Dr. David Harrington

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18 years, 97 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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I am a professor of chemistry at the University of Victoria, BC, Canada, where my research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are replies submitted by dharr

Perhaps if you upload a worksheet with the expression in to save someone typing it in you will get a response. Use the green up-arrow in the editor.

@ContrapuntoBrowniano Not clear if you want to use different arguments or the original parameters. Here are both ways.


F := proc (x, y, z) options operator, arrow; (r[x]+r[y])*r[z] end proc;

proc (x, y, z) options operator, arrow; (r[x]+r[y])*r[z] end proc

If you want the answer to depend on the arguments, then they need to be provided. Type function is a name with some arguments, such as f(x)

     if not type(f,function) then error "arguments need to be provided with %1",f end if;
   end proc:

G(F(a, b, c), 3)


G(F, 3)

Error, (in G) arguments need to be provided with F

If you want always to use the parameters that F was defined with

H:=proc(f::procedure,n) local params;
   end proc:

H(F, 3);




Download function.mw

Edit: For Maple 18 use add(i, i in op(f)) and add(i, i in params)

@yasi From your email: "How to show axis z representation as power numbers?(1.1E-7 to 1.1E-3) And the shape should be only lines and not colored". 

For the color, use color=white (and perhaps make the lines thicker with thickness=2). You need color to do hidden lines - if you use style=wireframe, all the lines show through. It looks a little glaring with color=white, perhaps because you chose lightmodel=none. You could perhaps change the lightmodel or choose an off-white color.

In my opinion, axes labels should never be written in this way, instead the plot should be scaled (as I explained earlier) so the numbers are human-friendly, say in the range 0..1, and the scaling should be explained in the axis label, e.g. 1000*error.

But you can individually choose a label for each tickmark, formatted with sprintf as requred, as shown here.

zticklabels := map(x->x = sprintf("%.0e", x), [seq(0.1e-3*i, i = 2 .. 10, 2)])

Default lightmodel.

(not all ticks rendered correctly here)


Or with lightmodel=none:



@nm The y range is the default one, which is what DEFAULT means. If another range has been chosen, then it will be there. 

p := plot(sin(x), x = -1 .. 10):

v := op(select(type, p, specfunc(VIEW)));

VIEW(-1. .. 10., DEFAULT, _ATTRIBUTE("source" = "mathdefault"))

p := plot(sin(x), x = -1 .. 10, view = [-2 .. 9, DEFAULT]); -1; v := op(select(type, p, specfunc(VIEW)))

VIEW(-2 .. 9, DEFAULT)

p := plot(sin(x), x = -1 .. 10, view = [-2 .. 9, 0 .. 1]); -1; v := op(select(type, p, specfunc(VIEW)))

VIEW(-2 .. 9, 0 .. 1)

p := plot(sin(x), x = -1 .. 10, 0 .. 1); -1; v := op(select(type, p, specfunc(VIEW)))

VIEW(-1. .. 10., 0. .. 1.)


Download view.mw

@mehdi jafari Your revised PDE is much simpler to solve, but it still needs one more boundary/initial condition as @tomleslie pointed out. As he also pointed out, if an analytical solution could be found, it would have one arbitrary constant. That is the case here, where your solution still contains _C1.

@dharr So I added in the surfdata, which I think is what you want. I really don't understand why the bottom of the z range wouldn't be zero. The managing of the ticks and ranges could be improved, since some are done locally and some passed to the procedure. surfdata needs to know what range the data is over, which has to be 0.1..0.9, even if the plot goes from zero to 1. You could read the 0.1..0.9 in from the Excel file. Note the three plot ranges can be given in the view option.



grids := proc (M, x_min, x_max, y_min, y_max) local z_min, z_max, plot_f, xz, yz, xy; z_min := 0.11e-6; z_max := 0.1e-2; plot_f := plots:-surfdata(M, .1 .. .9, .1 .. .9); xz := plot3d([x, y_min, z], x = x_min .. x_max, z = z_min .. z_max, style = line, color = blue, thickness = 0, grid = [6, 6]); yz := plot3d([x_min, y, z], y = y_min .. y_max, z = z_min .. z_max, style = line, color = blue, thickness = 0, grid = [6, 6]); xy := plot3d([x, y, z_min], x = x_min .. x_max, y = y_min .. y_max, style = line, color = blue, thickness = 0, grid = [6, 6]); plots:-display(plot_f, xz, yz, xy, lightmodel = none, tickmarks = [6, 6, 6], labels = [x, y, "Error"], labeldirections = [horizontal, horizontal, vertical], axes = frame, orientation = [45, 70, 0], view = [0 .. 1, 0 .. 1, 0.11e-6 .. 0.1e-2]) end proc:

dat := ExcelTools:-Import(cat(currentdir(), "/PlotTest.xlsx"), 1, "B2:J10"):

grids(dat, 0, 1, 0, 1);




Download Yasi2.mw

@yasi You can use display to combine the surfdata with your plot3d grids. If you want to scale the data by 1000, say, you can use 1e3*~dat.

If you aren't happy with any of the answers here, you need to explain exactly what you want - what is the image showing?

You should upload your worksheet using the green up-arrow in the editor.

@Rouben Rostamian  I missed your comment as I was writing my method. Perhaps my answer then is GIGO. It would be nice to have an analytical solution. I think in general an incompatiblilty in the initial condition can just indicate a step change (intended) in conditions at t=0, such as putting a uniform temperature object in contact with a freezer at t=0, forcing the boundary temperature immediately to a lower value. But I didn't think about the lack of a t derivative in this context.

@ijuptilk The theta(z,0) = thetab*sin(pi/2)*z IC is not compatible with the theta(d,t)=0 BC. Is this intended? And sin(pi/2)=1 so is this right?

@C_R Don't know much about Maplesim, but for electronics schematics (circuit diagrams), they moved away from arcs for non-crossing. They then went to a dot on the intersection if there was a connection, but the newer way is that connections are always at a T.

@nm Since v and theta are coupled, the time dependence of theta induces a time dependence of v. pdsolve (without any boundary/initial conditions) gives a general analytical solution where both theta and v are functions of z and t. Whether or not this solution can be made to fit the ic/bcs is not clear to me.

@mehdibgh Perhaps as a way to get initial guesses for fsolve, if the nonlinear problem is close to this...

@Christian Wolinski kk := Matrix(1, 8, {(1, 1) = .500000000000000+0.*I, (1, 2) = 0.1825477032e-15-.5879378012*I, (1, 3) = -.500000000000000+.555570233019603*I, (1, 4) = -0.2691433088e-15-.5879378012*I, (1, 5) = .500000000000000+0.*I, (1, 6) = -0.2691433088e-15+.587937801209679*I, (1, 7) = -.500000000000000-.5555702330*I, (1, 8) = 0.1825477032e-15+.587937801209679*I})

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