Prove by induction, for the interval case, that $latex \min\{ \mathscr E_m(u): u|_{\{0,1\}}=v\} = \mathscr E_0(v) = (v(0)-v(1))^2,$ with the minimizer satisfying $latex \displaystyle u\Big(\frac{2k+1}{2^m}\Big) = \frac{1}{2}\Big(u\Big(\frac{k}{2^{m-1}}\Big) + u\Big(\frac{k+1}{2^{m-1}}\Big)\Big).$
The minimum of $latex f(x,y,z) = (a-x)^2 + (x-y)^2 + (y-a)^2 + (x-b)^2 + (b-z)^2 + (z-x)^2 + (y-z)^2 + (z-c)^2 + (c-y)^2$ is attained at $latex \displaystyle x^* = \frac{2a+2b+c}{5},\; y^* = \frac{2a+b+2c}{5},\; z^* = \frac{a+2b+2c}{5},$ with $latex f(x^*, y^*, z^*) = \dfrac{3}{5}\big((a-b)^2 + (b-c)^2 + (c-a)^2\big).$
Prove that one can obtain the values $latex u(F_2(q_1)) = x, u(F_3(q_1)) = y$ of a harmonic function in terms of the values $latex a,b,c$ at the points $latex p_2, q_1, p_3$, respectively (as in the figure below).
Use the previous problem to show that, if $latex u$ is a harmonic function with boundary values $latex u(p_1) = u(p_2) = 0$ and $latex u(p_3) = 1$, then its restriction to the bottom side of the Sierpinski triangle is increasing.
If $latex u$ is harmonic, then it is uniformly continuous (use the fact $latex |u(x) - u(y)| \le c \Big(\dfrac{3}{5}\Big)^m$ if $latex x\sim y$ in $latex V_m$.)
Comentarios
Publicar un comentario