Question:
Consider the system
\begin{align} \frac{\partial u}{\partial x} + u \frac{\partial u}{\partial y} + 2\frac{\partial v}{\partial y} & = 1 \\ \frac{\partial v}{\partial x} + 2v\frac{\partial u}{\partial y} + u\frac{\partial v}{\partial y} & = \sqrt{v} \end{align}
with initial conditions $u(x,0) = x+\frac 12$ and $v(x,0) = \frac 14(x-1)^2$.
Find the solution $\big(u(x,y),v(x,y)\big)$.
Attempt:
So I started by writing the system in the form
$$\mathbf A \frac{\partial \mathbf u}{\partial x} + \mathbf B \frac{\partial \mathbf u}{\partial y} = \mathbf c$$
where
$$\mathbf A = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \qquad \mathbf B = \begin{pmatrix} u & 2 \\ 2v & u \end{pmatrix} \qquad \mathbf c = \begin{pmatrix} 1 \\ \sqrt{v} \end{pmatrix} \qquad \mathbf u = \begin{pmatrix} u \\ v \end{pmatrix}$$
By solving for $\lambda$ in $\det(\mathbf B - \lambda \mathbf A)=0 $, I found that the characteristic directions are
$$\frac{dy}{dx} = \lambda = u \pm 2\sqrt{v}$$
After a bit of algebra, I found that
\begin{align} u + 2\sqrt{v} - 2x = \text{constant} := \alpha \qquad \qquad & \text{on }\; \frac{dy}{dx} = u+2\sqrt{v} \\ u - 2\sqrt{v} = \text{constant} := \beta \qquad \qquad & \text{on } \; \frac{dy}{dx} = u-2\sqrt{v} \\ \end{align}
It follows that the characteristics are
$$\frac{dy}{dx} = \alpha + 2x \qquad \qquad \frac{dy}{dx} = \beta$$
i.e. a set of parabola and a set of straight lines.
However, how do I use the initial data? What should I do from here to find the solution?
Any hints would be much appreciated. Thanks!
(Of course, the solution is clearly
$$u(x,y) = x+\frac 12 \qquad v(x,y) = \frac 14(x-1)^2$$
but how do we get to this?)
Consider particular solutions $\big(u(x),v(x)\big)$ which do not depend on $y$. We have ${\bf A}\partial_x {\bf u} = {\bf c}$, which solutions are $$ u(x) = c_1 + x,\qquad v(x) = \tfrac14 \left( c_2 + x \right)^2 , $$ where $c_1$, $c_2$ are arbitrary constants. All these solutions are some kind of "steady-state solution", in the sense that they keep the above values for all $y$. Here, setting $c_1 = \frac12$, $c_2 = -1$ is such a solution which matches the boundary condition at $y=0$. Therefore, the solution keeps these values for all $y$.