# Is there any solution for this PDE system?

‎Let $(\mathbb{R}^2,\langle‎ .‎,.\rangle)$ be the Euclidean space and define the almost complex structure $J_{\delta,\beta}:TT\mathbb{R}^2\longrightarrow TT\mathbb{R}^2$ with‎ ‎\begin{align}‎ ‎J_{\delta,\beta}(X^h)=\beta X^h‎ +‎\alpha X^v\\‎ ‎J_{\delta,\beta}(X^v)=-\beta X^v‎ -‎\delta X^h‎, ‎\end{align}‎ ‎where $X^h,X^v$ are the horizontal and vertical lifts of the vector $X\in T\mathbb{R}^2$ and $\alpha‎ , ‎\delta‎, ‎\beta‎ : ‎T\mathbb{R}^2 \longrightarrow \mathbb{R}$ are functions satisfy in $\alpha \delta‎ - ‎\beta ^2=1$‎. ‎Then $J_{\delta‎, ‎\beta}$ is integrable if and only if $\delta‎ , ‎\beta$ satisfy in the expressed PDE when $d\beta \neq 0$‎.

Let $δ(x^1,x^2,y^1,y^2)$ and $β(x^1,x^2,y^1,y^2)$ be two functions. Are there $δ$ and $β$ which satisfy in the following PDE system?

\begin{align} \frac{\partial \beta}{\partial y^i}-\frac{\beta}{\delta}\frac{\partial \delta}{\partial y^i}=0 \hspace{1cm}i=1,2 \end{align}

\begin{align} \frac{\partial \delta}{\partial y^i}=-\delta ^2 \hspace{1cm}i=1,2 \end{align} \begin{align} \frac{\partial \beta}{\partial x^i}-\frac{\beta}{\delta}\frac{\partial \delta}{\partial x^i}=1 \hspace{1cm}i=1,2 \end{align} \begin{align} \frac{\partial \delta}{\partial x^i}=\beta \delta \hspace{1cm}i=1,2 \end{align}

• You should give some background as to why you want to solve this. Jan 22, 2016 at 15:48
• @ChrisRamsey Sure! I'll add some explanations ...
– Jack
Jan 22, 2016 at 15:50

Your system is inconsistent. For simplicity, fix $x_2, y_2$ and just consider the dependence on $x_1, y_1$ which I'll write as $x,y$. From equation (2), $$\delta(x, y) = \dfrac{1}{y + c(x)}$$ Then from equation (1), $$\beta(x,y) = \dfrac{d(x)}{y + c(x)}$$ From equation (4) we get $$d(x) = - c'(x)$$ and then equation (3) becomes $$- \dfrac{c''(x)}{y+c(x)} = 1$$ which is impossible.