Numerical integration of long fourth order tensor components containing singularities
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I need to evaluate a number of integrals over a unit circle, whereby the integrands are very long fourth order tensor components which are functions of phi but also of other tensor components, i.e. i have a tensor valued function of phi and another tensor:
B1111 (phi,C1312,etc.) = int_0^2*pi[((4*cos(phi)^2*((-1*C1312^2 + C1212*C1313)* cos(phi)^4 + A*(2*C1313*C2212 - 2*C1312*C2213 - 2*C1312*C2312 + 2*C1212*C2313)*cos(phi)^3*sin(phi) + A^3*(-2*C2213*C2223 - 2*C2223*C2312 + 2*C2222*C2313 + 2*C2212*C2323)*cos(phi)sin(phi)^3 + A^4(-1*C2223^2 + C2222*C2323)*sin(phi)^4))/(((C1313*cos(phi)^2 + A^2*C2323*sin(phi)^2)*((-1*C1112^2 + C1111*C1212)*cos(phi)^4 - (C1312*cos(phi)^2 + A^2*C2223*sin(phi)^2)*((-1*C1112*C1113 + C1111*C1312)* cos(phi)^4 + A*(C1113*(-1*C1122 - 1*C1212) + C1112*(-1*C1123 + C1312) + C1111*(C2213 + C2312))*cos(phi)^3*sin(phi) + A^4*(C1212*C2223 - 1*C2212*C2312)sin(phi)^4) - 1(C1113*cos(phi)^2 + A*(C1123 + C1312)*cos(phi)*sin(phi) + A^2*C2312*sin(phi)^2)*((C1113*C1212 - 1*C1112*C1312)cos(phi)^4 + A(C1123*C1212 - 1*C1122*C1312 + 2*C1113*C2212 - 1*C1112.*C2213 - 1*C1112*C2312)*cos(phi)^3*sin(phi) + A^3*((C1123 + C1312)*2 - 1*2*C2223 - 1*2*C2223 + C2212*(-1*C2213 + C2312))*cos(phi)*sin(phi)^3)))]dphi
Such integrals cannot be solved analytically, so I do it numerically. However, certain tensor arguments containing zero valued components lead to singularities in the integrals. I can get close to zero, but I'm interested in an integration solution at which the aforementioned components are exactly zero. An option would be to keep those components that are zero in symbolic form, integrate numerically and then see what the limit of the resulting sum is when the symbolic form components go to zero. There seems to be work on mixed numerical-symbolic integration methods, but I'm not inside those methods, and haven't thought yet of any other approach to solving this either. Can someone help?
integration numerical-methods singularity
asked Jan 21 at 22:44
EuclidesEuclides
11
$endgroup$
add a comment |
$begingroup$
I need to evaluate a number of integrals over a unit circle, whereby the integrands are very long fourth order tensor components which are functions of phi but also of other tensor components, i.e. i have a tensor valued function of phi and another tensor:
B1111 (phi,C1312,etc.) = int_0^2*pi[((4*cos(phi)^2*((-1*C1312^2 + C1212*C1313)* cos(phi)^4 + A*(2*C1313*C2212 - 2*C1312*C2213 - 2*C1312*C2312 + 2*C1212*C2313)*cos(phi)^3*sin(phi) + A^3*(-2*C2213*C2223 - 2*C2223*C2312 + 2*C2222*C2313 + 2*C2212*C2323)*cos(phi)sin(phi)^3 + A^4(-1*C2223^2 + C2222*C2323)*sin(phi)^4))/(((C1313*cos(phi)^2 + A^2*C2323*sin(phi)^2)*((-1*C1112^2 + C1111*C1212)*cos(phi)^4 - (C1312*cos(phi)^2 + A^2*C2223*sin(phi)^2)*((-1*C1112*C1113 + C1111*C1312)* cos(phi)^4 + A*(C1113*(-1*C1122 - 1*C1212) + C1112*(-1*C1123 + C1312) + C1111*(C2213 + C2312))*cos(phi)^3*sin(phi) + A^4*(C1212*C2223 - 1*C2212*C2312)sin(phi)^4) - 1(C1113*cos(phi)^2 + A*(C1123 + C1312)*cos(phi)*sin(phi) + A^2*C2312*sin(phi)^2)*((C1113*C1212 - 1*C1112*C1312)cos(phi)^4 + A(C1123*C1212 - 1*C1122*C1312 + 2*C1113*C2212 - 1*C1112.*C2213 - 1*C1112*C2312)*cos(phi)^3*sin(phi) + A^3*((C1123 + C1312)*2 - 1*2*C2223 - 1*2*C2223 + C2212*(-1*C2213 + C2312))*cos(phi)*sin(phi)^3)))]dphi
Such integrals cannot be solved analytically, so I do it numerically. However, certain tensor arguments containing zero valued components lead to singularities in the integrals. I can get close to zero, but I'm interested in an integration solution at which the aforementioned components are exactly zero. An option would be to keep those components that are zero in symbolic form, integrate numerically and then see what the limit of the resulting sum is when the symbolic form components go to zero. There seems to be work on mixed numerical-symbolic integration methods, but I'm not inside those methods, and haven't thought yet of any other approach to solving this either. Can someone help?
