Let u , v : [ 0 , ∞ ) → [ 0 , ∞ ) satsfying the following system of differential inequalities: u ′ ( t ) ≤ a 1 u ( t ) + a 2 v ( t ) + a 0 v ′ ( t ) ≤ b 1 u ( t ) + b 2 v ( t ) + b 0 for suitable coefficients a 0 , a 1 , a 2 , b 0 , b 1 , b 2 ∈ R . In particular I have a 1 , b 2 &lt; 0 and 0 &lt; b 1 &lt; | b 2 | Is there a Gronwall lemma for this system of linear differential inqualities? Namely an (optimal) inequality of type u ( t ) ≤ F ( t ) v ( t ) ≤ G ( t )where the functions F , G : [ 0 , ∞ ) → [ 0 , ∞ ) depend on u , v only through their initial values u ( 0 ) , v ( 0 )?

Question

Let   u  ,  v  :  [  0  ,  ∞  )  →  [  0  ,  ∞  ) satsfying the following system of differential inequalities:      u    ′    (  t  )  ≤      a    1      u  (  t  )  +      a    2      v  (  t  )  +      a    0            v    ′    (  t  )  ≤      b    1      u  (  t  )  +      b    2      v  (  t  )  +      b    0  for suitable coefficients       a    0    ,      a    1    ,      a    2    ,      b    0    ,      b    1    ,      b    2    ∈  R    . In particular I have       a    1    ,      b    2    &amp;lt;  0   and   0  &amp;lt;      b    1    &amp;lt;  |      b    2    |  Is there a Gronwall lemma for this system of linear differential inqualities? Namely an (optimal) inequality of type  u  (  t  )  ≤  F  (  t  )    v  (  t  )  ≤  G  (  t  )where the functions   F  ,  G  :  [  0  ,  ∞  )  →  [  0  ,  ∞  ) depend on   u  ,  v only through their initial values   u  (  0  )  ,  v  (  0  )?

Colin Moran · Accepted Answer

If   x  ∈        R    n  ,   A  ∈        M          n      ×      n        , and   a is a vector function of one variable with  X  (  t  )  ≤  a  (  t  )  +      ∫            t      0        t    A  X  (  s  )  d  sthen it is true that  X  (  t  )  ≤  a  (  t  )  +      ∫            t      0        t    V  (  t  ,  s  )  A  a  (  s  )  d  swhere the kernel is an   n  ×  n matrix that satisfies the relation  V  (  t  ,  s  )  =  I  +      ∫    s    t    A  V  (  y  ,  s  )  d  yHere the inequalities are taken component-wise. Basically what this theorem really says is that, indeed, a system of linear inequalities is upper bounded by the solution of the corresponding linear system of equalities. One can verify doing some trivial matrix algebra that whenever A is constant, the kernel has the form  V  (  t  ,  s  )  =      e          A      (      t      −      s      )      Finally, we turn to our specific problem. To transform it into the form required by the theorem, set  X  (  t  )  =      (                            u          (          t          )                                      v          (          t          )                      )       ,     A  =      (                                        a            1                                                a            2                                                            b            1                                                b            2                                )       ,     c  =      (                                        a            0                                                            b            0                                )  and integrate all inequalities once to obtain  X  (  t  )  ≤      X    0    +  c  t  +      ∫      0    t    A  X  (  s  )  d  sfor which the extended Gronwall&#039;s lemma implies that  X  (  t  )  ≤      X    0    +  c  t  +      ∫      0    t        e          A      (      t      −      s      )        A  (      X    0    +  c  s  )  d  swhich, noting that   A as defined above is invertible, with the help of the following integrals      ∫      0    t        e          −      A      s        d  s  =      A          −      1        (  I  −      e          −      A      t        )      ∫      0    t    s      e          −      A      s        d  s  =      A          −      2        (  I  −  (  I  +  A  t  )      e          −      A      t        )can be written in the simple form  X  (  t  )  ≤      X    0        e          A      t        +      A          −      1        (      e          A      t        −  1  )  cCalculating matrix exponentials for general coefficients requires diagonalization of the matrix   A, which is tractable since the matrix is only   2  ×  2

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