Levinson's theorem

Levinson's theorem - gives a condition guaranteeing that two systems are asymptotically equivalent .

Statement of the theorem

Let the system solutions

{\frac {dx}{dt}}=Ax,\quad (1)

{\ displaystyle {\ frac {dx} {dt}} = Ax, \ quad (1)}

{\frac {dx}{dt}}=Ax,\quad (1)

Where $A$ ${\ displaystyle A}$ $A$ - constant $(n\times n)$ ${\ displaystyle (n \ times n)}$ $(n\times n)$ - matrix limited to $[0,\infty )$ ${\ displaystyle [0, \ infty)}$ $[0,\infty )$ . Then the system

{\frac {dy}{dt}}=[A+B(t)]y,\quad (2)

{\ displaystyle {\ frac {dy} {dt}} = [A + B (t)] y, \ quad (2)}

{\frac {dy}{dt}}=[A+B(t)]y,\quad (2)

Where $B(t)\in C[0,\infty )$ ${\ displaystyle B (t) \ in C [0, \ infty)}$ $B(t)\in C[0,\infty )$ and $\int _{0}^{\infty }\lVert B(t)\rVert \,dt<\infty ,\quad (3)$ ${\ displaystyle \ int _ {0} ^ {\ infty} \ lVert B (t) \ rVert \, dt <\ infty, \ quad (3)}$ $\int _{0}^{\infty }\lVert B(t)\rVert \,dt<\infty ,\quad (3)$

asymptotically equivalent to the system $\quad (1)$ ${\ displaystyle \ quad (1)}$ $\quad (1)$ .

Proof

(The idea of the proof presented below belongs to Brouwer ^[1] )

Since system solutions $\quad (1)$ ${\ displaystyle \ quad (1)}$ $\quad (1)$ are limited then characteristic roots $\lambda \ (A)$ ${\ displaystyle \ lambda \ (A)}$ $\lambda \ (A)$ matrices $A$ ${\ displaystyle A}$ $A$ satisfy equality

Re\lambda \ (A)\leq \ 0,

{\ displaystyle Re \ lambda \ (A) \ leq \ 0,}

Re\lambda \ (A)\leq \ 0,

moreover, characteristic roots with zero real parts have simple elementary divisors.

Without loss of generality, suppose that the matrix $A$ ${\ displaystyle A}$ $A$ has a quasi-diagonal form

\quad A=diag(A_{1},A_{2}),\quad (4)

{\ displaystyle \ quad A = diag (A_ {1}, A_ {2}), \ quad (4)}

\quad A=diag(A_{1},A_{2}),\quad (4)

Where $\quad A_{1}$ ${\ displaystyle \ quad A_ {1}}$ $\quad A_{1}$ and $\quad A_{2}$ ${\ displaystyle \ quad A_ {2}}$ $\quad A_{2}$ - respectively $(p\times p)$ ${\ displaystyle (p \ times p)}$ $(p\times p)$ - and $(q\times q)$ ${\ displaystyle (q \ times q)}$ $(q\times q)$ matrices $\quad (p+q)$ ${\ displaystyle \ quad (p + q)}$ $\quad (p+q)$ such that

Re\lambda \ (A_{1})<-\alpha <\ 0,

{\ displaystyle Re \ lambda \ (A_ {1}) <- \ alpha <\ 0,}

Re\lambda \ (A_{1})<-\alpha <\ 0,

Re\lambda \ (A_{2})=0,\quad (5)

{\ displaystyle Re \ lambda \ (A_ {2}) = 0, \ quad (5)}

Re\lambda \ (A_{2})=0,\quad (5)

