Consider the differential system { <mtable columnalign="left left" rowspacing=".2em" co

Rachel Villa 2022-05-22 Answered
Consider the differential system
{ y ( t ) = a y ( t ) 3 + b z ( t ) z ( t ) = c z ( t ) 5 b y ( t )
with t > 0
y ( 0 ) = y 0 , z ( 0 ) = z 0 , a < 0 , c < 0 , b R
the question is to prouve that this system admits a unique solution on [ 0 , + ]?
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Answers (1)

Erick Blake
Answered 2022-05-23 Author has 10 answers
By standard ODE theorems, we know that if a solution doesn't exist, it will only not exist because the solution "blows up," that is, if there exists a T > 0 such that x ( t ) or y ( t ) is unbounded as t T .
Multiply the first equation by y ( t ), and the second solution by z ( t ), and then add them together, you obtain
1 2 d d t ( ( y ( t ) ) 2 + ( z ( t ) ) 2 ) = a ( y ( t ) ) 4 + c ( z ( t ) ) 6 .
Since a and c are negative, it follows that ( y ( t ) ) 2 + ( z ( t ) ) 2 is bounded. Hence a blow up can never happen.
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