### Problem with weak form formulation

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Hello everybody,

I have this PDE for the transversal vibration of a tensioned beam with variable axial tension:

EJ*v*u

where E,J,T

I am trying to integrate by parts but I do not know how to integrate the term: w*z*v*u

In particular, how should I handle the integration by parts if I have also the space variable 'z' which multiplies the trial and test function ?

Thank you very much for your help!

Diego

I have this PDE for the transversal vibration of a tensioned beam with variable axial tension:

EJ*v*u

_{zzz}- T

_{o}*v*u

_{zz}- w*z*v*u

_{zz}- w*v*u

_{z}+ m*v*u

_{tt}= f(z)*v

where E,J,T

_{o},w are constant, z is the space variable, t is the time variable,u(z,t) is the unknown displacement function and v is the test function.

I am trying to integrate by parts but I do not know how to integrate the term: w*z*v*u

_{zz}?

In particular, how should I handle the integration by parts if I have also the space variable 'z' which multiplies the trial and test function ?

Thank you very much for your help!

Diego

Community: FEniCS Project

### 2 Answers

1

Let \( \phi(z) = w z v(z)\) and \(\varGamma = \partial \varOmega\). Then

\[

\begin{align*}

\int_{\varOmega} \phi \frac{\partial^2 u}{\partial z^2} \mathrm{d}\varOmega &= - \int_{\varOmega} \frac{\partial \phi}{\partial z} \frac{\partial u}{\partial z} \mathrm{d}\varOmega + \int_{\varGamma} \phi \frac{\partial u}{\partial z} n_z \mathrm{d}\varGamma \\

&= - \int_{\varOmega} w \frac{\partial (z v)}{\partial z} \frac{\partial u}{\partial z} \mathrm{d}\varOmega + \int_{\varGamma} w z v \frac{\partial u}{\partial z} n_z \mathrm{d}\varGamma \\

&= - \int_{\varOmega} w \left[ v \frac{\partial z}{\partial z} + z \frac{\partial v}{\partial z} \right] \frac{\partial u}{\partial z} \mathrm{d}\varOmega + \int_{\varGamma} w z v \frac{\partial u}{\partial z} n_z \mathrm{d}\varGamma \\

&= - \int_{\varOmega} w \left[ v + z \frac{\partial v}{\partial z} \right] \frac{\partial u}{\partial z} \mathrm{d}\varOmega + \int_{\varGamma} w z v \frac{\partial u}{\partial z} n_z \mathrm{d}\varGamma,

\end{align*}

\]

where \(n_z\) is the \(z\) component of the normal vector \(\underline{n}\).

Thanks to Adam's good eye below.

\[

\begin{align*}

\int_{\varOmega} \phi \frac{\partial^2 u}{\partial z^2} \mathrm{d}\varOmega &= - \int_{\varOmega} \frac{\partial \phi}{\partial z} \frac{\partial u}{\partial z} \mathrm{d}\varOmega + \int_{\varGamma} \phi \frac{\partial u}{\partial z} n_z \mathrm{d}\varGamma \\

&= - \int_{\varOmega} w \frac{\partial (z v)}{\partial z} \frac{\partial u}{\partial z} \mathrm{d}\varOmega + \int_{\varGamma} w z v \frac{\partial u}{\partial z} n_z \mathrm{d}\varGamma \\

&= - \int_{\varOmega} w \left[ v \frac{\partial z}{\partial z} + z \frac{\partial v}{\partial z} \right] \frac{\partial u}{\partial z} \mathrm{d}\varOmega + \int_{\varGamma} w z v \frac{\partial u}{\partial z} n_z \mathrm{d}\varGamma \\

&= - \int_{\varOmega} w \left[ v + z \frac{\partial v}{\partial z} \right] \frac{\partial u}{\partial z} \mathrm{d}\varOmega + \int_{\varGamma} w z v \frac{\partial u}{\partial z} n_z \mathrm{d}\varGamma,

\end{align*}

\]

where \(n_z\) is the \(z\) component of the normal vector \(\underline{n}\).

Thanks to Adam's good eye below.

