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I need to solve a diferential equation of the form w'=g(t,w(t)) where g is defined as follows
g[t_, w_] := {f1[t, {w[[3]], w[[4]]}], f2[t, {w[[3]], w[[4]]}], w[[1]],w[[2]]};
and f1, f2 are
f1[t_, y_] := Sum[\[Mu][[i]] (s[[i]] - y)/Norm[s[[i]] - y]^2, {i, 1, 5}][[1]];
f2[t_, y_] := Sum[\[Mu][[i]] (s[[i]] - y)/Norm[s[[i]] - y]^2, {i, 1, 5}][[2]];
Everything else is defined properly and is not the cause of the error.
Yet when I use
sout = NDSolve[{y'[tvar] == g[tvar, y[tvar]],
y[0] == {Cos[Pi/6], Sin[Pi/6], 0, 0}}, y, {tvar, 0, 2}, Method -> "ExplicitRungeKutta"];
I get the error
Part::partw: Part 3 of y[tvar] does not exist.
Part::partw: Part 4 of y[tvar] does not exist.
I have looked in other questions and none of them solved this problem.
You want to find a function in $R^4$ satisfying a differential equation.
I don't think DSolve and NDSolve have a standard way of manipulating vectorial differential equations except by representing each component with an explicit name or an index for the dimensions.
Here is a working example, that can be executed without Method specification in dimension 4 with notations similar to your problem:
sout={w1[t],w2[t],w3[t],w4[t]} /. NDSolve[{
w1'[t]== t*w2[t],
w2'[t]== 2*t*w1[t],
w3'[t]==-2*w2[t]+w1[t],
w4'[t]== t*w3[t]-w1[t]+w2[t],
w1[0]==0,
w2[0]==1,
w3[0]==1,
w4[0]==0
},{w1[t],w2[t],w3[t],w4[t]},{t,0,2}]
ParametricPlot[{{sout[[1, 1]], sout[[1, 3]]}, {sout[[1, 2]], sout[[1, 4]]}}, {t, 0, 2}]
I think you will be able to adapt this working example to your needs.
I didn't use your original problem as I wanted to focus on the specification for Mathematica, not on the mathematics of your equation. There are constants involved such as Mu and s that you do not give.
I aim to calculate and preserve the results from the maximization of a function with two arguments and one exogenous parameter, when the maximum can not be derived (in closed form) by maximize. For instance, let
f[x_,y_,a_]=Max[0,Min[a-y,1-x-y]
be the objective function where a is positive. The maximization shall take place over [0,1]^2, therefore I set
m[a_]=Maximize[{f[x, y, a], 0 <= x <= 1 && 0 <= y <= 1 && 0 <= a}, {x,y}]
Obviously m can be evaluated at any point a and it is therefore possible to plot the maximizing x by employing
Plot[x /. m[a][[2]], {a, 0.01, 1}]
As I need to do several plots and further derivations containing the optimal solutions x and y (which of course are functions of a), i would like to preserve/save the results from the optimization for further use. Is there an elegant way to do this, or do I have to write some kind of loop to extract the values myself?
Now that I've seen the full text of your comment on my original comment, I suspect that you do understand the differences between Set and SetDelayed well enough. I think what you may be looking for is memoisation, sometimes implemented a bit like this;
f[x_,y_] := f[x,y] = Max[0,Min[a-y,1-x-y]]
When you evaluate, for example f[3,4] for the first time it will evaluate to the entire expression to the right of the :=. The rhs is the assignment f[3,4] = Max[0,Min[a-y,1-x-y]]. Next time you evaluate f[3,4] Mathematica already has a value for it so doesn't need to recompute it, it just recalls it. In this example the stored value would be Max[0,Min[a-4,-6]] of course.
I remain a little uncertain of what you are trying to do so this answer may not be any use to you at all.
