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

[Jacobpm64]

Homework Statement

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, $$x^2 + y^2 = 1$$.

Homework Equations

Listed above.

The Attempt at a Solution

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.

Thanks in advance.

 

[Dick]

Just use the double angle formulas for cos(2*theta) and sin(2*theta). I don't really see what the problem is.

 

[Jacobpm64]

All right, let me see what I can do. I had tried this, but it wasn't simplifying, maybe there's something I'm not seeing.

okay, so using those double angle identities:

$$cos^2(\theta) - sin^2(\theta) + 2sin(\theta)cos(\theta) = Acos^2(\theta) + Bcos(\theta)sin(\theta) + Csin^2(\theta)$$

Oh, and I can solve that for
$$A = 1,B = 2,C = -1$$

So my solution is :

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

This is the answer in the back of the book.

I was just wondering why they put the $$x^2 + y^2 = 1$$ in there? Is that something I didn't need to figure out the answer?

Thanks so much for that little nudge too. I really appreciate that.

 

[Dick]

No, you don't need it. If (x,y)=(cos(theta),sin(theta)) on the boundary, then of course x^2+y^2=1.

 

[Jacobpm64]

Thanks so much!