# 2nd Order Linear PDE - Harmonic Function

• April 29th 2011, 01:47 AM
bugatti79
2nd Order Linear PDE - Harmonic Function
Folks,

Can anyone assist in solving the attached question?

I believe the PDE is ellipitic becasue B^2-AC = 0 -(1)(1)<0 implies elliptic.
Therefore we can write

$\displaystyle \frac{dy}{dx}=\frac{0\pm \sqrt{0-AC}}{A}=\pm i$

Therefore I choose y-ix = Const. From this I choose new variables with s(x,y)= y and t(x,y)=-x

Not sure what to do after this
• April 29th 2011, 04:42 AM
Jester
Not sure what you're trying to achieve. You have Laplace's equation

$\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0$.

It won't reduce to more than that.
• April 29th 2011, 05:02 AM
bugatti79
Quote:

Originally Posted by Danny
Not sure what you're trying to achieve. You have Laplace's equation

$\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0$.

It won't reduce to more than that.

Well I am trying to answer part b i and b ii in the attachment. Obviously my attempt is wrong...

Any suggestions?
• April 29th 2011, 05:09 AM
Jester
Well, if you consider the Cauchy-Riemann equations

$u_x = v_y, u_y = - v_x$

then eliminating $v$ gives Laplace's equation. If you substitute your $u$ into these and solve for $v$, this is the imaginery part that you are seeking.
• April 30th 2011, 01:23 AM
bugatti79
Quote:

Originally Posted by Danny
Well, if you consider the Cauchy-Riemann equations

$u_x = v_y, u_y = - v_x$

then eliminating $v$ gives Laplace's equation. Is this for part i?

If you substitute your $u$ into these and solve for $v$, this is the imaginery part that you are seeking. I presume this for part ii

For part i, I attempted the following to find v(x,y)

$\displaystyle U_x=\frac{\partial }{\partial x} \left[\frac{x}{x^2+y^2}\right]=\frac{y^2-x^2}{(x^2+y^2)^2}=\upsilon_y (1)$

$\displaystyle U_y=\frac{\partial }{\partial y} \left[\frac{x}{x^2+y^2}\right]=\frac{-2xy}{(x^2+y^2)^2}=-\upsilon_x (2)$

I attempt to find v(x,y) from (1) and (2) by integrating and hence arriving at

from (2)

$\displaystyle v(x,y)=-\frac{y}{2(x^2+y^2)}+f(y)$
from (1)

$\displaystyle v(x,y)=-\frac{y}{(x^2+y^2)}+f(x)$

Not sure if this right approach or what to do next?

Thanks
• April 30th 2011, 05:26 AM
Jester
For part I, directly substitute $U = \dfrac{x}{x^2+y^2}$ into $U_{xx} + U_{yy} = 0$ to show it is identically satisfied. For part ii, yes what you have done is correct except for the 2 in the denominator of the first one and those arbitrary functions must both be a constant (the same constant as we want the same answer).

Another way to do this is to consider

$f(x+iy) = \dfrac{x}{x^2+y^2} + v(x,y) i$

Then set $y = 0$ giving, thus $f(x) = \dfrac{1}{x}$ (if $f(x)$ is real then $v(x,0)=0$). So $f(w) = \frac{1}{w}$.

Thus $f(x+iy) = \dfrac{1}{x+iy}$ simplify and the imaginery part is your $v$.
• May 4th 2011, 10:20 AM
bugatti79
Quote:

Originally Posted by Danny
Another way to do this is to consider

$f(x+iy) = \dfrac{x}{x^2+y^2} + v(x,y) i$

Then set $y = 0$ giving, thus $f(x) = \dfrac{1}{x}$ (if $f(x)$ is real then $v(x,0)=0$). So $f(w) = \frac{1}{w}$.

Thus $f(x+iy) = \dfrac{1}{x+iy}$ simplify and the imaginery part is your $v$.

Hmmm..Im afraid I dont follow this. How does one get the imaginary part v(x) from this last expression?
• May 5th 2011, 05:38 AM
Jester
Multiply top and bottom by the conjugate $x - iy$ and simplify.
• May 5th 2011, 10:56 AM
bugatti79
Quote:

Originally Posted by Danny
Multiply top and bottom by the conjugate $x - iy$ and simplify.

Perfect! Thank you!