Solve $u_{xx} + u_{ yy} = 0$ in the disk $\{r < a\}$ with the boundary condition $u = 1 + 3 \sin θ\,$ on $\,r = a.$
In this example, writer says that the full Fourier series for $h(\theta )=1+3\sin\theta$, $A_0=2, B_1=3/a$, and all other coefficients are $0$. How does he finds these?
Then he concludes that $u(r,\theta)=1+\frac 3 {a}r\sin\theta.$
In rectangular coordinates $u(x,y)=1+\frac {3y} {a}$
Here are the formulas we need (Poison's Formula):
$$u_{xx}+u_{yy}=0, \quad x^2+y^2\lt a^2$$ $$ u=h(\theta),\quad x^2+y^2= a^2 $$ $$ u=\frac 1 2A_0+\sum_{n=1}^\infty a_n [A_n\cos(n\theta)B_n\sin (n\theta)] $$
Indeed:
$$\frac 1 2A_0+\sum_{n=1}^\infty a^n (A_n\cos {n\theta}+B_n\sin {n\theta})=1+3\sin(\theta)$$
implies that:
$$\frac{A_0}{2}=1 \quad\text{and} \quad \sum_{n=1}^\infty a^nA_n\cos(n\theta)=0 \quad \text{and} \quad \sum_{n=1}^\infty a^nB_n\sin(n\theta)=3\sin(\theta)$$
Then $A_n=0 $ for all $n>0$. And
$$aB_1\sin(\theta)+a^2B_2\sin(2\theta)+a^3B_3\sin(3\theta)+\ldots=3\sin(\theta)$$
Hence $aB_1=3$ and the others terms are zero.