is there some analytical solution to the following problem:
Finding all $f:\mathbb{R}^2 \to \mathbb{R}$, $f\in C^1$ such that:
$$ x \cdot \nabla f(x) = f(x)$$
This Euler Relation means that $f$ is positively homogeneous of degree 1. In other words, for any $\alpha \geq 0$, $f(\alpha x) = \alpha f(x)$. My goal is to find all such $C^1$ functions in $\mathbb{R}^2$!
Thank you very much!
On
A partial answer, see the note at the bottom.
It is notoriously difficult to find the general solution to a PDE, even all the smooth continuous ones such as in this case.
In Cartesian coordinates, our equation reads:
$$x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=f$$
Let's consider the solutions in the form:
$$f(x,y)=\sum_{n,m} C_{n,m} x^n y^m$$
Substitution gives us the condition:
$$n+m=1$$
So we have either:
$$f(x,y)=x \sum_n A_n \frac{y^n}{x^n}$$
or:
$$f(x,y)=y \sum_m B_m \frac{x^m}{y^m}$$
Any convergent series in the form above will be a solution, and there's a lot of them.
But this kind of series doesn't describe all the possible solutions.
If we denote:
$$g= \frac{1}{f}$$
we obtain the following related equation:
$$x \frac{\partial g}{\partial x}+y \frac{\partial g}{\partial y}=-g$$
Which has series solutions in a similar form, but with the condition:
$$m+n=-1$$
One interesting particular solution is:
$$g_p(x,y)= C\int_0^\infty \frac{dt}{e^{xt}+e^{yt}}=\frac{C}{x} {_2 F_1 } \left(1, \frac{x}{x-y};1+\frac{x}{x-y};-1 \right)$$
Where $C$ is a constant.
We can also use other methods, like Green's function, which however will depend on the boundary conditions we assume.
Important note:
If the OP wants only solutions which are $C^1$ everywhere in $\mathbb{R}^2$, then the series solutions above don't all work.
Currently, I can't provide any example of a nontrivial solution which is $C^1$ everywhere. The trivial solution $f=0$ clearly works.
First we can change to polar coordinates to simplify the equation: if $x=r\cos{\theta}$ and $y=r\sin{\theta}$, then away from $r=0$, $$ (x,y) \cdot \nabla f = x \left( \frac{\partial r}{\partial x} \frac{\partial f}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial f}{\partial \theta} \right) + y \left( \frac{\partial r}{\partial y} \frac{\partial f}{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial f}{\partial \theta} \right) \\ = \frac{1}{r} x^2 \frac{\partial f}{\partial r} - \frac{xy}{r^2} \frac{\partial f}{\partial \theta} + \frac{y^2}{r} \frac{\partial f}{\partial r} + \frac{xy}{r^2} \frac{\partial f}{\partial \theta} \\ = r \frac{\partial f}{\partial r} , $$ and so the original equation becomes $$ r \frac{\partial f}{\partial r} = f . $$ Integrating this is easy, and gives $ f(r,\theta) = r g(\theta) $ for some function $g$. We note from the original equation that $f(0) = 0 \cdot \nabla f(0) = 0$, so this $f$ extends continuously to $0$. It remains to determine $g$. It is apparent from considering derivatives in the angular direction on $r= 1$ that $g$ must be $C^1$ for $f$ to be $C^1$. If we can live without $f$ being $C^1$ at $0$, this is all we can say. But for $f$ to be $C^1$ at $0$, it must have directional and partial derivatives at $0$. We can write the directional derivative at $0$ in two ways: as $ \frac{\partial f}{\partial r}(0,\theta) = g(\theta) $, and $$ \frac{\partial f}{\partial r}(0,\theta) = \frac{\partial x}{\partial r}(0,0) \frac{\partial f}{\partial x}(0,0) + \frac{\partial y}{\partial r}(0,0) \frac{\partial f}{\partial y}(0,0) \\ = A \cos{\theta} + B \sin{\theta} $$ for some fixed constants $A$ and $B$. Therefore we must have $g(\theta) = A\cos{\theta}+B\sin{\theta}$ for each $\theta$, and so $$ f(\mathbf{x}) = f(x,y) = rg(\theta) = Ar\cos{\theta} + B r\sin{\theta} = Ax+By = \mathbf{a} \cdot \mathbf{x} $$ for some $\mathbf{a} \in \mathbb{R}^2$ (abusing notation somewhat in this equation).
(And so we in fact find that any $C^1$ solution is actually $C^{\infty}$.)