Simple approach
results = Table[{x, y, a} /. m[a][[2]], {a, 0.01, 1, .01}]
ListPlot[{#[[3]], #[[1]]} & /# results, Joined -> True]
(The Set = is ok here so long as 'a' is not previosly defined )
If you want to utilise Plot[]s automatic evaluation take a look at Reap[]/Sow[]
{p, data} = Reap[Plot[x /. Sow[m[a]][[2]], {a, 0.01, 1}]];
Show[p]
(this takes a few minutes as the function output is a mess..).
hmm try this again: assuming you want x,y,a and the minimum value:
{p, data} = Reap[Plot[x /. Sow[{a, m[a]}][[2, 2]], {a, 0.01, .1}]];
Show[p]
results = {#[[1]], x /. #[[2, 2]], y /. #[[2, 2]], #[[2, 1]]} & /# data[[1]]
BTW Your function appears to be independent of x over some ranges which is why the plot is a mess..
I have a question about NDSolve function in Mathematica.
I have an oscillator defined by these two equations:
x' = v
v' = -x - u*v^3
where u is some constant.
How to create an NDSolve that resolves this? I tried following code (it has to depend on time) but it doesnt work:
eq1 = x'[t] == v;
eq2 = v' == -x[t] - u*v^3;
eq3 = x[0] == 2;
(initial displacement is 2m).
s = NDSolve[{eq1, eq2, eq3}, x, {t, 0, 30}]
Thank you very much...
You need to observe that the first equation once differentiated with respect to t can be used to substitute for v[t]. But then the second equation becomes a ODE of second order and requires to be supplied with another extra initial condition. We will give
v[0]==x'[0]==some number
Then after solving this ODE for x you can recover v[t]==x'[t]
I give you the solution in term of a Manipulate so that geometrically the situation becomes clear to you.
(* First equation *)
v[t] = x'[t];
(*
Differentiate this equation once and substitute
for v[t] in the second equation
*)
Manipulate[
With[{u = Constant, der = derval},
res = NDSolve[{x''[t] == -x[t] - u*x'[t]^3, x[0.] == 2,x'[0.] == der},
x, {t, 0., 30.}] // First;
Plot[Evaluate[{x[t], v[t]} /. res], {t, 0, 30}, PlotRange -> All,
Frame -> True,Axes -> None, ImageSize -> 600]
],
{{Constant, 0.,TraditionalForm#(u)}, 0.,3, .1},
{{derval, -3., TraditionalForm#(v[0] == x'[0])}, -3, 3, .1}
]
Hope this helps you but next time before you ask you need to brush up the theory first as you can see the question you asked concerns very basic and elementary Mathematics not Mathematica programming. Good luck!!
You need to specify a numeric value for your u as well as an initial condition for v[t] :
u=1.0;
solution=NDSolve[{x'[t]==v[t], v'[t]==-x[t]-u v[t]^3,x[0]==2,v[0]==-1},{x,v},{t,0,1}]
Plot[{solution[[1,1,2]][t],solution[[1,2,2]][t]},{t,0,1}]
I have a basic problem in Mathematica which has puzzled me for a while. I want to take the m'th derivative of x*Exp[t*x], then evaluate this at x=0. But the following does not work correct. Please share your thoughts.
D[x*Exp[t*x], {x, m}] /. x -> 0
Also what does the error mean
General::ivar: 0 is not a valid variable.
Edit: my previous example (D[Exp[t*x], {x, m}] /. x -> 0) was trivial. So I made it harder. :)
My question is: how to force it to do the derivative evaluation first, then do substitution.
As pointed out by others, (in general) Mathematica does not know how to take the derivative an arbitrary number of times, even if you specify that number is a positive integer.
This means that the D[expr,{x,m}] command remains unevaluated and then when you set x->0, it's now trying to take the derivative with respect to a constant, which yields the error message.
In general, what you want is the m'th derivative of the function evaluated at zero.
This can be written as
Derivative[m][Function[x,x Exp[t x]]][0]
or
Derivative[m][# Exp[t #]&][0]
You then get the table of coefficients
In[2]:= Table[%, {m, 1, 10}]
Out[2]= {1, 2 t, 3 t^2, 4 t^3, 5 t^4, 6 t^5, 7 t^6, 8 t^7, 9 t^8, 10 t^9}
But a little more thought shows that you really just want the m'th term in the series, so SeriesCoefficient does what you want:
In[3]:= SeriesCoefficient[x*Exp[t*x], {x, 0, m}]
Out[3]= Piecewise[{{t^(-1 + m)/(-1 + m)!, m >= 1}}, 0]
The final output is the general form of the m'th derivative. The PieceWise is not really necessary, since the expression actually holds for all non-negative integers.
