Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I am given a system first order differential equations: $x'=y$ and $y'=6x^2-a/2$, where $a$ is a constant and $'$ denotes $t$-derivatives.

I then make the substitution $(x,y)=(x_1y_1,y_1)$.

This implies that $x_1=\frac{x}{y}$ and $y_1=y$.

The $t$-derivatives are $$x_{1}'=\left(\frac{x}{y}\right)'=\frac{yx'-xy'}{y^2}=1-x_{1}y_{1}^{-1}(6x_{1}^2y_{1}^2-a/2)$$ and $$y_{1}'=6x_{1}^2y_{1}^2-a/2.$$

I need to continue in this fashion, that is, in the next step, for example, we let $(x_{1},y_{1})=(x_{2}y_{2},y_{2})$ and then find $x_{2}'$, $y_{2}'$, etc.

Is there a way I can compute the derivatives in Mathematica? For instance, how would I find $x_{1}'$ in Mathematica.

Thanks, Radz.

share|improve this question

migrated from math.stackexchange.com Jul 1 at 1:45

This question came from our site for people studying math at any level and professionals in related fields.

    
I am not sure if I properly understood. Suppose you differentiate the first equation and, in the result, replace $y'$ by the second equation. This gives a second order differential equation for $x$; solve it and compute $y$ from $x'$. –  Claude Leibovici Jun 27 at 11:03
    
Dear @Claude Leibovici, Yes. But I want to deal with only first order ODEs. I have to repeat the procedure a number of times (related to blow-ups). –  Radz Jun 27 at 11:23

5 Answers 5

up vote 2 down vote accepted
eq[0] = {Derivative[1][Subscript[x, 0]][t] == Subscript[y, 0][t], 
Derivative[1][Subscript[y, 0]][t] == 6 Subscript[x, 0][t]^2 - a/2};
ru[n_] := Block[
  {r = {Subscript[x, n - 1][t] :> 
      Subscript[x, n][t] Subscript[y, n][t], 
     Subscript[y, n - 1][t] :> Subscript[y, n][t]}},
  Flatten[{r, D[#, t] & /@ r}]]
eq[n_] := 
  Equal @@@ 
   Solve[eq[n - 1] /. ru[n], {Derivative[1][Subscript[x, n]][t], 
      Derivative[1][Subscript[y, n]][t]}][[1]];
Table[eq[n], {n, 0, 5}] // TableForm // TraditionalForm

enter image description here

share|improve this answer
    
You should try not to answer with images. Please try to put as much Mathematica code as possible or $\LaTeX$. –  Öskå Jul 1 at 12:36

Dt is useful for doing substutitions. You can think of Dt[x] as the differential of x. Substitutions can be performed with ReplaceAll (/.) and rules like {x -> x1 y1, y -> y1}. Then Solve can be used to solve for Dt[x1] and Dt[y1]. Solve returns a solution in the form of a Rule, so we replace Rule with Equal to get another differential equation.

eqns = {Dt[x] == y Dt[t], Dt[y] == (6 x^2 - a/2) Dt[t]};

eqns1 = First@
   Solve[eqns /. {x -> x1 y1, y -> y1}, {Dt[x1], Dt[y1]}] /. 
  Rule -> Equal
(*
  {Dt[x1] == -((-a x1 Dt[t] - 2 y1 Dt[t] + 12 x1^3 y1^2 Dt[t])/(2 y1)), 
   Dt[y1] == 1/2 (-a Dt[t] + 12 x1^2 y1^2 Dt[t])}
*)

You can repeat the process manually or do them all together with Fold or FoldList. FoldList will give the sequence of differential equations, which you might want.

