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February 27th, 2018, 02:02 PM  #1 
Newbie Joined: Feb 2018 From: Perú Posts: 3 Thanks: 1  How to prove the inequality of solutions for two Cauchy problems?
View image Thank you for your suggestion. 
March 2nd, 2018, 08:56 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 383 Thanks: 207 Math Focus: Dynamical systems, analytic function theory, numerics 
Just integrate to obtain $\Phi_i$ directly. You get \[ \Phi_i(t,x) = x + \int_0^t g_i(s,\phi_i(s,x)) \ ds \] so clearly $\Phi_1 \leq \Phi_2$ since this obviously holds for these integrals under the assumption that $g_1 \leq g_2$. 
March 3rd, 2018, 07:01 PM  #3 
Newbie Joined: Feb 2018 From: Perú Posts: 3 Thanks: 1  Why? Thank you

March 4th, 2018, 10:20 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 383 Thanks: 207 Math Focus: Dynamical systems, analytic function theory, numerics 
Your integrals make no sense. There should be no $t$ in the integrand. Also, you don't need to prove that \[ \int_0^1 g_1(s,\Phi_1(t,s)) \ ds \leq \int_0^1 g_2(s,\Phi_2(t,s)) \ ds \]. You only need to prove that the difference, $\Phi_2  \Phi_1$ is nonnegative whcih easily follows from the fact that its derivative is nonnegative. Specifically, write $h = \Phi_2  \Phi_1$ and $G = g_2  g_1$, then $\frac{dh}{dt} = G$ and $h(0,x) = 0$ for all $x$. Integrating the differential equation for $h$, you have \[h(t,x) = \int_0^t G(s, h(t,s)) \ ds \] so clearly $h \geq 0$. Is this more clear? 
March 5th, 2018, 01:56 PM  #5 
Newbie Joined: Feb 2018 From: Perú Posts: 3 Thanks: 1  Thank you very much
You're right. I have made a mistake.


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cauchy, inequality, problems, prove, solutions 
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