I am trying to evaluate the integral
\int_0^{2\pi}\frac{D_{11}^2+2D_{12}^2+D_{22}^2+D_{33}^2}{1+e\cos x(t)}dx
Where D_{ij} are third derivatives wrt t as defined below.
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];
To do this I have written the input
Assuming[Element[k1, Reals] && Element[e, Reals] &&
x'[t] == p2/(d[x[t]]^2) && Element[p2, Reals] &&
Element[p2, Reals],
Integrate[(D11^2 + 2*D12^2 + D22^2 + D33^2)/(1 + e*Cos[x[t]])^2, {x[
t], 0, 2*Pi}]]
This is taking a very long time to evaluate (has been running for hours), is it OK to specify to Mathematica that I have the condition \dot{x}=\frac{p_2}{d(x(t))^2} in this manner? How could this be sped up?
This integral arose in the study of the loss of energy of an orbiting mass due to gravitational radiation.
trounev
(Alexander Trounev)
October 14, 2018, 1:40pm
2
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)}
timmy2x
(Tim Laska)
October 14, 2018, 3:04pm
3
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)} *)
trounev
(Alexander Trounev)
October 14, 2018, 5:11pm
4
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.