Given the boundary value problem
$\Delta u = 0, 0 < x < 2\pi, 0 < y < \pi$
$u(x,0) = 0, u(x,\pi) = 0, 0 \leq x \leq 2 \pi$
$u(0,y) = y(\pi - y), u(2\pi,y) = 0, 0 \leq y \leq \pi$
I want to solve it by seperating into $u(x,y) = f(x) \cdot g(x)$ and find the minimum and maximum function value.
One boundary condition i cannot solve, but here is what i have so far:
$\Delta u = f''(x)\cdot g(y) + f(x) \cdot g''(x) = 0$
$\frac{f''(x)}{f(x)} = \frac{-g''(y)}{g(y)} := \lambda$
Which yields
$f''(x) - \lambda \cdot f(x) = 0$ and $g''(y) + \lambda \cdot g(y) = 0$
I started by defining $h_1(x) = 0, h_2(x) = 0, h_3(x) = y(\pi - y), h_4(x) = 0, a = 2 \pi$ and $b = \pi$.
Separation by product yields (for the boundaries):
$0 = h_1(x) = u(x, 0) = f(x) \cdot g(0) \Rightarrow g(0) = 0$
$0 = h_3(x) = u(x, \pi) = f(x) \cdot g(\pi) \Rightarrow g(\pi) = 0$
--
$g''(y) +\lambda \cdot g(y) = 0 \wedge g(0) = g(\pi) = 0 $ can only hold if $\lambda_k = k^ 2$ and $g_k(y) = sin(k\cdot y)$
With this the general solution of $f''(x)-k^2\cdot f(x) = 0$
$$f(x) = \overline c_k e^{kx} + \overline d_k e^{-kx} = c_k cosh(kx) + d_k \cdot sinh(k\cdot x)$$
With
$$0 = u(2\pi, y) = f( 2\pi)g(y) \Rightarrow f(2\pi) = 0 $$
And
$$sinh(s+t) = sinh(s) cosh(t) + cosh(s)sinh(t) $$
One can derive
$$f_k(x) = b_k \cdot sinh(k \cdot (x - 2 \pi))$$
The differential equation is linear, so we can make use of superposition principle:
$$ u(x,y) = \sum_{k=1}^{\infty} b_k sinh(k \cdot (x-2\pi)) \cdot sin(k \cdot y)$$
The last boundary condition missing is $y(\pi -y) = u(0,y) = \sum^\infty_{k=1} b_k \cdot sinh(-2\pi k)\cdot sin(k \cdot y) $
How do i continue from here? I am thinking the coefficient comparison, but i can not apply it here (i am aware that $sinh(-x) = -sinh(x)$):
$$\pi y - y ^2 = \sum^\infty_{k=1} - b_k \cdot sinh(2 \pi k ) \cdot sin(k \cdot y) $$