Question

(i) Use method of separation of variables in rectangular coordinates to obtain the following harmonic functions which are bounded in the upper half plane y > 0 of R2,
e−λy cosλx, e−λy sinλx; λ ≥0.
(ii) By integration with respect to the parameter λ, obtain the functions
x / x2 +y2, y / x2 +y2 which are harmonic in the upper half plane y > 0 of R2.

Answer 1

i) Let the required harmonic function be F such that,

F(x,y)=X(x,t).Y(y,t).....(1)

As per the question the boundary condition are y>0,(0,0),to $e^{-\lambda y}Cos\lambda x,e^{-\lambda y}{Sin\lambda x}$

from equation 1..

we have

$\frac{ d^2X}{ dx^2}+\frac{d^2Y}{dy^2}=0$

The above equation can be written according to separation of variable

$\frac{ d^2X}{ Xdx^2}=-\frac{d^2Y}{Y dy^2}$ = k$^2$

$\frac{d^2X}{dx^2}=k^2x, \frac{d^2Y}{dy^2}=-k^2y$

The solution of the above equation can be written as

X(x,t)=$c_1coskx+c_2sinkx,Y(y,t)=c_3e^{kx}+c_4e^{ky}$

Then the harmonic function is given by,

$F(x,y)=(c_1coskx+c_2sinkx)(c_3e^{kx}+c_4e^{ky})$

This is the required function.

ii)As $x=e^{-\lambda y}Cos\lambda x$

y=$x=e^{-\lambda y}Sin\lambda x$

then$\frac{x^2}{x^2+y^2}=\frac{e^{-2\lambda y} Cos^2\lambda x}{e^{-2\lambda y} Cos^2\lambda x+e^{-2\lambda y} Sin^2\lambda x}$

$\frac{x^2}{x^2+y^2}=\frac{e^{-2\lambda y} Cos^2\lambda x}{e^{-2\lambda y} (Cos^2\lambda x+ Sin^2\lambda x)}$

$\frac{x^2}{x^2+y^2}= Cos^2\lambda x$

similiarly$\frac{y^2}{x^2+y^2}=Sin^2\lambda x$

Now the integrate both of the function with respect to $\lambda$

$\int\frac{x^2}{x^2+y^2}d\lambda=\int Cos^2\lambda d\lambda=\int (\frac{1+Cos2\lambda x}{2})d\lambda$

=$\frac{\lambda +\frac{Sin2\lambda x}{2x}}{2}$

Now integrate

$\int\frac{y^2}{x^2+y^2}d\lambda=\int Sin^2\lambda d\lambda$

=$\int (\frac{1-Cos2\lambda x}{2})d\lambda$

=$\frac{\lambda -\frac{Sin2\lambda x}{2x}}{2}$

These are the required functions.