Indeed, this can be obtained using simple transformations. $\xi \ =S{\boldsymbol {x}},$ ${\ displaystyle \ xi \ = S {\ boldsymbol {x}},}$ $\xi \ =S{\boldsymbol {x}},$ and $\eta \ =S{\boldsymbol {y}},$ ${\ displaystyle \ eta \ = S {\ boldsymbol {y}},}$ $\eta \ =S{\boldsymbol {y}},$ Where $\quad S$ ${\ displaystyle \ quad S}$ $\quad S$ - constant $(n\times n)$ ${\ displaystyle (n \ times n)}$ $(n\times n)$ is a matrix , and a one-to-one correspondence between the new integral curves ${\boldsymbol {\xi }}(t)\Longleftrightarrow {\boldsymbol {\eta }}(t)$ ${\ displaystyle {\ boldsymbol {\ xi}} (t) \ Longleftrightarrow {\ boldsymbol {\ eta}} (t)}$ induces a one-to-one correspondence between the old integral curves ${\boldsymbol {x}}(t)=S^{-1}\xi (t)\Longleftrightarrow S^{-1}\eta (t)={\boldsymbol {y}}(t)$ ${\ displaystyle {\ boldsymbol {x}} (t) = S ^ {- 1} \ xi (t) \ Longleftrightarrow S ^ {- 1} \ eta (t) = {\ boldsymbol {y}} (t)}$ .

Also, from the marginal relationship ${\boldsymbol {\xi }}(t)-{\boldsymbol {\eta }}(t)\to 0,$ ${\ displaystyle {\ boldsymbol {\ xi}} (t) - {\ boldsymbol {\ eta}} (t) \ to 0,}$ at $t\to \infty$ ${\ displaystyle t \ to \ infty}$ obviously the limit relation follows

{\boldsymbol {x}}(t)-{\boldsymbol {y}}(t)\to 0,

{\ displaystyle {\ boldsymbol {x}} (t) - {\ boldsymbol {y}} (t) \ to 0,}

at

t\to \infty

{\ displaystyle t \ to \ infty}

.

$\quad 1)$ ${\ displaystyle \ quad 1)}$ Let be $\quad X(t)=diag(e^{tA_{1}},e^{tA_{2}})$ ${\ displaystyle \ quad X (t) = diag (e ^ {tA_ {1}}, e ^ {tA_ {2}})}$ is the fundamental matrix of the system $\quad (1),$ ${\ displaystyle \ quad (1),}$ normalized to zero: $\quad X(t)=E,$ ${\ displaystyle \ quad X (t) = E,}$ but $\quad I_{1}=diag(E_{p},0),$ ${\ displaystyle \ quad I_ {1} = diag (E_ {p}, 0),}$ and $\quad I_{2}=diag(0,E_{q}),$ ${\ displaystyle \ quad I_ {2} = diag (0, E_ {q}),}$ Where $\quad E_{q}$ ${\ displaystyle \ quad E_ {q}}$ and $\quad E_{p}$ ${\ displaystyle \ quad E_ {p}}$ are the identity matrices of the corresponding orders q and p, and, obviously, $\quad I_{1}+I_{2}=E_{n}.$ ${\ displaystyle \ quad I_ {1} + I_ {2} = E_ {n}.}$

Put $\quad X(t)=X_{1}(t)+X_{2}(t),$ ${\ displaystyle \ quad X (t) = X_ {1} (t) + X_ {2} (t),}$ Where $\quad X_{1}(t)=X(t)I_{1}\equiv diag(e^{tA_{1}},0),$ ${\ displaystyle \ quad X_ {1} (t) = X (t) I_ {1} \ equiv diag (e ^ {tA_ {1}}, 0),}$ and $\quad X_{2}(t)=X(t)I_{2}\equiv diag(0,e^{tA_{2}})$ ${\ displaystyle \ quad X_ {2} (t) = X (t) I_ {2} \ equiv diag (0, e ^ {tA_ {2}})}$ .

Hence the Cauchy matrix $\quad K(t,\tau )\equiv X(t)X^{-1}(\tau )=X(t-\tau )$ ${\ displaystyle \ quad K (t, \ tau) \ equiv X (t) X ^ {- 1} (\ tau) = X (t- \ tau)}$ can be represented as:

$\quad K(t,\tau )=X_{1}(t-\tau )+X_{2}(t-\tau ),$ ${\ displaystyle \ quad K (t, \ tau) = X_ {1} (t- \ tau) + X_ {2} (t- \ tau),}$

and provided $\quad (5)$ ${\ displaystyle \ quad (5)}$ we have

$\lVert X_{1}(t)\rVert =\lVert e^{tA_{1}}\rVert \leq ae^{-\alpha t},$ ${\ displaystyle \ lVert X_ {1} (t) \ rVert = \ lVert e ^ {tA_ {1}} \ rVert \ leq ae ^ {- \ alpha t},}$

at $0\leq t<\infty$ ${\ displaystyle 0 \ leq t <\ infty}$ $\quad (6)$ ${\ displaystyle \ quad (6)}$ and

