Solving $u_x+u_y=1$ with a special initial condition (Method of characteristics) I have tried to solve the following PDE using the method of characteristics:
$$
\begin{cases}
u_x+u_y=1 \\
u(x,x)=1
\end{cases}
$$
My solution: let the curve $(X(y),y)$ that verifies the differential equation:
$$
\begin{cases}
X'(y)=1 \\
X(x)=x
\end{cases}
$$
then we find that: the characterestics are: $X(y)=y$, and $\frac{\partial u(X(y),y)}{\partial y}=1$, then $u(X(y),y)=y+C(x)$. And using the initial condition, we find that:
$$u(X(x),x)=1=x+C(x)$$
then $C(x)=1-x$ which implies: $$u(X(y),y)=y+1-x$$
My problem is that I can't write $u$ with $X(y)$ and $y$, to give the final solution. 
 A: This Cauchy problem is unsolvable. This is, there are no points $x\in \mathbb{R}$ such that the problem 
$$(P)\begin{cases}
u_x+u_y=1 \\
u(x,x)=1
\end{cases}$$ 
has a local solution around $(x,x)$. To see this you can use the following results, which you can find in any PDE book:


*

*Definition: Consider the Cauchy problem 


$$(P)\begin{cases}
b_1 (x,y,u)u_x+b_2(x,y,u)u_y=c(x,y,u) \\
u(\sigma_1(s),\sigma_2(s))=\tau(s)
\end{cases}$$
We say that the curve $\sigma\equiv(\sigma_1 , \sigma_2)$ satisfies the transversality condition for $(P)$ at the point $(x_0,y_0)=(\sigma_1(s_0) , \sigma_2(s_0))$ if the following holds:
$$\begin{vmatrix}\sigma_1 '(s_0)&b_1(\sigma_1(s_0),\sigma_2(s_0),\tau(s_0))\\\\\sigma_2'(s_0)&b_2(\sigma_1(s_0),\sigma_2(s_0),\tau(s_0))\\ \end{vmatrix} \neq 0$$


*

*Theorem 1: If the transversality condition doesn't hold at $(x_0,y_0)$ then the Cauchy problem has either no solution or infinitely many around $(x_0,y_0)$.

*Theorem 2: Suppose that the transversality condition doesn't hold at $(x_0,y_0)$. If the vectors $\bigl(\sigma_1'(s_0),\sigma_2'(s_0),\tau'(s_0)\bigr)$ and $\Bigl(b_1 \bigl(x_0,y_0,\tau(s_0)\bigr),b_2\bigl(x_0,y_0,\tau(s_0)\bigr),c\bigl(x_0,y_0,\tau(s_0)\bigr)\Bigr)$ are linearly dependent, then $(P)$ has infinitely many solutions around $(x_0,y_0)$. Otherwise it has no solution around $(x_0,y_0)$.

In your example, for any $s\in \mathbb{R}$ you have: 
$$\sigma(s)=(s,s) \implies \sigma'(s)=(1,1)$$
$$\tau(s)=1 \implies \tau'(s)=0$$
$$b_1(x,y,z)=b_2(x,y,z)=c(x,y,z)=1$$ 
So if $x\in \mathbb{R}$:
$$\begin{vmatrix}\sigma_1 '(s_0)&b_1(\sigma_1(s_0),\sigma_2(s_0),\tau(s_0))\\\\\sigma_2'(s_0)&b_2(\sigma_1(s_0),\sigma_2(s_0),\tau(s_0))\\ \end{vmatrix} =\begin{vmatrix}1&1\\ 1&1\\ \end{vmatrix} = 0$$
So $\sigma$ doesn't satisfy the transversality condition. Moreover: 
$$\bigl(\sigma_1'(s_0),\sigma_2'(s_0),\tau'(s_0)\bigr)=(1,1,0)$$
$$\Bigl(b_1 \bigl(x_0,y_0,\tau(s_0)\bigr),b_2\bigl(x_0,y_0,\tau(s_0)\bigr),c\bigl(x_0,y_0,\tau(s_0)\bigr)\Bigr)=(1,1,1)$$
So there doesn't exist a solution to  $(P)\begin{cases}
u_x+u_y=1 \\
u(x,x)=1
\end{cases}$ in any neighbourhood of $(x,x)$.
A: There is something missing or wrong in the problem statement, because the solution that you achieve, i.e. $u(X(y),y) = y + 1 - x$, don't meet the PDE. Your problem is
$$\left\{ \begin{array}{ll} & \displaystyle \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1,\\ & u(x,y) = 1\quad\text{on}\quad y = x.\end{array} \right.$$
The mapping 
$$(x',y') = (x+y,x-y),$$
transform your original PDE in
$$\frac{\partial u}{\partial x'} = \frac{1}{2}.$$
The general solution is
$$u(x',y') = \frac{1}{2}x' + C(y'),$$
or 
$$u(x,y) = \frac{1}{2}\Big[x + y + f(x-y)\Big],$$
where $f(x-y)$ is an arbitrary function. Bear in mind that the boundary condition is applied on the line $y = x$, which means $y' = 0$ in the mapping. On this line, the problem set that $u$ must be constant, but the PDE has a source, that is $1/2$ in the $(x',y')$-plane. Therefore, $u$ must be a function of $x'$ on $y' = 0$ and any other. Either the boundary condition is wrong or your problem is  
$$\left\{ \begin{array}{ll} & \displaystyle \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0,\\ & u(x,y) = 1\quad\text{on}\quad y = x.\end{array} \right.$$
In such a case, your solution is  $u(x,y) = 1 - (x - y)$.
