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Let $u:\mathbb{R}^n\setminus \{0\} \rightarrow \mathbb{R}$ be a $C^1$-function which is positive homogeneous of degree $k \neq 0$, then by Euler's theorem $x \cdot \nabla u = k u$. Homogeneous polynomials of degree $k$ satisfy this equation, can I evoke a uniqueness theorem to conclude that the these homogeneous functions are necessarily sum of Homogeneous polynomials of degree $k$? In case this is true, is there any other simple proof of this fact?

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Uniqueness theorems always come with some requirements, with their exact nature depending on the nature of the PDE. In the case of a first-order equation like $x \cdot \nabla u =k u$, this is provided by the method of characteristics: solutions to the PDE are uniquely determined by their values on a surface that intersects each characteristic exactly once. Since the characteristics here are just the rays from the origin, the most natural surface to use is the unit sphere $$\mathbb S^{n-1} = \{ x \in \mathbb R^n : |x| = 1\}.$$

Indeed, any function $u$ defined on the sphere can be extended to $\mathbb R^n \setminus \{0\}$ by $$u(x) = |x|^k u\left(\frac{x}{|x|}\right),$$ which is homogeneous of degree $k$; so there are as many $k$-homogeneous functions as there are functions on the sphere. In particular, there is an infinite-dimensional vector space of such functions, while the homogeneous polynomials of degree $k$ form a finite-dimensional space.

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The function $f: \mathbb{R}^n \backslash \{0\} \to \mathbb{R}$ given by $f(x) = |x|^k$ is homogeneous of degree $k$ but not a polynomial.

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  • $\begingroup$ Of course! thanks. But then there is no uniqueness to the PDE problem? $\endgroup$ – inquisitor Dec 1 '17 at 16:44

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