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February 27th, 2018, 03:02 PM   #1
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How to prove the inequality of solutions for two Cauchy problems?

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March 2nd, 2018, 09:56 PM   #2
SDK
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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 3rd, 2018, 08:01 PM   #3
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Why?

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Originally Posted by SDK View Post
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, 11:20 PM   #4
SDK
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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?
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March 5th, 2018, 02:56 PM   #5
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Thank you very much

You're right. I have made a mistake.
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