Finite Differencing Dynamic Boundary

[member 428835]

Hi PF!

I'm using a finite differencing scheme to solve the following $$h_t = h h_{zz} + 2h_z^2$$ where the subscripts denote partial derivatives. The difficulty I'm facing is the boundary conditions are dynamic, and move with time $t$. This makes choosing a $\Delta z$ very difficult and unintuitive for me.

I have already developed a MatLab code that works pretty well but I feel my $\Delta z$ should change in time. If anyone knows any references or has advice on this issue of dealing with the $\Delta z$ please let me know.

Thanks so much!

Josh

 

[soarce]

Try to change the time variable in order to achieve time independent boundary conditions.

 

[member 428835]

I've already performed a similarity transform, which can actually be solved analytically under certain physical situations! But I'm trying to solve before the transform, so moving boundary conditions is all I can work with.

 

[soarce]

Then you must look into Free Boundary problems.

 

[member 428835]

Thanks!

 

[Chestermiller]

On a problem like this, what I would do is change variables, so that the z variable is as follows:

$$z^*=\frac{(z-z_{left}(t))}{(z_{right}(t)-z_{left}(t))}$$

You would have to transform the differential equation so that the partial derivative with respect to t is at constant z* versus z. But the advantage would be that the boundary conditions could always be applied at z* = 0 and z* = 1. Basically what you are doing is taking the complexity out of the boundary conditions and putting it into the differential equation where it can be handled more easily.

Chet

 

[member 428835]

Chet, this is exactly what the department chair recommended. We then have $$\frac{\partial}{\partial z} = \frac{1}{z_r(t)-z_l(t)}\frac{\partial}{\partial z^*}$$Unfortunately at the moment I need to solve this problem in the $z$ space, not the $z^*$ space. This being said, later I'll look into this method. Would you mind if I asked you for advice then if I run into any trouble?

I do have a well-prepared strategy to proceed, but I'm wondering how to use finite differencing techniques to solve for the last spatial node in my domain? Currently I'm using a forward time centered space scheme. During each time loop, I use FD to recalculate each $h$ value, which gives me something like this $$\frac{h_i^{j+1}-h_i^j}{\Delta t} \approx h_i^j \left( \frac{h_{i+1}^j-2h_i^j+h_{i-1}^j}{\Delta z^2} \right) + 2\left( \frac{h_{i+1}^j-h_{i-1}^j}{2 \Delta z} \right)^2$$

where the subscript denotes space and the superscript denotes time. To find $h^{j+1}_{right}$ should I allow $h_{right}^j=0$ since that is what the right end was at time $j$ and then use a backward space technique for the first derivative term?

 

[Chestermiller]

Chet, this is exactly what the department chair recommended. We then have $$\frac{\partial}{\partial z} = \frac{1}{z_r(t)-z_l(t)}\frac{\partial}{\partial z^*}$$Unfortunately at the moment I need to solve this problem in the $z$ space, not the $z^*$ space. This being said, later I'll look into this method. Would you mind if I asked you for advice then if I run into any trouble?

I do have a well-prepared strategy to proceed, but I'm wondering how to use finite differencing techniques to solve for the last spatial node in my domain? Currently I'm using a forward time centered space scheme. During each time loop, I use FD to recalculate each $h$ value, which gives me something like this $$\frac{h_i^{j+1}-h_i^j}{\Delta t} \approx h_i^j \left( \frac{h_{i+1}^j-2h_i^j+h_{i-1}^j}{\Delta z^2} \right) + 2\left( \frac{h_{i+1}^j-h_{i-1}^j}{2 \Delta z} \right)^2$$

where the subscript denotes space and the superscript denotes time. To find $h^{j+1}_{right}$ should I allow $h_{right}^j=0$ since that is what the right end was at time $j$ and then use a backward space technique for the first derivative term?