1

I think there should also be the term

\[ - w \int_\Omega \frac{\partial u}{\partial z} v \,\mathrm{d} \Omega, \]

arising from the derivative of \(z\).

\[ - w \int_\Omega \frac{\partial u}{\partial z} v \,\mathrm{d} \Omega, \]

arising from the derivative of \(z\).

written
9 months ago by
Adam Janecka

I suppose you integrate by parts this way, using \(\phi(z) = w z v \):

\[

\begin{align*}

\int_{\Omega} \phi \frac{\partial^2 u}{\partial z^2} d\Omega &= - \int_{\Omega} \frac{\partial \phi}{\partial z} \frac{\partial u}{\partial z} d\Omega + \int_{\partial \Omega} \phi \frac{\partial u}{\partial z} n_z d\partial \Omega \\

&= - \int_{\Omega} w \frac{\partial (z v)}{\partial z} \frac{\partial u}{\partial z} d\Omega + \int_{\partial \Omega} w z v \frac{\partial u}{\partial z} n_z d\partial \Omega \\

&= - \int_{\Omega} w v \frac{\partial z}{\partial z} \frac{\partial u}{\partial z} d\Omega - \int_{\Omega} w z \frac{\partial v}{\partial z} \frac{\partial u}{\partial z} d\Omega + \int_{\partial \Omega} w z v \frac{\partial u}{\partial z} n_z d\partial \Omega \\

&= - \int_{\Omega} w v \frac{\partial u}{\partial z} d\Omega - \int_{\Omega} w z \frac{\partial v}{\partial z} \frac{\partial u}{\partial z} d\Omega + \int_{\partial \Omega} w z v \frac{\partial u}{\partial z} n_z d\partial \Omega,

\end{align*}

\]

but... nevermind, you're right.

\[

\begin{align*}

\int_{\Omega} \phi \frac{\partial^2 u}{\partial z^2} d\Omega &= - \int_{\Omega} \frac{\partial \phi}{\partial z} \frac{\partial u}{\partial z} d\Omega + \int_{\partial \Omega} \phi \frac{\partial u}{\partial z} n_z d\partial \Omega \\

&= - \int_{\Omega} w \frac{\partial (z v)}{\partial z} \frac{\partial u}{\partial z} d\Omega + \int_{\partial \Omega} w z v \frac{\partial u}{\partial z} n_z d\partial \Omega \\

&= - \int_{\Omega} w v \frac{\partial z}{\partial z} \frac{\partial u}{\partial z} d\Omega - \int_{\Omega} w z \frac{\partial v}{\partial z} \frac{\partial u}{\partial z} d\Omega + \int_{\partial \Omega} w z v \frac{\partial u}{\partial z} n_z d\partial \Omega \\

&= - \int_{\Omega} w v \frac{\partial u}{\partial z} d\Omega - \int_{\Omega} w z \frac{\partial v}{\partial z} \frac{\partial u}{\partial z} d\Omega + \int_{\partial \Omega} w z v \frac{\partial u}{\partial z} n_z d\partial \Omega,

\end{align*}

\]

but... nevermind, you're right.

written
9 months ago by
pf4d

Exactly, omitting \(w\),

\[ \int_{\partial \Omega} z \frac{\partial u}{\partial z} n_z v \,\mathrm{d} a = \int_{\Omega} \frac{\partial}{\partial z} \left( z \frac{\partial u}{\partial z} v \right) \,\mathrm{d} v = \int_\Omega \frac{\partial u}{\partial z} v \,\mathrm{d}v + \int_{\Omega} z \frac{\partial^2 u}{\partial z^2} v \,\mathrm{d} v + \int_{\Omega} z \frac{\partial u}{\partial z} \frac{\partial v}{\partial z} \,\mathrm{d} v. \]

I do not see, how you can 'move' the derivative just to the test function a leave \(z\) untouched.

\[ \int_{\partial \Omega} z \frac{\partial u}{\partial z} n_z v \,\mathrm{d} a = \int_{\Omega} \frac{\partial}{\partial z} \left( z \frac{\partial u}{\partial z} v \right) \,\mathrm{d} v = \int_\Omega \frac{\partial u}{\partial z} v \,\mathrm{d}v + \int_{\Omega} z \frac{\partial^2 u}{\partial z^2} v \,\mathrm{d} v + \int_{\Omega} z \frac{\partial u}{\partial z} \frac{\partial v}{\partial z} \,\mathrm{d} v. \]

I do not see, how you can 'move' the derivative just to the test function a leave \(z\) untouched.

written
9 months ago by
Adam Janecka

Yes, you are correct.

written
9 months ago by
pf4d

I like how you use mathrm on those d's, but I dislike your choice of "dv" and "da" that is confusing.

written
9 months ago by
pf4d

Thanks. 'dv' should indicate volume, while 'da' area -- maybe 'ds' as for surface would be better.

written
9 months ago by
Adam Janecka

personally, I prefer the explicit or the use of \(\varGamma = \partial \varOmega\).

written
9 months ago by
pf4d

I'm fine with it (or anything that makes sense) ;-)

written
9 months ago by
Adam Janecka

I edited my answer once again after you kindled my obsessive-math-symbol-notation disorder.

written
9 months ago by
pf4d

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