$\lVert X_{2}(t)\rVert =\lVert e^{tA_{2}}\rVert \leq b,$ ${\ displaystyle \ lVert X_ {2} (t) \ rVert = \ lVert e ^ {tA_ {2}} \ rVert \ leq b,}$

at $-\infty <t<\infty$ ${\ displaystyle - \ infty <t <\ infty}$ $\quad (7),$ ${\ displaystyle \ quad (7),}$ Where $\quad a,b$ ${\ displaystyle \ quad a, b}$ are some positive constants. Using the method of variation of arbitrary constants, the differential equation can be written in integral form

$\quad y(t)=X(t-t_{0}){\boldsymbol {y}}(t_{0})+\int _{t_{0}}^{t}X_{1}(t-\tau )B(\tau ){\boldsymbol {y}}(\tau )\,d\tau +$ {\ displaystyle \ quad y (t) = X (t-t_ {0}) {\ boldsymbol {y}} (t_ {0}) + \ int _ {t_ {0}} ^ {t} X_ {1} (t- \ tau) B (\ tau) {\ boldsymbol {y}} (\ tau) \, d \ tau +} $\int _{t_{0}}^{t}X_{2}(t-\tau )B(\tau ){\boldsymbol {y}}(\tau )\,d\tau ,$ ${\ displaystyle \ int _ {t_ {0}} ^ {t} X_ {2} (t- \ tau) B (\ tau) {\ boldsymbol {y}} (\ tau) \, d \ tau,}$ Where $t\in [0,\infty )$ ${\ displaystyle t \ in [0, \ infty)}$ arbitrary.

Since the matrix $\quad B(t)$ ${\ displaystyle \ quad B (t)}$ completely integrated on $[0,\infty ),$ ${\ displaystyle [0, \ infty),}$ then all decisions ${\boldsymbol {y}}(t)$ ${\ displaystyle {\ boldsymbol {y}} (t)}$ the system $\quad (2)$ ${\ displaystyle \ quad (2)}$ limited to $[0,\infty ),$ ${\ displaystyle [0, \ infty),}$

and therefore the improper integral $\int _{t_{0}}^{\infty }X_{2}(t-\tau )B(\tau ){\boldsymbol {y}}(\tau )\,d\tau$ ${\ displaystyle \ int _ {t_ {0}} ^ {\ infty} X_ {2} (t- \ tau) B (\ tau) {\ boldsymbol {y}} (\ tau) \, d \ tau}$ is convergent.

Hence, given that $\quad X_{2}(t-\tau )=X(t-\tau )I_{2}=X(t-t_{0})X(t_{0}-\tau )I_{2}=X(t-t_{0})X_{2}(t_{0}-\tau ),$ ${\ displaystyle \ quad X_ {2} (t- \ tau) = X (t- \ tau) I_ {2} = X (t-t_ {0}) X (t_ {0} - \ tau) I_ {2 } = X (t-t_ {0}) X_ {2} (t_ {0} - \ tau),}$ our integral equation can be represented as

$\quad y(t)=X(t-t_{0})\left\lbrack {\boldsymbol {y}}(t_{0})+\int _{t_{0}}^{\infty }X_{2}(t_{0}-\tau )B(\tau ){\boldsymbol {y}}(\tau )\,d\tau \right\rbrack +$ ${\ displaystyle \ quad y (t) = X (t-t_ {0}) \ left \ lbrack {\ boldsymbol {y}} (t_ {0}) + \ int _ {t_ {0}} ^ {\ infty } X_ {2} (t_ {0} - \ tau) B (\ tau) {\ boldsymbol {y}} (\ tau) \, d \ tau \ right \ rbrack +}$ $+\int _{t_{0}}^{t}X_{1}(t-\tau )B(\tau ){\boldsymbol {y}}(\tau )\,d\tau -\int _{t}^{\infty }X_{2}(t-\tau )B(\tau ){\boldsymbol {y}}(\tau )\,d\tau (8)$ ${\ displaystyle + \ int _ {t_ {0}} ^ {t} X_ {1} (t- \ tau) B (\ tau) {\ boldsymbol {y}} (\ tau) \, d \ tau - \ int _ {t} ^ {\ infty} X_ {2} (t- \ tau) B (\ tau) {\ boldsymbol {y}} (\ tau) \, d \ tau (8)}$

