Solving an ODE from a spherical laplacian

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JHZR2

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So I'm always big on saying how important calculus and all is. Of course even those of us who respect and understand the need for it, dont always remember all of the basics. I havent done this one in a while, and Im doing something wrong.

Any math whizzes who can help me out here?

The basis is a laplacian in spherical coordinates.

8275dcad2a7a2f9587c996beb5cc8c37.png


This would be a PDE, with a time derivative on the LHS, but everything simplifies out to be 1D, so I guess it is really an ODE.

My equation is this, at steady state:

Sql Query:
0 = 1/r^2 ∂/∂r (D*r^2 ∂c/∂r)


Where D is some numerical Scalar, it does not imply a directional derivative or anything.

So the thing that I'm just not remembering is how to deal with the basic function of the form:

Sql Query:
∂/∂r (r^2 ∂c/∂r)
in terms of how to deal with it to make it solvable.

I was kind of under the impression that I should do a chain rule to make it a real ODE, which gives me:

Sql Query:
0 = D*( ∂2c/∂r2 + 2/r ∂c/∂r)


Problem is that the 2/r would integrate into log(r), right? Implementing the BC for the sphere that at r=0, dc/dr=0, I would have an undefined answer since I'd have Log(0).

So something is wrong.

I just dont recall how to solve. I want to actually go through the math to prove it to myself. I know there are solutions for this mind of form, but it annoys me that Im forgetting the basic step to solve this form of equation... so I need to go through the motions with the right start.

Thanks!
 
Haven't done calculus since my first year of my first undergrad degree
smile.gif


That was back in the early 90-s...
 
Originally Posted By: CivicFan
Haven't done calculus since my first year of my first undergrad degree
smile.gif


That was back in the early 90-s...


Me too and I added a Math minor(few classes) to my engineering major and have not used it once.
 
All I care to know is how in the world did you manage to enter those equations! I have had trouble figuring out how to enter symbol for PI in this forum :-)

And what is ODE??

- Vikas

P.S. There was a time in my life when I could solve this but now I am happy if I can recognize all the Greek letters from that equations.
 
Originally Posted By: rjundi
Originally Posted By: CivicFan
Haven't done calculus since my first year of my first undergrad degree
smile.gif


That was back in the early 90-s...


Me too and I added a Math minor(few classes) to my engineering major and have not used it once.


They make you take those courses just to provided job security for math PhDs because lets face it what else would they do with their advanced math knowledgde lol. I could run the question by a math PhD I know but it doesn't seem like it's worth the bother.
 
I mean no disrespect to you or you skills...but of what value is this ? I am asking as a serious question, not as a smart %&#. I have always considered myself to be good at everyday math,but this leaves me in the dust !
 
Originally Posted By: Vikas
All I care to know is how in the world did you manage to enter those equations! I have had trouble figuring out how to enter symbol for PI in this forum :-)

And what is ODE??

- Vikas

P.S. There was a time in my life when I could solve this but now I am happy if I can recognize all the Greek letters from that equations.


ODE = Ordinary Differential Equation
PDE = Partial Differential Equation
LHS = Left Hand Side of the equation

But while I remember what they are, I can't remember how to do this stuff...so I am no help...
 
I'm not following the equation. I think it might be formatting that makes it look strange to me.

Maybe some explanation would help. It's been alot of years since I took differential equations too, so... it may or may not come back to me.
 
C is concentration in my equations, also shown as f in the fancy one I posted as the definition of the laplacian. The laplacian is a second derivative (essentially the rate of change of the rate of change).

The basis of the equation is to understand the rate of diffusion in a spherical body.

This is the basis of how things like time release drugs work in the body, or, one could figure out based upon the rate of production, what the concentration of a substance is at some distance.

One could figure, for example, if a droplet of water fell into a vat of oil (there is some solubility!), what would the concentration of water be in the oil.

Folks likely wouldn't solve this for that, but it is an example that we could likely identify with...
 
Quote:
Problem is that the 2/r would integrate into log(r), right? Implementing the BC for the sphere that at r=0, dc/dr=0, I would have an undefined answer since I'd have Log(0).

So something is wrong.


Quote:
C is concentration in my equations, also shown as f in the fancy one I posted as the definition of the laplacian. The laplacian is a second derivative (essentially the rate of change of the rate of change).

The basis of the equation is to understand the rate of diffusion in a spherical body.



I think you answer yourself there already. I too forgot most of my math but without spending too much time finding my Calculus textbook my gut feeling would tell me that at r = 0, dc/dr would not be 0 as it is the point origin of the diffusion. If you are looking at the slope of the diffusion (dc/dr I assume), you will see a vertical line descending from 0 and gradually flatten out as it go further away.
 
Last edited:
Originally Posted By: JHZR2
C is concentration in my equations, also shown as f in the fancy one I posted as the definition of the laplacian. The laplacian is a second derivative (essentially the rate of change of the rate of change).

The basis of the equation is to understand the rate of diffusion in a spherical body.

This is the basis of how things like time release drugs work in the body, or, one could figure out based upon the rate of production, what the concentration of a substance is at some distance.

One could figure, for example, if a droplet of water fell into a vat of oil (there is some solubility!), what would the concentration of water be in the oil.

Folks likely wouldn't solve this for that, but it is an example that we could likely identify with...


Thanks for that. It actually gives a lot of context to what you are showing. I remember asking a math professor showing similar type equations and what was a practical example of them. He knew of none. The funny thing he is still at my school, looked old when I was there and now still does not have an email address.
 
Originally Posted By: rjundi
Originally Posted By: CivicFan
Haven't done calculus since my first year of my first undergrad degree
smile.gif


That was back in the early 90-s...


Me too and I added a Math minor(few classes) to my engineering major and have not used it once.


Same here!
 
Originally Posted By: PandaBear
Quote:
Problem is that the 2/r would integrate into log(r), right? Implementing the BC for the sphere that at r=0, dc/dr=0, I would have an undefined answer since I'd have Log(0).

So something is wrong.


Quote:
C is concentration in my equations, also shown as f in the fancy one I posted as the definition of the laplacian. The laplacian is a second derivative (essentially the rate of change of the rate of change).

The basis of the equation is to understand the rate of diffusion in a spherical body.



I think you answer yourself there already. I too forgot most of my math but without spending too much time finding my Calculus textbook my gut feeling would tell me that at r = 0, dc/dr would not be 0 as it is the point origin of the diffusion. If you are looking at the slope of the diffusion (dc/dr I assume), you will see a vertical line descending from 0 and gradually flatten out as it go further away.



maybe. Thing is, at the sphere's center, r=0, there is no diffucion and all changes occur outward from that point. So at that singular point, there cannot be any change itself as everything only changes riadially outward from there in all directions.
 
Originally Posted By: JHZR2
maybe. Thing is, at the sphere's center, r=0, there is no diffucion and all changes occur outward from that point. So at that singular point, there cannot be any change itself as everything only changes riadially outward from there in all directions.



if you are only interested in the end result, you likely have to multiply a volume or surface area to the "rate" of dc/dr anyways, and it is probably more practical to cancel out some terms rather than plugging in a zero (as the size of the source of diffusion shrink in volume and surface down to zero) multiply by an infinite large or infinite small (dc/dr being undefined). Then the term after cancelling these terms out may approach something that makes more sense (likely a zero).
 
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