# Math Help - Laplace's equation

1. ## 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

2. 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:

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

3. 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.

4. 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.

5. Cheers guys got the answer to the first part now.

Not to difficult once you get your head round what all the words mean.

6. 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.