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On solution of a nonlinear differential inequality
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I have the following differential inequality:
$$f'(x)geq cf(x)^a,quad forall xin[0,1]$$
where $0<a<1,, f(x)geq 0$.
I'm taking the following approach to solve the problem:
$$f'(x)geq cf(x)^aRightarrow fracdfdxf(x)^-ageq c$$
Integrating from $alpha$ to $t$ gives:
$$int_x=alpha^t f(x)^-adf-c, dxgeq 0$$
Then:
$$fracf(t)^1-a1-a-fracf(alpha)^1-a1-ageq c(t-alpha)$$
$$ f(t)geq left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a$$
This answer seems different from the one obtain in my previous question. Is this answer correct?
Also there exist some sort of initial condition in this solution. I mean to obtain a lower bound on $f(x)$ one should know the value of $f$ at some point $alpha$. But the derivation seems to be true for all $alpha$-s. Hence I conclude:
$$f(t)geq left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a forall alphain[0,1]$$
$$f(t)geq max_alphain[0,1]left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a$$
If this conclusion is true there doesn't exist any closed form lower-bound on $f(x)$ given initial condition for a single point (since the lower-bound depends on $f$ over the interval $[0,1]$). With this respect I think I didn't actually solve the problem. Because the solution to $f$ depends on $f$ over the entire interval. so
Is it possible to find a closed form necessary and sufficient condition on $f(x)$ to be a solution of $f'(x)geq cf(x)^a$?
calculus differential-equations integral-inequality nonlinear-analysis
asked Jul 30 at 7:00
SMA.D
422420
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up vote
2
down vote
favorite
I have the following differential inequality:
$$f'(x)geq cf(x)^a,quad forall xin[0,1]$$
where $0<a<1,, f(x)geq 0$.
I'm taking the following approach to solve the problem:
$$f'(x)geq cf(x)^aRightarrow fracdfdxf(x)^-ageq c$$
Integrating from $alpha$ to $t$ gives:
$$int_x=alpha^t f(x)^-adf-c, dxgeq 0$$
Then:
$$fracf(t)^1-a1-a-fracf(alpha)^1-a1-ageq c(t-alpha)$$
$$ f(t)geq left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a$$
This answer seems different from the one obtain in my previous question. Is this answer correct?
Also there exist some sort of initial condition in this solution. I mean to obtain a lower bound on $f(x)$ one should know the value of $f$ at some point $alpha$. But the derivation seems to be true for all $alpha$-s. Hence I conclude:
$$f(t)geq left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a forall alphain[0,1]$$
$$f(t)geq max_alphain[0,1]left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a$$
If this conclusion is true there doesn't exist any closed form lower-bound on $f(x)$ given initial condition for a single point (since the lower-bound depends on $f$ over the interval $[0,1]$). With this respect I think I didn't actually solve the problem. Because the solution to $f$ depends on $f$ over the entire interval. so
Is it possible to find a closed form necessary and sufficient condition on $f(x)$ to be a solution of $f'(x)geq cf(x)^a$?
calculus differential-equations integral-inequality nonlinear-analysis
asked Jul 30 at 7:00
SMA.D
422420
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following differential inequality:
$$f'(x)geq cf(x)^a,quad forall xin[0,1]$$
where $0<a<1,, f(x)geq 0$.
I'm taking the following approach to solve the problem:
$$f'(x)geq cf(x)^aRightarrow fracdfdxf(x)^-ageq c$$
Integrating from $alpha$ to $t$ gives:
$$int_x=alpha^t f(x)^-adf-c, dxgeq 0$$
Then:
$$fracf(t)^1-a1-a-fracf(alpha)^1-a1-ageq c(t-alpha)$$
$$ f(t)geq left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a$$
This answer seems different from the one obtain in my previous question. Is this answer correct?
Also there exist some sort of initial condition in this solution. I mean to obtain a lower bound on $f(x)$ one should know the value of $f$ at some point $alpha$. But the derivation seems to be true for all $alpha$-s. Hence I conclude:
$$f(t)geq left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a forall alphain[0,1]$$
$$f(t)geq max_alphain[0,1]left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a$$
If this conclusion is true there doesn't exist any closed form lower-bound on $f(x)$ given initial condition for a single point (since the lower-bound depends on $f$ over the interval $[0,1]$). With this respect I think I didn't actually solve the problem. Because the solution to $f$ depends on $f$ over the entire interval. so
Is it possible to find a closed form necessary and sufficient condition on $f(x)$ to be a solution of $f'(x)geq cf(x)^a$?
calculus differential-equations integral-inequality nonlinear-analysis
asked Jul 30 at 7:00
SMA.D
422420
I have the following differential inequality:
$$f'(x)geq cf(x)^a,quad forall xin[0,1]$$
where $0<a<1,, f(x)geq 0$.
I'm taking the following approach to solve the problem:
$$f'(x)geq cf(x)^aRightarrow fracdfdxf(x)^-ageq c$$
Integrating from $alpha$ to $t$ gives:
$$int_x=alpha^t f(x)^-adf-c, dxgeq 0$$
Then:
$$fracf(t)^1-a1-a-fracf(alpha)^1-a1-ageq c(t-alpha)$$
$$ f(t)geq left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a$$
This answer seems different from the one obtain in my previous question. Is this answer correct?
Also there exist some sort of initial condition in this solution. I mean to obtain a lower bound on $f(x)$ one should know the value of $f$ at some point $alpha$. But the derivation seems to be true for all $alpha$-s. Hence I conclude:
$$f(t)geq left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a forall alphain[0,1]$$
$$f(t)geq max_alphain[0,1]left(c(1-a)t+f(alpha)^1-a-cfracalpha1-aright)^frac11-a$$
If this conclusion is true there doesn't exist any closed form lower-bound on $f(x)$ given initial condition for a single point (since the lower-bound depends on $f$ over the interval $[0,1]$). With this respect I think I didn't actually solve the problem. Because the solution to $f$ depends on $f$ over the entire interval. so
Is it possible to find a closed form necessary and sufficient condition on $f(x)$ to be a solution of $f'(x)geq cf(x)^a$?
calculus differential-equations integral-inequality nonlinear-analysis
asked Jul 30 at 7:00
SMA.D
422420
edited Jul 30 at 14:27
asked Jul 30 at 7:00
SMA.D
422420
asked Jul 30 at 7:00
SMA.D
422420
asked Jul 30 at 7:00
SMA.D
422420
422420
add a comment |Â
add a comment |Â
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