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Aslam Kittur

Why does the DSolve not solve the PDE giving the 'Arbitrary functions'?

Aslam Kittur, University of Alberta

Posted 1 month ago

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Hello, I have two PDEs (strainDisp11 & strainDisp22) in 2 variables x1 and x2. strainDisp11 is a PDE with the partial differential term in x1 whereas, strainDisp22 is a PDE with the partial differential term in x2 I am trying to solve these two PDEs separately using DSolve (Last two command lines in the attached file), however, the solution is not generated along with the required arbitrary functions C1[1] which should be f1[x2] and C1[1] which should be f2[x1] in the respective solutions of the PDEs. Attached is Notebook for your reference. Appreciate your help. In[]:= Clear["Global`"](ExactAnalyticalusingAiry(Real)StressPotentialFunctions))(Case-1:End-LoadedCantileverin2DPlane-Stressb<<a)endLoad= P end ;height=2a;width=b;length=L; σ 13 =0; σ 23 =0; σ 33 =0; σ 31 =0; σ 32 =0;(MaterialConstantsApplicabletoLinearElastic,Isotropic)poissonR=ν;modulusE=;λ=(ν)/((1+ν)(1-2ν));μ=/(2(1+ν));(BodyForce)Ω=0;(Airy'sFunction)ϕ=phi;phi=-(3endLoad x 1 * x 2 /(4(height/2)width))+(endLoad* x 1 ( x 2 ^3)/(4((height/2)^3)width))(StessesDefinedFromAiry'sFunction) σ 11 =D[ϕ,{ x 2 ,2}]-Ω; σ 22 =D[ϕ,{ x 1 ,2}]-Ω; σ 12 =D[ϕ, x 1 , x 2 ]; σ 21 =D[ϕ, x 1 , x 2 ];(TheStressTensor)stress={{ σ 11 , σ 12 , σ 13 },{ σ 21 , σ 22 , σ 23 },{ σ 31 , σ 32 , σ 33 }};(TheConstitutiveEquations-Hooke'sLawForIsotropicSolid)strain=((1+poissonR/modulusE))stress-(poissonR/modulusE)( σ 11 + σ 22 + σ 33 )IdentityMatrix[3]; ϵ 11 =strain[[1,1]] ϵ 22 =strain[[2,2]] ϵ 33 =strain[[3,3]]; ϵ 12 =strain[[1,2]] ϵ 23 =strain[[2,3]]; ϵ 13 =strain[[1,3]]; ϵ 21 =strain[[2,1]]; ϵ 32 =strain[[3,2]]; ϵ 31 =strain[[3,1]];(CheckForTheStrainCompatabilityIn2D)(TheStrain-DisplacementPDEquationsForLinearElasticity)strainDisp11=D[u1[ x 1 , x 2 ], x 1 ]- ϵ 11 strainDisp22=D[u2[ x 1 , x 2 ], x 2 ]- ϵ 22 strainDisp12=1/2( ∂ x 2 u1[ x 1 , x 2 ]+ ∂ x 1 u2[ x 1 , x 2 ])- ϵ 12 (SolvingtheDifferentialEquationforDisplacements*)solDisp11=DSolve[strainDisp110,u1[ x 1 , x 2 ],{ x 1 , x 2 }]solDisp22=DSolve[strainDisp220,u2[ x 1 , x 2 ],{ x 1 , x 2 }] Out[]= - 3 P end x 1 x 2 4ab + P end x 1 3 x 2 4 3 a b Out[]= - 3ν P end x 1 x 2 2 3 a b + 31+ ν   P end x 1 x 2 2 3 a b Out[]= - 3ν P end x 1 x 2 2 3 a b Out[]= 1+ ν  - 3 P end 4ab + 3 P end 2 x 2 4 3 a b Out[]= 3ν P end x 1 x 2 2 3 a b - 31+ ν   P end x 1 x 2 2 3 a b + (1,0) u1 [ x 1 , x 2 ] Out[]= 3ν P end x 1 x 2 2 3 a b + (0,1) u2 [ x 1 , x 2 ] Out[]= -1+ ν  - 3 P end 4ab + 3 P end 2 x 2 4 3 a b + 1 2  (0,1) u1 [ x 1 , x 2 ]+ (1,0) u2 [ x 1 , x 2 ] DSolve :To avoid possible ambiguity, the arguments in  3ν P end x 1 x 2 2 3 a b - 31+ ν   P end x 1 x 2 2 3 a b + (1,0) u1 [ x 1 , x 2 ]0 should literally match the independent variables. Out[]= DSolve 3ν P end x 1 x 2 2 3 a b - 31+ ν   P end x 1 x 2 2 3 a b + (1,0) u1 [ x 1 , x 2 ]0,u1[ x 1 , x 2 ],{ x 1 , x 2 } DSolve :To avoid possible ambiguity, the arguments in  3ν P end x 1 x 2 2 3 a b + (0,1) u2 [ x 1 , x 2 ]0 should literally match the independent variables. Out[]= DSolve 3ν P end x 1 x 2 2 3 a b + (0,1) u2 [ x 1 , x 2 ]0,u2[ x 1 , x 2 ],{ x 1 , x 2 }

POSTED BY: Aslam Kittur

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Mariusz Iwaniuk

Mariusz Iwaniuk, engineer,hobbyist.