Thanks to your update, it's clear what's happening here. Mathematica doesn't actually calculate the derivative; you then replace x with 0, and it ends up looking at this:
D[Exp[t*0],{0,m}]
which obviously is going to run into problems, since 0 isn't a variable.
I'll assume that you want the mth partial derivative of that function w.r.t. x. The t variable suggests that it might be a second independent variable.
It's easy enough to do without Mathematica: D[Exp[t*x], {x, m}] = t^m Exp[t*x]
And if you evaluate the limit as x approaches zero, you get t^m, since lim(Exp[t*x]) = 1. Right?
Update: Let's try it for x*exp(t*x)
the mth partial derivative w.r.t. x is easily had from Wolfram Alpha:
t^(m-1)*exp(t*x)(t*x + m)
So if x = 0 you get m*t^(m-1).
Q.E.D.
Let's see what is happening with a little more detail:
When you write:
D[Sin[x], {x, 1}]
you get an expression in with x in it
Cos[x]
That is because the x in the {x,1} part matches the x in the Sin[x] part, and so Mma understands that you want to make the derivative for that symbol.
But this x, does NOT act as a Block variable for that statement, isolating its meaning from any other x you have in your program, so it enables the chain rule. For example:
In[85]:= z=x^2;
D[Sin[z],{x,1}]
Out[86]= 2 x Cos[x^2]
See? That's perfect! But there is a price.
The price is that the symbols inside the derivative get evaluated as the derivative is taken, and that is spoiling your code.
Of course there are a lot of tricks to get around this. Some have already been mentioned. From my point of view, one clear way to undertand what is happening is:
f[x_] := x*Exp[t*x];
g[y_, m_] := D[f[x], {x, m}] /. x -> y;
{g[p, 2], g[0, 1]}
Out:
{2 E^(p t) t + E^(p t) p t^2, 1}
HTH!
I'm trying to solve a really simple problem of finding object position under force {k1+k2 * y, k3*t}. Here's what I'm entering into Mathematica 7:
DSolve[{
x''[t]*m == k1 + k2*y[t],
y''[t]*m == k3*t,
y'[0] == 0,
y[0] == 0,
x'[0] == 0,
x[0] == 0
}, {x[t], y[t]}, t]
and I get this error:
DSolve::deqn: Equation or list of equations expected instead of True in the first argument {-C m (x^[Prime])[t]^2==k1+k2 y[t],m (y^[Prime][Prime])[t]==k3 t,True,y[0]==0,True,x[0]==0}.
It seems that Mathematica is unhappy about boundary conditions x'[0] == 0. Why is that?
It worked as you typed it ... try to do it in a fresh notebook
When I cut and paste the code you've posted into M'ma 7.0.1 and evaluate, I get the result
{{x[t] -> (60*k1*m*t^2 + k2*k3*t^5)/(120*m^2),
y[t] -> (k3*t^3)/(6*m)}}
Your M'ma error message tells me you actually have only one prime on x (i.e. x'[t]) in your actual M'ma input. The equation it cites, -C m (x^[Prime])[t]^2==k1+k2 y[t], does not match the first line of your code above.
I also suspect that x'[0] and y'[0] have been assigned to zero previously, which is causing x'[0]==0, ..., y'[0]==0 to both collapse to True. Best way to test: kill your kernel and re-evaluate the input above (after fixing typos).
Both, belisarius and Eric Towers have suggested killing the kernel and re-evaluating. They're most likely correct in that something has a prior definition. You can check if that is true via
?<variable name>
As an alternative to killing the kernel, I'd suggest clearing their values via
Clear[x, y, k1, k2, k3, m]
Or, if you really want to rid yourself of any definition of a variable there's Remove. This way, you won't have to recalculate anything else from your current session.