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

Anthony Kramer

Anthony Kramer

Answered question

2022-05-19

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 )?

Answer & Explanation

asafand2c

asafand2c

Beginner2022-05-20Added 11 answers

However, the answer is "no" 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.

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