Understanding the definition of second order partial differential equations Classification: we consider an open two-dimensional domain $(x,y)\in\Omega\subset\mathbb{R}^2$ with boundary $\Gamma = \partial\Omega$ as the domain of the second order linear partial differential equation (PDE) for the unknown field $u = u(x,y)$ and the source function $f(x,y)$ that can be written as
$$\mathcal{L}(u) = f \,\,\text{on} \,\,\Omega$$
where the operator $\mathcal{L}$ has constant coefficients $a_{ij}, b_i,$ and $c$
$$\mathcal{L}(u) = a_{11}\frac{\partial^2 u(x,y)}{\partial x^2} + 2a_{12}\frac{\partial^2 u(x,y)}{\partial x\partial y} + a_{22}\frac{\partial^2 u(x,y)}{\partial y^2} + b_1\frac{\partial u(x,y)}{\partial x} + b_2\frac{\partial u(x,y)}{\delta y} + cu(x,y).$$
Questions: 


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*What does the boundary do for a second order partial differential equation? I read on partial differential equations that boundary functions are used to give the constraint $u(x,y) = g(x,y)$ on $\partial\Omega$. What does this mean; isn't $u(x,y)$ unknown (according to introduction to partial differential equations it is)?  

*Why is the term unknown field $u = u(x,y)$ used instead of just the function $u(x,y)$? Is this because we want to speak about the range of $u$?

*In an exercise I have to solve the following is stated: "... given the domain $\Omega = (0,1)$ with boundary $\partial\Omega$ and outward normal $n$...". What does outward normal $n$ mean? 


Thanks in advance!
 A: In 2D, the boundary of a domain is typically described by a curve, either in parametric form
$$ \textbf{r}(t) = (x(t), y(t))$$
or implicit form
$$ F(x,y) = 0 $$
"normal" in this context means the vector in question is perpendicular to the tangent of the boundary curve
$$ \hat{\textbf{n}}\cdot \textbf{r}'(t) = 0 $$
or in the direction of the gradient
$$ \hat{\textbf{n}} = \frac{\nabla F}{|\nabla F|} $$

"Outward pointing" is a little harder to put in mathematical terms. It's a convention to define a unique normal, as there's always 2 possible directions for a perpendicular vector. If you place the tail of the unit vector its corresponding point the curve, then the head of the vector is somewhere outside the enclosed domain.
If you had the boundary curve, you can think of it as some level curve (contour) of a 3d surface, then the normal points to the direction of greatest increase. If you take the negative of the outward vector, then the result is another normal vector that points "inwards"
In higher dimensions, this extends to a surface with dimension 1 less then that of the domain.

In 1D, the domain is reduced down to a segment of the real line, so the boundary is defined by its endpoints. The "normal" becomes the first derivative evaluated at these points
$$ n(x) = \frac{du}{dx} $$

As to why there are constraints given for boundary functions, they denote that $u(x,y)$ takes on some specific values in the boundaries, but unknown everywhere else. This is what "boundary value problem" means, and is similar to the initial conditions you may encounter in ODES.
For example, if your domain is a rectangle on the $xy$-plane with $0 < x < a$, $0 < y < b$ then the boundary conditions may be given as
$$ u(0,y) = f(y), \ u(a,y) = g(y) \\ u(x,0) = h(x), \ u(x,b) = k(x) $$ 
Or in terms of the normal derivatives
$$ \frac{\partial u}{\partial x}(0,y) = f(y) \\ 
\frac{\partial u}{\partial y}(x,0) = h(x) $$
Note that these boundary conditions are not sufficient to determine the unknown function on the interior. You have to solve differential equation for that reason.
On the other hand, if no boundary conditions are given, then the equation might still be solved, but the solution will not be unique and depend upon some arbitrary constant. Again, this is just how ODEs carry over to higher dimensions.

In general, the shape of the boundary influences how the equation is solved. For the rectangular boundary above, the solution is usually in terms of rectangular coordinates. However, this is not always the case, as the boundary may be of any shape, such as a circle
$$ \Omega: x^2 + y^2 \le 4 $$
then you'll have to perform a change of variables into polar coordinates. 

For more information, look for online resources such as Wikipedia or this tutorial