Decision $\quad {\boldsymbol {y}}(t)$ ${\ displaystyle \ quad {\ boldsymbol {y}} (t)}$ the system $\quad (2)$ ${\ displaystyle \ quad (2)}$ with initial condition $\quad {\boldsymbol {y}}(t_{0})={\boldsymbol {y_{0}}}$ ${\ displaystyle \ quad {\ boldsymbol {y}} (t_ {0}) = {\ boldsymbol {y_ {0}}}}$ comparable solution $\quad {\boldsymbol {x}}(t)$ ${\ displaystyle \ quad {\ boldsymbol {x}} (t)}$ the system $\quad (1)$ ${\ displaystyle \ quad (1)}$ with initial condition

$\quad {\boldsymbol {x}}(t_{0})={\boldsymbol {y_{0}}}(t_{0})+\int _{t_{0}}^{\infty }X_{2}(t-\tau )B(\tau ){\boldsymbol {y}}(\tau )\,d\tau (9)$ ${\ displaystyle \ quad {\ boldsymbol {x}} (t_ {0}) = {\ boldsymbol {y_ {0}}} (t_ {0}) + \ int _ {t_ {0}} ^ {\ infty} X_ {2} (t- \ tau) B (\ tau) {\ boldsymbol {y}} (\ tau) \, d \ tau (9)}$

Since decisions $\quad {\boldsymbol {x}}(t)$ ${\ displaystyle \ quad {\ boldsymbol {x}} (t)}$ and $\quad {\boldsymbol {y}}(t)$ ${\ displaystyle \ quad {\ boldsymbol {y}} (t)}$ completely determined by their initial conditions, then the formula $\quad (9)$ ${\ displaystyle \ quad (9)}$ establishes a unique correspondence between the set of all decisions $\lbrace {\boldsymbol {y}}(t)\rbrace$ ${\ displaystyle \ lbrace {\ boldsymbol {y}} (t) \ rbrace}$ the system $\quad (2)$ ${\ displaystyle \ quad (2)}$ and many decisions $\lbrace {\boldsymbol {x}}(t)\rbrace$ ${\ displaystyle \ lbrace {\ boldsymbol {x}} (t) \ rbrace}$ (or part of) the system $\quad (1)$ ${\ displaystyle \ quad (1)}$ . Note that the relation $\quad (9)$ ${\ displaystyle \ quad (9)}$ continuous relative to the initial value $\quad {\boldsymbol {y}}(t_{0})={\boldsymbol {y_{0}}}.$ ${\ displaystyle \ quad {\ boldsymbol {y}} (t_ {0}) = {\ boldsymbol {y_ {0}}}.}$

$\quad 2)$ ${\ displaystyle \ quad 2)}$ We show that the correspondence between the solutions $\quad {\boldsymbol {x}}(t)$ ${\ displaystyle \ quad {\ boldsymbol {x}} (t)}$ and $\quad {\boldsymbol {y}}(t),$ ${\ displaystyle \ quad {\ boldsymbol {y}} (t),}$ which is determined by the formula $\quad (9),$ ${\ displaystyle \ quad (9),}$ is one-to-one and extends to the whole set of decisions $\lbrace {\boldsymbol {x}}(t)\rbrace$ ${\ displaystyle \ lbrace {\ boldsymbol {x}} (t) \ rbrace}$ .

Let be $\quad Y(t)$ ${\ displaystyle \ quad Y (t)}$ is the fundamental matrix of the system $\quad (1)$ ${\ displaystyle \ quad (1)}$ such that $\quad Y(t_{0})=E$ ${\ displaystyle \ quad Y (t_ {0}) = E}$ . We have

$Y(t)=X(t-t_{0})+\int _{t_{0}}^{t}X(t-\tau )B(\tau )Y(\tau )\,d\tau .$ ${\ displaystyle Y (t) = X (t-t_ {0}) + \ int _ {t_ {0}} ^ {t} X (t- \ tau) B (\ tau) Y (\ tau) \, d \ tau.}$

