CODEWITHSUNDEEP
wolfram-mathematica
saimon
saimon

Reputation: 57

Mathematica 3d plot of a function of a function

I would like to make a 3d plot of a function that takes as a variable a function of another variable.

The whole thing is complicated by the fact that my functions are nested piecewise functions

here is my code:

phi0=Function[u,1.21*10^-6/((u/10^44.25)^1.01 + (u/10^44.25)^2.38)][Lx]
zc=Function[v,Piecewise[{{2.49,v>=10^45.74},{2.49*(v/10^45.74)^0.2,v<10^45.74}}]][Lx]

e=Function[uu,Piecewise[{{(1+uu)^4.62,uu<=zc},{(1+zc)^4.62*((1+uu)/(1+zc))^-1.15,uu>zc}}]][z]

Plot3D[e[z,Lx],{z,0,7},{Lx,10^42,10^47}, PlotRange->Full]

but it is not plotting anything, and I am not sure of what to do

EDIT:

thanks, for the hint, I think I solved it this way. It is not giving me any error, but it is taking a lot of time to evaluate the result even in one single point... do you think it is normal?

phizero[Lx_] := 1.21/10^6/((Lx/10^44.25)^1.01 + (Lx/10^44.25)^2.38)

zc[Lx_] := 
 Piecewise[{{2.49, Lx >= 10^45.74}, {2.49*(Lx/10^45.74)^0.2, 
    Lx < 10^45.74}}]

e[z_, Lx_] := 
 Piecewise[{{(1 + z)^4.62, 
    z <= zc[Lx]}, {(1 + zc[Lx])^4.62*((1 + z)/(1 + zc[Lx]))^-1.15, 
    z > zc[Lx]}}]

phi[z_, Lx_] := phizero[Lx]*e[z, Lx]

(*D[phi[z,Lx],Lx]:=Lx*phi[z,Lx]*)

p[z_, Lx_] = Integrate[Lx*phi[z, Lx], Lx]
p[4, 10^44]

Upvotes: 0

Views: 590

Answers (2)

agentp
agentp

Reputation: 6999

It is taking a long time because it works hard to symbolically integrate the function. It takes a while just to learn it cant be done. -- use NIntegrate for numerical integration. Here is a stab at it.. notice as a good habit i leave out the floating point values from the expressions until really needed:

 constants = {c1 -> 10^45.74, c2 -> 10^44.25, c3 -> 2.49, c4 -> 4.62, 
   c5 -> 2.38, c6 -> 0.2, c7 -> 1.21/10^6, c8 -> -1.15, c9 -> 1.01}
phizero[Lx_] := c7/((Lx/c2)^c9 + (Lx/c2)^c5)
zc[Lx_] := Piecewise[{{c3, Lx >= c1}, {c3 (Lx/c1)^c6, Lx < c1}}]
e[z_, Lx_] := 
      Piecewise[{{(1 + z)^c4, 
           z <= zc[Lx]}, {(1 + zc[Lx])^c4 ((1 + z)/(1 + zc[Lx]))^c8, 
           z > zc[Lx]}}]
phi[z_, Lx_] := phizero[Lx] e[z, Lx]
p[z_, L1_, L2_] := NIntegrate[Lx phi[z, Lx] /. constants, {Lx, L1, L2}]
p[4, 10^42, 10^47]

quickly returns:

   (* ~7 10^84 *)

Upvotes: 0

Bill
Bill

Reputation: 3977

First, Function works like this:

In[1]:= q = Function[x, x^2];
In[2]:= q[4]
Out[2]= 16

so lose the [var] that you have at the end of each of your Function definitions. You could also do

q[x_]:= x^2

and skip the use of Function[] if that would be simpler.

Next, you define the function e to accept a single argument, but then you give it two arguments when you use it inside your Plot3D. So you need to figure out what your definition of the function e should be and I can't even guess how to do that.

Upvotes: 1

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