Posted 1 month ago

A Tip: Don't use Subscript , because causes problems.

POSTED BY: Mariusz Iwaniuk

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Aslam Kittur

Aslam Kittur, University of Alberta

Posted 1 month ago

Thanks! Very much appreciated.

POSTED BY: Aslam Kittur

Answer

Aslam Kittur

Aslam Kittur, University of Alberta

Posted 11 days ago

Hello, I have two PDEs in 2 variables 'r' and 'theta'. I am trying to solve these two PDEs separately using DSolve (The last two command lines in the attached file). The solution is generated as expected for the 1st PDE (Integration with respect to variable 'r'), however, the solution is not generated for the 2nd PDE (Integration with respect to 'theta'). I cannot understand why Mathematica does not solve all the terms and has replaced 'theta' by K[1] in the unsolved integral with limits? Attached is Notebook for your reference. Appreciate your help.

POSTED BY: Aslam Kittur

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Mariusz Iwaniuk

Mariusz Iwaniuk, engineer,hobbyist.

Posted 11 days ago

Maybe: solDispRR = DSolve[strainDispRR == 0, uR, {r, \[Theta]}] // Flatten; solDisp\[Theta]\[Theta] = DSolve[strainDisp\[Theta]\[Theta] == 0, u\[Theta], {r, \[Theta]}] // Flatten; uRFunctionTemp = uR[r, \[Theta]] /. solDispRR[[1]] u\[Theta]FunctionTemp = (u\[Theta][r, \[Theta]] /. solDisp\[Theta]\[Theta][[1]] /. solDispRR[[1]]) // Activate // ExpandAll Looks like MMA can't integrate, a workaround: u\[Theta]FunctionTemp = (Integrate[#, {K[1], 1, \[Theta]}] & /@ (u\[Theta]FunctionTemp[[1, 1]])) + u\[Theta]FunctionTemp[[2]] (Integrate[-C[1][K[1]], {K[1], 1, \[Theta]}] + (2P\[Nu]^2Log[r](Sin[1] - Sin[\[Theta]]))/(Pi\[DoubleStruckCapitalE]) + (2P\[Nu](-Sin[1] + Sin[\[Theta]]))/(Pi\[DoubleStruckCapitalE]) + (2P\[Nu]^2(-Sin[1] + Sin[\[Theta]]))/(Pi\[DoubleStruckCapitalE]) + (2PLog[r](-Sin[1] + Sin[\[Theta]]))/(Pi\[DoubleStruckCapitalE]) + C[1][r]*) In this line: Integrate[-C[1][K[1]], {K[1], 1, \[Theta]}] what answer do you expect?

Maybe:

solDispRR = DSolve[strainDispRR == 0, uR, {r, \[Theta]}] // Flatten;
solDisp\[Theta]\[Theta] = 
  DSolve[strainDisp\[Theta]\[Theta] == 0, u\[Theta], {r, \[Theta]}] //
    Flatten;
uRFunctionTemp = uR[r, \[Theta]] /. solDispRR[[1]]
u\[Theta]FunctionTemp = (u\[Theta][r, \[Theta]] /. 
      solDisp\[Theta]\[Theta][[1]] /. solDispRR[[1]]) // Activate // 
  ExpandAll

Looks like MMA can't integrate, a workaround:

u\[Theta]FunctionTemp = (Integrate[#, {K[1], 
       1, \[Theta]}] & /@ (u\[Theta]FunctionTemp[[1, 1]])) + 
  u\[Theta]FunctionTemp[[2]]

 (*Integrate[-C[1][K[1]], {K[1], 1, \[Theta]}] + (2*P*\[Nu]^2*Log[r]*(Sin[1] - Sin[\[Theta]]))/(Pi*\[DoubleStruckCapitalE]) + 
  (2*P*\[Nu]*(-Sin[1] + Sin[\[Theta]]))/(Pi*\[DoubleStruckCapitalE]) + (2*P*\[Nu]^2*(-Sin[1] + Sin[\[Theta]]))/(Pi*\[DoubleStruckCapitalE]) + 
  (2*P*Log[r]*(-Sin[1] + Sin[\[Theta]]))/(Pi*\[DoubleStruckCapitalE]) + C[1][r]*)

In this line:

Integrate[-C[1][K[1]], {K[1], 1, \[Theta]}]

what answer do you expect?

POSTED BY: Mariusz Iwaniuk

Answer

Aslam Kittur

Aslam Kittur, University of Alberta

Posted 11 days ago

The expression is not complicated So as I want the DSolve (Integration with respect to Theta) to result in three terms as follows; But, I cannot understand why is MMA causing a definite integral with limits [1 to Theta], besides changing the variable 'Theta' to 'K[1]'

POSTED BY: Aslam Kittur

Answer

Aslam Kittur

Aslam Kittur, University of Alberta

Posted 11 days ago

Actually, to not confuse... the below is problem defined as PDE The required solution for uTheta, upon Integrating with respect to Theta, we should get

POSTED BY: Aslam Kittur

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