But from the inequalities $\quad (6),(7)$ ${\ displaystyle \ quad (6), (7)}$ should $\lVert X(t-t_{0})\rVert \leq max(a,b)=c,$ ${\ displaystyle \ lVert X (t-t_ {0}) \ rVert \ leq max (a, b) = c,}$ at $t\geq t_{0}$ ${\ displaystyle t \ geq t_ {0}}$ ; so

$\lVert Y(t)\rVert \geq c+\int _{t_{0}}^{t}c\lVert B(\tau )\rVert \lVert Y(\tau )\rVert \,d\tau$ ${\ displaystyle \ lVert Y (t) \ rVert \ geq c + \ int _ {t_ {0}} ^ {t} c \ lVert B (\ tau) \ rVert \ lVert Y (\ tau) \ rVert \, d \ tau}$

and by virtue of the Gronwall-Bellman lemma we find

$\lVert Y(t)\rVert \geq c\,\exp(\int _{t_{0}}^{t}c\lVert B(\tau )\rVert \,d\tau )\geq c\,\exp(c\int _{0}^{\infty }\lVert B(\tau )\rVert \,d\tau )=k,\quad$ ${\ displaystyle \ lVert Y (t) \ rVert \ geq c \, \ exp (\ int _ {t_ {0}} ^ {t} c \ lVert B (\ tau) \ rVert \, d \ tau) \ geq c \, \ exp (c \ int _ {0} ^ {\ infty} \ lVert B (\ tau) \ rVert \, d \ tau) = k, \ quad}$

at

t_{0}\geq t<\infty \qquad (10),

{\ displaystyle t_ {0} \ geq t <\ infty \ qquad (10),}

moreover, the constant $\quad k$ ${\ displaystyle \ quad k}$ at the rate $\quad (10)$ ${\ displaystyle \ quad (10)}$ independent of the choice of the initial moment $t_{0}(t_{0}\leq 0).$ ${\ displaystyle t_ {0} (t_ {0} \ leq 0).}$

Obviously, we have ${\boldsymbol {y}}(t)=Y(t){\boldsymbol {y}}(t_{0}).$ ${\ displaystyle {\ boldsymbol {y}} (t) = Y (t) {\ boldsymbol {y}} (t_ {0}).}$

Therefore, from the formula $\quad (9)$ ${\ displaystyle \ quad (9)}$ we get ${\boldsymbol {y}}(t_{0})=\lbrack E+Z(t_{0})\rbrack {\boldsymbol {y}}(t_{0}),\quad$ ${\ displaystyle {\ boldsymbol {y}} (t_ {0}) = \ lbrack E + Z (t_ {0}) \ rbrack {\ boldsymbol {y}} (t_ {0}), \ quad}$ Where $Z(t_{0})=\int _{t_{0}}^{\infty }X_{2}(t_{0}-\tau )B(\tau )Y(\tau )\,d\tau ,\quad$ ${\ displaystyle Z (t_ {0}) = \ int _ {t_ {0}} ^ {\ infty} X_ {2} (t_ {0} - \ tau) B (\ tau) Y (\ tau) \, d \ tau, \ quad}$ based on $\quad (7),(10)$ ${\ displaystyle \ quad (7), (10)}$ we deduce

$\lVert Z(t_{0})\rVert \geq \int _{t_{0}}^{\infty }\lVert X_{2}(t_{0}-\tau )\rVert \lVert B(\tau )\rVert \lVert Y(\tau )\rVert \,d\tau \geq bk\int _{t_{0}}^{\infty }\lVert B(\tau )\rVert \,d\tau \quad (12).$ ${\ displaystyle \ lVert Z (t_ {0}) \ rVert \ geq \ int _ {t_ {0}} ^ {\ infty} \ lVert X_ {2} (t_ {0} - \ tau) \ rVert \ lVert B (\ tau) \ rVert \ lVert Y (\ tau) \ rVert \, d \ tau \ geq bk \ int _ {t_ {0}} ^ {\ infty} \ lVert B (\ tau) \ rVert \, d \ tau \ quad (12).}$

