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I would like to evaluate the integral $\int (\Delta u) v d\Omega$, where the domain $\Omega$ is a cylinder. On the boundaries, either the normal derivative $\partial_n u$ is zero or $v$ is zero. An additional assumption is that $u$ and $v$ are axial symmetric, ie. $\frac{\partial u}{\partial \phi} = \frac{\partial v}{\partial \phi} = 0$.

The two steps in the derivation below are partial integration of the second derivative, and using cylindrical coordinates for the nabla operator. However, the order seems to matter.

Method 1

$\int (\Delta u) v d\Omega$

Using Green's first identity this becomes

$- \int_\Omega \nabla u \cdot \nabla v + \oint_{\partial \Omega} (\partial_n u)v$

where the second term disappears because of the boundary conditions assumed. With the expression for the gradient in cylindrical coordinates, $\nabla = \left(\frac{\partial}{\partial \rho} \boldsymbol{\hat{\rho}}, \frac1{\rho}\frac{\partial}{\partial \phi} \boldsymbol{\hat{\phi}}, \frac{\partial}{\partial z} \mathbf{\hat{z}}\right)$, and the dot product in the coordinates defined by the orthogonal (local) unit vectors $\boldsymbol{\hat{\rho}}, \boldsymbol{\hat{\phi}}, \boldsymbol{\hat{z}}$: $(\rho_1, \phi_1, z_1) \cdot (\rho_2, \phi_2, z_2) = \rho_1 \rho_2 + \phi_1 \phi_2 + z_1 z_2$, the first term becomes:

$-2 \pi \int \int d\rho dz \left[ \frac{\partial u}{\partial \rho} \frac{\partial v}{\partial \rho} + \frac1{\rho^2} \frac{\partial u}{\partial \phi} \frac{\partial v}{\partial \phi} + \frac{\partial u}{\partial z} \frac{\partial u}{\partial z}\right] $

or due to axial symmetry:

$-2 \pi \int \int d\rho dz \left[ \frac{\partial u}{\partial \rho} \frac{\partial v}{\partial \rho} + \frac{\partial u}{\partial z} \frac{\partial u}{\partial z}\right]$

Or:

Method 2

$\int (\Delta u) v d\Omega$

With the Laplacian in cylindrical coordinates, $\Delta = \frac1{\rho}\frac{\partial}{\partial \rho} \left(\rho \frac{\partial}{\partial \rho}\right) + \frac1{\rho^2} \frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2}$ this gives

$2 \pi \int \int d\rho dz \left[ \frac1{\rho}\frac{\partial u}{\partial \rho} + \frac{\partial^2 u}{\partial \rho^2} + \frac1{\rho^2} \frac{\partial^2 u}{\partial \phi^2} + \frac{\partial^2 u}{\partial z^2}\right] v$

Applying now partial integration to the terms with a second derivative, eg. $\int \int d\rho dz \frac{\partial^2 u}{\partial \rho^2}v = \int dz \left. \frac{\partial u}{\partial \rho}v \right|_0^{\rho_{max}} - \int \int d\rho dz \frac{\partial u}{\partial \rho} \frac{\partial v}{\partial \rho} = - \int d\rho dz \frac{\partial u}{\partial \rho} \frac{\partial v}{\partial \rho}$ where the last step is due to axial symmetry and assumed boundary conditions. The resulting equation then is

$2 \pi \int \int d\rho dz \left[\frac1{\rho} \frac{\partial u}{\partial \rho} v - \frac{\partial u}{\partial \rho} \frac{\partial v}{\partial \rho} - \frac{\partial u}{\partial z} \frac{\partial u}{\partial z}\right]$

Where now there is an additional term compared to method 1, and I don't see what I did wrong?

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  • $\begingroup$ FWIW, I used both versions in a finite element implementation and the results are very similar, though not identical. $\endgroup$
    – Maarten
    Sep 22, 2014 at 9:58

1 Answer 1

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Found it eventually, I missed the Jacobian everywhere. All double integrals should be multiplied by $\rho$, which makes no difference between method 1 and 2 except for the partial integration term as written below.

$\int \int \frac{\partial^2 u}{\partial \rho^2} v \rho d\rho dz = \int \left. \frac{\partial u}{\partial \rho}v \rho \right|_0^{\rho_{max}} dz - \int \int \frac{\partial u}{\partial \rho} \frac{\partial (\rho v)}{\partial \rho} d\rho dz = - \int \int \frac{\partial u}{\partial \rho} \rho\frac{\partial v}{\partial \rho} d\rho dz - \int \int \frac{\partial u}{\partial \rho} v d\rho dz$

where this last term thus cancels with the 'mystery term' in the original question (after multiplication by $\rho$, as explained above)

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