integration numerical-methods singularity
asked Jan 21 at 22:44
EuclidesEuclides
11
$endgroup$
2
$begingroup$
Why don't you write out the integrals? As it stands, it is very difficult to understand your exact problem.
$endgroup$
– Mattos
Jan 21 at 22:50
$begingroup$
It appears that the integrand is a fraction between two polynomials in $sin(phi)$ and $cos(phi)$. I am voting for closure because the reader should not have to struggle to interpret the text.
$endgroup$
– Carl Christian
Jan 23 at 11:43
$begingroup$
yes, I know it is a fraction of two polynomials in sin(phi) and cos(phi), but I guess these are not textbook polynomials as in here: math.stackexchange.com/questions/1182910/…, meaning we can't get antiderivatives by using standard methods for improper integral solving. I tried to solve this symbolically on Matlab, but the machine ran out of memory. So, I guess I always have to integrate numerically (please correct me if I'm wrong). If I'm not wrong, the question of how to get a sum when certain coefficients of such polynomials are exactly zero remains.
$endgroup$
– Euclides
Jan 24 at 13:07
add a comment |
$begingroup$
I need to evaluate a number of integrals over a unit circle, whereby the integrands are very long fourth order tensor components which are functions of phi but also of other tensor components, i.e. i have a tensor valued function of phi and another tensor:
B1111 (phi,C1312,etc.) = int_0^2*pi[((4*cos(phi)^2*((-1*C1312^2 + C1212*C1313)* cos(phi)^4 + A*(2*C1313*C2212 - 2*C1312*C2213 - 2*C1312*C2312 + 2*C1212*C2313)*cos(phi)^3*sin(phi) + A^3*(-2*C2213*C2223 - 2*C2223*C2312 + 2*C2222*C2313 + 2*C2212*C2323)*cos(phi)sin(phi)^3 + A^4(-1*C2223^2 + C2222*C2323)*sin(phi)^4))/(((C1313*cos(phi)^2 + A^2*C2323*sin(phi)^2)*((-1*C1112^2 + C1111*C1212)*cos(phi)^4 - (C1312*cos(phi)^2 + A^2*C2223*sin(phi)^2)*((-1*C1112*C1113 + C1111*C1312)* cos(phi)^4 + A*(C1113*(-1*C1122 - 1*C1212) + C1112*(-1*C1123 + C1312) + C1111*(C2213 + C2312))*cos(phi)^3*sin(phi) + A^4*(C1212*C2223 - 1*C2212*C2312)sin(phi)^4) - 1(C1113*cos(phi)^2 + A*(C1123 + C1312)*cos(phi)*sin(phi) + A^2*C2312*sin(phi)^2)*((C1113*C1212 - 1*C1112*C1312)cos(phi)^4 + A(C1123*C1212 - 1*C1122*C1312 + 2*C1113*C2212 - 1*C1112.*C2213 - 1*C1112*C2312)*cos(phi)^3*sin(phi) + A^3*((C1123 + C1312)*2 - 1*2*C2223 - 1*2*C2223 + C2212*(-1*C2213 + C2312))*cos(phi)*sin(phi)^3)))]dphi
Such integrals cannot be solved analytically, so I do it numerically. However, certain tensor arguments containing zero valued components lead to singularities in the integrals. I can get close to zero, but I'm interested in an integration solution at which the aforementioned components are exactly zero. An option would be to keep those components that are zero in symbolic form, integrate numerically and then see what the limit of the resulting sum is when the symbolic form components go to zero. There seems to be work on mixed numerical-symbolic integration methods, but I'm not inside those methods, and haven't thought yet of any other approach to solving this either. Can someone help?
