Is $U_{x} = - U_{yy}$ solvable? Can someone solve this pde?  
$U_{x} = - U_{yy}$ with $x>0$ and $-\infty < y < \infty$
Initial condition: $U(0,y) = f(y)$
This follows from a mathematical model and I am not interested in the physical interpretation.
 A: Since we are dealing with an infinite domain, I hope you are familiar with Fourier transform. Let 
$$\hat{u}(k,x) = \mathcal{F}\{u\}=\int_{-\infty}^\infty u(x,y) e^{-iky} dy.$$
Using the properties of Fourier transform we arrive to
$$
\frac{\partial}{\partial x} \hat{u} = k^2 \hat{u},
$$
whose solution is
$$
\hat{u} = C(k) e^{k^2x}.
$$
From the initial condition, we have $\hat{u} = \hat{f}(k) e^{k^2x}$, and the solution for $u$ is
$$
u(x,y) = -\frac{i}{\sqrt{4\pi x}} \int_{-\infty}^\infty f(y') e^{\frac{(y-y')^2}{4x}} dy'.
$$
For the particular case in which the initial condition is the delta function $u(0,y) = \delta(y)$, the solution is
$$
u(x,y) = -i\frac{e^{\frac{y^2}{4x}}}{\sqrt{4\pi x}}.
$$
What's the meaning of this? The physical interpretation of the problem is precisely the heat diffusion but with time going backward. Let $-x=t$, and the equation is
$$
\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial y^2},
$$
in the domain $t<0$. Therefore, the solution 'explodes' because it is the opposite of the phenomena of diffusion, it is, the heat will 'concentrate' in a region instead of to dissipate to the entire domain. Furthermore, the solution is imaginary because there is no real solution of which could be before the Dirac delta. If it was a regular heat diffusion problem, the heat would be totally concentred in the origin and then dissipate to the rest of the domain. If the opposite happens, it will concentrate from the entire domain to a single point, but it can't go further than this, and this fact is reflected in the inexistence of a real solution.
Instead of a non-smooth function, let's assume that the initial condition is a gaussian $u(0,y)=\exp -y^2$. In that case,
$$
\int_{-\infty}^\infty \exp -y'^2 e^{\frac{(y-y')^2}{4x}} dy' = \sqrt{\frac{\pi x}{4x-1}} \exp \frac{y^2}{4x-1} \mathrm{erf} \left(\frac{(4x-1)y'+y}{2 \sqrt{x(4x-1)}} \right)_{-\infty}^\infty=
$$
$$
-2i\sqrt{\frac{\pi x}{1-4x}} \exp \frac{y^2}{4x-1},
$$
and our solution is
$$
u(x,y) = \frac{e^{\frac{y^2}{4x-1}}}{\sqrt{1-4x}}.
$$
See that this is a real solution, which can be explained by the fact that the initial condition can 'un-diffuse' until a singular distribution, that occurs at $x=1/4$.
Apparently, if the initial condition is smooth (perhaps a mathematician could say how much smooth) you will have a valid solution for $0<x<x_s$ (in which $x_s$ is the $x$ in which the singularity occurs) for this case in which we assumed (implicitly with the Fourier transform) that $u=0$ for $|y| \to \infty$. As Robert Israel pointed, solutions like $u(x,y)=\exp(x)\cos(y)$ or $u=\exp(y-x)$ also satisfy the equation and do not have singularity.
