Why is initial curve not assumed to be coincident with characteristics I was reading the method of characteristic for first order p.d.e's.
$a\frac{\partial u}{\partial x}+b\frac{\partial u}{\partial y}=c$.
Now it said to find out the solution we try to find out a specific curve along which the pde becomes an ode. This part I get, since the curve is where the first order derivatives get discontinuous. But why does it say that the initial data curve should not coincide with this chatacteristic. What will happen if the initial conditions are specified on the boundary. Can somebody give an example.?
Say i have the equation as $$yp+q=2$$ with $u$ known on the initial curve $y=0, 0\leq x\leq1$. So i get the characteristics and solution as
$y^2=2(x-x_R)$ and $u= 2y +u_R$

You ask to solve (for example) :
$$yp+q=y\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=2\tag 1$$
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dx}{y}=\frac{dy}{1}=\frac{du}{2}$$
A first characteristic equation comes from solving $\frac{dx}{y}=\frac{dy}{1}$ :
$$x-\frac12 y^2=c_1$$
A second characteristic equation comes from solving $\frac{dy}{1}=\frac{du}{2}$ :
$$u-2y=c_2$$
The general solution of the PDE on implicit form $c_2=F(c_1)$ is :
$$u-2y=F\left(x-\frac12 y^2\right)$$
$F$ is an arbitrary function (to be determined according to some boundary condition).
$$\boxed{u(x,y)=2y+F\left(x-\frac12 y^2\right)} \tag 2$$
Then, all depends on the given boundary condition which can a continuous function or not, a picewise function or not.
Depending of the boundary condition the solution might exists or not or be not unique, might be continuous or not, piecewise or not.
In the question you wrote "with $u$ known on the initial curve $y=0$ , $0\leq x\leq 1."$ I suppose that you mean : $u(x,0)=f(x)$ with a given (known) function $f(x)$. If not let us know what you mean.
So with the condition : $u(x,0)=f(x)$ that we put into the general solution :
$$f(x)=u(x,0)=0+F(x-0)\quad\implies\quad F(x)=f(x)$$
Now the function $F$ is known since the function $f$ is given :
$$F(X)=f(X)\quad\text{any } X$$
We put it into the above general solution where $X=x-\frac12 y^2$
$$u(x,y)=2y+f\left(x-\frac12 y^2\right) \tag 3$$
The difference with equation $(2)$ is that $f$ is known instead of arbitrary $F$ in Eq.$(2)$.
Now we have to discuss about the "given" function $f$ in your question.
You didn't say what is $u(x,0)$ on $0\leq x \leq 1.$  So the boundary condition is hill posed and one cannot say if a solution exists or not or if they are many solutions.
You wrote : "So i get the characteristics and solution as $y^2=2(x−x_R)$ and $u=2y+u_R$ ".
What do you mean with $x_R$ and $u_R$ which appears like fly in the ointment ?
SEVERAL DIFFERENT EXAMPLES OF BOUNDARY CONDITION.
First example :
With the condition $u(x,0)=x^2$. The boundary $y=0$ is not a characteristic curve.
$$u(x,0)=x^2=0+F(x)\quad\implies\quad F(x)=x^2$$
$$u(x,y)=2y+\left(x-\frac12 y^2\right)^2$$
Second example :
With condition $u(x,y)=0$ on $y^2=2x$. The boundary is on a characteristic curve.
$$u(x,y)=0=2y+F(0) \quad\implies\quad y=\text{constant, which is not true any }y.$$
$$\text{There exists no solution for }u(x,y).$$
Third example :
With condition $u(x,y)=2y$ on $x-\frac12 y^2=1$. The boundary is on a characteristic curve.
$$u(x,y)=2y=2y+F(1)\quad\implies\quad F(1)=0$$
There exist an infinity of functions which are nul at argument=$1$. Thus :
$$\text{They are an infinity of solutions for }u(x,y)\: :$$
$F(X)=X-1 \quad\to\quad u(x,y)=2y+\left(x-\frac12 y^2\right)-1$
$F(X)=X^4-1 \quad\to\quad u(x,y)=2y+\left(x-\frac12 y^2\right)^4-1$
$F(X)=\sin(X-1) \quad\to\quad u(x,y)=2y+\sin\left(x-\frac12 y^2-1\right)$
Etc.
