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Speed up evaluation of this integral?

tom ri

Posted 7 years ago

POSTED BY: tom ri

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Alexander Trounev

Alexander Trounev, A&E Trounev IT Consulting

Posted 7 years ago

It takes less than a minute. But first you need to simplify the expression a[psi_] := k1/(1 + eCos[psi]); d[psi_] := p1/(1 + eCos[psi]); D11 = D[a[x[t]](Cos[x[t]]^2 - 1/3), {t, 3}]; D12 = D[a[x[t]]Cos[x[t]]Sin[x[t]], {t, 3}]; D22 = D[a[x[t]](Sin[x[t]]^2 - 1/3), {t, 3}]; D33 = D[-a[x[t]]/3, {t,3}]; x'[t] = p2/d[x[t]]^2 x''[t] = D[x'[t], t] x'''[t] = D[x'[t], t, t] Et = (D11^2 + 2D12^2 + D22^2 + D33^2)/(1 + eCos[x[t]])^2 //FullSimplify Integrate[Et, {x[t], 0, 2*Pi}] //AbsoluteTiming {2.57266, ((1536 + 20320 e^2 + 35616 e^4 + 10988 e^6 + 315 e^8) k1^2 p2^6 \[Pi])/(24 p1^12)}

It takes less than a minute. But first you need to simplify the expression

a[psi_] := k1/(1 + e*Cos[psi]);
d[psi_] := p1/(1 + e*Cos[psi]);
D11 = D[a[x[t]]*(Cos[x[t]]^2 - 1/3), {t, 3}];
D12 = D[a[x[t]]*Cos[x[t]]*Sin[x[t]], {t, 3}];
D22 = D[a[x[t]]*(Sin[x[t]]^2 - 1/3), {t, 3}];
D33 = D[-a[x[t]]/3, {t,3}];


x'[t] = p2/d[x[t]]^2

x''[t] = D[x'[t], t]

x'''[t] = D[x'[t], t, t]

Et = (D11^2 + 2*D12^2 + D22^2 + D33^2)/(1 + e*Cos[x[t]])^2 //FullSimplify
Integrate[Et, {x[t], 0, 2*Pi}] //AbsoluteTiming
{2.57266, ((1536 + 20320 e^2 + 35616 e^4 + 10988 e^6 + 
        315 e^8) k1^2 p2^6 \[Pi])/(24 p1^12)}

POSTED BY: Alexander Trounev

Tim Laska

Tim Laska, A Priori, Ltd

Posted 7 years ago

Very insightful. There is a problem that the OP's numerator is not dimensionally consistent. The D33 term is the partial derivative with respect to nothing, which bothered me. Presuming that D33 should also be the third derivative with respect to t, your approach is even faster and contains no conditional expression. a[psi_] := k1/(1 + eCos[psi]); d[psi_] := p1/(1 + eCos[psi]); D11 = D[a[x[t]](Cos[x[t]]^2 - 1/3), {t, 3}]; D12 = D[a[x[t]]Cos[x[t]]Sin[x[t]], {t, 3}]; D22 = D[a[x[t]](Sin[x[t]]^2 - 1/3), {t, 3}]; (D33=D[-a[x[t]]/3])(* From the OP ) D33 = D[-a[x[t]]/3, {t, 3}]; x'[t] = p2/d[x[t]]^2; x''[t] = D[x'[t], t]; x'''[t] = D[x'[t], t, t]; Et = (D11^2 + 2D12^2 + D22^2 + D33^2)/(1 + eCos[x[t]])^2 // FullSimplify Integrate[Et, {x[t], 0, 2Pi}] // AbsoluteTiming (* {2.819966958116478`,((1536+20320 e^2+35616 e^4+10988 e^6+315 e^8) \ k1^2 p2^6 \[Pi])/(24 p1^12)} *)

Very insightful. There is a problem that the OP's numerator is not dimensionally consistent. The D33 term is the partial derivative with respect to nothing, which bothered me. Presuming that D33 should also be the third derivative with respect to t, your approach is even faster and contains no conditional expression.

a[psi_] := k1/(1 + e*Cos[psi]);
d[psi_] := p1/(1 + e*Cos[psi]);
D11 = D[a[x[t]]*(Cos[x[t]]^2 - 1/3), {t, 3}];
D12 = D[a[x[t]]*Cos[x[t]]*Sin[x[t]], {t, 3}];
D22 = D[a[x[t]]*(Sin[x[t]]^2 - 1/3), {t, 3}];
(*D33=D[-a[x[t]]/3]*)(* From the OP *)
D33 = D[-a[x[t]]/3, {t, 3}];

x'[t] = p2/d[x[t]]^2;

x''[t] = D[x'[t], t];

x'''[t] = D[x'[t], t, t];

Et = (D11^2 + 2*D12^2 + D22^2 + D33^2)/(1 + e*Cos[x[t]])^2 // 
  FullSimplify
Integrate[Et, {x[t], 0, 2*Pi}] // AbsoluteTiming

(* {2.819966958116478`,((1536+20320 e^2+35616 e^4+10988 e^6+315 e^8) \
k1^2 p2^6 \[Pi])/(24 p1^12)} *)

POSTED BY: Tim Laska

Alexander Trounev

Alexander Trounev, A&E Trounev IT Consulting

Posted 7 years ago

Thank you, I did not even notice the error in the definition of D3. After correcting the code, the result has changed. I also changed my post.

POSTED BY: Alexander Trounev

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