Choose parameters to make a harmonic function

Question

Let $B(0;1)=\{x\in \mathbb{R}^N ;|x|\leq 1\}$, the ball of $\mathbb{R}^N$ equipped with the euclidian scalar product $$x \cdot y=x_1y_1+...+x_Ny_N,\ \ \ x=(x_1,...,x_N),\ \ y=(y_1,...,y_N)\ \ \ |x|=\sqrt{x\cdot x}.$$ Let $\alpha\mbox{ , } \beta \mbox{ two real numbers, } \alpha\geq 0\mbox{ , } a \in \partial B(0,1)$ fix. $$h(x)=\log\left((1-x\cdot a)^{\alpha}+|x-a|^{\beta}\right)$$ Can we find $\alpha$ and $\beta$ such that $h$ is harmonic in $B(0,1)$?

Answer 1

Without loss of generality, choose a unit vector $a=(a_1,\dots,a_n)$ with components $a_1=1$, $a_j=0\;\,\forall\,j\geqslant 2$.  This may be done by rotation $\,y=Sx\,$ with an orthogonal matrix $S$ such that $(Sa)_1=1$, $(Sa)_j=0\;\,\forall\,j \geqslant 2$.  A rather trivial check shows that function $$ h(x)=\ln\Bigl(1-x_1+\sqrt{(1-x_1)^2+x_2^2+x_3^2}\Bigr) $$ is harmonic inside a unit ball $\,|x|<1\,$ in $\mathbb{R}^3$, i.e., $\,\alpha=\beta=1\,$ is the answer to the question for space dimension $n=3$.  For all $n\geqslant 2$,  there are no other $\,\alpha,\beta\in\mathbb{R}$  with nonconstant harmonic $h$.  Restriction $\,\alpha\geqslant 0\,$ proves absolutely unimportant.

Interestingly enough, all more or less sophisticated approaches fail to give a complete answer to the question. Thus conformal mappings are accompanied by restriction $n=2$ on space dimension, a maximum principle for a subharmonic function $|\nabla\,h|^2$ works very nice with any $\,n\geqslant 2\,$ but only for $\,\alpha=0\,$ and $\,\beta>0\,$, and so on, so on, so on.  Instead, absolutely unsophisticated, naive Taylor expansion around the origin $\,x=0\,$ gives a complete answer to the question for all required values of $n\geqslant 2$, $\,\alpha,\beta\in\mathbb{R}\,$ along with a bonus option $\,\alpha<0\,$ in addition.  The Taylor series
$$ h(x)=\sum_{k=0}^{\infty}P_k(x), \quad x\in B=\{x\in\mathbb{R}^n\,\colon\,|x|<1\}, $$ is absolutely convergent inside the unit ball $B$, with homogeneous polynomials $\,P_k\,$ of degree $k$ being harmonic for any $k\geqslant 0$, i.e., $$ \Delta\,P_k(x)=0,\quad \forall\, x\in B\;\Rightarrow\;\forall\, x\in \mathbb{R}^n\;\; \forall\, k\geqslant 0. $$ To establish that either $\,\alpha=\beta=0\,$ or $\,\alpha=\beta=1\,$, it suffices to examine just three cases of $\,k=2,3,4\,$, i.e., examine the implications of identities $$ \Delta\,P_2(x)=\Delta\,P_3(x)=\Delta\,P_4(x)=0\quad \forall\, x\in\mathbb{R}^n. \tag{$\ast$} $$ To find polynomials $\,\Delta P_k(x)\,$, one can apply the Taylor series expansion to the function $$ g(x_1,r)=\ln\biggl((1-x_1)^{\alpha}+\bigl(1-2x_1+x_1^2+r^2\bigr)^{\beta/2}\biggr) $$ with respect to variables $x_1$ and $r$, applying herewith the Laplace operator in the form $$ \Delta =\frac{\partial^2\,}{\partial x_1^2}+\frac{\partial^2\,}{\partial r^2}+ \frac{n-2}{r}\!\cdot\!\frac{\partial\,}{\partial r}\,, $$ and substituting thereafter $\,r=|x'|^2$, $x'\in\mathbb{R}^{n-1}$.  Running the available symbolic calculations software readily reduces the identities $\,(\ast)\,$ to four algebraic equations, the three of which: $$ \begin{gather} \alpha^2-2\alpha\beta+\beta^2+2\beta n-2\alpha-4\beta=0, \\ \alpha\beta n-\beta^2 n+3\alpha^2-7\alpha\beta+4\beta^2+4\beta n-4\alpha-8\beta=0,\\ \beta(\beta n+5\alpha-4\beta-4n+8)=0, \end{gather} $$ are more than enough to show that either $\,\alpha=\beta=0\,$ or $\,\alpha=\beta=1\,$.