$(frac{d}{dt} + 2a)int w^2 dx le 0 implies w=0$ where $a>0$
$begingroup$
$F(t) := int w^2 dx$ where x is a vector and $a,t>0$
Given that $(frac{d}{dt} - 2a)F(t) le 0$
How do I deduce that $w=0$?
My idea was to do:
$(frac{d}{dt} + 2a)F(t) le 0$
$implies $$frac{d}{dt}F(t) le 2aF(t)$
$implies $$int frac{d}{dt}F(t) le int 2aF(t)$
$implies $$F(t) le int 2aF(t)$
Since $a, F(t) > 0 $ $implies $$int 2aF(t)ge0$
$ implies F(t) le 0$ just realised this line is not true either
Since $F(t) ge 0 implies F(t)=0 implies w=0$
The only line am i not sure about is:
$int frac{d}{dt}F(t) le int 2aF(t)$ $implies $$F(t) le int 2aF(t)$
because there should be a constant of integration introduced somewhere
The soln i was given was that $F(t)$ is bdd above by $G(t)$ where $(frac{d}{dt}-2a)G(t) = 0$ and $G(0)=0$ so $implies G=0$
but since $F(t) le 0 implies F=0 implies w=0$
I cant seem to understand this since answer since G isnt defined explicitly
ordinary-differential-equations multivariable-calculus
asked Dec 15 '18 at 13:45
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
$F(t) := int w^2 dx$ where x is a vector and $a,t>0$
Given that $(frac{d}{dt} - 2a)F(t) le 0$
How do I deduce that $w=0$?
My idea was to do:
$(frac{d}{dt} + 2a)F(t) le 0$
$implies $$frac{d}{dt}F(t) le 2aF(t)$
$implies $$int frac{d}{dt}F(t) le int 2aF(t)$
$implies $$F(t) le int 2aF(t)$
Since $a, F(t) > 0 $ $implies $$int 2aF(t)ge0$
$ implies F(t) le 0$ just realised this line is not true either
Since $F(t) ge 0 implies F(t)=0 implies w=0$
The only line am i not sure about is:
$int frac{d}{dt}F(t) le int 2aF(t)$ $implies $$F(t) le int 2aF(t)$
because there should be a constant of integration introduced somewhere
The soln i was given was that $F(t)$ is bdd above by $G(t)$ where $(frac{d}{dt}-2a)G(t) = 0$ and $G(0)=0$ so $implies G=0$
but since $F(t) le 0 implies F=0 implies w=0$
I cant seem to understand this since answer since G isnt defined explicitly
ordinary-differential-equations multivariable-calculus
asked Dec 15 '18 at 13:45
pablo_mathscobarpablo_mathscobar
996
$endgroup$
$begingroup$
$G(t)$ is defined expticitly: $G(t) = 0$ for all $t$. What you need to do is to make use of the fact that $G$ is the solution of the differential equation $G' = 2aG$ with $G(0) = 0$ to show that it must also be an upper bound for $F$.
$endgroup$
– Paul Sinclair
Dec 15 '18 at 21:58
$begingroup$
@PaulSinclair but how does G being a soln to that diff eq make it an upper bound for f
$endgroup$
– pablo_mathscobar
Dec 15 '18 at 21:59
add a comment |
$begingroup$
$F(t) := int w^2 dx$ where x is a vector and $a,t>0$
Given that $(frac{d}{dt} - 2a)F(t) le 0$
How do I deduce that $w=0$?
My idea was to do:
$(frac{d}{dt} + 2a)F(t) le 0$
$implies $$frac{d}{dt}F(t) le 2aF(t)$
$implies $$int frac{d}{dt}F(t) le int 2aF(t)$
$implies $$F(t) le int 2aF(t)$
Since $a, F(t) > 0 $ $implies $$int 2aF(t)ge0$
$ implies F(t) le 0$ just realised this line is not true either
Since $F(t) ge 0 implies F(t)=0 implies w=0$
The only line am i not sure about is:
$int frac{d}{dt}F(t) le int 2aF(t)$ $implies $$F(t) le int 2aF(t)$
because there should be a constant of integration introduced somewhere
The soln i was given was that $F(t)$ is bdd above by $G(t)$ where $(frac{d}{dt}-2a)G(t) = 0$ and $G(0)=0$ so $implies G=0$
but since $F(t) le 0 implies F=0 implies w=0$
I cant seem to understand this since answer since G isnt defined explicitly
ordinary-differential-equations multivariable-calculus
asked Dec 15 '18 at 13:45
pablo_mathscobarpablo_mathscobar
996
$endgroup$
$F(t) := int w^2 dx$ where x is a vector and $a,t>0$
Given that $(frac{d}{dt} - 2a)F(t) le 0$
How do I deduce that $w=0$?
