Consider the first order PDE given by $$\frac{\partial u}{\partial x}+2y^{\frac{1}{2}}\frac{\partial u}{\partial y}=xy,$$ with the Cauchy data for $0\leq x \leq 2$ given by $$u(x,0)=f(x)=\begin{cases} x^2, & \text{for }\quad 0 \leq x \leq 1 \\4-x^2, & \text{for }\quad 1<x\leq 2\end{cases}.$$
I've used the method of characteristics to find the solution of this, and this turns out to be $$u(x,y)=\begin{cases}\frac{1}{4}y^2+\frac{1}{3}y^{\frac{3}{2}}(x-\sqrt{y})+(x-\sqrt{y})^2, & \text{for } \quad 0\leq x-\sqrt{y}\leq 1\\\frac{1}{4}y^2+ \frac{1}{3}y^{\frac{3}{2}}(x-\sqrt{y})+4-(x-\sqrt{y})^2, & \text{for } \quad 1<x-\sqrt{y}\leq2\end{cases}.$$ I'm asked to find the value of $y$ for which the partial derivative $\frac{\partial u}{\partial x}$ is discontinuous along $x=2$.
I can see that the solution $u(x,y)$ is discontinuous along $x-\sqrt{y}=1,$ so the solution is discontinuous at $(x,y)=(2,1)$. Is this enough to show that $\frac{\partial u}{\partial x}$ is discontinuous at this point as well though?
For $0\leq x-\sqrt{y}\leq 1$ : $$\frac{\partial u}{\partial x}=\frac13y^{3/2}+2x-2\sqrt{y}$$ Along $x=2\quad$ and insofar $\quad 0\leq 2-\sqrt{y}\leq 1\quad$ that is $\quad 1\leq y\leq 4$ : $$\left(\frac{\partial u}{\partial x}\right)_{x=2}=\frac13y^{3/2}+4-2\sqrt{y}$$ This function of $y$ is continuous.
$\left(\frac{\partial u}{\partial x}\right)_{x=2\:,\:y=1}=\frac13+4-2=\frac73$ $$ $$
For $1< x-\sqrt{y}\leq 2$ : $$\frac{\partial u}{\partial x}=\frac13y^{3/2}-2x+2\sqrt{y}$$ Along $x=2\quad$ and insofar $\quad 1< 2-\sqrt{y}\leq 2\quad$ that is $\quad 0\leq y<1$ : $$\left(\frac{\partial u}{\partial x}\right)_{x=2}=\frac13y^{3/2}-4+2\sqrt{y}$$ This function of $y$ is continuous.
$\left(\frac{\partial u}{\partial x}\right)_{x=2\:,\:y=1}=\frac13-4+2=-\frac53$ $$ $$
Thus the discontinuity point might lay on the interface between the two domains, that is at $y=1$.
The discontinuity of $\frac{\partial u}{\partial x}$ at $(2,1)$ is confirmed by the two different above values $\frac73$ and $-\frac53$ .