# A symmetric parabolic second order PDE

Here I want to solve a second order PDE symmetrically depending on two variables $$3(\partial_{\alpha\alpha}f+\partial_{\beta\beta}f-2\partial_{\alpha\beta}f)+2(\cot\alpha)\partial_\alpha f +2(\cot\beta)\partial_\beta f=cf$$ where $$c$$ is a constant.

Please do you see any way of solving it for all $$c$$?

I tried the change of variable $$x=\sin((\alpha-\beta)/2),\, y=\sin((\theta+\delta)/2)$$ and it's transformed into the following PDE $$3(1-x^2)\partial_{xx}g-5x\partial_x g+\frac{2(1-y^2)}{y^2-x^2}(y\partial_y g-x\partial_x g)=cg.$$ It now does not have second order term in $$y$$, but is no longer symmetric. I am not sure how to continue.

• MSE is a right place for such type questions. – user64494 Oct 3 '19 at 15:50

pdsolve(3*(diff(f(alpha, beta), alpha $$2) + diff(f(alpha, beta), beta$$ 2) - 2*diff(diff(f(alpha, beta), alpha), beta)) + 2*cot(alpha)*diff(f(alpha, beta), alpha) + 2*cot(beta)*diff(f(alpha, beta), beta) = c*f(alpha, beta), explicit);