@tomleslie 

Weirdly your worksheet works fine- my guess is that this was something odd that happened when i pasted it.

Thanks

@tomleslie 

weirdly still an error. I am using version

`Maple 2019.1, X86 64 WINDOWS, May 21 2019, Build ID 1399874`

@tomleslie 
Weirdly it produces an error when i copy/paste it
York_Identifiability_images.mw

@acer 

thanks, I had to spend a fair bit of time digesting all that you sent. Here is a script file illustrating the problem (the last line is the one that takes up all the memory, if you want something that will process faster try putting in %[1..6] or similar rather than %)

Linda_script_full.mpl

I think its a simple enough problem to get a course solution given my universities resources, but I would rather find the least expensive way of doing this kind of solve action.

@tomleslie Thats close! what I meant was I wanted at mutliples of 24 to add 150 (i think this represents a drug) and then it decays back close to zero, until another 150 is added.

I updated the question to make that a little clearer

@Annonymouse a work around could be to use the solutions you got and plot each of them in 3d space where z is tswitch.

I tried this but couldn't get it to work

@tomleslie I would really prefer a 3d plot; a 3d plot can be coloured in various ways to illustrate the map tswitch-> solution curve.

Significantly I have a differential polynomial that is solved on the diaganal of the surface (t=tswitch) for the solution and its derivatives. If i can get code for the 3d plot i can also make 3d plots of the derivatives - and then see graphs of all the variables that are being fed into the differential polynomial.

@Annonymouse i needed to take absolute values

@Carl Love For some reason when i do that my bar chart ends up missing bars

the numbers

             [                          -14  
             [y(t) = 1.73194791841524 10   ,

                                             -11  
               y[0](t) = -1.98951966012828 10   ,

               y[1](t) = -0.00304885510956351,

                                            -8  
               y[2](t) = 3.07160927189243 10  ,

                                             -12  
               y[3](t) = -4.88302273709004 10   ,

                                            -14  
               y[4](t) = 9.20694116760555 10   ,

                                             -15]
               y[5](t) = -1.85145481241699 10   ]

end up looking like

could you look at my worksheet?
(just press enter on every line)
Visualising_numerous_derivatives_of_the_L2_model.mw

In the end I just used a program that i had previusly written to iteratively take the lie-derivatives of the variable i am interested in, and then wrote some ugly plotting commands to get the graphs i wanted.

Visualising_numerous_derivatives_of_the_L2_model.mw

This at least solves half of my problem; but is probably not useful for anyone who finds this page through google

@vv 

"I don't understand why not simply plot the solutions of the system."

i thought that was what i was doing - could you explain?

@Kitonum Thats often true, but not always true, and generally unhelpful. Consider the function:

f(x,y)=x^2 +y^2;

the level set for  f(x,y)=0 is a  point
whereas the other level sets are all closed curves circles- by contrast

f(x,y)=x*y;

has all of its level sets as curves or sets of curves; none of which  are closed.

These two cases illustrate thaty we can understand the shape of functions by understanding their level sets.

If someone were to ask about finding the level sets of a scalar function from R^2 it would therefore be both inaccurate and unhelpful to say that they were all 1 dimensional surfaces.

Perhaps a better way of visualising it would be as a set of triangles in 2d space; with the x coordinates of each triangle corner being the dependant variables, and the corresponding y co ordinates being the independant variables.

@Annonymouse is there a way of making the explore polts bigger?

@acer Thats really helpful