R$, respectively. The question is whether there exists a solution of $\mathcal P$ that is continuous and nonvanishing on the entire $\Omega$. Suppose that this is the case, and that $u=\varphi(t,x)$ is such a solution. Define the function $\zeta(t):=\varphi(t,-1)$, $t\in\mathbb R$. We have $\zeta\in {\rm C}(\mathbb R)$ and $$\varphi(t,x)=
\begin{cases}
\zeta(t)\exp\left(\dfrac1t\arctan\dfrac xt+\dfrac\pi{2|t|}-\dfrac{\arctan t}t\right) \quad\mbox{if}\quad t\ne0,\ x\in\mathbb R,
\\[2.5ex]
\zeta(0)\exp\left(-\dfrac1x-1\right) \quad\mbox{if}\quad t=0,\ x\in\mathbb R_-.
\end{cases}$$ Here we use the equality $$\begin{gathered}
\label{eq:IdentityWithArctan}
\dfrac1t\arctan\dfrac xt-\dfrac\pi{2|t|}{\mathop{\rm sgn}\nolimits}x=-\dfrac1t\arctan\dfrac tx \quad\mbox{if}\quad t\ne0,\ x\ne0.\end{gathered}$$ The right hand side function is continuous on $\mathbb R^2\setminus\big(\{0\}\times[0,+\infty)\big)$ but cannot be continuously extended to $\Omega$ if $\zeta(0)\ne0$ since for $x>0$ and $t\to0$ we obtain $$\zeta(t)\exp\left(\dfrac1t\arctan\dfrac xt+\dfrac\pi{2|t|}-\dfrac{\arctan t}t\right)=
\zeta(t)\exp\left(-\dfrac1t\arctan\dfrac tx+\dfrac\pi{ |t|}-\dfrac{\arctan t}t\right)\to\infty,$$ where the sign of infinity coincides with the sign of $\zeta(0)$. In other words, the equation $\mathcal P$ has *no (continuous) solution that is nonzero on the entire domain* $\Omega$. Moreover, any solution of this equation on $\Omega$ vanishes on the half-axis $\{0\}\times(-\infty,0)$, and the corresponding function $\zeta$ is $O(e^{-\pi/|t|})$ as $t\to 0$. Consider the solution $$\varphi^1(t,x)=
\begin{cases}
\exp\left(\dfrac1t\arctan\dfrac xt-\dfrac\pi{2|t|}+\dfrac{\arctan t}t\right) \quad\mbox{if}\quad t\ne0,\ x\in\mathbb R,
\\[2.5ex]
\exp\left(\dfrac1x-1\right) \quad\mbox{if}\quad t=0,\ x\in\mathbb R_+,
\\[2.5ex]
0 \quad\mbox{if}\quad t=0,\ x\in\mathbb R_-.
\end{cases}$$ Since $\varphi^1(t,1)=1$, any solution $\varphi$ of the equation $\mathcal P$ on 