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Mathematics Calculus Equation Solving Wolfram Language

How to solve this equation?

Jacques Ou

Posted 6 years ago

Would you like to explain whether this equation can be solved and how? In[1308]:= eq1 = \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T] == 1 - C2[R1, T] + \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T]/R1 + (Rv \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T])/R1 + \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T] Out[1308]= \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T] == 1 - C2[R1, T] + \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T]/R1 + (Rv \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T])/R1 + \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T] In[1309]:= eq2 = Rv C2[1, T] + \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[1, T] == 0 Out[1309]= Rv C2[1, T] + \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[1, T] == 0

Would you like to explain whether this equation can be solved and how?

In[1308]:= eq1 = 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T] == 1 - C2[R1, T] + 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]/R1  + (Rv 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T])/R1 + 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]

Out[1308]= 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T] == 1 - C2[R1, T] + 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]/R1 + (Rv 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T])/R1 + 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]

In[1309]:= eq2 = Rv C2[1, T] + 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[1, T] == 0

Out[1309]= Rv C2[1, T] + 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[1, T] == 0

POSTED BY: Jacques Ou

9 Replies

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Jacques Ou

Posted 6 years ago

I absolutely agree. This one is just a little different from that one which can be solved by Maple 2020 beta. With respect to the mathematical structure, they are same. The numerical solutions have been done earl

POSTED BY: Jacques Ou

Jacques Ou

Posted 6 years ago

How about this one? eq1 = \!\(\SuperscriptBox[\(C1\), TagBox[ RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}], Derivative], MultilineFunction->None]\)[R, T] == 1 - C1[R, T] + \!\(\SuperscriptBox[\(C1\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R, T]/R + (Rv \!\(\SuperscriptBox[\(C1\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R, T])/R + \!\(\SuperscriptBox[\(C1\), TagBox[ RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R, T] eq2 = Rv C1[Rx, T] + Rx \!\(\*SuperscriptBox[\(C1\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[Rx, T] == 0

How about this one?

eq1 = 
\!\(\*SuperscriptBox[\(C1\), 
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[R, T] == 1 - C1[R, T] + 
\!\(\*SuperscriptBox[\(C1\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R, T]/R + (Rv 
\!\(\*SuperscriptBox[\(C1\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R, T])/R + 
\!\(\*SuperscriptBox[\(C1\), 
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R, T]

eq2 = Rv C1[Rx, T] + Rx 
\!\(\*SuperscriptBox[\(C1\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[Rx, T] == 0

POSTED BY: Jacques Ou

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 6 years ago

Unfortunately Maple does not solve this system analytically. Do you have any reason to think there is a closed form? Most PDE's don't have one. Maybe the best you can do is numerical methods. All computer algebra systems, including Mathematica,Maple, are limited in their capabilities.

POSTED BY: Mariusz Iwaniuk

Jacques Ou

Posted 6 years ago

I appreciate your help and wish you happy new year 2020.

POSTED BY: Jacques Ou

Jacques Ou

Posted 6 years ago

Dear Mariusz, I have no Maple 2020 beta version. Would you like to run the original problem in this version? Thanks a lot. eq1 = \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T] == 1 - C2[R1, T] + \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T]/(Rd (R0 + R1 Rd)) + (Rv \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T])/(Rd (R0 + R1 Rd)) + \!\(\SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"2", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[R1, T]/Rd^2 eq2 = Rv C2[1, T] + (Rx \!\(\*SuperscriptBox[\(C2\), TagBox[ RowBox[{"(", RowBox[{"1", ",", "0"}], ")"}], Derivative], MultilineFunction->None]\)[1, T])/Rd == 0

Dear Mariusz, I have no Maple 2020 beta version. Would you like to run the original problem in this version? Thanks a lot.

eq1 = 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T] == 1 - C2[R1, T] + 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]/(Rd (R0 + R1 Rd)) + (Rv 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T])/(Rd (R0 + R1 Rd)) + 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]/Rd^2

eq2 = Rv C2[1, T] + (Rx 
\!\(\*SuperscriptBox[\(C2\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[1, T])/Rd == 0

POSTED BY: Jacques Ou

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 6 years ago

Unfortunately Maple 2020 does not solve.

