Toronto Math Forum

Writing the change of coordinates Sheng applied to the boundary conditions I found,  $ w|_{\pi=0}=0$ and $  w|_{r=1}=2\sin^2\theta $ I can apply separation of variables to the function $ w = P(\theta)R(r) $ which I can solve for this half disk as
 $$ w = \sum A r^n \sin(n\theta) $$ and therefore
 $$ A = \frac{4}{\pi}\int_{0}^{\pi} \sin^2(\theta) \sin(n\theta) d \theta  =\frac{ (4 (2 \cos(π n) - 2))}{(π (n^3 - 4 n)) }$$
and this is $ \frac{-8}{(π (n^3 - 4 n)) } $ when n is odd. So I have
$$ w =  \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) $$
and therefore I can write $$ u = w + v =  \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) + r^2 \cos^2(\theta) - r^2 sin^2(\theta) $$ which can be written most concisely as
$$ u = r^2 \cos(2 \theta) +  \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) $$

 

Begin via separation of variables.
$$ X''+\lambda X = 0 $$
$$ Y''-\lambda Y=0 $$
with this the solution to the  new Dirichlet Boundary conditions gives  $ \lambda = n^2  $  $n = 1,2,... $
so
$$  X= A_n \sin(2nx) $$ where we don't have any constants that might be forgotten...
Now the solution to the Y ODE are exponential, where we discard the infinite terms so our general solution is of the form
$$ u(x,y)=\sum_{n=1}^\infty B_ne^{-2ny}\sin(2nx) $$ Then the Robin Boundary Condition gives us
$$ - \sum _{n=1}^\infty B_n (2n+1)\sin(2nx)=1. $$ This allows us to solve the traditional method that
 $$ B_n = -\int_{0}^{\frac{\pi}{2}}  (2n+1)^{-1}\sin(2nx)  \text{dx} $$ Which has the nice solution that
$$ = B_n ((2n(2n+1))^{-1}(\cos(n\pi) - 1)$$
$$ B_n = -(n(2n+1))^{-1} $$ for n odd

and therefore we have what would have been a much nicer solution of
$$ u(x,y)=-\sum_{n \text{  odd}}^\infty  (n(2n+1))^{-1} e^{-2ny}\sin(2nx)  $$

Only considering (1) and (2),
My candidate is
$$ v = x^2 -y^2 $$
which is harmonic, (has zero laplacian) and $$ v(x,0) = x^2 $$
By homogeneous (2) do you mean?
$$ w|_{y=0}=0  $$
How do I modify (3) ?

The solution is spherically symmetric because the question has that symmetry.
Then using the hint I reduce the Laplacian to $$(ru)"  = v" $$ then do separation of variables on v and I get the solution to the eigenvalue problem of v is that  $ R = \sin(r) $ because the eigenvalue must be one and the cos term has a singularity at r = 0
So now my general solution is that $$ u = \frac{\sin(r)}{r} (A\cos(t) + Bsin(t)) $$ To satisfy the boundary conditions, A = 0 and
$$ u = \frac{\sin(r)}{r} \sin(t) \text{        for               } r< \theta $$ B = 1.
For the other boundary condition, the only conceivable option is that u is defined as the trivial solution outside of this domain, B = 0

a) separation of variables gives us that $$ P'' + \lambda P = 0 $$
$$ r^2R''+ rR' -\lambda R = 0 $$
The boundary conditions translate to Dirichlet Boundary Conditions $ P(0) = P(\theta) = 0  $
b) With the eigenvalue problem of  $$ P  \lambda = (n)^2 $$ gives $$ P = \sin(n\theta) $$
c) and then solving the ODE for R we know $$ R = Ar^n + Br^{-n} $$
d) Now solving for coefficients $$ u = \sin(n\theta)(Ar^n + Br^{-n} ) $$
Now we know $ A = -B $  because  of (2),(3) and so
$$A_n =\frac{2}{\pi} \int_{0}^{\pi} \frac{1}{2^{-n}-2^{n}} \sin(n\theta) $$ and via computation
becomes $$ u =\sum_{n \text{  odd}} \frac{4}{n\pi(2^n-2^{-n})} \sin(n\theta) ( r^n + r^{-n}) $$

d) *from Gel'fand*
Equations 5 and 6 come from the natural log of  the generalized expression
$$ (x+iy)^{\nu} =  \exp[\nu \log(x+iy) + iArg(x+iy)]
\tag{A}$$  which is just the representation on the complex plane
Then we can find
$$ (x+i0)^{-\nu} = x^{-\nu} - \frac{i\pi(-1)^{\nu-1}}{(\nu -1)!}\delta^ {(nu -1)}(x)
\tag{B}$$
$$ (x-i0)^{-\nu} = x^{-\nu} + \frac{i\pi(-1)^{\nu-1}}{(\nu -1)!}\delta^ {(nu -1)}(x) \tag{C}$$
and by combining these and setting $ \nu =1 $ we arrive at
$$ (x-i0)^{-1} + (x+i0)^{-1}=2 x^{-1}  \tag{D}$$
$$(x-i0)^{-1}- (x+i0)^{-1}=2\pi i \delta(x)  \tag{E}$$

a) Consider
$$ (x\pm i0)^{\nu}= f'$$
Now via definitions $$ (f',\theta) = - (f,\theta') $$ and then f is defined as
 $$ \frac{1}{\nu+1}(x \pm i0)^{\nu+1}  = f $$ This  exists for  $ \Re\nu >-1 $. Therefore the limit exists for this condition.
b) Extending the proof from part a, define $$ (x\pm i0)^{\nu}= f^{(n)}$$ then with the same rules of distribution derivatives I can write $$  [(x\pm \varepsilon i)^{\nu}]^{(n)}= (\nu ! )(x\pm \varepsilon i)^{\nu-n}$$ which is then defined for corresponding $ \Re\nu \neq -n $ where  $ n = 2, 3,\ldots $

Above is incomplete.
we begin with separation of variables $ U = X(x)Y(y) $
The equation simplifies to
$$ \frac{X"}{X} +  \frac{Y"}{Y}  = -\lambda $$

We know that these fractions must remain constant and so we have corresponding $  -\lambda_x + -\lambda_y = -\lambda $
We now have  Not enough letters? V.I. I'm sorry I don't follow
$$ X" + \lambda_xX = 0 $$
$$ Y" + \lambda_yY = 0 $$
Furthermore boundary conditions state that  $$ X(0) = X'(a) $$ and $$ Y(0) = Y'(b) $$ These are mixed boundary conditions (Dirichlet and Nuemann). They dictate the eigenvalues and corresponding eigenfunctions for the ODEs.
\begin{align*}
&\lambda_x =( \frac{\pi(2n+1)}{2a})^2 , &&X_n = \sin(\frac{\pi(2n+1)}{2a}) && n = 0,1,2,...\\
&\lambda_y =( \frac{\pi(2m+1)}{2b})^2  &&X_n = \sin(\frac{\pi(2m+1)}{2b}) &&m = 0,1,2,...
\end{align*}
We can and must then conclude that
$$
\lambda  =\pi^2( (\frac{(2n+1)}{2a})^2+ (\frac{(2m+1)}{2b})^2)$$
ERROR above. V.I. corrected
and our eigenfunctions are of the formula
$$ U_{n,m}(x,y) =  \sin(\frac{\pi(2n+1)}{2a})\sin(\frac{\pi(2m+1)}{2b}) $$

This is the complete answer. No need to go on a tangent.