Fold[
 First@Solve[#1 /. #2, Dt /@ Variables[Last /@ #2]] /. Rule -> Equal &,
 eqns,
 {{x -> x1 y1, y -> y1},
  {x1 -> x2 y2, y1 -> y2}}
 ]
(*
  {Dt[x2] == -((-Dt[t] - a x2 Dt[t] + 12 x2^3 y2^4 Dt[t])/y2), 
   Dt[y2] == 1/2 (-a Dt[t] + 12 x2^2 y2^4 Dt[t])}
*)

Warning: Depending on the form of the substitutions you use, Variables might not work to perfection. Manual intervention might be necessary. Or you might write a more complicated procedure for Fold in which the new variables are explicitly listed with each substitution.

share|improve this answer

It's a little hard to tell what you are looking for, but here is a try. The first step can be written in almost exactly the same way you have above:

x1[t] = x[t]/y[t];
y1[t] = y[t];
D[x1[t], t] /. {D[x[t], t] -> y[t], D[y[t], t] -> 6 x[t]^2 - a/2}
D[y1[t], t] /. {D[x[t], t] -> y[t], D[y[t], t] -> 6 x[t]^2 - a/2}

This returns the same equations you have for the derivatives. But now you can continue in exactly the same manner:

x2[t] = x1[t]/y1[t];
y2[t] = y1[t];
D[x2[t], t] /. {D[x1[t], t] -> y1[t], D[y1[t], t] -> 6 x1[t]^2 - a/2}
D[y2[t], t] /. {D[x1[t], t] -> y1[t], D[y1[t], t] -> 6 x1[t]^2 - a/2}

enter image description here

and so on. The next one would be:

x3[t] = x2[t]/y2[t];
y3[t] = y2[t];
D[x3[t], t] /. {D[x2[t], t] -> y2[t], D[y2[t], t] -> 6 x2[t]^2 - a/2}
D[y3[t], t] /. {D[x2[t], t] -> y2[t], D[y2[t], t] -> 6 x2[t]^2 - a/2}
share|improve this answer

The idea:

eq = {x'[t] == y[t],
      y'[t] == 6 x[t]^2 - a/2}

eq2 = eq /. x -> (x1[#] y1[#] &) /. y -> y1
{
 y1[t] x1'[t] + x1[t] y1'[t] == y1[t],
 y1'[t] == -(a/2) + 6 x1[t]^2 y1[t]^2
}
  Thread[
    {x1'[t], y1'[t]} == First[ {x'[t], y'[t]} /. Solve[eq, {x1'[t], y1'[t]}]] ]
 x1'[t] == 1 + (a x1[t])/(2 y1[t]) - 6 x1[t]^3 y1[t] 
 y1'[t] == -(a/2) + 6 x1[t]^2 y1[t]^2

General approach

You don't have to even change the names!

and if you need to do this couple of times then just use:

eq = {x'[t] == y[t], y'[t] == 6 x[t]^2 - a/2}
Do[
 eq = eq /. x -> (x[#] y[#] &);
 eq = Thread[
        {x'[t], y'[t]} == First[ {x'[t], y'[t]} /. Solve[eq, {x'[t], y'[t]}]] ]
 , {5}]
share|improve this answer

Here is a fully automated way of solving your problem. To streamline the notation, I redefine your initial equations using index 0.

Define the initial equations.

eqns = {Derivative[1][x[0]][t] == y[0][t], Derivative[1][y[0]][t] == 6 x[0][t]^2 - a/2};

Define the transformation rule.

transfrule = {x[n_] :> (x[n + 1][#] y[n + 1][#] &), y[n_] :> (y[n + 1][#] &)};

Iterate the transformation, and construct each new set of equations with the derivatives on the left hand side as you go along. I iterate 5 times, but choose your own number as required.

NestList[
  (
    eqnsnew = # /. transfrule;
    derivs = eqnsnew // Cases[#, Derivative[_][_][_], \[Infinity]] & // DeleteDuplicates;
    Thread[derivs == Simplify[derivs /. Solve[eqnsnew, derivs][[1]]]]
  ) &,
  eqns,
  5
]
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.