Answer to Question #131400 in Differential Equations for Meenu Rani

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.