My idea was to do:
$(frac{d}{dt} + 2a)F(t) le 0$
$implies $$frac{d}{dt}F(t) le 2aF(t)$
$implies $$int frac{d}{dt}F(t) le int 2aF(t)$
$implies $$F(t) le int 2aF(t)$
Since $a, F(t) > 0 $ $implies $$int 2aF(t)ge0$
$ implies F(t) le 0$ just realised this line is not true either
Since $F(t) ge 0 implies F(t)=0 implies w=0$
The only line am i not sure about is:
$int frac{d}{dt}F(t) le int 2aF(t)$ $implies $$F(t) le int 2aF(t)$
because there should be a constant of integration introduced somewhere
The soln i was given was that $F(t)$ is bdd above by $G(t)$ where $(frac{d}{dt}-2a)G(t) = 0$ and $G(0)=0$ so $implies G=0$
but since $F(t) le 0 implies F=0 implies w=0$
I cant seem to understand this since answer since G isnt defined explicitly
ordinary-differential-equations multivariable-calculus
ordinary-differential-equations multivariable-calculus
asked Dec 15 '18 at 13:45
pablo_mathscobarpablo_mathscobar
996
asked Dec 15 '18 at 13:45
pablo_mathscobarpablo_mathscobar
996
edited Dec 16 '18 at 11:54
pablo_mathscobar
asked Dec 15 '18 at 13:45
pablo_mathscobarpablo_mathscobar
996
asked Dec 15 '18 at 13:45
pablo_mathscobarpablo_mathscobar
996
asked Dec 15 '18 at 13:45
pablo_mathscobarpablo_mathscobar
996
996
$begingroup$
$G(t)$ is defined expticitly: $G(t) = 0$ for all $t$. What you need to do is to make use of the fact that $G$ is the solution of the differential equation $G' = 2aG$ with $G(0) = 0$ to show that it must also be an upper bound for $F$.
$endgroup$
– Paul Sinclair
Dec 15 '18 at 21:58
$begingroup$
@PaulSinclair but how does G being a soln to that diff eq make it an upper bound for f
$endgroup$
– pablo_mathscobar
Dec 15 '18 at 21:59
add a comment |
$begingroup$
$G(t)$ is defined expticitly: $G(t) = 0$ for all $t$. What you need to do is to make use of the fact that $G$ is the solution of the differential equation $G' = 2aG$ with $G(0) = 0$ to show that it must also be an upper bound for $F$.
$endgroup$
– Paul Sinclair
Dec 15 '18 at 21:58
$begingroup$
@PaulSinclair but how does G being a soln to that diff eq make it an upper bound for f
$endgroup$
– pablo_mathscobar
Dec 15 '18 at 21:59
$begingroup$
$G(t)$ is defined expticitly: $G(t) = 0$ for all $t$. What you need to do is to make use of the fact that $G$ is the solution of the differential equation $G' = 2aG$ with $G(0) = 0$ to show that it must also be an upper bound for $F$.
$endgroup$
– Paul Sinclair
Dec 15 '18 at 21:58
$begingroup$
$G(t)$ is defined expticitly: $G(t) = 0$ for all $t$. What you need to do is to make use of the fact that $G$ is the solution of the differential equation $G' = 2aG$ with $G(0) = 0$ to show that it must also be an upper bound for $F$.
$endgroup$
– Paul Sinclair
Dec 15 '18 at 21:58
$begingroup$
@PaulSinclair but how does G being a soln to that diff eq make it an upper bound for f
$endgroup$
– pablo_mathscobar
Dec 15 '18 at 21:59
$begingroup$
@PaulSinclair but how does G being a soln to that diff eq make it an upper bound for f
$endgroup$
– pablo_mathscobar
Dec 15 '18 at 21:59
add a comment |
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$begingroup$
$G(t)$ is defined expticitly: $G(t) = 0$ for all $t$. What you need to do is to make use of the fact that $G$ is the solution of the differential equation $G' = 2aG$ with $G(0) = 0$ to show that it must also be an upper bound for $F$.
$endgroup$
– Paul Sinclair
Dec 15 '18 at 21:58
$begingroup$
@PaulSinclair but how does G being a soln to that diff eq make it an upper bound for f
$endgroup$
– pablo_mathscobar
Dec 15 '18 at 21:59