I want to find solutions $u$ to the equation $$b(x,y,z)\cdot\nabla u(x,y,z)=0\quad (x,y,z)\in D$$ $$u(1,0,z)=1,\, u(0,y,0)=0$$ where $D$ is the domain (a simplex) defined by $$D=\{(x,y,z)\in[0,1]^3: 0\leq x+y\leq 1,\quad 0\leq z\leq x\}$$ The vector field $b$ is given by $$b(x,y,z)=\begin{pmatrix}\delta z(1-x)-\beta xy\\\alpha(1-x)-(\alpha+\beta)y\\\gamma x-(\gamma+\delta)z\end{pmatrix}$$ where $0<\alpha,\beta,\gamma,\delta<1$ are constants. Using the method of characteristic, one has to solve the system $$(\dot{x}(s),\dot{y}(s),\dot{z}(s))^T=b(x(s),y(s),z(s))$$ which was the content of my related question. This system has two fixed points $(0,\frac{\alpha}{\alpha+\beta},0)$ and $(1,0,\frac{\gamma}{\gamma+\delta})$. Any interesting solution of this ODE system can be extended to move from one fixed point to the other. Now the method of characteristics would give $\dot{u}(x(s),y(s),z(s))=0$, so $u$ is constant along characteristic curves.
Questions: This seems to contradict the boundary conditions. Can we conclude that this equation has no (continuous) solution? Or does this simply mean that characteristics don't start and end at the parts of the boundary where $u$ is given? Do I need to find conditions on other parts of the boundary or is there another way to solve this equation? I'm sorry if this is kind of vague, but I never encountered such a problem before.
Any help or comment is greatly appreciated!