Can anyone help me to understand the PDE problem? Consider the IVP $$xu_x +tu_t=u+1 , x\in\mathbb {R},t>0$$
$$u(x,t)=x^2, t=x^2$$. Then 
1.the solution is a singular at (0,0).
2.the given space curve $(x,u,t)=(w,w^2,w^2)$ is not a characteristic curve at (0,0)
3.there is no base characteristic curve in $(x,t)$ plane passing through $(0,0)$
4.a necessary condition for the IVP to have a unique $C^1$ solution at $(0,0)$ doesn't hold.
I'm not understanding that  In the first option , what is the meaning of singular solution at (0,0) while we can see at (0,0) the solution is trivial.Can anyone help to understand all options? Thanks in advance.
 A: $$xu_x+tu_t=u+1$$
Charpit-Lagrange system of ODEs : $\quad \frac{dx}{x}=\frac{dt}{t}=\frac{du}{u+1}$
First characteristic equation from $\frac{dx}{x}=\frac{dt}{t}$ :
$$\frac{t}{x}=c_1$$
Second characteristic equation from $\frac{dx}{x}=\frac{du}{u+1}$ :
$$\frac{u+1}{x}=c_2$$
General solution of the PDE on the form of implicit equation $c_2=F(c_1)$ :
$$\frac{u+1}{x}=F\left(\frac{t}{x}\right)$$
where $F$ est an arbitrary function to be determined according to the specified condition.
$$u(x,t)=-1+xF\left(\frac{t}{x}\right)$$
CONDITION : 
$u(x,x^2)=x^2=-1+xF\left(\frac{x^2}{x}\right)=-1+xF(x)$
$F(x)=\frac{1}{x}(x^2+1)$
So, the function $\quad F(X)=X+\frac{1}{X}\quad$ is determined. We put it into the above general solution where $X=\frac{t}{x}$ :
$u(x,t)=-1+x\left(\frac{t}{x}+\frac{x}{t}\right)$
$$u(x,t)=-1+t+\frac{x^2}{t}$$
Of course, $u(x,t)$ is singular at $(0,0)$ . But why considering this point when asking about unicity ? What this point has more special than any other point ?
The important is to check if the condition is satisfied or not on a characteristic curve. The condition is specified on $t=x^2$ which is not a characteristic curve :
$\frac{t}{x}=\frac{x^2}{x}=x\neq c_1$
$\frac{u+1}{x}=\frac{x^2+1}{x}\neq c_2$
Thus the solution is unique.