Since the matrix $\quad B(t)$ ${\ displaystyle \ quad B (t)}$ completely integrated on $\quad [0,\infty )$ ${\ displaystyle \ quad [0, \ infty)}$ then $\int _{t_{0}}^{\infty }\lVert B(\tau )\rVert \,d\tau \to 0$ ${\ displaystyle \ int _ {t_ {0}} ^ {\ infty} \ lVert B (\ tau) \ rVert \, d \ tau \ to 0}$ at $t_{0}\to \infty$ ${\ displaystyle t_ {0} \ to \ infty}$ therefore by virtue $\quad (12)$ ${\ displaystyle \ quad (12)}$ starting point $\quad t_{0}$ ${\ displaystyle \ quad t_ {0}}$ can be chosen so large that takes place $\det \lbrack E+Z(t_{0})\rbrack >0.(13)$ ${\ displaystyle \ det \ lbrack E + Z (t_ {0}) \ rbrack> 0. (13)}$ Further $\quad t_{0}\quad$ ${\ displaystyle \ quad t_ {0} \ quad}$ we will consider fixed and assume the presence of inequality $\quad (13)$ ${\ displaystyle \ quad (13)}$ . From here and from the formula $\quad (11)$ ${\ displaystyle \ quad (11)}$ we deduce

${\boldsymbol {y}}(t_{0})=\lbrack E+Z(t_{0})\rbrack <sup>-1</sup>{\boldsymbol {x}}(t_{0}).\qquad (14)$ ${\ displaystyle {\ boldsymbol {y}} (t_ {0}) = \ lbrack E + Z (t_ {0}) \ rbrack <sup> -1 </sup> {\ boldsymbol {x}} (t_ {0 }). \ qquad (14)}$

Since the formulas $\quad (11)$ ${\ displaystyle \ quad (11)}$ and $\quad (14)$ ${\ displaystyle \ quad (14)}$ are equivalent then for each solution $\quad {\boldsymbol {x}}(t)$ ${\ displaystyle \ quad {\ boldsymbol {x}} (t)}$ the system $\quad (1)$ ${\ displaystyle \ quad (1)}$ with initial condition ${\boldsymbol {x}}(t_{0})={\boldsymbol {x_{0}}}\quad$ ${\ displaystyle {\ boldsymbol {x}} (t_ {0}) = {\ boldsymbol {x_ {0}}} \ quad}$ there is only one solution ${\boldsymbol {y}}(t)\quad$ ${\ displaystyle {\ boldsymbol {y}} (t) \ quad}$ the system $\quad (2),$ ${\ displaystyle \ quad (2),}$ which corresponds to the relation established above, namely, this solution, the initial condition ${\boldsymbol {y}}(t_{0})\quad$ ${\ displaystyle {\ boldsymbol {y}} (t_ {0}) \ quad}$ which is determined by the formula $\quad (14).$ ${\ displaystyle \ quad (14).}$

Matching Decisions ${\boldsymbol {x}}(t)$ ${\ displaystyle {\ boldsymbol {x}} (t)}$ and ${\boldsymbol {y}}(t)$ ${\ displaystyle {\ boldsymbol {y}} (t)}$ set by formulas $\quad (11)$ ${\ displaystyle \ quad (11)}$ and $\quad (14),\quad$ ${\ displaystyle \ quad (14), \ quad}$ - one-to-one, i.e. every decision ${\boldsymbol {y}}(t)$ ${\ displaystyle {\ boldsymbol {y}} (t)}$ one and only one solution ${\boldsymbol {x}}(t)\quad$ ${\ displaystyle {\ boldsymbol {x}} (t) \ quad}$ , and vice versa.

Note that the trivial solution ${\boldsymbol {y}}\equiv 0\quad$ ${\ displaystyle {\ boldsymbol {y}} \ equiv 0 \ quad}$ corresponds to a trivial solution ${\boldsymbol {x}}\equiv 0\quad$ ${\ displaystyle {\ boldsymbol {x}} \ equiv 0 \ quad}$ and due to the linearity of the relations $\quad (11)$ ${\ displaystyle \ quad (11)}$ and $\quad (14)$ ${\ displaystyle \ quad (14)}$ various solutions ${\boldsymbol {y_{1}}}(t)$ ${\ displaystyle {\ boldsymbol {y_ {1}}} (t)}$ and ${\boldsymbol {y_{2}}}(t)\quad$ ${\ displaystyle {\ boldsymbol {y_ {2}}} (t) \ quad}$ the system $\quad (2),$ ${\ displaystyle \ quad (2),}$ different solutions ${\boldsymbol {x_{1}}}(t)\quad$ ${\ displaystyle {\ boldsymbol {x_ {1}}} (t) \ quad}$ and ${\boldsymbol {x_{2}}}(t)\quad$ ${\ displaystyle {\ boldsymbol {x_ {2}}} (t) \ quad}$ the system $\quad (1),$ ${\ displaystyle \ quad (1),}$ and vice versa.

