# PDE's - Finding a certain solution to Laplace's equation on a circle

• Sep 2nd 2008, 02:44 PM
Jacobpm64
PDE's - Finding a certain solution to Laplace's equation on a circle
Find a solution of Laplace's equation $u_{xx} + u_{yy} = 0$ of the form $u(x,y) = Ax^2 + Bxy + Cy^2 \ (A^2 + B^2 + C^2 \not= 0 )$ which satisfies the boundary condition $u(cos(\theta),sin(\theta)) = cos(2\theta) + sin(2\theta)$ for all points $(cos(\theta),sin(\theta))$ on the circle, $x^2 + y^2 = 1$.

first, I found $u_{xx}$ and $u_{yy}$

$u_{xx} = 2A$
$u_{yy} = 2C$

From $u_{xx} + u_{yy} = 0$ and the above results, I can get $2A + 2C = 0$.

Now, I plugged in the boundary condition:
$cos(2\theta) + sin(2\theta) = Acos^2(\theta) + Bcos(\theta) sin(\theta) + Csin^2(\theta).$

I tried various trig substitutions here and couldn't seem to get anywhere. However, with this equation and the one above, I have two equations (but there are three unknowns). I am pretty sure I have to use the $x^2 + y^2 = 1$ to write another equation so that I can solve for A, B, and C, but I do not know how to use the circle information.

Any help would be greatly appreciated.

• Sep 2nd 2008, 03:26 PM
sire, it would seem to me that theres an angle theta you can arbitrarily choose to eliminate one of the variables from the equation, thus allowing you to solve for the other variables.

I hope this helps

awfully nice of them to already give u an intended form of solution (Rofl)
• Sep 2nd 2008, 03:29 PM
ThePerfectHacker
Quote:

Originally Posted by Jacobpm64
Now, I plugged in the boundary condition:
$cos(2\theta) + sin(2\theta) = Acos^2(\theta) + Bcos(\theta) sin(\theta) + Csin^2(\theta).$

$\cos 2\theta = \cos ^2 \theta - \sin ^2 \theta$ and $\sin 2\theta = 2\sin \theta \cos \theta$.

Thus, we get,
$\cos^2 \theta - 2\sin \theta \cos \theta - \sin^2 \theta = A \cos^2 \theta + B\sin \theta \cos \theta + C\sin^2 \theta$

Now compare coefficients.
• Sep 2nd 2008, 03:30 PM
Jacobpm64
haha thanks, I got it.

$u(x,y) = x^2 +2xy -y^2$

It's the beginning of a PDE course.

All we've done so far is review ODE's and start "checking" to see if a solution satisfies a PDE.

You're interested in helping me with part b? It's a tad bit harder.
• Sep 2nd 2008, 03:32 PM
i like hacker's approach better, but u said you were failing at the trig identities... either way should yield the correct solution , pls correct me of im off base here

ahh i remember my first undergrad PDE course:D

Just wait til u'r solvin for green's functions muahah
• Sep 2nd 2008, 03:41 PM
Jacobpm64
Yeah I used hacker's approach. It worked very smoothly (I just substituted the wrong identity.. I used a different one for cos(2theta) and that gave me some problems because i couldn't equate the sides)

Anyway,
Here's part (b) of that problem:

Show that the graph of any solution u(x,y) of Laplace's equation of the form $u(x,y) = Ax^2 + Bxy + Cy^2$ $(A^2 + B^2 + C^2 \not= 0 )$ intersects the xy-plane in a pair of perpendicular lines through (0,0).

Obviously, I'll have to plug (0, 0) in somewhere. The confusion I get is when I see xy-plane. (i'm guessing that means I have to set u(x,y) = 0).

Doing that, I get $0 = Ax^2 + Bxy + Cy^2$

AH, I'm confused (Worried)
• Sep 2nd 2008, 03:54 PM
you are right about the u(x,y) = z = 0 ie xy plane thing

im wondering if you cant assume a linear relation between x and y when z = 0 and show it to be true
.. err went thru that in my head no lightbulbs went off, maybe scratch that approach
• Sep 2nd 2008, 04:01 PM
Jacobpm64
well, for the lines to be perpendicular, we know we'll have to find some derivatives to show that the slopes are opposite reciprocals at that point (0,0) i suppose.

Hmm so just calculating these .. because we might need them:

$u_{x} = 2Ax + By$
$u_{y} = Bx +2Cy$

Hmm, not sure what to do with these.

(I also don't know if we can assume the linear relationship.)