If you can scrape the essential part of your code from your notebook and edit your post above to paste that onto the screen with four leading blanks (white space) in front of each line then that will let the posting process that you mean the paste to be code instead of interpreting some of that as formatting directions.

If you can include enough of the code that it should be able to be run by someone else who doesn't have anything other than what they read on the screen and you include the input, what the output is and what the output should be instead then perhaps someone can spot it issue and help you get it working.

Dr. Dolhaine

Thank you very much for your help.

You mentioned something about a notebook that you sent to me. I looked for it but could not find it and your code has been very valuable to me. Oops I see it thanks! Briefly looking at it there are some striking similarities but I think I have my work cut out for me.

I would not have guessed using a diffusion and drift equation for your tower and would have tried something along the Stokes flow approach but I think friction is involved there maybe.. And since the tower was high and narrow you can use a one dimensional flow with your version of the diffusion equation which is like a blessing from heaven.

Sorry but what is "barometric flow"? Is it the opposite of Stokes flow? I haven't taken a physics class in a while. Googling did not help very much.

Dr. Dolhaine ,as a matter of fact I do have a copy of Carslaw and Jaeger. I could not have solved my differential equation without it. Today I looked for the transformation equation that takes the very same differential equation I am using and turns it into a simple diffusion equation but could not find it today. However I did write it down and use it often..
. However the most elegant form of what I'm working on is the simple ODE. It was not my idea and was so simple that I could kick myself.

The ordinary differential equation is

D C"(x) - u C'(x) - k C(x) ==0

and of course it has cases depending on the coefficients. This model was custom made for Uranium Roll front deposits. Because of the cases associated with the constants each combination of constants D, u, k is a period in the life time of the deposit. It was hinted to me that if D and u are known then the reaction rate can be estimated-- something not known back in 1982.

Once again, Thanks very much.

Very truly yours,

Mike Bernthal

I made a mistake In my haste to post this blog I should have mentioned. I got it right 26 years ago but did not get it right ironically as I was checking it for "bugs".The outflow condition should be as Dr. Dolhaine wrote it. Tan(x) - x = 0 not Tan(x) + x.=0

Mike Bernthal

Whether the zeros approach really (2 j + 1 ) Pi / 2 is an open question. At least the 1st hundred lie near to these values

ListPlot[Table[ (2 j + 1) Pi/2 - (x /. FindRoot[Tan[x] - x == 0, {x, (2 j + 1) Pi/2 - .1}, 
      WorkingPrecision -> 40]), {j, 1, 100}]]

Guten Tag Michael.

Try (you can transfer the following line to a notebook via copy and paste)

Plot[Tan[x] - x, {x, 0, 15}]

From the plot you can see that the zeros are in the vicinity of startvalues (given below. You may well use other startvalues). Then apply Findroot to each of them to get the zeros (here FindRoot is incorporated in a Function which is applied to the list of startvalues).

In[19]:= startvalues = {4, 7, 10.5, 14};
FindRoot[Tan[x] - x == 0, {x, #}, WorkingPrecision -> 20] & /@ startvalues

Out[20]= {{x -> 4.4934094579090641753}, {x -> 7.7252518369377071642},
{x -> 10.904121659428899827}, {x -> 14.066193912831473480}}

For example

In[22]:= Tan[10.9041216594288] - 10.9041216594288

Out[22]= -1.19265*10^-11

From a Plot like

Plot[Tan[x] - x, {x, 15, 35}, Epilog -> {PointSize[.02], Red, Point /@ Table[{(2 j - 1) Pi/2, 0}, {j, 20}]}]

one may conclude that the zeros of Tan[x]-x approach (2 n - 1) Pi / 2.

I hope this helps.

Hi Bill, It's a tricky problem for me. I am pretty sure the code is wrong. However nothing like this has ever happened to me even with a flawed program.

I am 61 and am a slow worker but I am patient and don't give up very easily . I'm still working on the right algorithm. There has to be a simple answer.

I will post it as soon as I know I've got the right code.

Thanks for your patience.

Mike

I