For relevant decisions ${\boldsymbol {x}}(t)\quad$ ${\ displaystyle {\ boldsymbol {x}} (t) \ quad}$ and ${\boldsymbol {y}}(t)\quad$ ${\ displaystyle {\ boldsymbol {y}} (t) \ quad}$ estimate the norm of their difference. Since, it is obvious that

{\boldsymbol {x}}(t)=X(t-t_{0}){\boldsymbol {x}}(t_{0}),\qquad

{\ displaystyle {\ boldsymbol {x}} (t) = X (t-t_ {0}) {\ boldsymbol {x}} (t_ {0}), \ qquad}

Where

{\boldsymbol {x}}(t_{0})

{\ displaystyle {\ boldsymbol {x}} (t_ {0})}

defined by the formula

\quad (9)

{\ displaystyle \ quad (9)}

then from the formula

\quad (8)

{\ displaystyle \ quad (8)}

we have

${\boldsymbol {y}}(t)-{\boldsymbol {x}}(t)=\int _{t_{0}}^{t}X_{1}(t-\tau )B(\tau ){\boldsymbol {y}}(\tau )\,d\tau -\int _{t}^{\infty }X_{2}(t-\tau )B(\tau ){\boldsymbol {y}}(\tau )\,d\tau .$ ${\ displaystyle {\ boldsymbol {y}} (t) - {\ boldsymbol {x}} (t) = \ int _ {t_ {0}} ^ {t} X_ {1} (t- \ tau) B ( \ tau) {\ boldsymbol {y}} (\ tau) \, d \ tau - \ int _ {t} ^ {\ infty} X_ {2} (t- \ tau) B (\ tau) {\ boldsymbol { y}} (\ tau) \, d \ tau.}$

Hence, given that

$\lVert {\boldsymbol {y}}(t)\rVert =\lVert Y(t){\boldsymbol {y}}(t_{0})\lVert \leq \lVert Y(t)\rVert \lVert {\boldsymbol {y}}(t_{0})\rVert \leq k\,\lVert {\boldsymbol {y}}(t_{0})\rVert ,$ ${\ displaystyle \ lVert {\ boldsymbol {y}} (t) \ rVert = \ lVert Y (t) {\ boldsymbol {y}} (t_ {0}) \ lVert \ leq \ lVert Y (t) \ rVert \ lVert {\ boldsymbol {y}} (t_ {0}) \ rVert \ leq k \, \ lVert {\ boldsymbol {y}} (t_ {0}) \ rVert,}$ at $t\geq t_{0},$ ${\ displaystyle t \ geq t_ {0},}$

based on ratings $\quad (6)$ ${\ displaystyle \ quad (6)}$ and $\quad (7)$ ${\ displaystyle \ quad (7)}$ we get

\lVert {\boldsymbol {y}}(t)-{\boldsymbol {x}}(t)\rVert \leq \int _{t_{0}}^{t}\lVert X_{1}(t-\tau )\rVert \lVert B(\tau )\rVert \lVert {\boldsymbol {y}}(\tau )\,d\tau +\int _{t}^{\infty }\lVert X_{2}(t-\tau )\rVert \lVert B(\tau )\rVert \lVert {\boldsymbol {y}}(\tau )\,d\tau \leq

{\ displaystyle \ lVert {\ boldsymbol {y}} (t) - {\ boldsymbol {x}} (t) \ rVert \ leq \ int _ {t_ {0}} ^ {t} \ lVert X_ {1} ( t- \ tau) \ rVert \ lVert B (\ tau) \ rVert \ lVert {\ boldsymbol {y}} (\ tau) \, d \ tau + \ int _ {t} ^ {\ infty} \ lVert X_ { 2} (t- \ tau) \ rVert \ lVert B (\ tau) \ rVert \ lVert {\ boldsymbol {y}} (\ tau) \, d \ tau \ leq}

