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Is there any way I could use the MaxStepFraction (or grid size) as used in NDSolve in the example below as ticks on the 3d Plot?

That was I would be able to plot the grid points on the X and Y axis.

I tried creating a table and using that as the argument in Ticks but that didn't work.

Minimum working example:

Clear[u, L, t, x, y, sol, Eq]
L = 4;
Eq = -D[u[t, x, y], t, t] + D[u[t, x, y], x, x] + 
   D[u[t, x, y], y, y] + Sin[u[t, x, y]];
uSol = u /. NDSolve[{
     Eq == 0, u[t, -L, y] == u[t, L, y], 
     u[t, x, -L] == u[t, x, L], 
     u[0, x, y] == Exp[-(x^2 + y^2)], 
     Derivative[1, 0, 0][u][0, x, y] == 0
     }, 
    u,
     {t, 0, L/2}, {x, -L, L}, {y, -L, L},
    MaxStepFraction -> 1/11
    ][[1]]




tt = 1.2;
Plot3D[ uSol[tt, x, y], {x, 0, L}, {y, 0, L}, 
 Ticks -> {{0, 2/5, 4/5, 6/5, 8/5, 2, 12/5, 14/5, 16/5, 18/5, 4}, {0, 
    2/5, 4/5, 6/5, 8/5, 2, 12/5, 14/5, 16/5, 18/5, 4}}]

enter image description here

The ticks in the above figure were created using, Table[i 4/10, {i, 10}] and then manually pasted into the curly brackets. Obviously, that isn't the most efficient way.

I tried pasting the Table command inside Plot3D but that didn't work.

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9
  • $\begingroup$ Try Ticks -> Evaluate[{Table[(* stuff *)], (* other ticks *)}]; Plot3D[] is HoldAll, see... $\endgroup$ Sep 25, 2012 at 13:45
  • $\begingroup$ @J.M. Evaluate.... hmm... I never thought of that. This isn't the first time I've come across this Evaluate... $\endgroup$
    – dearN
    Sep 25, 2012 at 13:49
  • $\begingroup$ @J.M. Are my questions really poor? I thought I'd ask. $\endgroup$
    – dearN
    Sep 25, 2012 at 13:49
  • $\begingroup$ Well, they aren't poor per se... I suppose the HoldAll attribute of plotting functions isn't really obvious at first glance. $\endgroup$ Sep 25, 2012 at 14:00
  • 3
    $\begingroup$ @drN Can you point out, why Table didn't work? For me something like Plot3D[uSol[tt, x, y], {x, 0, L}, {y,0, L}, Ticks -> Table[x, {3}, {x, 0, 4, 1/3}]] does work, even with the HoldAll attribute of Plot3D. $\endgroup$
    – halirutan
    Sep 25, 2012 at 19:14

1 Answer 1

2
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tt = 1.2;
Plot3D[uSol[tt, x, y], {x, 0, L}, {y, 0, L}, 
      Ticks -> {Table[i 4/10, {i, 10}], Table[i 4/10, {i, 10}]}
]

or the slightly more concise version of mr.wizard:

Plot3D[uSol[tt, x, y], {x, 0, L}, {y, 0, L}, 
  Ticks -> ({#, #} & @ Table[i 4/10, {i, 10}])
]

yield:

Mathematica graphics

or, since you wanted to use your setting for MaxStepFraction (1/11), you could use FindDivisions with that value:

tt = 1.2;
Plot3D[uSol[tt, x, y], {x, 0, L}, {y, 0, L},
      Ticks -> {FindDivisions[{0, L, 1/11}, 10],FindDivisions[{0, L, 1/11}, 10]}
]

Mathematica graphics

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7
  • $\begingroup$ Looks like I lost. If you add the concise {#, #} form I'll delete my answer. $\endgroup$
    – Mr.Wizard
    Sep 26, 2012 at 9:18
  • $\begingroup$ @Mr.Wizard ;-P Looks like it. Perhaps time to go to bed? $\endgroup$ Sep 26, 2012 at 9:20
  • 1
    $\begingroup$ Why do you need such a big Function construct? Was Table[i 4/10, {2}, {i, 10}] too simple? ;-) $\endgroup$
    – halirutan
    Sep 26, 2012 at 10:03
  • $\begingroup$ This is definitely a good answer like the other one :P. However, haliurtan's answer/comment is simplicity itself. I don't quite understand why it didn't work previously. $\endgroup$
    – dearN
    Sep 26, 2012 at 13:59
  • $\begingroup$ @drN No! A single Table doesn't give the ticks as you showed in the figure in your question. Besides, you'd need to place {} around it, otherwise you'd get errors. (that might even have been the error that plagued you) $\endgroup$ Sep 26, 2012 at 14:03

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