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Math 4200
Maple play, November 21
Work for the Poisson integral formula for harmonic functions on the unit disk
with(plots):
g:=(x,y)->(1-x^2-y^2)/(1-2xcos(theta)-2ysin(theta)
+x^2+y^2);
g :=( x y , )→
1 − x −
2 y
2
1 − 2 x cos( ) θ − 2 y sin( ) θ + x +
2 y
2
diff(g(x,y),x,x)+diff(g(x,y),y,y);
1 − 2 x cos( ) θ − 2 y sin( ) θ + x +
2 y
2
4 x ( − 2 cos( ) θ + 2 x )
( 1 − 2 x cos( ) θ − 2 y sin( ) θ + x + )
2 y
2
2
2 ( 1 − x − )
2 y
2 ( − 2 cos( ) θ + 2 x )
2
( 1 − 2 x cos( ) θ − 2 y sin( ) θ + x + )
2 y
2
3
4 ( 1 − x − )
2 y
2
( 1 − 2 x cos( ) θ − 2 y sin( ) θ + x + )
2 y
2
2
4 y ( − 2 sin( ) θ + 2 y )
( 1 − 2 x cos( ) θ − 2 y sin( ) θ + x + )
2 y
2
2
2 ( 1 − x − )
2 y
2 ( − 2 sin( ) θ + 2 y )
2
( 1 − 2 x cos( ) θ − 2 y sin( ) θ + x + )
2 y
2
3
simplify(%); #gettin zero here shows the sense in which
#the Poisson integral formula represents
#a harmonic function as an integral superposition
#of harmonic functions.
Here’s an example where the boundary values are 1 on a quarter circle, and zero on the
rest of the circle: (Could you write this function using the imaginary part of some complex logarithm
expression?)
u:=(rho,phi)->1/(2Pi)evalf(int((1-rho^2)/(1-2rhocos(theta-phi)
+rho^2),theta=0..Pi/2));
u :=( ρ φ, )→
evalf d
0
π
2
1 −ρ
2
1 − 2 ρ cos( θ −φ )+ρ
2
θ
π
plot3d([rhocos(phi),rhosin(phi),u(rho,phi)],rho=0.01..(.99),phi=
0..2.05*Pi,grid=[40,40],axes=boxed,style=wireframe,
color=black,title=‘equilibrium heat distribution on the disk‘);
equilibrium heat distribution on the disk
–0.
0
1
0
1
0
1
So, without any work, what is the value of this harmonic function at the origin - precisely?
