# How can I solve a certain complicated second-order PDE?

I am trying to solve the following complicated second-order PDE

eqn =
(D[D[U[x, y], x], x] - 1/x D[U[x, y], x] - 1/x D[D[U[x, y], y], y])^2 +
4(-D[1/x D[U[x, y], y], x])^2
==
(2 C q + (D[D[U[x, y], x], x] + 1/x D[U[x, y], x] +
1/x^2 D[D[U[x, y], y], y]) p)^2;

DSolve[eqn, U[x, y], {x, y}]


where p, q and C are the constant values.

Unfortunately, Mathematica doesn't solve the equation and just returns the DSolve expression unevaluated. Any idea or comments to solve the equation would be appreciated.

• As per Wolfram tutorial, DSolve can find the general solution for a RESTRICTED type of homogeneous linear second-order PDEs, so I guess Mathematica can only solve the three basic types of PDEs (elliptic, hyperbolic, and parabolic). – Kevin Aug 28 '18 at 2:58
• More precisely, DSolve attempts to solve differential equations by applying known methods. It is not able to develop and employ previously unknown methods. Sadly, it also sometimes fails to solve differential equations even when they can be solved by known methods. – bbgodfrey Sep 1 '18 at 1:35

Here is one solution, although admittedly not the most general one. Suppose that U is a function of x only:

eqn /. U -> Function[{x, y}, f[x]];
DSolve[%, f[x], x]
(* {{f[x] -> -((C q x^2)/(2 p)) +
1/2 (1 - p) x (x - p x)^(-((1 + p)/(-1 + p))) C + C},
{f[x] -> (-C q x^2 + p ((1 + p) x)^(2/(1 + p)) C)/(2 p) + C}} *)


Similarly, a solution can be obtained depending on y only,

eqn /. U -> Function[{x, y}, g[y]];
DSolve[%7, g[y], y]


The result is a bit long to be reproduced here.

• Hi bbgodfrey, thank you so much for your solution and valuable time; appreciate that! U in the PDE is a biharmonic stress function around a tunnel for a cavity expansion in compressible soil under biaxial stress field, so U cannot be assumed to be only a function of x. Assuming U as a function of x leads to shear stresses of zero, which is not correct for the compressible materials. Any other idea or comments would be appreciated! Thanks again bbgodfrey! – Kevin Aug 30 '18 at 0:40
• Would a solution of the form f[x] + g[y] be helpful? – bbgodfrey Sep 1 '18 at 1:43