Understanding whether or not a curve is a characteristic projection

Question

I am trying to familiarise myself with the method of characteristics in solving first-order PDEs. I have come across the following example:

Solve the first-order PDE $$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1$$ given the initial data:

1. $u = 0$ on $x + y = 0$
2. $u = 0$ on $x = y$
3. $u = x$ on $x = y$.

I am perfectly OK with showing that $$u(x,y) = \frac{x+y}{2} + F(x - y)$$

where $F$ is an arbitrary function and I can find the solution in the first case and show that in case 2 there is no solution and in case 3 the solution is non-unique. However, I am having difficulty following the geometric reasoning that the lecture notes follow - they assert that in case 1 the initial curve is normal to the characteristic projection (and therefore nowhere tangent to it) but that in the other two cases the initial curves are both characteristic projections.

Could someone please explain to me why this is the case and why this means from a geometric point of view that the PDE does not have a unique solution?

Answer 1

Let us plot the characteristic curves passing at the points with coordinates $(x_0, y_0)$ in the $x$-$y$ plane. These curves satisfy $$ \begin{aligned} {\text d y}/{\text d s} &= 1 \, ,\\ {\text d x}/{\text d s} &= 1 \, ,\\ {\text d u}/{\text d s} &= 1 \, , \end{aligned} \qquad\text{i.e.}\qquad \begin{aligned} x &= x_0+y-y_0 \, ,\\ u &= u_0 + y-y_0 \, , \end{aligned} $$ where $u_0 = u(x_0,y_0)$. Thus, along the characteristics, $u-x$ is constant.

We plot the lines $y=-x$ (red) and $y=x$ (blue).

1. The value $u_0$ is obtained by following the characteristic line until it intersects the red line. This occurs at the orthogonal projection $(x_p,-x_p)$ of $(x_0,y_0)$, where $u$ is equal to zero. Therefore, we have $$ \left\lbrace \begin{aligned} x_p &= x_0-x_p-y_0 \, ,\\ 0 &= u_0-x_p-y_0\, , \end{aligned} \right. \qquad \text{i.e.}\qquad u_0 = \frac{x_0+y_0}{2} \, . $$ Hence, $u(x,y)=\frac{1}{2}(x+y)$ everywhere.
2. The value of $u$ at $(x_0,y_0)$ is given by following the characteristic line until it intersects the blue line. If $y_0 = x_0$, the characteristic line intersects the blue line along its whole length, where $u$ is equal to zero. If $y_0\neq x_0$, the value of $u$ is undetermined (no intersection occurs).
3. If $y_0 = x_0$, the characteristic line intersects the blue line along its whole length, where $u$ is not uniquely defined. If $y_0\neq x_0$, the value of $u$ is still undetermined.