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Plot a stream function?

Nurul Ainina

Posted 8 years ago

Hi, I have problem to plot this equation \[Psi](y,z)=( k[y]' z^2 (k[y]-z))/(8k[y] ); where k[y]'=Sqrt[(1-k[y]/p)(1-k[y] p Cos[\[Alpha]]- k[y] Log[k[y]/p])] and I am expected to get this kind of graph I get k[y] from this f[k_?NumericQ]:=p*NIntegrate[1/Sqrt[(1-t)(1-t p^2 Cos[\[Alpha]])-3 M p t Log[t]],{t,k/p,1}] ky=FindRoot[y==f[k],{k,0}]; Table[{y,ky[[1,2]]},{y,0,a1+\[Delta],\[Delta]}]//N where p=1.57516;a1=1.7824; Can anyone help me with this problem? Thanks in advance!

Hi, I have problem to plot this equation

\[Psi](y,z)=( k[y]' z^2 (k[y]-z))/(8k[y] );
 where
k[y]'=Sqrt[(1-k[y]/p)(1-k[y] p Cos[\[Alpha]]- k[y] Log[k[y]/p])]

and I am expected to get this kind of graph

enter image description here

I get k[y] from this

f[k_?NumericQ]:=p*NIntegrate[1/Sqrt[(1-t)(1-t p^2 Cos[\[Alpha]])-3 M p t Log[t]],{t,k/p,1}]
ky=FindRoot[y==f[k],{k,0}];
Table[{y,ky[[1,2]]},{y,0,a1+\[Delta],\[Delta]}]//N

where
p=1.57516;a1=1.7824;

Can anyone help me with this problem? Thanks in advance!

POSTED BY: Nurul Ainina

3 Replies

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 8 years ago

You can use `Interpolation`: k = Interpolation[ Chop[{{0., 1.57516 + 0. I}, {0.05, 1.57289 + 2.4447310^-16 I}, {0.1, 1.56608 - 5.1521510^-17 I}, {0.15, 1.55475 + 3.2843110^-14 I}, {0.2, 1.53893 - 1.518210^-13 I}, {0.25, 1.51865 - 2.7144310^-15 I}, {0.3, 1.49396 + 2.7485410^-16 I}, {0.35, 1.46494}, {0.4, 1.43164}, {0.45, 1.39415}, {0.5, 1.35259}, {0.55, 1.30705}, {0.6, 1.25767}, {0.65, 1.20459}, {0.7, 1.14797}, {0.75, 1.088}, {0.8, 1.02488}, {0.85, 0.958821}, {0.9, 0.890087}, {0.95, 0.818952}, {1., 0.74573}, {1.05, 0.670772}, {1.1, 0.594476}, {1.15, 0.517297}, {1.2, 0.439757}, {1.25, 0.362468}, {1.3, 0.286157}, {1.35, 0.211719}, {1.4, 0.140297}, {1.45, 0.073469}, {1.5, 0.0138314}, {1.55, -0.0326749 + 1.6869810^-15 I}, {1.6, -0.0692389 - 3.0689610^-15 I}, {1.65, -0.136826 - 1.6695210^-15 I}, {1.7, -0.186826 - 2.8714110^-15 I}, {1.75, -0.236826 - 2.1663310^-15 I}, {1.8, -0.286826 - 3.0206410^-15 I}}]]; Plot[k[y], {y, 0, 1.8}]

