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I am wondering why the following does not plot the way it's written and exactly how does the Do[] function operate? I think I could be using it incorrectly. 

In[1]:= u[x_, t_] := (1/2) (Exp[-(x + t)^2] + Exp[-(x - t)^2]) Do[ Plot[u[x, k], {x, -5, 5}, AxesLabel -> {x, u}, PlotRange -> {0, 1}], {k, 0, 3} ]

EDIT

Clear["Global`*"] a = 1; b = 1; ϕ[ξ_, ν_, t_] := Sum[(((5*(-1 + (-1)^j)*(-1 + (-1)^i))/(j*i*Pi^2))* Cos[(1/Pi^4)*Sqrt[i^4*Pi^4 + j^4*Pi^4]*t])*Sin[i*Pi*ξ]*Sin[j*Pi*ν], {i, 1, 10}, {j, 1, 10}]; Do[Print[Plot3D[ϕ[ξ, ν, m], {ξ, 0, a}, {ν, 0, b}, Mesh :> Automatic, PlotLabel :> StringJoin["Surface Plot of Solution at t = ", ToString[m]], ColorFunction :> {"BlueGreenYellow"}]], {m, 0, 10, 2}]
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3 Answers 3

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The reason is that since version 6 of Mathematica, the Plot returns its output just like any other function. It does not print is as a side effect. Do is meant to be used with operations that have side effects and simply discards the output, so you never see the plot.

You can either use Table, which collects the values in a list:

Mathematica graphics

Note how you get a list of graphics objects. Graphics are expressions like any other, they're simply shown using a special formatting. You can always see their internal structure by using InputForm on them or pressing Ctrl-Shift-I on the containing cell.

Or you can use Print explicitly:

Do[Print@Plot[u[x, k], {x, -5, 5}, AxesLabel -> {x, u}, PlotRange -> {0, 1}], {k, 0, 3}]

Note: If as @JM noted, you are trying to animate the plot, use Animate:

Animate[Plot[u[x, k], {x, -5, 5}, AxesLabel -> {x, u}, PlotRange -> {0, 1}], {k, 0, 3, 1}]
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  • $\begingroup$ I have a strong feeling OP's code was taken from an old Mathematica book, and the intention was to animate the plot; you might thus want to mention Animate[]. $\endgroup$ Commented Jun 22, 2012 at 8:27
  • $\begingroup$ @J.M.: No this was not the intention, I was trying to find a more convenient way of making multiple plots when necessary by the the computer do the work for me in a more programmatic fashion rather than copy/paste several Plot commands or use Manipulate over a certain range (parameter). But that does sounds interesting now that you bring that idea up. Thanks :) $\endgroup$ Commented Jun 22, 2012 at 9:56
  • $\begingroup$ @nightowl You can use Row/Column/Grid to arrange the plots in a nicer layout $\endgroup$ Commented Jun 22, 2012 at 9:57
  • $\begingroup$ @night owl: I see; as I said, it's exactly how animations were done in old Mathematica, so that's what I thought you were doing. Then indeed, Table[] is what you want (though that too was used for producing animations as well for subsequent exporting). $\endgroup$ Commented Jun 22, 2012 at 10:07
  • $\begingroup$ @Szabolcs: Thanks! I will look into how to arrange them nicely using the suggestions you've given. $\endgroup$ Commented Jun 23, 2012 at 2:20
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Another way is to use DisplayFunction, something before the time of graphic input/output of version 6. The following line prints the same way as Szabolcs's 2nd example.

Do[Plot[u[x, k], {x, -5, 5}, AxesLabel -> {x, u}, PlotRange -> {0, 1}, DisplayFunction -> Print], {k, 0, 3}]

DisplayFunction -> f wraps the actual output of any Graphics function (Plot here) into f. This way one can directly send the plot to any pipe, for example to a stream or file. If you are version 6+ though, the best solution is to use Table, just as Szabolcs has said.

