Solving $u_x+u_y=1$ with a special initial condition (Method of characteristics)

Question

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.

Answer 1

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 $(x_0,y_0)$.

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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 $(x,x)$.

Answer 2

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