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Mathematics Calculus Equation Solving Wolfram Language

What is the Syntax error for ....is incomplete; more input is needed

Nomsa Ledwaba

Posted 11 months ago

Input: DSolve[-Sqrt[r] v h[t] - 2 r v Derivative[1][Xi[2]][t] == 0, h[t], t] Output: {{h[t] -> -2 Sqrt[r] Derivative[1][Xi[2]][t]}} Input: LXi = {rD[Xi[2][t], t] + Sqrt[r](-2 Sqrt[r] D[Xi[2][t], t])(Xi[1][r,t]), Xi[2][t]}, LPhi = {f[r, t] v + g[r, t]}; EDsI = SubstInfinitesimals[EDs, LXi, LPhi] Output: Syntax error: ....is incomplete; more input is needed. I'm expecting the Output to be a list of {0, 0, 0, 0, 0, 0, mathematical expression }. I cannot identify what is not there on input to complete the expression to compute the correct answer. This is the Mathematica package. Please advise further.

Input:  DSolve[-Sqrt[r] v h[t] - 2 r v Derivative[1][Xi[2]][t] == 0, h[t], t]
    Output: {{h[t] -> -2 Sqrt[r] Derivative[1][Xi[2]][t]}}

Input: LXi = {r*D[Xi[2][t], t] + Sqrt[r]*(-2 *Sqrt[r] * D[Xi[2][t], t])(*Xi[1][r,t]*), Xi[2][t]},
LPhi = {f[r, t] v + g[r, t]};
EDsI = SubstInfinitesimals[EDs, LXi, LPhi] 
Output: Syntax error:  ....is incomplete; more input is needed.

I'm expecting the Output to be a list of {0, 0, 0, 0, 0, 0, mathematical expression }. I cannot identify what is not there on input to complete the expression to compute the correct answer. This is the Mathematica package. Please advise further.

POSTED BY: Nomsa Ledwaba

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Rohit Namjoshi

Posted 11 months ago

Unlikely that anyone can answer the question unless the definition for `SubstInfinitesimals` is provided.

POSTED BY: Rohit Namjoshi

Nomsa Ledwaba

Posted 11 months ago

Indeed SubstInfinitesimals was provided. That being said the previous input and output whereby its was a list of {0, 0, 0, 0, 0, mathematical expression, mathematical expression } were: Input: LXi = {rD[Xi[2][t], t] + Sqrt[r] h[t], Xi[2][t]}; LPhi = {f[r, t] v + g[r, t]}; EDsI = SubstInfinitesimals[EDs, LXi, LPhi] Output: {0, 0, 0, 0, 0, -r g[r, t] - Sqrt[r] v h[t] - 2 r v Derivative[1][Xi[2]][t] + v \!\(\SuperscriptBox[\(f\), TagBox[RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + \!\(\SuperscriptBox[\(g\), TagBox[RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}],Derivative], MultilineFunction->None]\)[r, t] + v \[Rho] \!\(\SuperscriptBox[\(f\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] - v \[Alpha] \[Lambda][r, t] \!\(\SuperscriptBox[\(f\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative], MultilineFunction->None]\)[r, t] + r v \[Beta] \[Lambda][r, t] \!\(\SuperscriptBox[\(f\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + \[Rho] \!\(\SuperscriptBox[\(g\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative], MultilineFunction->None]\)[r, t] - \[Alpha] \[Lambda][r, t] \!\(\SuperscriptBox[\(g\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + r \[Beta] \[Lambda][r, t] \!\(\SuperscriptBox[\(g\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative], MultilineFunction->None]\)[r, t] + 1/2 r v \[Sigma]^2 \!\(\SuperscriptBox[\(f\), TagBox[RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + 1/2 r \[Sigma]^2 \!\(\SuperscriptBox[\(g\), TagBox[RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t], -((\[Rho] h[t])/( 2 Sqrt[r])) + (\[Sigma]^2 h[t])/( 8 Sqrt[r]) + (\[Alpha] h[t] \[Lambda][r, t])/(2 Sqrt[r]) + 1/2 Sqrt[r] \[Beta] h[t] \[Lambda][r, t] - Sqrt[r] Derivative[1][h][t] + r \[Beta] \[Lambda][r, t] Derivative[1][Xi[2]][t] - r (Xi[2]^\[Prime]\[Prime])[t] - \[Alpha] Xi[2][t] \!\(\SuperscriptBox[\(\[Lambda]\),TagBox[RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + r \[Beta] Xi[2][t] \!\(\SuperscriptBox[\(\[Lambda]\), TagBox[RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}],Derivative], MultilineFunction->None]\)[r, t] + r \[Sigma]^2 \!\(\SuperscriptBox[\(f\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] - Sqrt[r] \[Alpha] h[t] \!\(\SuperscriptBox[\(\[Lambda]\), TagBox[RowBox[{"(",RowBox[{"1", ",", "0"}], ")"}],Derivative], MultilineFunction->None]\)[r, t] + r^(3/2) \[Beta] h[t] \!\(\SuperscriptBox[\(\[Lambda]\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] - r \[Alpha] Derivative[1][Xi[2]][t] \!\(\SuperscriptBox[\(\[Lambda]\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + r^2 \[Beta] Derivative[1][Xi[2]][t] \!\(\SuperscriptBox[\(\[Lambda]\), TagBox[RowBox[{"(",RowBox[{"1", ",", "0"}], ")"}],Derivative], MultilineFunction->None]\)[r, t]} Hope this is clearer.