integration numerical-methods singularity
asked Jan 21 at 22:44
EuclidesEuclides
11
$endgroup$
I need to evaluate a number of integrals over a unit circle, whereby the integrands are very long fourth order tensor components which are functions of phi but also of other tensor components, i.e. i have a tensor valued function of phi and another tensor:
B1111 (phi,C1312,etc.) = int_0^2*pi[((4*cos(phi)^2*((-1*C1312^2 + C1212*C1313)* cos(phi)^4 + A*(2*C1313*C2212 - 2*C1312*C2213 - 2*C1312*C2312 + 2*C1212*C2313)*cos(phi)^3*sin(phi) + A^3*(-2*C2213*C2223 - 2*C2223*C2312 + 2*C2222*C2313 + 2*C2212*C2323)*cos(phi)sin(phi)^3 + A^4(-1*C2223^2 + C2222*C2323)*sin(phi)^4))/(((C1313*cos(phi)^2 + A^2*C2323*sin(phi)^2)*((-1*C1112^2 + C1111*C1212)*cos(phi)^4 - (C1312*cos(phi)^2 + A^2*C2223*sin(phi)^2)*((-1*C1112*C1113 + C1111*C1312)* cos(phi)^4 + A*(C1113*(-1*C1122 - 1*C1212) + C1112*(-1*C1123 + C1312) + C1111*(C2213 + C2312))*cos(phi)^3*sin(phi) + A^4*(C1212*C2223 - 1*C2212*C2312)sin(phi)^4) - 1(C1113*cos(phi)^2 + A*(C1123 + C1312)*cos(phi)*sin(phi) + A^2*C2312*sin(phi)^2)*((C1113*C1212 - 1*C1112*C1312)cos(phi)^4 + A(C1123*C1212 - 1*C1122*C1312 + 2*C1113*C2212 - 1*C1112.*C2213 - 1*C1112*C2312)*cos(phi)^3*sin(phi) + A^3*((C1123 + C1312)*2 - 1*2*C2223 - 1*2*C2223 + C2212*(-1*C2213 + C2312))*cos(phi)*sin(phi)^3)))]dphi
Such integrals cannot be solved analytically, so I do it numerically. However, certain tensor arguments containing zero valued components lead to singularities in the integrals. I can get close to zero, but I'm interested in an integration solution at which the aforementioned components are exactly zero. An option would be to keep those components that are zero in symbolic form, integrate numerically and then see what the limit of the resulting sum is when the symbolic form components go to zero. There seems to be work on mixed numerical-symbolic integration methods, but I'm not inside those methods, and haven't thought yet of any other approach to solving this either. Can someone help?
integration numerical-methods singularity
integration numerical-methods singularity
asked Jan 21 at 22:44
EuclidesEuclides
11
asked Jan 21 at 22:44
EuclidesEuclides
11
edited Jan 21 at 23:48
Euclides
asked Jan 21 at 22:44
EuclidesEuclides
11
asked Jan 21 at 22:44
EuclidesEuclides
11
asked Jan 21 at 22:44
EuclidesEuclides
11
11
2
$begingroup$
Why don't you write out the integrals? As it stands, it is very difficult to understand your exact problem.
$endgroup$
– Mattos
Jan 21 at 22:50
$begingroup$
It appears that the integrand is a fraction between two polynomials in $sin(phi)$ and $cos(phi)$. I am voting for closure because the reader should not have to struggle to interpret the text.
$endgroup$
– Carl Christian
Jan 23 at 11:43
$begingroup$
yes, I know it is a fraction of two polynomials in sin(phi) and cos(phi), but I guess these are not textbook polynomials as in here: math.stackexchange.com/questions/1182910/…, meaning we can't get antiderivatives by using standard methods for improper integral solving. I tried to solve this symbolically on Matlab, but the machine ran out of memory. So, I guess I always have to integrate numerically (please correct me if I'm wrong). If I'm not wrong, the question of how to get a sum when certain coefficients of such polynomials are exactly zero remains.
