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I've followed the method explained in Numerical Recipes in C, chapter 19, to solve a elliptic equation:

http://www.capca.ucalgary.ca/top/teaching/phys499+535/PHYS535/nrf90/pde-c19-0.pdf

I'm currently working with polar coordinates, because the problem I'm modelling is a circular furnace. So the values in the band matrix are not constant. Depend on the radius and the angle.

How can I proof that there'll always be a solution, analytically?

I've been thinking of demonstrate that it's strictly diagonally dominant, by showing that

$|-2 r_j^2 b^2 + r_j a b^2 - 2 a^2| > |a^2| + |a^2| + |r_j^2 b^2| + |r_j^2 b^2 - a r_j b^2|$

where $r_j$ is the length of a radius, $a = \Delta r$ and $b = \Delta \theta$ (both two are constant)


The subject is Numerical methods

A circular furnace is used to melt metal. The temperature inside the furnice is 1500°C. The inner side (suppose it's circular) is also 1500°. The temperature of the outer side of the furnice is registered by a set of $n$ thermocouples.

The problem consists in to find the 400°C - isotherm.


We were given this formula*

$\frac{\partial^2T(r,\theta)}{\partial r^2} + \frac{1}{r} \frac{\partial T(r,\theta)}{\partial r} + \frac{1}{r^2} \frac{\partial ^2 T(r,\theta)}{\partial \theta ^2} = 0 $

which is satisfied for every point $(r,\theta)$ in polar coordinates, for $0 < r_{inner} < r < r_{outer}$ and $0 \leq \theta \leq 2\pi$

$T(r,\theta)$ is the temperature in $(r,\theta)$

The idea is to discretize the domain (a grid of size $n \times (m+1)$), as it's shown in the pdf file.

(*)NOTE: Math in my career does not include Differential Equations (except for Optative subjects)


As a result of replacing differentials by finite differences, and adding the boundary conditions, my problem is now to solve a linear equation system of size $(m+1)n \times (m+1)n$. Each variable is the value of the temperature at a point of the grid ("the solution")

( The Dirichlet's Problem http://es.wikipedia.org/wiki/Ecuaci%C3%B3n_de_Laplace#Problema_de_Dirichlet )

The matrix ("the band matrix") I've obtained is similar to the one in page 832. But the non-zero values are not constant (they are the ones I've written at the very first)

It's not clear for me if the matrix is invertible

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    Thanks for clarifying. Your question was probably transferred by a moderator of the other forum. The answer to your question is yes, it is always solvable. This can be seen using Hilbert space methods. I'll type up a detailed answer when I get a moment.2011-10-18

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I will outline the typical method of showing that elliptic problems are solvable, applied to your particular PDE. Since you mentioned you have not taken any PDE classes, I have included links to the facts I will use. In particular, if you are not familiar with weak derivatives and Sobolev Spaces, you might want to take a glance at the respective Wikipedia pages.

The operator you have mentioned is just the Laplacian expressed in polar coordinates. So you are trying to show that $-\Delta u = f$ is solvable with Dirichlet boundary conditions (also assume $f$ is square-integrable). Let $\Omega$ denote your region in question (all that is really important about $\Omega$ is that it is open and bounded).

In the weak sense, this equation is equivalent to the statement $\int_\Omega -u \Delta v dx = \int_\Omega fv dx \text{ for all } v \in D(\Omega)$ If we apply Green's identity then this is equivalent to $\int_\Omega \nabla u \cdot \nabla v dx = \int_\Omega fv dx \text{ for all } v \in D(\Omega)$

Here is used $D(\Omega) = C^{\infty}_0(\Omega)$ to denote the space of test functions.

But the term on the left hand side defines an inner product on $H^1_0(\Omega)$ (because of Friedrich's inequality)and the term on the right is a bounded linear functional $F$ on $H^1_0(\Omega)$ by first Holder's and then Friedrich's inequality. So in the weak form, our equation is $\langle u,v \rangle = F(v) \text{ for all } v \in D(\Omega)$ But by the Riesz Representation Theorem, there exists a $u \in H^1_0(\Omega)$ such that the above equation is satisfied for all $v \in H^1_0(\Omega)$, so in particular it is satisfied for all $v \in D(\Omega)$. So we have proved the existence of a weak solution.

As of now we can only ensure that our solution $u$ lies in $H^1_0(\Omega)$ but we would like to know that at least $u \in H^2_0(\Omega)$. For this we will need the Elliptic Regularity Theorem. For our purposes, the following simplified version will suffice:

Let $k \geq 0$. If $f \in H^k(\Omega)$ and $\Delta u = f$, then $u \in H^{k+2}(\Omega)$.

In particular, if $f$ is smooth, then $f \in H^{k}(\Omega)$ for every $k$ so $u \in H^{k+2}(\Omega)$ for every $k$ which by the Sobolev Embedding Theorem implies that $u$ is smooth.

So we have proven that your PDE has a unique smooth classical solution whenever the data is smooth.

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    @A.J. Oh, sorry I though you wanted to know that the equation is solvable. However, I think for Poisson's equation the finite difference method and finite element method are equivalent if you take a grid. The above guarantees that the linear system obtained from FEM will be invertible (positive definite, in fact). So the same applies to the FDM system.2011-10-18