We have seen the following problem in our class: $$ \begin{cases} \Delta u=0 \text{ in } B_R (0) \\ u(x)=\phi(x) \text{ in } \partial B_R (0) \end{cases} $$ where $B_R (0)$ the $n$-dimensional ball centered in $0$ with radius $R$. The solution is given by $$ u(x)=\frac{R^2 - |x|^2}{R |\partial B_1|} \int_{\partial B_R (0)} \frac{\phi(y)}{|x-y|^n} d \sigma_y $$ where $|\partial B_1|$ is the measure of the surface of the $n$-dimensional ball with radius $R=1$
Now, for finding this magical formula, we have first of all searched for radial harmonic functions. It turns out that, if the dimension is $n=2$, the radial harmonic function is $\log(r)$ while for $n \geq 3$ it has the form $\frac{1}{|x|^{n-2}}$. Then we have seen how to derive the solution starting from these radial harmonic functions. The only problem is that we have focused on the case $n\geq 3$, so I was wondering both how to deal with the case $n=2$ and if the result was the same.
For $n=2$ the formula stays the same, e.g., Wikipedia article on Poisson kernel states it as such for all $n$. It's often written using polar coordinates $r, \theta$ when $n=2$, and takes the form $$ u(re^{i\theta})=\frac{R^2 - r^2}{2\pi } \int_{0}^{2\pi} \frac{\phi(Re^{it})}{R^2-2Rr\cos(t-\theta)+r^2} \,d t $$ But this is only a cosmetic difference.
Even for $n=1$ the formula works. In this case $\partial B_R(0)$ is $\{-R,R\}$, so the formula becomes
$$ u(x)=\frac{R^2 - x^2}{2R} \left(\frac{\phi(R)}{R-x} + \frac{\phi(-R)}{R+x} \right) = \frac{\phi(R)(R+x)+\phi(-R)(R-x)}{2R} $$ which indeed describes a harmonic (=affine when $n=1$) function with the desired boundary values.
The proof of the Poisson formula is the same in all dimensions: in involves applying Green's formula to $u$ and the fundamental solution on a suitable domain. It's true that the fundamental solution $\phi$ looks different when $n=2$, but in the process of using Green's formula we end up working with its gradient $\nabla \phi(x)$, which is $x/|x|^n$ (with some constant factor) in all dimensions.