Is it possible to find the value of scalar function u(x, y) anywhere in the region $\Omega$, given the following:
$\nabla \cdot$ ($\nabla$u) = f,
u(x,y) = g when y = 0, L$_2$
($\nabla$u)$_x$ = h when x = 0, L$_1$
where f, g and h are given scalar valued functions
I ask because it is possible to solve the equivalent 1D problem:
$\frac{d^2u}{dx^2}$ = f,
u(0) = g$_1$,
$\frac{du}{dx}$ = g$_2$ when x = L
by stating: u(x) = g$_1$ + $\int_0^x\bigl(-\int_x^L$f dx + g$_2\bigl)$ dx.
And both the 1D and 2D problems are solvable using the finite element method. But I can’t figure out a way on paper to derive u(x,y).
I tried reasoning:
u(x,y) = u(x,0) + $\int_0^y$ ($\nabla$u)$_y$ dy.
where ($\nabla$u)$_y$ = ($\nabla$u(x,0))$_y$ + $\int_0^y \frac{\delta ^2u}{\delta y^2} $ dy.
But ($\nabla$u(x,0))$_y$ and $\frac{\delta ^2u}{\delta y^2}$ are unknown so I can’t get any further.
Alternatively, using the divergence theorem:
$\int_{\Omega_0}$ f dA = -$\int_{L_a}^{L_b}$ ($\nabla$u(0,y))$_x$ dy + $\int_0^{L_1}$ ($\nabla$u(x,$L_b$))$_y$ dx + $\int_{L_a}^{L_b}$ ($\nabla$u($L_1$,y))$_x$ dy - $\int_0^{L_1}$ ($\nabla$u(x,$L_a$))$_y$ dx.
Where $L_a$ and $L_b$ are two values of y between 0 and $L_2$ and $\Omega_0$ is the rectangular region bounded by 0 and $L_1$ on the x axis and $L_a$ and $L_b$ on the y axis.
So $\int_0^{L_1}$ ($\nabla$u(x,$L_b$))$_y$ dx - $\int_0^{L_1}$ ($\nabla$u(x,$L_a$))$_y$ dx = A.
where A = $\int_\Omega$ f dA + $\int_{L_a}^{L_b}$ ($\nabla$u(0,y))$_x$ dy - $\int_{L_a}^{L_b}$ ($\nabla$u($L_1$,y))$_x$ dy.
Therefore ($\nabla$u(x,$L_b$))$_y$ - ($\nabla$u(x,$L_a$))$_y$ = $\frac{\delta A}{\delta x}$,
but there are parts of A for which we don’t know $\frac{\delta}{\delta x}$.
So my question is, am I missing something and it’s possible to find u(x,y) or is there not enough information?
Many thanks and sorry it’s so long winded.
In your problem
We get two problems
$$ \nabla^{2}u =0 \tag{1}$$
on a rectangle with $ x \in [0,0.3]$ and $ y\in [0.8] $ subject to the boundaries $$ \textrm{ Left } \frac{\partial u}{\partial x}(0,y) = 0 \\ \textrm{ Right } \frac{\partial u}{\partial x}(0.3,y) = 0 \\ $$ $$ \textrm{Bottom } u(x,0) =A(x) \\ \textrm{ Top } u(x,0.8) = B(x) $$
in the end it is a function of $x,y$ and we solve the bottom.
$$ u(x,y) = v(x,y) + r(x,y) \tag{2} $$
$$ r(x,0) = A(x) = 300(1+\frac{x}{3}) \tag{3} $$ $$ r(x,0.8) = B(x) = 310(1+8x^{2}) \tag{4} $$
$$ \frac{d^{2}}{dx^{2}}r(x,0) = 0 \tag{5} $$ $$ \frac{d^{2}}{dx^{2}}r(x,.8) = 4960 \tag{6} $$