# 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.

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 $$(x,x)$$.

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

$$(x',y') = (x+y,x-y),$$

$$\frac{\partial u}{\partial x'} = \frac{1}{2}.$$

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 $$u(x,y) = 1 - (x - y)$$.