POSTED BY: Mariusz Iwaniuk

Updating Name

Posted 6 years ago

Dear Mariusz, I used the PDEtools of MAPLE and I didn't get the result. Would you like share the codes with me? Thanks a lot.

POSTED BY: Updating Name

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 6 years ago

Oh, my mistake PDE is solved in Maple 2020 beta version. I'm a beta tester. Yes Maple 2019.2 can't solve. Regards M.I. I added a Maple file PDE solution.mw you must only chage a file extension to *.mw Attachments: PDE solution.nb

POSTED BY: Mariusz Iwaniuk

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 6 years ago

Equation can be solved.I used Maple 2019.2 LateX: $$\text{C2}(\text{R1},T)=1-\frac{\text{Rv} \text{R1}^{-\frac{\text{Rv}}{2}} I_{\frac{\text{Rv}}{2}}(\text{R1})}{\text{Rv} I_{\frac{\text{Rv}}{2}}(1)+I_{\frac{\text{Rv}}{2}+1}(1)}$$ Wolfram language: C2[R1, T] == 1 - (Rv R1^(-(Rv/2))BesselI[Rv/2, R1])/(RvBesselI[Rv/2, 1] + BesselI[Rv/2 + 1, 1]) Check: Derivative[0, 1][C2][R1, T] == 1 - C2[R1, T] + Derivative[1, 0][C2][R1, T]/R1 + (RvDerivative[1, 0][C2][R1, T])/ R1 + Derivative[2, 0][C2][R1, T] /. C2 -> Function[{R1, T}, 1 - (Rv R1^(-(Rv/2))BesselI[Rv/2, R1])/( RvBesselI[Rv/2, 1] + BesselI[Rv/2 + 1, 1])] // FullSimplify (True) RvC2[1, T] + Derivative[1, 0][C2][1, T] == 0 /. C2 -> Function[{R1, T}, 1 - (Rv R1^(-(Rv/2))BesselI[Rv/2, R1])/( RvBesselI[Rv/2, 1] + BesselI[Rv/2 + 1, 1])] // FullSimplify (True) Regards M.I.

Equation can be solved.I used Maple 2019.2

LateX:

$$\text{C2}(\text{R1},T)=1-\frac{\text{Rv} \text{R1}^{-\frac{\text{Rv}}{2}} I_{\frac{\text{Rv}}{2}}(\text{R1})}{\text{Rv} I_{\frac{\text{Rv}}{2}}(1)+I_{\frac{\text{Rv}}{2}+1}(1)}$$

Wolfram language:

C2[R1, T] == 1 - (Rv R1^(-(Rv/2))*BesselI[Rv/2, R1])/(Rv*BesselI[Rv/2, 1] + BesselI[Rv/2 + 1, 1])

Check:

Derivative[0, 1][C2][R1, T] == 
1 - C2[R1, T] + 
 Derivative[1, 0][C2][R1, T]/R1 + (Rv*Derivative[1, 0][C2][R1, T])/
 R1 + Derivative[2, 0][C2][R1, T] /. 
 C2 -> Function[{R1, T}, 
1 - (Rv R1^(-(Rv/2))*BesselI[Rv/2, R1])/(
 Rv*BesselI[Rv/2, 1] + BesselI[Rv/2 + 1, 1])] // FullSimplify
 (*True*)

 Rv*C2[1, T] + Derivative[1, 0][C2][1, T] == 0 /. 
  C2 -> Function[{R1, T}, 
  1 - (Rv R1^(-(Rv/2))*BesselI[Rv/2, R1])/(
  Rv*BesselI[Rv/2, 1] + BesselI[Rv/2 + 1, 1])] // FullSimplify
 (*True*)

Regards M.I.

POSTED BY: Mariusz Iwaniuk

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