Use of variational operator, in Mathematica

Question

I am working in Hamilton's principle. Part of deriving the equation of motion is to use the delta operator () which can be operated just like a differential operator. It is not the function VariationalD in package VariationalMethods.

A post was posted regarding the same subject but it seems that who asked the question found the answer somewhere else. The previous question is here.

Example on the equation is:

T= 0.5 Mu'^2

applying the delta operator

T=M u' u'

I need to find the first and second variation of this function w.r.t the dependent variable.

regards

Natas · Answer 1 · 2020-11-03T07:03:22.640

4

Indeed, VariationalD does perform integration by parts which might be unwanted.

For this simple univariat case, one can hack together a quick function myVariationalD:

Clear[δ, myVariationalD]
myVariationalD[expr_, f_[x_]] := Module[{maxDerivative},
  maxDerivative = Max @ Cases[expr, Derivative[m_][f][x] :> m, Infinity];
  D[expr, f[x]] δ[f[x]] + 
   Sum[D[expr, Derivative[m][f][x]] δ[Derivative[m][f][x]], {m,
      1, maxDerivative}]
  ]

The example you have given can then be computed as

expr = 1/2 \[DoubleStruckCapitalM] u'[t]^2;
myVariationalD[expr, u[t]]
(* \[DoubleStruckCapitalM] δ[u'[t]] u'[t] *)

edited Nov 03 '20 at 07:03

answered Nov 02 '20 at 17:33

Natas

2,310
4
14

Thank you Natas,
What about below equation:

T = 0.5 M D[u[t], t]^2 [Delta]T = D[T /. {u -> u + s [Delta]u}, s] /. s -> 0

– ayman zayed Nov 13 '20 at 20:06
@aymanzayed Sorry, but I don't quite understand what you are asking. Perhaps you can modify your question and explain in detail the expected output for a given input and I can then edit my answer? – Natas Nov 14 '20 at 07:53

score 0 · Answer 2 · answered Jan 06 '22 at 13:35

I am also working on Hamilton's Principle to derive the governing equation of beam. After evaluating the strain and kinetic energy, I am combining both the energies. Now I want to apply the variational operator on my Hamilton's principle. The strain energy and kinetic energy terms contain the five variables such as u[x,t], v[x,t], w[x,t],Theta[x,t] and Phi[x,t].

How I can get the governing equation with these five variables? The Strain energy terms are as follows:

D00 ks \[Theta][x, t]^2 + D00 \[Phi][x, t]^2 + B00 
\!\(\*SuperscriptBox[\(u\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 - 2 D00 \[Phi][x, t] 
\!\(\*SuperscriptBox[\(v\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] + D00 
\!\(\*SuperscriptBox[\(v\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 - 2 D00 ks \[Theta][x, t] 
\!\(\*SuperscriptBox[\(w\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] + D00 ks 
\!\(\*SuperscriptBox[\(w\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 - 2 B10 
\!\(\*SuperscriptBox[\(u\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] 
\!\(\*SuperscriptBox[\(\[Theta]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] + B20 
\!\(\*SuperscriptBox[\(\[Theta]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 - 2 B01 
\!\(\*SuperscriptBox[\(u\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] - 2 B01 \[Theta][x, t] 
\!\(\*SuperscriptBox[\(u\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] + 2 B10 \[Phi][x, t] 
\!\(\*SuperscriptBox[\(u\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] + B11 
\!\(\*SuperscriptBox[\(\[Theta]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] + B20 
\!\(\*SuperscriptBox[\(\[Theta]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] + B11 \[Theta][x, t] 
\!\(\*SuperscriptBox[\(\[Theta]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] + B20 \[Theta][x, t] 
\!\(\*SuperscriptBox[\(\[Theta]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] - 2 B20 \[Phi][x, t] 
\!\(\*SuperscriptBox[\(\[Theta]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t] + B11 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 + 2 B11 \[Theta][x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 + B11 \[Theta][x, t]^2 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 - B11 \[Phi][x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 - B20 \[Phi][x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 - B11 \[Theta][x, t] \[Phi][x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 - B20 \[Theta][x, t] \[Phi][x, t] 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2 + B20 \[Phi][x, t]^2 
\!\(\*SuperscriptBox[\(\[Phi]\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, t]^2

This doesn't appear to be an answer to the original question, rather a question in its own right. You should post it as a question and delete this 'answer'. — High Performance Mark, Jan 06 '22 at 15:13
This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. To get notified when this question gets new answers, you can follow this question. Once you have enough reputation, you can also add a bounty to draw more attention to this question. - From Review — Hausdorff, Jan 06 '22 at 16:54

Use of variational operator, in Mathematica

2 Answers2