which by replacing (
13) is polynomial (
11). Thus, the equilibrium
\(T^*\) of (
10) is asymptotically stable if and only if
\(\forall i=1,2,\ldots ,\omega\) and for
\(\phi (\lambda )=0\)
$$\begin{array}{cc}\left| \lambda \right| <1,&\lambda \in \mathbb {C}.\end{array}$$
For the proof of (b), since governmental expenditure is a fully controlled variable for each country, equation (
10) will take the form
$$\begin{aligned} T_k&=\bar{G}_k+\left[ C^{(1)}+A^{(1)}+BA^{(1)}+(E-I_\omega )M^{(1)}\right] T_{k-1}\\&\quad + \sum _{q=2}^{p-1}\left\{ \left[ C^{(q)}+A^{(q)}+B\left( A^{(q)}-A^{(q-1)}\right) +(E-I_\omega )M^{(q)}\right] T_{k-q}\right\} \\&\quad + \left[ C^{(p)}-BA^{(p-1)}+(E-I_\omega )M^{(p)}\right] T_{k-p}, \quad k \ge 0,\quad p \ge 3. \end{aligned}$$
By adopting (
14) and applying it into the above expression, we arrive at (
13) which is a linear discrete-time control system with input vector
\(G_k\). Linear control involves modification of the behaviour of a given system by applying state feedback. It is known Azzo and Houpis (
1995), Dorf (
1983), Kuo (
1996), Ogata (
1987) that this is possible if and only if the system is completely controllable. The necessary and sufficient condition for complete controllability is that
$${\text {rank}}\left[ \begin{array}{ccccc}Q&FQ&F^2Q&\cdots&F^{p-1}Q\end{array}\right] =p\omega .$$
Since
$${\text {det}}\left[ \begin{array}{ccccc}Q&FQ&F^2Q&\cdots&F^{p-1}Q\end{array}\right] =(-1)^{p\omega }\ne 0,$$
which means system (
13) is controllable and the state feedback replaces the input
\(G_k\) by
where
$$K=\left[ \begin{array}{cccc}K_1&K_2&\cdots&K_p\end{array}\right]$$
and
\(K_1,K_2,\ldots ,K_p\in \mathbb {C^{\omega \times \omega }}\). The system then takes the form
where
\(QK=\left[ \begin{array}{c}0_{\omega ,\omega }\\ \vdots \\ 0_{\omega ,\omega }\\ K \end{array}\right]\). The basic problem is that of choosing a state feedback
K such that the resulting (closed-loop equation) is stable. The stabilisation in the time-invariant case is via results on eigenvalue placement in the complex plane. In our situation, eigenvalues of the closed-loop system are specified to have modulus less than unity for stability. Thus, we have
$$\begin{aligned} F-QK=\left[ \begin{array}{cccccc} 0_{\omega ,\omega }&\quad I_\omega &\quad 0_{\omega ,\omega }&\quad \cdots &\quad 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }\\ 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }&\quad I_\omega &\quad \cdots &\quad 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }\\ \vdots &\quad \vdots &\quad \vdots &\quad \ddots &\quad \vdots &\quad \vdots \\ 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }&\quad 0_{\omega ,\omega }&\quad \cdots &\quad 0_{\omega ,\omega }&\quad I_\omega \\ v_1-K_1&\quad v_2-K_2&\quad v_3-K_3&\quad \cdots &\quad v_{p-1}-K_{p-1}&\quad v_p-K_p\end{array}\right] , \end{aligned}$$
where
\(v_i\),
\(i=1,2,\ldots ,\omega\) are given by (
14). Then, the characteristic equation of
\(F-QK\) will be
\(\tilde{\phi }(\tilde{\lambda })\), with
\(\tilde{\lambda }\) the desired eigenvalues. The proof is completed.
\(\square\)