Laplace's equation

• Dec 23rd 2009, 03:43 AM
jezzyjez
Laplace's equation
I understand using Laplace transforms but I think this is something different the question reads.

Show that the following functions $u(x,y)$ are harmonic (ie satisfy Laplace's equation) and in each case find $f(z)$ such that $f(z) = u + iv$ is regular.
(a) $u = x^4 - 6x^2y^2 + y^4$

(b) $u = sinxsinhy$
Any help would be great.

Jez
• Dec 23rd 2009, 03:58 AM
AnonymitySquared
Quote:

Originally Posted by jezzyjez
I understand using Laplace transforms but I think this is something different the question reads.

Show that the following functions $u(x,y)$ are harmonic (ie satisfy Laplace's equation) and in each case find $f(z)$ such that $f(z) = u + iv$ is regular.
(a) $u = x^4 - 6x^2y^2 + y^4$

(b) $u = sinxsinhy$
Any help would be great.

Jez

This is nothing to do with Laplace transforms.

Laplace's equation is this:

$\nabla^2 \phi = 0$

That is:

$\frac{\partial \phi}{\partial u} + \frac{\partial \phi}{\partial v} = 0$
• Dec 23rd 2009, 05:45 AM
shawsend
Quote:

Originally Posted by jezzyjez
I understand using Laplace transforms but I think this is something different the question reads.

Show that the following functions $u(x,y)$ are harmonic (ie satisfy Laplace's equation) and in each case find $f(z)$ such that $f(z) = u + iv$ is regular.
(a) $u = x^4 - 6x^2y^2 + y^4$

(b) $u = sinxsinhy$
Any help would be great.

Jez

If $u = x^4 - 6x^2y^2 + y^4$

is harmonic then:

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

You can take the second partials and check that right? Same dif for the other one. Now, in order to find a complex-analytic function for which the real part is $u(x,y)$, then you need to find the harmonic conjugate of $u(x,y)$ using the Cauchy-Riemann equations and a little bit of integration.
• Dec 23rd 2009, 06:06 AM
HallsofIvy
Quote:

Originally Posted by AnonymitySquared
This is nothing to do with Laplace transforms.

Laplace's equation is this:

$\nabla^2 \phi = 0$

That is:

$\frac{\partial \phi}{\partial u} + \frac{\partial \phi}{\partial v} = 0$

You've missed the "second order"! Laplace's equation is
$\nabla^2u= \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2u}{\partial y^2}= 0$
(I have changed AnonimitySquared's " $\phi$" to "u" since the functions are given as "u(x,y)".)
Also, jezzyjez, a function, u(x,y)+ iv(x,y), is "regular" if it satifies the Cauchy-Riemann equations:
$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$
and
$\frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}$
Since you are given u is both cases, you can find the right hand sides of these two equations and use them to find v.
• Dec 23rd 2009, 06:31 AM
jezzyjez
Cheers guys got the answer to the first part now.

Not to difficult once you get your head round what all the words mean.
• Dec 23rd 2009, 08:09 AM
shawsend
You can find the harmonic conjugate by the following expression:

$v(x,y)=\int\frac{\partial u}{\partial x}dy+\int\left(-\frac{\partial u}{\partial y}-\frac{\partial}{\partial x}\int\frac{\partial u}{\partial x}dy\right)dx$

Yeah, I know it looks intimidating. Start from the inside and work out and when you do an integration, drop the constant of integration. I'll do the first part for $u(x,y)=u = x^4 - 6x^2y^2 + y^4$ then:

$\frac{\partial u}{\partial x}=4 x^3-12 x y^2$

see, I'm starting from the most inner level. Now, I want to partially integrate that with respect to y:

$\int 4x^3-12xy^2 \partial y=4x^3 y-12/3 xy^3$

Now, I want to take the partial of that with respect to x. See, that's the first three steps inside the parantheses. Now get the partial of u with respect to y, negate them, then integrate the bunch with respect to x. See, got that whole parentheses thing now. Do that last integral, bingo-bango.

Edit: Oh yeah, I forgot to mention don't use that expression unless you can go through the steps in deriving it. Any Complex Analysis text book goes over finding the complex conjugate. I just combined all the steps into one expression for fun.