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  • $\begingroup$ Thanks for that alternative method. Could you read the last comment under Szabolcs post. I want to try and use GraphicsGrid with the Do command. Also, another concern came up. What is the most efficient way of using the iterator for the Do command. Because at most it take three arguments, e.g. Do[expr,{k,min,max,step-size}]. But what if you have a function such as $\phi[\xi_,\nu_,t]$, and you want to plot for value of time say: $t=0,0.3,0.7,0.10,0.14,0.16$. As you can notice, the time instances (steps) are not by the same amount, so how could you account for this when using $\endgroup$ Commented Jun 23, 2012 at 8:10
  • $\begingroup$ the Do command. Or is there something you can add.. Or would you just have to input the times in manually for a case like this? $\endgroup$ Commented Jun 23, 2012 at 8:11
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    $\begingroup$ @nightowl the syntax you are looking for is Do[f[x], {x, {0, 0.3, 0.7, 0.10, 0.14, 0.16}}] -- this also works for Table. Also realize you could use Scan or Map, e.g. Scan[f, {0, 0.3, 0.7, 0.10, 0.14, 0.16}] $\endgroup$ Commented Jun 23, 2012 at 8:29
  • $\begingroup$ @Mr.Wizard: Oh my! Thank You. This helps alot! :). So could you ("usually") input list in place of an iterator if you have several values to input with staggering increments of time? Now only to get GraphicsGrid to work in junction with it and that will complete my handy toolbox for the moment. :) Do each graphic have a internal name associated with them or could one be assigned to them. Because I know that GraphicsGrid[{{g_11,g_12,...},...}] needs to have variables (functions) hence -> ($g_{11},g_{12}$) assigned for input to create the 2-D grid. $\endgroup$ Commented Jun 23, 2012 at 8:57
  • $\begingroup$ @nightowl I'm not sure what you're trying to do but it's quite possible you don't realize you can pass a table of graphics objects directly to GraphicsGrid: GraphicsGrid[ Table[Plot[k Sin[x] + x, {x, 0, 3 Pi}], {k, 4}] ~Partition~ 2 ] (Here I generate four plots with Table, split it into a 2x2 table with Partition, and pass the result to GraphicsGrid.) $\endgroup$ Commented Jun 23, 2012 at 9:38
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Since this is getting a bit lengthy for comments:

What is the most efficient way of using the iterator for the Do command. Because at most it take three arguments, e.g. Do[expr,{k,min,max,step-size}]. But what if you have a function such as ϕ[ξ,ν,t], and you want to plot for value of time say: t=0,0.3,0.7,0.10,0.14,0.16. As you can notice, the time instances (steps) are not by the same amount, so how could you account for this

The syntax you are looking for is:

Do[f[x], {x, {0, 0.3, 0.7, 0.10, 0.14, 0.16}}]

This also works for Table. In simple cases such as this example you could use Scan or Map:

Scan[f, {0, 0.3, 0.7, 0.10, 0.14, 0.16}]

Do each graphic have a internal name associated with them or could one be assigned to them. Because I know that GraphicsGrid[{{g_11,g_12,...},...}] needs to have variables (functions) hence -> (g11,g12) assigned for input to create the 2-D grid.

No, the elements of the table given to GraphicsGrid do not need to be assigned to symbols first. Here is an example using the function given in your question:

u[x_, t_] := (1/2) (Exp[-(x + t)^2] + Exp[-(x - t)^2]) plots = Table[ Plot[u[x, k], {x, -5, 5}, AxesLabel -> {x, u}, PlotRange -> {0, 1}], {k, 0, 3} ]; Partition[plots, 2] // GraphicsGrid

Mathematica graphics

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  • $\begingroup$ Okay, Thanks. I don't see what Scan does. It must be something internally but does not return anything to the screen. And Map just seems to do something weird like {Cos[x][0],Cos[x][0.3],...} with f being Cos[x], from your example shown. And I'm glad to see that it works for Table, but my question was how to get it to work with the example I provided in the question underneath EDIT using the Do command. $\endgroup$ Commented Jun 23, 2012 at 21:45
  • $\begingroup$ @nightowl Scan is to Map as Do is to Table. Try: Scan[Print, {1, 2, 3, 4, 5}] $\endgroup$ Commented Jun 23, 2012 at 23:33

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