# Matrix representation of the radial Laplace operator isn't symmetric as supposed

I'm working with the cylindrical coordinates. I'm using the central difference to convert the radial part of Laplace operator into a matrix.

$\nabla^2 u = \frac{\partial ^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}$

which in the discretized form gives:

$\nabla^2 u_i = \frac{u_{i+1} - 2u_i +u_{i-1}}{h^2}+\frac{1}{r_i}\frac{u_{i+1} -u_{i-1}}{2h}$

where "h" is the step size and $\ r_i = i*h$. Thus:

$\nabla^2 u_i = \frac{u_{i+1} - 2u_i +u_{i-1}}{h^2}+\frac{1}{i*h}\frac{u_{i+1} -u_{i-1}}{2h}=\frac{1}{h^2}[u_{i+1}(1+\frac{1}{2i})-2u_i+u_{i-1}(1-\frac{1}{2i})]$

This means that the off diagonals: coefficients of $(u_{i+1})$ and $(u_{i-1})$ will have the $1/r_i$ dependence, so that would affect the matrix symmetry. This is because in the matrix, the element "ij" will depend on $1/r_i$ while its transpose "ji" will depend on $1/r_j$. Hence, the first few elements of the matrix will look like:

\begin{matrix} -2 & 1.5 & 0 & \cdots \\ 0.75 & -2 & 1.25 & 0 & \cdots\\ 0 & \frac{5}{6} & -2 & \frac{7}{6} & \cdots\\ \cdots & \cdots & \cdots & \cdots \\ \end{matrix}

I know that the Laplace operator is hermitian, why isn't the matrix symmetric? What am I missing?

• You are missing the angular term in your Laplacian $$\nabla^2 u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \phi^2}$$ – nicoguaro Apr 25 '16 at 21:48
• @nicoguaro it looks like this question is only in reference to the radial component. – Charles Apr 26 '16 at 18:59

What Charlie is saying is basically that the discrete Laplacian is Hermitian, but not with regard to the Cartesian inner product.

When defined in polar coordinates, the continuous Laplacian $\nabla^2$ satisfies $$\int u \nabla^2 v rdrd\varphi = \int \nabla^2 v u r dr d\varphi$$ for any $u(r, \varphi)$ and $v(r, \varphi)$. For the discrete Laplacian, the integrals become sums weighted by $hr_i$, or in matrix notation $$\langle u, WAv \rangle = \langle A u,Wv \rangle,$$ where $W_{ii} = hr_i$ and $\langle x, y \rangle = x^Ty$. Hence, $WA$, or in Charlie's notation $dVA$, is Hermitian but $A$ is not.

I don't see a problem in solving the eigenvalue problem for $WA$ instead of $A$: if $x$ is an eigenvector of $WA$, then $W^{-1}x$ is an eigenvector of $A$ with the same eigenvalue.

Start with the original definition and discretize consistently: \begin{align*} \frac{1}{r}\frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r} \right) &= \frac{1}{r_i \thinspace h} \left( r_{i+\frac{1}{2}} \left[ \frac{u_{i+1} - u_i}{h} \right] - r_{i-\frac{1}{2}} \left[ \frac{u_i - u_{i-1}}{h} \right] \right) \\ &= \frac{1}{i \thinspace h^2} \left( (i+\tfrac{1}{2}) \left[ u_{i+1} - u_i \right] - (i-\tfrac{1}{2}) \left[ u_i - u_{i-1} \right] \right) \\ &= \frac{1}{i \thinspace h^2} \left( (i+\tfrac{1}{2}) u_{i+1} - 2 \thinspace i \thinspace u_i + (i-\tfrac{1}{2}) u_{i-1} \right) \end{align*}

That should result in the symmetric matrix you're looking for.

• Whatever definition used, both your expression and mine are the same. As an ex., compare the coefficient of $u_{i+1}$ in yours and mine. Yours: $\frac{1}{i}*(i+\frac{1}{2})$, mine:$1+\frac{1}{2i}$. So, the matrix is still not hermitian. – user334106 Apr 25 '16 at 1:32
• My bad ... back to square one. Don't know the answer yet. – NameRakes Apr 25 '16 at 3:17

I think the answer might be in the equation you're looking at. Consider solving the equation $Ax=b$, where $A=\nabla^2$, $x$ is the unknown and $b$ is the right hand side. Multiplying the left and right hand side by the volume ($r dr d\theta dz$), will result in a symmetric matrix $A$ on the left hand side, right? Then you'd be solving the modified system $Dx = f$ where $D = dV A$ and $f = dV b$.

I'm not sure if there is a way to symmetrize the operator without modifying other parts of the equation.

As @nameRakes pointed out, this would become

\begin{align*} dV \frac{1}{r}\frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r} \right) &= \frac{k}{h} \left( r_{i+\frac{1}{2}} \left[ \frac{u_{i+1} - u_i}{h} \right] - r_{i-\frac{1}{2}} \left[ \frac{u_i - u_{i-1}}{h} \right] \right) \\ &= \frac{k}{\thinspace h} \left( (i+\tfrac{1}{2}) \left[ u_{i+1} - u_i \right] - (i-\tfrac{1}{2}) \left[ u_i - u_{i-1} \right] \right) \\ &= \frac{k}{\thinspace h} \left( (i+\tfrac{1}{2}) u_{i+1} - 2 \thinspace i \thinspace u_i + (i-\tfrac{1}{2}) u_{i-1} \right) \end{align*}

Where $dV=rk,k=drd\theta dz$. You'll notice that the equation is slightly different now and you're only dividing by $h$ instead of $h^2$, but this is okay so long you treat your RHS the same (by multiplying by $V$).

I think this is symmetric, I'm not sure about Hermitian though.

• The point is that I'm using the operator in an eigen-value problem: $Ax=\lambda x$. So multiplying both sides by $dV$ will change the equation to $Dx=\lambda y$ where $y=dV.x$. That's why I'm looking for a soultion to the LHS without affecting the RHS. In addition, it's strange that the matrix of a hermitian operator not hermitian. – user334106 Apr 27 '16 at 0:35
• Ahh, I see... I just wasn't familiar with what hermitian meant, so I apologize if I misspoke about it. – Charles Apr 30 '16 at 8:20