$\leq ak\,\lVert {\boldsymbol {y}}(t_{0})\lVert \int _{t_{0}}^{t}e^{-\alpha (t-\tau )}\lVert B(\tau )\rVert \,d\tau \,+\,bk\,\lVert {\boldsymbol {y}}(t_{0})\rVert \int _{t}^{\infty }\lVert B(\tau )\rVert \,d\tau .(15)$ ${\ displaystyle \ leq ak \, \ lVert {\ boldsymbol {y}} (t_ {0}) \ lVert \ int _ {t_ {0}} ^ {t} e ^ {- \ alpha (t- \ tau) } \ lVert B (\ tau) \ rVert \, d \ tau \, + \, bk \, \ lVert {\ boldsymbol {y}} (t_ {0}) \ rVert \ int _ {t} ^ {\ infty } \ lVert B (\ tau) \ rVert \, d \ tau. (15)}$

Given the absolute integrability of the matrix $\quad B(t)$ ${\ displaystyle \ quad B (t)}$ at $t\geq 2t_{0}$ ${\ displaystyle t \ geq 2t_ {0}}$ we have $\int _{t_{0}}^{t}e^{-\alpha (t-\tau )}\lVert B(\tau )\rVert \,d\tau =\int _{t_{0}}^{\frac {t}{2}}e^{-\alpha (t-\tau )}\lVert B(\tau )\rVert \,d\tau \,+\,\int _{\frac {t}{2}}^{t}e^{-\alpha (t-\tau )}\lVert B(\tau )\rVert \,d\tau \leq$ ${\ displaystyle \ int _ {t_ {0}} ^ {t} e ^ {- \ alpha (t- \ tau)} \ lVert B (\ tau) \ rVert \, d \ tau = \ int _ {t_ { 0}} ^ {\ frac {t} {2}} e ^ {- \ alpha (t- \ tau)} \ lVert B (\ tau) \ rVert \, d \ tau \, + \, \ int _ { \ frac {t} {2}} ^ {t} e ^ {- \ alpha (t- \ tau)} \ lVert B (\ tau) \ rVert \, d \ tau \ leq}$

$\leq e^{-{\frac {\alpha t}{2}}}\int _{0}^{\infty }\lVert B(\tau )\rVert \,d\tau \,+\,\int _{\frac {t}{2}}^{t}\lVert B(\tau )\rVert \,d\tau \,<\varepsilon \,,$ ${\ displaystyle \ leq e ^ {- {\ frac {\ alpha t} {2}}} \ int _ {0} ^ {\ infty} \ lVert B (\ tau) \ rVert \, d \ tau \, + \, \ int _ {\ frac {t} {2}} ^ {t} \ lVert B (\ tau) \ rVert \, d \ tau \, <\ varepsilon \ ,,}$ if a $\quad t>T.$ ${\ displaystyle \ quad t> T.}$

So,

$\lim \limits _{t\to \infty }\int _{t_{0}}^{t}e^{-\alpha (t-\tau )}\lVert B(\tau )\rVert \,d\tau =0.$ ${\ displaystyle \ lim \ limits _ {t \ to \ infty} \ int _ {t_ {0}} ^ {t} e ^ {- \ alpha (t- \ tau)} \ lVert B (\ tau) \ rVert \, d \ tau = 0.}$

Thus, from the inequality $\quad (15)$ ${\ displaystyle \ quad (15)}$ we deduce $\lim \limits _{t\to \infty }[x(t)-y(t)]=0,$ ${\ displaystyle \ lim \ limits _ {t \ to \ infty} [x (t) -y (t)] = 0,}$ i.e. systems $\quad (1)$ ${\ displaystyle \ quad (1)}$ and $\quad (2)$ ${\ displaystyle \ quad (2)}$ asymptotically equivalent. It is proved.

Notes

↑ Brauer, Nonlinear differential equations with forcing terms, Proc. Amer. Math. Soc. 15, 5 (1964), 758-765

Sources

Demidovich B.P. Lectures on the mathematical theory of stability. M .: "Science", 1967. (Russian) (Russian)

[1] Brauer, Nonlinear differential equations with forcing terms, Proc. Amer. Math. Soc. 15, 5 (1964), 758-765

Levinson's theorem

Statement of the theorem

Proof

See also

Notes

Sources

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