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February 27th, 2018, 02:02 PM   #1
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How to prove the inequality of solutions for two Cauchy problems?

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 March 2nd, 2018, 08:56 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 620 Thanks: 391 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
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Why?

Quote:
 Originally Posted by SDK 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 $g_1 \leq g_2$.
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 March 4th, 2018, 10:20 PM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 620 Thanks: 391 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 non-negative whcih easily follows from the fact that its derivative is non-negative. 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? Thanks from Mategatico
 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. Thanks from Joppy

 Tags cauchy, inequality, problems, prove, solutions

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