$f$ is function that $f(0)=1$ and $f:[0,\infty)\to\mathbb{R}$. It's continuous on $[0,\infty)$ and is $C^2$ on $(0,\infty)$. Let $u$ be another function defined on $\mathbb{R}^2$ and $u(x,y)=f(\sqrt{x^2+y^2})$. In $\mathbb{R}^2$ we have the equation $\Delta u=0$.
By polar coodinates I knew $f$ satisfies the following ODE: $$\Delta u=f''(r)+\frac{1}{r}f'(r)=0$$ Then easily got the general solution $f(r)=C_{1}\log r+C_{2}$ on 2D Laplacian equation. By condition $f(0)=1$ and its continuity, meaning that $$f(0)=\lim_{r\to0}C_{1}\log r+C_{2}=1$$ must satisfy $C_{1}=0$ and $C_{2}=1$. Does the solution $f\equiv1$ just the unique one solution?
I have met this kind of question in test for many times. Another version is, $f$ is replaced as two variable function $F(r,\theta)$, the condition change to $\lim_{r\to0}F(r,0)=1$ and $F_{\theta}(r,\theta)$, which still lead to the same result $F(r,\theta)\equiv1$.