You can use Interpolation:

k = Interpolation[
   Chop[{{0., 1.57516 + 0. I}, {0.05, 
      1.57289 + 2.44473*10^-16 I}, {0.1, 
      1.56608 - 5.15215*10^-17 I}, {0.15, 
      1.55475 + 3.28431*10^-14 I}, {0.2, 
      1.53893 - 1.5182*10^-13 I}, {0.25, 
      1.51865 - 2.71443*10^-15 I}, {0.3, 
      1.49396 + 2.74854*10^-16 I}, {0.35, 1.46494}, {0.4, 
      1.43164}, {0.45, 1.39415}, {0.5, 1.35259}, {0.55, 
      1.30705}, {0.6, 1.25767}, {0.65, 1.20459}, {0.7, 
      1.14797}, {0.75, 1.088}, {0.8, 1.02488}, {0.85, 0.958821}, {0.9,
       0.890087}, {0.95, 0.818952}, {1., 0.74573}, {1.05, 
      0.670772}, {1.1, 0.594476}, {1.15, 0.517297}, {1.2, 
      0.439757}, {1.25, 0.362468}, {1.3, 0.286157}, {1.35, 
      0.211719}, {1.4, 0.140297}, {1.45, 0.073469}, {1.5, 
      0.0138314}, {1.55, -0.0326749 + 
       1.68698*10^-15 I}, {1.6, -0.0692389 - 
       3.06896*10^-15 I}, {1.65, -0.136826 - 
       1.66952*10^-15 I}, {1.7, -0.186826 - 
       2.87141*10^-15 I}, {1.75, -0.236826 - 
       2.16633*10^-15 I}, {1.8, -0.286826 - 3.02064*10^-15 I}}]];
Plot[k[y], {y, 0, 1.8}]

POSTED BY: Gianluca Gorni

Marco Thiel

Marco Thiel, University of Aberdeen - Dept. of Physics/Mathematics

Posted 8 years ago

Dear Nurui, this is a bit difficult to answer like this. The code fragments are too buggy and it is hard to see what exactly you want to achieve. Here are some observations: You do not stick to Mathematica's syntax, I suppose. For example you "define" \[Psi](y,z)=( k[y]' z^2 (k[y]-z))/(8k[y] ); which I assume is supposed to be a function of y and z. The thing is that functions should have the arguments in square brackets. Also Phi does not appear to be used anywhere in the code that follows. 3. What do you want to achieve with this line: k[y]'=Sqrt[(1-k[y]/p)(1-k[y] p Cos[\[Alpha]]- k[y] Log[k[y]/p])] k[y] seems to be the derivative of a function. You do not define it with a delayed assignment (:=), and you never ask Mathematica to solve that ODE. I do not think that this way of defining Psi will/should work. You do not define all parameters. You use numerical integration but do not define. e.g. Alpha. I suppose that ky would need to break defined in a delayed way (:=), because it cannot be solved numerically before you give it the value of y, which is a line later. You code does not appear to be meant to run as you define parameters after everything. You should aim at posting code that at least has an order to it. Independent variables in functions should also be defined with an underscore. I think that all in all, you will have to work your way through a book with syntax of the Wolfram Language. Here, construct the function psi "by foot". First I need to solve for k[y]. sol = NDSolveValue[{D[k[y], y] == Sqrt[(1 - k[y]/p) (1 - k[y] p Cos[\[Alpha]] - k[y] Log[k[y]/p])],k[0] == 0.1} /. {\[Alpha] -> 0/10., p -> 1.57516, a1 -> 1.7824, \[Alpha] -> 2 Pi}, k, {y, 0, 10}] Now, I can define Psi. \[Psi][y_, z_] := Re[(Sqrt[(1 - sol[y]/p) (1 - sol[y] p Cos[\[Alpha]] - sol[y] Log[sol[y]/p])] z^2 (sol[y] - z))/(8 sol[y]) /. {p -> 1.57516, a1 -> 1.7824, \[Alpha] -> 0.0}]; You can plot that function like so: ContourPlot[\[Psi][y, z] /. {p -> 1.57516, a1 -> 1.7824, \[Alpha] -> 0.0}, {y, 0, 1.8}, {z, 0, 15}, PlotPoints -> 50] I don't think that this is what you want. Also, I don't have the parameters that I need to generate your image. I hope this gives you some starting point. Cheers, Marco

POSTED BY: Marco Thiel

Nurul Ainina

Posted 8 years ago

POSTED BY: Nurul Ainina

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