solving a first order PDE: $u_x+2x u_t=2x$ I'm back with another PDE problem! :D 
I want to solve the following:
\begin{cases}
u_x+2x u_t=2x & x>0,t>0\\
u(0,t)=u(x,0)=1
\end{cases}
I changed the variables:
\begin{cases}
s=x\\
r=x^2-t
\end{cases}
and then,
$$
u_x=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}=u_s+2x u_r
$$
$$
u_t=\frac{\partial u}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial u}{\partial r}\frac{\partial r}{\partial t}=-u_r
$$
so we have
$$
u_s+2x u_r-2xu_r=2x \quad\Rightarrow\quad u_s=2s
$$
$$
\Longrightarrow \qquad u(s,r)=s^2+\phi(r)
$$
Now with $x,t$ :
$$
u(x,t)=x^2+\phi(x^2-t)
$$
$$
u(0,t)=\phi(-t)=1 \quad,\quad u(x,0)=x^2+\phi(x^2)=1
$$
How to find $\phi$?Is this solution correct?
 A: Let me rewrite the problem several times, first, I'd like to call $u_t = p, u_x = q, u = z,$ and then write the PDE as
$$F(p,q,z,x,t) := p + \frac{q}{2x} - 1 = 0$$
We can blindly apply the method of characteristics to solve along the $t = 0$ boundary first:
\begin{align}
\frac{dx}{ds} = \partial_q F &= \frac{1}{2x}; \;\; x(r,s=0) = r\\
\frac{dt}{ds} = \partial_p F &= 1; \;\;t(r,s = 0) = 0\\
\frac{dz}{ds} = p\partial_p F + q\partial_q F &= 1; \;\;z(r,s = 0) = 1\\
\end{align}
We can solve these without much trouble, $t = s, \;  x^2 - t = r^2$ so $u(x,t) = z(r,t) = t + 1,$ for characteristics originating at the $t = 0$ boundary.
We can do the same exact thing for the characteristics beginning at the $x = 0$ boundary, namely
\begin{align}
\frac{dx}{ds} = \partial_q F &= \frac{1}{2x}; \;\; x(r,s = 0) = 0\\
\frac{dt}{ds} = \partial_p F &= 1; \;\;t(r,s = 0) = r\\
\frac{dz}{ds} = p\partial_p F + q\partial_q F &= 1; \;\;z(r,s = 0) = 1\\
\end{align}
Solving this, again we have $x^2 = s, \; t = r + s,$ and $z(r,s) = s + 1,$ so $u(x,t) = x^2 + 1$ for characteristics originating from the $x = 0$ boundary.
The answer seems a bit strange, the value of $u$ grows with respect to time or space depending on which characteristic it is on.  The equations for the characteristics coming from $x= 0$ are given by $r = x^2 - t$ and from $t = 0$, $r^2 = x^2 - t.$ Since the answer appears to not depend on $r$ in either instance, it seems like the only curve that the PDE has a solution on is when $x^2 - t = 0,$ where the initial data agrees. Anywhere else, talking about "the solution" doesn't make sense, at least to me.
