Is there a $\mathcal{C}^1$ solution for this Cauchy problem?

Question

I have the following Cauchy problem: $$\left\{ \begin{array}{l} 2u_x+(x+u)u_y=2 \\ u\left(x,\frac{x^2}{2}\right)=x \end{array}\right. $$ I want to study whether it has $\mathcal{C}^1$ solutions in a neighborhood of the initial curve $$\Gamma(s)=\left(s, \frac{s^2}{2},s\right)$$ First I check whether the transversality condition is fulfilled. I have that $$\begin{vmatrix} 2 & 2s \\ 1 & s \end{vmatrix}=0 \quad \forall s$$ Therefore I check the rank of the extended matrix: $$\text{rank}\begin{pmatrix}2 & 2s & 2\\ 1 & s & 1 \end{pmatrix}=\left\{\begin{array}{lcr}2 & \text{if} & s\neq 1\\ 1 & \text{if} & s=1 \end{array}\right.$$ Therefore if $s\neq1$, the problem has no $\mathcal{C}^1$ solution. However I am having trouble deciding what happens when $s=1$. If I solve the system, for the particular case $s=1$ I get that the solution is $(t+1, 3t^2/2+2t+1/2,2t+1)$. This is not parallel nor the same as my initial curve, so there aren't infinite solutions. What happens here?

Answer 1

Below, consider this as information (or comment) instead of a direct answer.

$$2u_x+(x+u)u_y=2$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{2}=\frac{dy}{x+u}=\frac{du}{2}$$ A first characteristic equation comes from solving $\frac{dx}{2}=\frac{du}{2}$ : $$u-x=c_1$$ A second characteristic equation comes from solving $\frac{dx}{2}=\frac{dy}{x+(c_1+x)}$ : $$c_1x+x^2-2y=c_2$$ General solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ : $$(u-x)x+x^2-2y=F(u-x)$$ $F$ is an arbitrary function. $$\boxed{xu-2y=F(u-x)}$$ Condition : $$u(x,\frac12 x^2)=x\quad\implies\quad x*x-2(\frac12 x^2)=F(x-x)=F(0)$$ $$F(0)=0$$ They are infinity many functions which are equal to zero when the argument is zero. Thus the PDE with the specified condition has infinity many solutions.

For examples :

With the function $F(X)=0$ we get the solution $u(x,y)=x$ (in agreement with jan's comment).

With F(X)=X we get the solution $u(x,y)=\frac{2y-x}{x-1}$ .

More complicated, with $F(X)=(\exp(X)-1)$ we get the solution $u(x,y)=\frac{2y-1}{x}-W\left(-\frac{1}{x}e^{-(x^2-2y+1)/x} \right)$ involving the $W$ Lambert's function.

But generally one cannot express $u(x,y)$ on explicit form. We have to be satisfied with the implicit equation $xu-2y=F(u-x)$ with functions $F$ such as $F(0)=0$ for the infinity many solutions.