You need actual boundary conditions at the two ends. What are the boundary conditions?

chet

 

[Strum]

On a problem like this, what I would do is change variables, so that the z variable is as follows:

$$z^*=\frac{(z-z_{left}(t))}{(z_{right}(t)-z_{left}(t))}$$

You would have to transform the differential equation so that the partial derivative with respect to t is at constant z* versus z. But the advantage would be that the boundary conditions could always be applied at z* = 0 and z* = 1. Basically what you are doing is taking the complexity out of the boundary conditions and putting it into the differential equation where it can be handled more easily.

Chet

I am not quite sure I understand this method. What is $z_{left}$ and $z_{right}$? What happens if the moving boundary is part of the solution to the problem, as in a Stefan problem?

 

[Chestermiller]

I am not quite sure I understand this method. What is $z_{left}$ and $z_{right}$? What happens if the moving boundary is part of the solution to the problem, as in a Stefan problem?

Josh,

Do you feel like you can explain this to Strum?

Chet

 

[Strum]

So Chet. It doesn't look like Josh wants to chip in. Would you mind explaining how this would generalize to moving boundary problems?

 

[Chestermiller]

So Chet. It doesn't look like Josh wants to chip in. Would you mind explaining how this would generalize to moving boundary problems?

z_left(t) and z_right(t) are the time-dependent locations of the left- and right boundaries of the region being analyzed. Both boundaries are assumed to be moving. To complete the problem statement, Josh would have to provide the boundary conditions on h at these locations.

We would like to convert the problem statement from h = h (z,t) to h = h(z*,t) so that, in the z* - t framework, the boundaries are not moving. The transformation that I gave in post #6 will do this. But, one has to transform the differential equation properly. We start out by writing:

$$dh=\left(\frac{\partial h}{\partial t}\right)_{z^*}dt+\left(\frac{\partial h}{\partial z^*}\right)_{t}dz^*$$
Then we evaluate the partial derivatives in the differential equation as follows:
$$\left(\frac{\partial h}{\partial z}\right)_t=\left(\frac{\partial h}{\partial z^*}\right)_{t}\left(\frac{\partial z^*}{\partial z}\right)_t=\left(\frac{\partial h}{\partial z^*}\right)_{t}\frac{1}{(z_{right}(t)-z_{left}(t))}$$
$$\left(\frac{\partial h}{\partial t}\right)_z=\left(\frac{\partial h}{\partial t}\right)_{z^*}
+\left(\frac{\partial h}{\partial z^*}\right)_{t}
\left(\frac{\partial z*}{\partial t}\right)_z=
\left(\frac{\partial h}{\partial t}\right)_{z^*}
-\frac{z^*}{(z_{right}(t)-z_{left}(t))}\left(\frac{\partial h}{\partial z^*}\right)_{t}
\left(\frac{dz_{right}}{dt}-\frac{dz_{left}}{dt}\right)$$

This is then substituted into the differential equation.

Hope this makes sense.

Chet

 

[member 428835]

Thanks Chet! Sorry it has taken so long for me to thank you, but thanks so much!

 

[member 428835]

And sorry Strum, I recently changed my email address, and that coupled with a summer vacation meant I was scarce; I'm sorry and didn't intend to be rude.

At the risk of sounding redundant to what Chet said, $z_{left}$ is the left boundary position of $z$ in the problem. If you imagine $z$ as the length of a candle, then perhaps $z_{left}$ is the height, and since the candle is presumable burning, $z_{left}$ would be getting smaller in time. Obviously the differential equation above, while it can be re-cast as the heat equation, is definitely not a candle; in fact, it's capillary corner flow! The same logic extends to $z_{right}$.

Chet's proposed change of variables implies $z^* \in [0,1]$. This is a common substitution in non-dimentionalizing pressure terms in navier-stokes equations, if you've studies those at all.

Again, I'm terribly sorry for the late delays, and the only reason I was able to respond to Chet was via my phone, which took some work.

And Chet, $h(z_{left}) = 1$ and $h(z_{right}) = 0$. I'll work with this some and let you know what I get, although it seems in your previous post you already did the heavy lifting.