Indeed SubstInfinitesimals was provided. That being said the previous input and output whereby its was a list of {0, 0, 0, 0, 0, mathematical expression, mathematical expression } were:

Input:  LXi = {r*D[Xi[2][t], t] + Sqrt[r] h[t], Xi[2][t]};
LPhi = {f[r, t] v + g[r, t]};
EDsI = SubstInfinitesimals[EDs, LXi, LPhi]

Output:   {0, 0, 0, 0, 0, -r g[r, t] - Sqrt[r] v h[t] - 2 r v Derivative[1][Xi[2]][t] + v \!\(\*SuperscriptBox[\(f\), 
TagBox[RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + 
\!\(\*SuperscriptBox[\(g\), TagBox[RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}],Derivative],
MultilineFunction->None]\)[r, t] + v \[Rho] \!\(\*SuperscriptBox[\(f\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] - v \[Alpha] \[Lambda][r, t] \!\(\*SuperscriptBox[\(f\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],
MultilineFunction->None]\)[r, t] + r v \[Beta] \[Lambda][r, t] \!\(\*SuperscriptBox[\(f\), TagBox[
RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + \[Rho] 
\!\(\*SuperscriptBox[\(g\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],
MultilineFunction->None]\)[r, t] - \[Alpha] \[Lambda][r, t] \!\(\*SuperscriptBox[\(g\), TagBox[
RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + r \[Beta] \[Lambda][r, t] \!\(\*SuperscriptBox[\(g\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],
MultilineFunction->None]\)[r, t] + 1/2 r v \[Sigma]^2 \!\(\*SuperscriptBox[\(f\), TagBox[RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + 1/2 r \[Sigma]^2 
\!\(\*SuperscriptBox[\(g\), TagBox[RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t], -((\[Rho] h[t])/(   2 Sqrt[r])) + (\[Sigma]^2 h[t])/(
  8 Sqrt[r]) + (\[Alpha] h[t] \[Lambda][r, t])/(2 Sqrt[r]) + 1/2 Sqrt[r] \[Beta] h[t] \[Lambda][r, t] - 
  Sqrt[r] Derivative[1][h][t] +   r \[Beta] \[Lambda][r, t] Derivative[1][Xi[2]][t] - 
  r (Xi[2]^\[Prime]\[Prime])[t] - \[Alpha] Xi[2][t] \!\(\*SuperscriptBox[\(\[Lambda]\),TagBox[RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + r \[Beta] Xi[2][t] 
\!\(\*SuperscriptBox[\(\[Lambda]\), TagBox[RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}],Derivative],
MultilineFunction->None]\)[r, t] + r \[Sigma]^2 \!\(\*SuperscriptBox[\(f\), TagBox[RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] - Sqrt[r] \[Alpha] h[t] 
\!\(\*SuperscriptBox[\(\[Lambda]\), TagBox[RowBox[{"(",RowBox[{"1", ",", "0"}], ")"}],Derivative],
MultilineFunction->None]\)[r, t] + r^(3/2) \[Beta] h[t] \!\(\*SuperscriptBox[\(\[Lambda]\), TagBox[
RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] - r \[Alpha] Derivative[1][Xi[2]][t] \!\(\*SuperscriptBox[\(\[Lambda]\), TagBox[RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}],Derivative],MultilineFunction->None]\)[r, t] + r^2 \[Beta] Derivative[1][Xi[2]][t] 
\!\(\*SuperscriptBox[\(\[Lambda]\), TagBox[RowBox[{"(",RowBox[{"1", ",", "0"}], ")"}],Derivative],
MultilineFunction->None]\)[r, t]}

Hope this is clearer.

POSTED BY: Nomsa Ledwaba

Nomsa Ledwaba

Posted 11 months ago

I have attached a .nb file to show how evaluation of the inputs results in outputs, particularly for EDsI[[6]]. I am not sure what is incorrect from the syntax. Attachments: Practice1.nb

POSTED BY: Nomsa Ledwaba

Eric Rimbey

Posted 11 months ago

You still haven't shown us the definitions for `SubstInfinitesimals` and `EDs`

POSTED BY: Eric Rimbey

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