# Is it trivial that I will always find a solution to Laplace's equation via finite-difference method

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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## migrated from stackoverflow.comOct 17 '11 at 21:13

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This question is unreadable. Please give an exact description of your set up and question. Just referring to "the band matrix" or "the solution" is extremely ambiguous. Also use TEX, please. – user12014 Oct 17 '11 at 21:45
Why is my question in math.se? :S – A.J. Oct 17 '11 at 22:48
@PZZ: right, I'll be completing in the next hours. – A.J. Oct 17 '11 at 22:52
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. – user12014 Oct 18 '11 at 0:54

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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Thank you, PZZ. It's still too advanced for me Does this mean the system of equations has a solution? I mean, doesn't the discretization affect? – A.J. Oct 18 '11 at 3:42
@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. – user12014 Oct 18 '11 at 8:03