Suppose this code in Mathematica:
w[w1_] := NDSolve[{y''[x] + y[x] == 2, y'[0] == 0, y[0] == w1},y[x], {x, 0, 30}];
Manipulate[Plot[Evaluate[{y[x], y'[x]} /. w[w1]], {x, 0, 30}], {w1, -2, 3}]
The code works this way: A slidebar (in Manipulate) controls one initial condition (value of y[0]) of differential equation saved in variable w and when it is changed, the solution is plotted. This works fine, but I have a problem with plotting the first derivative of the solution (y'[x]). Instead of a function graph there is just simply blank space and nothing is plotted, even I checked it in Mathematica help, where the example code
s = NDSolve[{y''[x] + Sin[y[x]] y[x] == 0, y[0] == 1, y'[0] == 0}, y, {x, 0, 30}]
Plot[Evaluate[{y[x], y'[x], y''[x]} /. s], {x, 0, 30},PlotStyle -> Automatic]
works. Does anybody know where's the problem?
I found the problem, it is just in NDSolve it is needed to write
w[w1_] := NDSolve[{y''[x] + y[x] == 2, y'[0] == 0, y[0] == w1},y, {x, 0, 30}]
y instead of y[x] and everything works.
Related
I'm solving and plotting the equations of motion for the double pendulum using Mathematica's NDSolve.
I've successfully plotted the Angular position using a standard plot. But when I come to use the parametric plot for the position of each mass. I get no errors but simply no plot.
eqn1 = 2 th''[t] + Sin[th[t] - ph[t]] (ph'[t])^2 + Cos[th[t] - ph[t]] (ph''[t]) + (2 g/l) Sin[th[t]]
eqn2 = ph''[t] + Sin[th[t] - ph[t]] (th'[t])^2 + Cos[th[t] - ph[t]] (th''[t]) + (g/l) Sin[th[t]]
eqnA = eqn1 /. {g -> 10, l -> 1}
eqnB = eqn2 /. {g -> 10, l -> 1}
sol = NDSolve[{eqnA == 0, eqnB == 0, th[0] == 0.859, th'[0] == 0, ph[0] == 0.437, ph'[0] == 0}, {th, ph}, {t, 0, 10}]
Plot[{th[t], ph[t]} /. sol, {t, 0, 10}]
r1 = {lSin[th[t]] + lSin[ph[t]], -lCos[th[t]] - lCos[ph[t]]} /. {l -> 1, g -> 10}
ParametricPlot[r1 /. sol, {t, 0, 10}]
Replace
r1 = {lSin[th[t]] + lSin[ph[t]], -lCos[th[t]] - lCos[ph[t]]} /. {l->1, g->10}
with
r1 = {l*Sin[th[t]] + l*Sin[ph[t]], -l*Cos[th[t]] - l*Cos[ph[t]]} /. {l->1, g->10}
and your ParametricPlot should appear.
One useful trick you might remember, when any plot doesn't appear you can try replacing the plot with Table and see what it shows. Often the table of data provides the needed hint about why the plot isn't appearing.
I want to prove the expression Limit[Sin[x*x] *Exp[-x*x]*x, x -> Infinity] ==0
So I do this Normal[ Series[ Sin[x*x] *Exp[-x*x]*x, {x, 0, 40}]].And the result
imply the expression will be infinity.
That is odd, and I make some change. Let t=x^2, then the expression will be Limit[Sin[t] *Exp[-t]*Sqrt[t], t -> Infinity]. I try again Normal[ Series[Sin[x] *Exp[-x]*Sqrt[x], {x, 0, 40}]]. That's the answer I need.
The right series
I can't figure out what's wrong. Why I can't generate series directly?
I wnat to say thay f, f1 and f2 should be the same, but they looks like this:
f=f1,but f2 diffuse
Normal[Series[Sin[x*x]*Exp[-x*x]*x, {x, 0, 40}]]
with t = x^2 is equivalent to
Normal[Series[Sin[t]*Exp[-t]*x, {x, 0, 40}]]
because x != Sqrt[x^2] e.g. for `x = -2
Results for negative t are not plotted.
Plot[Sin[t]*Exp[-t]*Sqrt[t], {t, -8, 8}]
I have a plot
Plot[40500*x^(-0.1), {x, 1, 100}, PlotRange -> {0, 50000}]
I'm trying to plot the cumulative of these y values. I'll try to explain with an example:
I'm trying to get
for x=1: 40500*1^(-0.1)
for x=2: 40500*(2^(-0.1)+1^(-0.1))
for x=3: 40500*(3^(-0.1)+2^(-0.1)+1^(-0.1))
and so on up to x=100.
