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Mathematics Calculus Equation Solving Wolfram Language

Avoid different solutions of two equivalent systems of equations?

Jacques Ou

Posted 6 years ago

Consider the following code: In[401]:= eq1 = DSolve[{ \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R2, T] == 0, (Jrm R0^2 + Rr C20[0, T]) \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, Assumptions -> {T > 0}, GeneratedParameters -> F1] Out[401]= {{C20 -> Function[{R2, T}, F1[1][T]]}, {C20 -> Function[{R2, T}, (-Jrm R0^2 + R2 Rr F1[2][T])/Rr]}} In[402]:= eq2 = DSolve[{ -2 R0 \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R2, T] == 0, 4 R0 (Jrm R0^2 + Rr C20[0, T]) \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, Assumptions -> {T > 0}, GeneratedParameters -> F1] Out[402]= {{C20 -> Function[{R2, T}, -((Jrm R0^2)/Rr)]}} eq1 and eq2 are equivalent, but their solutions are different. Why?

Consider the following code:

In[401]:= eq1 = DSolve[{ 
    \!\(\*SuperscriptBox[\(C20\), 
    TagBox[
    RowBox[{"(", 
    RowBox[{"2", ",", "0"}], ")"}],
    Derivative],
    MultilineFunction->None]\)[R2, T] == 0, (Jrm R0^2 + Rr C20[0, T]) 
    \!\(\*SuperscriptBox[\(C20\), 
    TagBox[
    RowBox[{"(", 
    RowBox[{"1", ",", "0"}], ")"}],
    Derivative],
    MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, 
      Assumptions -> {T > 0}, GeneratedParameters -> F1]



Out[401]= {{C20 -> Function[{R2, T}, F1[1][T]]}, {C20 -> 
   Function[{R2, T}, (-Jrm R0^2 + R2 Rr F1[2][T])/Rr]}}

In[402]:= eq2 = DSolve[{ -2 R0 
\!\(\*SuperscriptBox[\(C20\), 
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R2, T] == 0, 4 R0 (Jrm R0^2 + Rr C20[0, T]) 
\!\(\*SuperscriptBox[\(C20\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, 
  Assumptions -> {T > 0}, GeneratedParameters -> F1]

Out[402]= {{C20 -> Function[{R2, T}, -((Jrm R0^2)/Rr)]}}

eq1 and eq2 are equivalent, but their solutions are different. Why?

POSTED BY: Jacques Ou

3 Replies

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 6 years ago

The first solution comes from setting `(Jrm R0^2 + Rr C20[0, T]) == 0`. The second solution is a special case of the first, with `F1[2][T]=0`. They both miss the solution `C20 -> Function[{R2,T}, F1[2][T]]` that comes from setting `Derivative[1,0][C20][0, T] == 0`.

POSTED BY: Gianluca Gorni

Jacques Ou

Posted 6 years ago

The unique difference between 2 systems of equations is the constant coefficients. The possible problem is that one is taken as PDEs and the other as ODEs. eq1 = DSolve[{ \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R2, T] == 0, (Jrm R0^2 + Rr C20[0, T]) \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, Assumptions -> {T > 0}, GeneratedParameters -> F1] ({{C20\[Rule]Function[{R2,T},F1[1][T]]},{C20\[Rule]Function[{R2,T},(-\ Jrm R0^2+R2 Rr F1[2][T])/Rr]}}) eq2 = DSolve[{ -2 \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R2, T] == 0, 4 (Jrm R0^2 + Rr C20[0, T]) \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, Assumptions -> {T > 0}, GeneratedParameters -> F1] ({{C20\[Rule]Function[{R2,T},-((Jrm R0^2)/Rr)]}}) eq3 = DSolve[{ -2 R0 \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R2, T] == 0, 4 R0 (Jrm R0^2 + Rr C20[0, T]) \!\(\SuperscriptBox[\(C20\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, Assumptions -> {T > 0}, GeneratedParameters -> F1] ({{C20\[Rule]Function[{R2,T},-((Jrm R0^2)/Rr)]}})

