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Question: I am trying to solve a set of simultaneous pde and am getting solutions in the form of unevaluated integrals. I can manually integrate them but was wondering if there was a way to do that automatically.

Question:I am trying to solve a set of simultaneous pde and am getting solutions in the form of unevaluated integrals. I can manually integrate them but was wondering if there was a way to do that automatically.

Rakshak 150 Maple
integral pde pdsolve differential-equations duplicate-question
October 14 2022
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eq1 := -6*sin(theta)*(cos(theta)^2*(diff(n(r, theta), r))*a^3+cos(theta)^2*a^4+3*cos(theta)^2*a^2*r^2+(diff(n(r, theta), r))*a*r^2+2*r^4-n(r, theta)*a*r)/(r^2+cos(theta)^2*a^2)^(3/2) = 0

-6*sin(theta)*(cos(theta)^2*(diff(n(r, theta), r))*a^3+cos(theta)^2*a^4+3*cos(theta)^2*a^2*r^2+(diff(n(r, theta), r))*a*r^2+2*r^4-n(r, theta)*a*r)/(r^2+cos(theta)^2*a^2)^(3/2) = 0

 (1)
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eq2 := -(6*(cos(theta)^2*sin(theta)*(diff(n(r, theta), theta))*a^3+a^4*r*cos(theta)+2*cos(theta)*a^2*r^3+cos(theta)*r^5+n(r, theta)*a^3*cos(theta)+cos(theta)*n(r, theta)*a*r^2+sin(theta)*(diff(n(r, theta), theta))*a*r^2))/(r^2+cos(theta)^2*a^2)^(3/2) = 0

-6*(cos(theta)^2*sin(theta)*(diff(n(r, theta), theta))*a^3+a^4*r*cos(theta)+2*cos(theta)*a^2*r^3+cos(theta)*r^5+n(r, theta)*a^3*cos(theta)+cos(theta)*n(r, theta)*a*r^2+sin(theta)*(diff(n(r, theta), theta))*a*r^2)/(r^2+cos(theta)^2*a^2)^(3/2) = 0

 (2)
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pdsolve([eq1, eq2])

{n(r, theta) = (1/2)*(4*r^2+2*a^2*cos(2*theta)+2*a^2)^(1/2)*((Int(-2*((a^4+3*a^2*r^2)*cos(2*theta)+a^4+3*r^2*a^2+4*r^4)/((4*r^2+2*a^2*cos(2*theta)+2*a^2)^(1/2)*(a^2*cos(2*theta)+a^2+2*r^2)*a), r))*sin(theta)+Int(5*(a^2*(-(1/5)*(a^2+4*r^2)^2*cos(3*theta)+(-(3/5)*a^4-(8/5)*r^2*a^2)*cos(5*theta)-(1/5)*a^4*cos(7*theta)+cos(theta)*(a^4+(16/5)*r^2*a^2+(16/5)*r^4))*(4*r^2+2*a^2*cos(2*theta)+2*a^2)^(1/2)*(Int(16*((a^2+3*r^2)*cos(2*theta)+a^2-r^2)/(4*r^2+2*a^2*cos(2*theta)+2*a^2)^(5/2), r))-(16/5)*(a^2+r^2)*((-(1/2)*a^2-2*r^2)*cos(3*theta)-(1/2)*a^2*cos(5*theta)+cos(theta)*(a^2+2*r^2))*r)*a/((4*r^2+2*a^2*cos(2*theta)+2*a^2)^(1/2)*((32*a^4+64*a^2*r^2)*cos(2*theta)+8*a^4*cos(4*theta)+24*a^4+64*r^2*a^2+64*r^4)), theta)+_C1)/sin(theta)}, {n(r, theta) = -(a^2*r+r^3)/a}

 (3)
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