Is there a way to do that?
Running some examples for x = 3
for x=3: 40500*(3^(-0.1)+2^(-0.1)+1^(-0.1))
114574.
This can be found using Sum:
Sum[40500*i^(-0.1), {i, 3}]
or using Fold
Fold[#1 + 40500*#2^(-0.1) &, 0, {1, 2, 3}]
114574.
FoldList outputs the intermediate steps.
FoldList[#1 + 40500*#2^(-0.1) &, 0, {1, 2, 3}]
{0, 40500., 78287.8, 114574.}
Accumulating to 100 and discarding the initial zero value:
ListLinePlot[Rest[FoldList[#1 + 40500*#2^(-0.1) &, 0, Range[100]]]]
I am working on a solution to solve a Partial Differential Equation, Fick's Second Law of Diffusion to be exact.
I was able to produce a 3D Plot using the NDSolve and Plot3D functions.
Code used:
NDSolve[{D[c[t, h], t] == 1*D[c[t, h], h, h],
c[0, h] == Erfc[h/(2*81.2)],
c[t, 0] == 1,
c[t, 4000] == 3.08*^-18}, c, {t, 0, 900}, {h, 0, 274}]
Instead of a graphical representation, I would like to find numerical points of the graph at t = 900.
I would like to know how to put in t = 900 into NDSolve (or other functions) so as to generate detailed numerical points of the solution.
Try saving the solution in a variable first:
e = NDSolve[{D[c[t, h], t] == 1*D[c[t, h], h, h], c[0, h] == Erfc[h/(2*81.2)], c[t, 0] == 1, c[t, 4000] == 3.08*^-18}, c, {t, 0, 900}, {h, 0, 274}]
Then we can Evaluate this expression for our desired variables:
Evaluate[c[900, 10] /. e]
(*{0.914014}*)
Or to make it more versatile, we can use Manipulate:
Manipulate[Evaluate[c[t, h] /. e], {t, 0, 900}, {h, 0, 274}]
Update:
Considering the information I received from the comments below; we can define a function like q[t,h] which will give us the solution as a function:
q[t_, h_] := Evaluate[c[t, h] /. e]
q[900, 10]
(*{0.914014}*)
I wrote a program that defines two piecewise functions "gradino[x_]" and "gradino1[x_]", where x is a vector of m components.
I'm not able to write these functions explicitly using the x_i, I need to keep x as a vector.
I need to measure the distance between these two function doing:
Integrate[Abs[gradino[x]-gradino1[x]],{x[[1]],0,100},{x[[2],0,100},{x[[3]],0,100}...{x[[m]],0,100}]
but it's not working.
Any idea how to do this? Remembering that I can't simply express gradino[x1_,x2_ etc...].
re: "its not working" posting the actual error message is usually a good idea,
in this case "Part specification x[[1]] is longer than depth of object.".. tells you exactly what the problem is. If x is not already defined as a list you cannot use list elements as integration variables.
f[y_] := y[[1]] y[[2]];
Integrate[ f[x] , {x[[1]], 0, 1}, {x[[2]], 0, 1}]
(* error Part specification x[[1]] is longer than depth of object. *)
If you first define x as a list, then it works:
x = Array[z, 2];
Integrate[ f[x] , {x[[1]], 0, 1}, {x[[2]], 0, 1}]
(*1/4*)
Note you can not do this with nintegrate:
NIntegrate[ f[x] , {x[[1]], 0, 1}, {x[[2]], 0, 1}]
(*error Tag Part in x[[1]] is Protected *)
you need to use the explicit elements:
NIntegrate[ f[x] , {z[1], 0, 1}, {z[2], 0, 1}]
(* 0.25 *)
According to the model above, with
x = Array[z, 2];
why the following is ok:
f[y_] := NIntegrate[y[[1]] y[[2]] t, {t, 0, 1}];
NIntegrate[f[x], {z[1], 0, 1}, {z[2], 0, 1}]
but the following is not:
f[y_] := NIntegrate[y[[1]] y[[2]] Exp[t], {t, 0, 1}];
NIntegrate[f[x], {z[1], 0, 1}, {z[2], 0, 1}]
The only difference is changing t in the inner integration into Exp[t].