Suppose we have a 1-dimensional differential inequality d x d t ≤ x − x 3 We can apply the Comparison principle to claim that if y ( t ) is the solution to d y d t = y − y 3 , then x ( t ) ≤ y ( t ) (assuming x ( 0 ) ≤ y ( 0 ). Can we extend similar argument to a 2-dimensional system? For example, let us consider the following system of equations d x 1 d t = x 1 − x 2 , d x 2 d t ≤ x 1 + x 2 − x 2 4 x 1 2 Is the solution to following d y 1 d t = y 1 − y 2 , d y 2 d t = y 1 + y 2 − y 2 4 y 1 2 related with the solution of the original problem. Specifically, can we apply the Comparison principle to first say that x 2 ( t ) ≤ y 2 ( t ) and then subsequently use it to claim x 1 ( t ) ≥ y 1 ( t )?

Question

Suppose we have a   1-dimensional differential inequality            d      x              d      t        ≤  x  −      x    3  We can apply the Comparison principle to claim that if   y  (  t  ) is the solution to             d      y              d      t        =  y  −      y    3  , then   x  (  t  )  ≤  y  (  t  ) (assuming   x  (  0  )  ≤  y  (  0  ). Can we extend similar argument to a   2-dimensional system? For example, let us consider the following system of equations            d              x        1                    d      t        =      x    1    −      x    2    ,                  d              x        2                    d      t        ≤      x    1    +      x    2    −            x      2      4              x      1      2      Is the solution to following            d              y        1                    d      t        =      y    1    −      y    2    ,                  d              y        2                    d      t        =      y    1    +      y    2    −            y      2      4              y      1      2      related with the solution of the original problem. Specifically, can we apply the Comparison principle to first say that       x    2    (  t  )  ≤      y    2    (  t  ) and then subsequently use it to claim       x    1    (  t  )  ≥      y    1    (  t  )?

asafand2c · Accepted Answer

However, the answer is &quot;no&quot; for general problems, even if they are linear. Consider the structure:                              x          ′                                    =        f        (        x        ,        y        )                                      y          ′                                    =        g        (        x        ,        y        )        ≤        h        (        x        ,        y        )            Specifically, compare the following two systems:System 1                              x          ′                                    =        x        −        y                                      y          ′                                    =                                                    2              x              −              y              −              1                        ⏟                                g            (            x            ,            y            )                              System 2                              w          ′                                    =        w        −        z                                      z          ′                                    =                                                    2              w              −              z                        ⏟                                h            (            w            ,            z            )                              Suppose initial conditions are  (  x  (  0  )  ,  y  (  0  )  )  =  (  w  (  0  )  ,  z  (  0  )  )  =  (  0  ,  0  ).Then   (  w  (  t  )  ,  z  (  t  )  )  =  (  0  ,  0  ) for all   t  ≥  0. However,   (  x  (  t  )  ,  y  (  t  )  ) has solution:                    x        (        t        )                            =        1        −        cos        ⁡        (        t        )                            y        (        t        )                            =        1        −        sin        ⁡        (        t        )        −        cos        ⁡        (        t        )            and indeed   y  (  t  ) takes both positive and negative values over   t  ∈  [  0  ,  ∞  ). So we cannot say that   y  (  t  )  ≤                      z        (        t        )            ⏟          0   for all   t.

Suppose we have a 1 -dimensional differential inequality d x </m

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