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How should I approach to obtain an analytical or approximate analytical solution of an equation of the form $$a_1f_{xxy} + a_2f_{xx} + a_3f_{yy} + a_4f_y + a_5f + a_6 = 0$$ with $a_1, a_2, a_3, a_4, a_5, a_6 \in\Bbb{R}$ ?

Please note that I do not require a numerical solution. Thank you

Edit: Boundary conditions $$y=0,f=0$$ $$y=0,\frac{\partial f}{\partial y}=0$$ $$x=0, f=1$$ $$x=1, f=0$$

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  • $\begingroup$ What are the boundary conditions? $\endgroup$
    – user121049
    Aug 2, 2017 at 18:54

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Solving it for general solution is not difficult, even with huge formulas.

As usual the big difficulty comes after, when the arbitrary parameters and/or functions involved in the formulas have to be determined according to some boundary conditions. Fortunately, no boundary condition is specified in the wording of the question, which now avoid a lot of work to fully solve the problem.

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  • $\begingroup$ I have added the boundary conditions as suggested by you. Can you go through it once? Any help would be largely appreciated $\endgroup$
    – Apurba Roy
    Aug 3, 2017 at 16:55
  • $\begingroup$ I never suggested to add the boundary conditions. On the contrary, without boundary condition, it was possible to answer to your question which was : "How should I approach to obtain an analytical or approximate analytical solution?". The answer provides not only an analytical solution, but an infinity of exact solutions. If you add some boundary conditions, you change a lot the question and make it more difficult, but also more interesting. $\endgroup$
    – JJacquelin
    Aug 4, 2017 at 4:21
  • $\begingroup$ With the specified boundary conditions, the problem becomes even more interesting since the conditions $f(x,0)=0$ and $f(0,y)=1$ introduce a special point $f(0,0)=0=1$. To overcome the apparent contradiction, one have to consider complex values of $\lambda$ so that sinusoidal terms appear into the solutions. This allows to express them in term of Fourier series, likely to fit the boundary conditions on a limited range. I let to someone else the pleasure to solve such a so interesting problem. Good luck. $\endgroup$
    – JJacquelin
    Aug 4, 2017 at 4:30
  • $\begingroup$ Thank you for all your help and time. $\endgroup$
    – Apurba Roy
    Aug 4, 2017 at 4:59

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