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.

When working with differential equations with independent (e.g. t )and dependent (e.g., x[t]) variables Mathematica expresses partial and 'regular' derivatives nicely, but not the derivative itself. Can anybody manage to do this?

I give an example: Let f depend on x, y and t. x and y depend on t. So, the partial derivative

enter image description here

with e.g. respect to x is in Matheamtica expressed by

f^(1,0,0)[x,y,t]

, the derivative with respect to the independent variable

enter image description here

as an example by

x'[t]

. But how can I make Mathematica to show the derivative e.g.

enter image description here

explicitly?

share|improve this question
    
Does DifferentialD[x] work for your purposes? –  Timothy Wofford Oct 10 '13 at 15:09
    
Using Solve[Dt[w[l2[t], lp[t], t], t] == 0, Dt[lp[t], t]], I would then for instance like to see dlp. I tried to use your (adjusted) example instead of Dt[lp[t], t]. It did not work. What do you think? –  Andreas Oct 10 '13 at 15:43
    
Using the Notation package we can make Dt appear as DifferentialD, and both Dt[w[x[t],y[t],t]] and Dt[w[x,y,t]] seem to expand as I would expect. If you are after the functionality of exterior derivatives, you may need to define it. Maybe one of these packages can help. 1, 2, 3 –  Timothy Wofford Oct 10 '13 at 22:09
add comment

1 Answer 1

up vote 3 down vote accepted

Dt is one way to work with differentials, unless you're after the more advanced functionality Timothy Wofford linked to in his comment. I'm not sure whether you want to do calculus with differentials, for instance in differential equations, or just want them to be formatted in a certain way.

Formatting

For formatting derivatives and differentials, these may be of help:

The derivative and Dt

The differential of a multivariable function:

w = f[x, y, t];
Dt[w]
(* Dt[t] * Derivative[0, 0, 1][f][x, y, t] + 
    Dt[y] * Derivative[0, 1, 0][f][x, y, t] + 
    Dt[x] * Derivative[1, 0, 0][f][x, y, t]   *)

The (matrix) derivative of f may be obtained with D

D[w, {{x, y, t}}]
(* {Derivative[1, 0, 0][f][x, y, t], 
    Derivative[0, 1, 0][f][x, y, t], 
    Derivative[0, 0, 1][f][x, y, t]}

The output of Dt[w] above may also be obtained with

D[w, {{x, y, t}}] . Dt[{x, y, t}]

One can substitute functions for x and y to make them explicitly depending on t, and the Chain Rule is automatically applied (of course):

Dt[w] /. {x -> x[t], y -> y[t]} // Factor // InputForm
(* Dt[t]*
    (Derivative[0, 0, 1][f][x[t], y[t], t] +
      y'[t]*Derivative[0, 1, 0][f][x[t], y[t], t] +
      x'[t]*Derivative[1, 0, 0][f][x[t], y[t], t])  *)

Differential equations and Dt

One can turn expressions in terms of Dt[x] etc. into differential equations:

dtToDE[diffForm_, vars_List, var_] /; MemberQ[vars, var] := 
  With[{indep = DeleteCases[vars, var]},
   0 == diffForm /. 
     Flatten[{Thread[Dt /@ indep -> D[Through[indep[var]], var]], 
       Thread[indep -> Through[indep[var]]]}] /. Dt[var] -> 1
   ];
dtToDE[diffForm_, vars_List, var_] := 
  D[Through[vars[var]], var] == 
   Extract[CoefficientList[diffForm, Dt /@ vars] /. 
     Thread[vars -> Through[vars[var]]], 
    1 + IdentityMatrix[Length[vars]]];

orthogonalTrajectory[diffForm_, vars_List] := 
  diffForm /. Thread[# -> Cross[#]] &[Dt /@ vars];

Examples

An exact differential.

Dt[x^2 + y^2 x - y^2]                            (* the differential of f *)
ode = dtToDE[Dt[x^2 + y^2 x - y^2], {x, y}, x]   (* as a derivative equation *)
DSolve[ode, y[x], x]                             (* solution to the ODE *)
Solve[x^2 + y^2 x - y^2 == C[1]/2, y]            (* solution to f = constant *)

(* 2 x Dt[x] + y^2 Dt[x] - 2 y Dt[y] + 2 x y Dt[y] *)
(* 0 == 2 x + y[x]^2 - 2 y[x] Derivative[1][y][x] + 2 x y[x] Derivative[1][y][x] *)
(* {{y[x] -> -(Sqrt[-2 x^2 - C[1]]/(Sqrt[2] Sqrt[-1 + x]))},
    {y[x] ->   Sqrt[-2 x^2 - C[1]]/(Sqrt[2] Sqrt[-1 + x])}}  *)
(* {{y    -> -(Sqrt[-2 x^2 + C[1]]/(Sqrt[2] Sqrt[-1 + x]))},
    {y    ->   Sqrt[-2 x^2 + C[1]]/(Sqrt[2] Sqrt[-1 + x])}}  *)

A differential with an integrating factor (pasted from an old test's TeX file, converted to a differential form with Dt):

ode = dtToDE[x dy + y dx + 4 x^3 y^4 dy /. {dx -> Dt[x], dy -> Dt[y]}, {x, y}, x]
sol = DSolve[ode, y[x], x];
ode /. {y'[x] -> Hold[D[y[x], x]]} /. sol // ReleaseHold // Simplify
Times @@ Subtract @@@ DeleteDuplicates@Simplify[x y^2 == x y[x]^2 /. sol] == 0 // Simplify

(* 0 == y[x] + x y'[x] + 4 x^3 y[x]^4 y'[x]   -- ODE *)
(* {True, True, True, True}                   -- all branches of sol satisfy the ODE *)
(* x^2 (4 y^4 - 2 y^2 C[1]) == 1              -- the integral curve (hacked from sol) *)

A system of equations, following three orthogonal trajectories to the gradient of a function.

sols = NDSolveValue[
         {dtToDE[orthogonalTrajectory[Dt[(x^2 + y^2)^2 - 4 x y], {x, y}], {x, y}, t],
          x[0] == #, y[0] == #},
         {x, y}, {t, -1.35, 1.35}] & /@ {0.99, 1., 1.01};
ParametricPlot @@ {Through[#[t]] & /@ sols, Flatten @ {t, sols[[1,1]]["Domain"]}}

Mathematica graphics

share|improve this answer
add comment

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.