The unique difference between 2 systems of equations is the constant coefficients. The possible problem is that one is taken as PDEs and the other as ODEs.

eq1 = DSolve[{ 

\!\(\*SuperscriptBox[\(C20\), 
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R2, T] == 0, (Jrm R0^2 + Rr C20[0, T])  
\!\(\*SuperscriptBox[\(C20\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, 
        Assumptions -> {T > 0}, GeneratedParameters -> F1]


(*{{C20\[Rule]Function[{R2,T},F1[1][T]]},{C20\[Rule]Function[{R2,T},(-\
Jrm R0^2+R2 Rr F1[2][T])/Rr]}}*)

eq2 = DSolve[{ 
    -2  
\!\(\*SuperscriptBox[\(C20\), 
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R2, T] == 0, 
   4 (Jrm R0^2 + Rr C20[0, T])  
\!\(\*SuperscriptBox[\(C20\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, 
        Assumptions -> {T > 0}, GeneratedParameters -> F1]


(*{{C20\[Rule]Function[{R2,T},-((Jrm R0^2)/Rr)]}}*)

eq3 = DSolve[{
    -2 R0  
\!\(\*SuperscriptBox[\(C20\), 
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R2, T] == 0, 4 R0 (Jrm R0^2 + Rr C20[0, T]) 
\!\(\*SuperscriptBox[\(C20\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[0, T] == 0}, C20, {R2, T}, 
    Assumptions -> {T > 0}, GeneratedParameters -> F1]

(*{{C20\[Rule]Function[{R2,T},-((Jrm R0^2)/Rr)]}}*)

POSTED BY: Jacques Ou

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 6 years ago

eq1 and eq2 are not equivalent,not the same.Mathematica gives a correct solution. Check for eq1: eq1 = DSolve[{Derivative[2, 0][C20][R2, T] == 0, (JrmR0^2 + RrC20[0, T])Derivative[1, 0][C20][0, T] == 0}, C20, {R2, T}, Assumptions -> {T > 0}, GeneratedParameters -> F1] Derivative[2, 0][C20][R2, T] == 0 /. eq1 ( {True, True}) (JrmR0^2 + RrC20[0, T])Derivative[1, 0][C20][0, T] == 0 /. eq1 (* {True, True}) Check for eq2: eq2 = DSolve[{-2R0Derivative[2, 0][C20][R2, T] == 0, 4R0(JrmR0^2 + RrC20[0, T])Derivative[1, 0][C20][0, T] == 0}, C20, {R2, T}, Assumptions -> {T > 0}, GeneratedParameters -> F1] -2R0Derivative[2, 0][C20][R2, T] == 0 /. eq2 (True) 4R0(JrmR0^2 + RrC20[0, T])Derivative[1, 0][C20][0, T] == 0 /. eq2 (True*)

eq1 and eq2 are not equivalent,not the same.Mathematica gives a correct solution.

Check for eq1:

 eq1 = DSolve[{Derivative[2, 0][C20][R2, T] == 
     0, (Jrm*R0^2 + Rr*C20[0, T])*Derivative[1, 0][C20][0, T] == 0}, 
   C20, {R2, T}, 
         Assumptions -> {T > 0}, GeneratedParameters -> F1]

Derivative[2, 0][C20][R2, T] == 0 /. eq1
(* {True, True}*)
(Jrm*R0^2 + Rr*C20[0, T])*Derivative[1, 0][C20][0, T] == 0 /. eq1
(* {True, True}*)

Check for eq2:

eq2 = DSolve[{-2*R0*Derivative[2, 0][C20][R2, T] == 0, 
4*R0*(Jrm*R0^2 + Rr*C20[0, T])*Derivative[1, 0][C20][0, T] == 0}, 
 C20, {R2, T}, 
 Assumptions -> {T > 0}, GeneratedParameters -> F1]

 -2*R0*Derivative[2, 0][C20][R2, T] == 0 /. eq2
 (*True*)
  4*R0*(Jrm*R0^2 + Rr*C20[0, T])*Derivative[1, 0][C20][0, T] == 0 /. eq2
 (*True*)

POSTED BY: Mariusz Iwaniuk

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