gugo82 ha scritto:1. Let \( \Omega \subseteq \mathbb{R}^2\setminus \{\mathbf{o}\} \) be an open set made up of rays from \( \mathbf{o}=(0,0) \), let \( f,g:\Omega \to \mathbb{R} \) be two smooth \( \alpha \)-homogeneous functions with \( \alpha\neq -1 \) and \( (x_0,y_0)\in \Omega \) s.t. \( g(x_0,y_0)\neq 0 \).
Prove that if:
\[ \tag{1} f_y (x,y) = g_x (x,y) \]
in \( \Omega \), then any solution of the Cauchy's problem:
\[ \tag{P} \begin{cases} g(x,y(x))\ y^\prime (x) = - f(x,y(x))\\ y(x_0) = y_0 \end{cases} \]
is implicitly defined by the following equation:
\[ \tag{2} x\ f(x,y) + y\ g(x,y) = x_0\ f(x_0,y_0) + y_0\ g(x_0 ,y_0) \]
in a suitable neighbourhood of the initial data \( (x_0,y_0) \).
Let:
\[
\Phi (x,y) := x\ f(x,y) + y\ g(x,y)\; .
\]
Since the functions \((x,y)\mapsto x\) and \((x,y)\mapsto y\) are \(1\)-homogeneous, both products \(x\; f(x,y)\) and \(y\; g(x,y)\) are \(\alpha +1\)-homogeneous functions; hence, \(\Phi\) is an \(\alpha +1\)-homogenous function which is differentiable in \(\Omega\) whose derivatives satisfy:
\[
\begin{split}
\Phi_x (x,y) &= f(x,y) + x\ f_x(x,y) + y\ g_x(x,y)\\
&= f(x,y) + x\ f_x(x,y) + y\ f_y(x,y)\\
&= f(x,y) + \alpha\ f(x,y)\\
&= (\alpha +1)\ f(x,y)\\
\Phi_y(x,y) &= g(x,y) + x\ f_y(x,y) + y\ g_y(x,y))\\
&= g(x,y) + x\ g_x(x,y) + y\ g_y(x,y)\\
&= g(x,y) + \alpha\ g(x,y)\\
&= (\alpha +1)\ g(x,y)
\end{split}
\]
with \(\alpha +1\neq 0\), because of
Euler's Theorem on positively homogeneous functions. Therefore the ODE in (
P) rewrites:
\[\tag{I}
\Phi_x(x,y(x))+ \Phi_y(x,y(x))\ y^\prime (x) =0\; .
\]
Now, assume that (
P) has at least a solution (which is indeed the case, because
Peano's Existence Theorem applies).
Upon setting:
\[
\phi (x) := \Phi (x,y(x))\; ,
\]
from (
I) we infer:
\[
\phi^\prime (x) =0
\]
and finally:
\[
\tag{II}
\phi (x) = \phi (x_0)
\]
everywhere in the interval where the maximal solution of (
P) is defined. But (
II) is in fact equivalent to (
2), so we have the claim.