<blockquote>
  <p>The problem with my derivation is I am not sure if it correct and I do not have a way to check it.</p>
</blockquote>

<p>Try dimensional analysis to check your solution. Is your solution consistent with dimensional analysis? Estimate a solution to a real example, based upon you knowledge of geophysics and geology. Does your solution even come close to a reasonable estimate? If you are unfamiliar with dimensional analysis and estimation methods, I recommend the online book (MIT course) <a href="http://ocw.mit.edu/courses/mathematics/18-098-street-fighting-mathematics-january-iap-2008/readings/sf_math.pdf" rel="nofollow">Street Fighting Mathematics</a> for some possible ways to stimulate ideas of ways to check your derivation.   </p>

<p>For example, as you have written your solution, the $b$ and the $\alpha$ are constant over time ($t$) and location ($x)$. First substitute your solution into the differential equation, and carry out the differentiation to confirm that it is in fact a solution. If it can be verified as a solution, turn your attention to the dimensions of b and alpha. The dimension of b must be time and the dimension of alpha must be length. You are modeling the system as a viscoelastic solid, so the the initial condition ($t=0$) for your displacements $w(0,x)$ are just the elastic solution, and the time-dependent term is unity. Use the elastic solution to solve for alpha (the characteristic wavelength) because this does not change over time. Name the purely elastic displacement '$e$' and it has dimension of length. </p>

<p>You know you want to include viscosity somewhere, so let's try the kinematic viscosity $v$ (dimensions $M*M/s$). So try $b = -(L * L)/v$, or $b = -(L * 1)/v$, because this is a 2D problem and can be one unit deep. You'll need the negative sign to make the displacement grow faster for a smaller viscosity.</p>

<p>What could $L$ be? I would try $L=e$, or $L=e/2$. Because you really don't know the kinematic viscosity of the lithosphere, the factor is not critical. </p>

<p>Substitute $b$ and $\alpha$ into your solution. Does the result seem to make physical and geological sense? </p>

CC BY-SA 3.0
2015-05-13T04:25:09.557Z
2016-03-12T03:37:01.243Z
2016-03-12T03:37:01.243Z
Mark Rovetta
<p>I have been trying to figure out solution for the fourth order differential equation of flexure for a viscous plate. </p>

<p>The fourth order differential equation for a viscous plate is given as</p>

<p>$$Dv*\dfrac{\partial^4}{\partial x^4}\bigg(\dfrac{\partial w}{\partial t}\bigg) + kw = 0,$$</p>

<p>here, $Dv$ is a function of viscosity and elastic thickness, and $k$ is the specific weight. 
My solution so far is 
$$w = w_0*\exp(t/b)*\exp(-x/\alpha)*\cos(x/\alpha),$$ 
where, $w_0$ is the maximum deflection, $b$ is function of $Dv$, $k$ and $\alpha$; and $\alpha$ is function of $Dv$ and $k$.</p>

<p>The problem with my derivation is I am not sure if it correct and I do not have a way to check it. Moreover, the $b$ term is an issue as it resolves to constant integer, which is not physically likely.
Anyone out there, who could help me get on the right path? Any papers, any links?</p>

CC BY-SA 3.0
2015-05-12T13:28:22.267Z
Please review my changes to your markup to make sure I haven't botched up the equations as you intended.
casey
CC BY-SA 3.0
2016-03-13T01:17:08.457Z
Can you post your derivation?
Antonio
<geophysics><geodynamics><lithosphere>
Flexure of a viscous lithosphere - derivation
geodyn_usr
# Flexure of a viscous lithosphere - derivation

(Asked by: username_0 on 2015-05-11)

I have been trying to figure out solution for the fourth order differential equation of flexure for a viscous plate.

The fourth order differential equation for a viscous plate is given as

$$Dv\*\dfrac{\partial^4}{\partial x^4}\bigg(\dfrac{\partial w}{\partial t}\bigg) + kw = 0,$$

here, $Dv$ is a function of viscosity and elastic thickness, and $k$ is the specific weight.
My solution so far is
$$w = w\_0\*\exp(t/b)\*\exp(-x/\alpha)\*\cos(x/\alpha),$$
where, $w\_0$ is the maximum deflection, $b$ is function of $Dv$, $k$ and $\alpha$; and $\alpha$ is function of $Dv$ and $k$.

The problem with my derivation is I am not sure if it correct and I do not have a way to check it. Moreover, the $b$ term is an issue as it resolves to constant integer, which is not physically likely.
Anyone out there, who could help me get on the right path? Any papers, any links?

**Tags:** geophysics, geodynamics, lithosphere

> **Comments:**
> **username_1**: Please review my changes to your markup to make sure I haven't botched up the equations as you intended.  
> **username_2**: Can you post your derivation?  

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## 1 answers

### Answer by username_3 on 2015-05-13. Score: 4

> The problem with my derivation is I am not sure if it correct and I do not have a way to check it.

Try dimensional analysis to check your solution. Is your solution consistent with dimensional analysis? Estimate a solution to a real example, based upon you knowledge of geophysics and geology. Does your solution even come close to a reasonable estimate? If you are unfamiliar with dimensional analysis and estimation methods, I recommend the online book (MIT course) [Street Fighting Mathematics](http://ocw.mit.edu/courses/mathematics/18-098-street-fighting-mathematics-january-iap-2008/readings/sf_math.pdf) for some possible ways to stimulate ideas of ways to check your derivation.

For example, as you have written your solution, the $b$ and the $\alpha$ are constant over time ($t$) and location ($x)$. First substitute your solution into the differential equation, and carry out the differentiation to confirm that it is in fact a solution. If it can be verified as a solution, turn your attention to the dimensions of b and alpha. The dimension of b must be time and the dimension of alpha must be length. You are modeling the system as a viscoelastic solid, so the the initial condition ($t=0$) for your displacements $w(0,x)$ are just the elastic solution, and the time-dependent term is unity. Use the elastic solution to solve for alpha (the characteristic wavelength) because this does not change over time. Name the purely elastic displacement '$e$' and it has dimension of length.

You know you want to include viscosity somewhere, so let's try the kinematic viscosity $v$ (dimensions $M\*M/s$). So try $b = -(L \* L)/v$, or $b = -(L \* 1)/v$, because this is a 2D problem and can be one unit deep. You'll need the negative sign to make the displacement grow faster for a smaller viscosity.

What could $L$ be? I would try $L=e$, or $L=e/2$. Because you really don't know the kinematic viscosity of the lithosphere, the factor is not critical.

Substitute $b$ and $\alpha$ into your solution. Does the result seem to make physical and geological sense?


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