# Does a particular iteration produce a weak solution to a non linear pde?

Consider the following non linear pde in the unknown $v(x,y)$: $$\frac{\partial v(x,y)}{\partial x} + \Big(\frac{\partial v(x,y)}{\partial x} \Big)^2 = e^{2 ty}-1$$ where $t$ is some fixed small non zero real number. Now define the following sequence of functions: $$v_{0}(x,y) : \equiv 0$$ $$v_{n+1}(x,y):= \int_{0}^{x} (e^{2 ty}-1)- \Big(\frac{\partial v_n(x^{\prime},y)}{\partial x^{\prime}} \Big)^2 d x^{\prime}$$

My question is the following: is there any hope of saying that this sequence "converges" in any reasonable sense, for small $t$? More precisely, the idea is as follows: Is there some appropriate "weighted Sobolev space" in which these functions $v_{n}(x,y)$ live; and hopefully with respect to that weighted norm, the sequence is a Cauchy sequence?

Note that it can not be the usual Sobolev space, because the function $e^{2t y}$ does not have finite $L^p$ norm.

For the time being I am not asking whether the "limit" $v$, if it exists is smooth or not. Right now all I am looking for is a "weak" solution. Eventually of course I would like to know about the regularity of $v$. But that is a meaningful question only if $v$ exists.

Every thing is over the real numbers, the functions $v(x,y)$ and $v_{n}(x,y)$ are supposed to be defined on the whole of $\mathbb{R}^2$ (at least they should be measurable with respect to the right weight). I am not imposing any boundary conditions; I am simply looking for some solution of the pde.

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Just a remark: Any reasonable solution should look like $f(y) + a_yx$ where $a_y$ is a solution to $z^2 + z = e^{2ty}-1$, since the equation only imposes a condition on the $x$ derivative. –  Connor Mooney Aug 20 '13 at 12:58

This is $$(\partial_x\nu+\frac12)^2=\frac14+e^{2ty}-1=e^{2ty}-\frac34.$$ The parameter $ty$ should be chosen so that $2ty\ge \ln 3-\ln 4$, and then a solution is given by $$\partial_x\nu+\frac12=\alpha,\quad \alpha^2=e^{2ty}-\frac34,$$ e.g. $\nu=(\alpha -\frac12)x+\phi(y).$