$endgroup$
– Euclides
Jan 24 at 13:07
add a comment |
2
$begingroup$
Why don't you write out the integrals? As it stands, it is very difficult to understand your exact problem.
$endgroup$
– Mattos
Jan 21 at 22:50
$begingroup$
It appears that the integrand is a fraction between two polynomials in $sin(phi)$ and $cos(phi)$. I am voting for closure because the reader should not have to struggle to interpret the text.
$endgroup$
– Carl Christian
Jan 23 at 11:43
$begingroup$
yes, I know it is a fraction of two polynomials in sin(phi) and cos(phi), but I guess these are not textbook polynomials as in here: math.stackexchange.com/questions/1182910/…, meaning we can't get antiderivatives by using standard methods for improper integral solving. I tried to solve this symbolically on Matlab, but the machine ran out of memory. So, I guess I always have to integrate numerically (please correct me if I'm wrong). If I'm not wrong, the question of how to get a sum when certain coefficients of such polynomials are exactly zero remains.
$endgroup$
– Euclides
Jan 24 at 13:07
2
2
$begingroup$
Why don't you write out the integrals? As it stands, it is very difficult to understand your exact problem.
$endgroup$
– Mattos
Jan 21 at 22:50
$begingroup$
Why don't you write out the integrals? As it stands, it is very difficult to understand your exact problem.
$endgroup$
– Mattos
Jan 21 at 22:50
$begingroup$
It appears that the integrand is a fraction between two polynomials in $sin(phi)$ and $cos(phi)$. I am voting for closure because the reader should not have to struggle to interpret the text.
$endgroup$
– Carl Christian
Jan 23 at 11:43
$begingroup$
It appears that the integrand is a fraction between two polynomials in $sin(phi)$ and $cos(phi)$. I am voting for closure because the reader should not have to struggle to interpret the text.
$endgroup$
– Carl Christian
Jan 23 at 11:43
$begingroup$
yes, I know it is a fraction of two polynomials in sin(phi) and cos(phi), but I guess these are not textbook polynomials as in here: math.stackexchange.com/questions/1182910/…, meaning we can't get antiderivatives by using standard methods for improper integral solving. I tried to solve this symbolically on Matlab, but the machine ran out of memory. So, I guess I always have to integrate numerically (please correct me if I'm wrong). If I'm not wrong, the question of how to get a sum when certain coefficients of such polynomials are exactly zero remains.
$endgroup$
– Euclides
Jan 24 at 13:07
$begingroup$
yes, I know it is a fraction of two polynomials in sin(phi) and cos(phi), but I guess these are not textbook polynomials as in here: math.stackexchange.com/questions/1182910/…, meaning we can't get antiderivatives by using standard methods for improper integral solving. I tried to solve this symbolically on Matlab, but the machine ran out of memory. So, I guess I always have to integrate numerically (please correct me if I'm wrong). If I'm not wrong, the question of how to get a sum when certain coefficients of such polynomials are exactly zero remains.
$endgroup$
– Euclides
Jan 24 at 13:07
add a comment |
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$begingroup$
Why don't you write out the integrals? As it stands, it is very difficult to understand your exact problem.
$endgroup$
– Mattos
Jan 21 at 22:50
$begingroup$
It appears that the integrand is a fraction between two polynomials in $sin(phi)$ and $cos(phi)$. I am voting for closure because the reader should not have to struggle to interpret the text.
$endgroup$
– Carl Christian
Jan 23 at 11:43
$begingroup$
yes, I know it is a fraction of two polynomials in sin(phi) and cos(phi), but I guess these are not textbook polynomials as in here: math.stackexchange.com/questions/1182910/…, meaning we can't get antiderivatives by using standard methods for improper integral solving. I tried to solve this symbolically on Matlab, but the machine ran out of memory. So, I guess I always have to integrate numerically (please correct me if I'm wrong). If I'm not wrong, the question of how to get a sum when certain coefficients of such polynomials are exactly zero remains.
$endgroup$
– Euclides
Jan 24 at 13:07