I wish to solve the following system of hyperbolic PDEs: $\frac{\partial y}{\partial u}=\alpha \sqrt{2y-y^2}$, $\frac{\partial y}{\partial v}=-\frac{1}{\alpha}\frac{\partial\alpha}{\partial v}\:\left(1-y\right)$, for the functions $\alpha=\alpha(u,v)$ and $y=y(u,v)$ on a rectangle aligned with the axes on the first quadrant, and boundary conditions $\alpha (u,0)=1$, $y(u,0)=\sin u+1$, $y(0,v)=1$. However, I posed the problem as follows
eq1 = D[y[u, v], u] == \[Alpha][u, v] Sqrt[2 y[u, v] - y[u, v]^2];
eq2 = D[y[u, v], v] \[Alpha][u,v] == -D[\[Alpha][u, v], v] (1 - y[u, v]);
bc = {\[Alpha][u, 0] == 1, y[u, 0] == 1 + Sin[u], y[0, v] == 1};
region = Rectangle[{0, 0}, {1.5, 1.5}];
sol = NDSolve[{{eq1, eq2}, bc}, {\[Alpha], y}, {u, v} \[Element]region]
but I am not obtaining any solution, just a string of errors starting with
NDSolve`FEM`InitializePDECoefficients::femcnsd: The PDE coefficient -(((2-2 y[u,v]) \[Alpha][u,v])/(2 Sqrt[2 y[u,v]-y[u,v]^2])) does not evaluate to a numeric scalar at the coordinate {1.2075,1.2075}; it evaluated to Indeterminate instead.
and ending with
NDSolve::fempslf: The linearization process in PDESolve failed.
Anybody could please guide me about what is going wrong?
bc
atu=0, v=0
are formulated outside ofregion
. Also yourregion
is not "a rectangle aligned with the axes on the first quadrant". $\endgroup$