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?

  • 1
    $\begingroup$ You are missing the angular term in your Laplacian $$\nabla^2 u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \phi^2}$$ $\endgroup$ – nicoguaro Apr 25 '16 at 21:48
  • $\begingroup$ @nicoguaro it looks like this question is only in reference to the radial component. $\endgroup$ – 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 \begin{equation} \int u \nabla^2 v rdrd\varphi = \int \nabla^2 v u r dr d\varphi \end{equation} 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 \begin{equation} \langle u, WAv \rangle = \langle A u,Wv \rangle, \end{equation} 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.

  • $\begingroup$ 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. $\endgroup$ – user334106 Apr 25 '16 at 1:32
  • $\begingroup$ My bad ... back to square one. Don't know the answer yet. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – user334106 Apr 27 '16 at 0:35
  • $\begingroup$ Ahh, I see... I just wasn't familiar with what hermitian meant, so I apologize if I misspoke about it. $\endgroup$ – Charles Apr 30 '16 at 8:20

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