Condition for finite number of solutions for Cauchy problem of first order PDE

Question

I am studying the result in pde, which says that

Theorem: Consider the Cauchy problem for Partial differential equation of first order $$P(x,y,z)p+Q(x,y,z)q=R(x,y,z)$$ With the initial curve $\Gamma:(x_0(t),y_0(t),z_0(t))$, and let us denote \begin{align} \Delta= \begin{vmatrix} P(x_0(t),y_0(t),z_0(t)) & Q(x_0(t),y_0(t),z_0(t))\\ \frac{dx_0}{dt} & \frac{dy_0}{dt} \\ \end{vmatrix}, \end{align} \begin{align} \Delta_1= \begin{vmatrix} R(x_0(t),y_0(t),z_0(t)) & Q(x_0(t),y_0(t),z_0(t))\\ \frac{dz_0}{dt} & \frac{dy_0}{dt}\\ \end{vmatrix}, \end{align} \begin{align} \Delta_2= \begin{vmatrix} P(x_0(t),y_0(t),z_0(t)) & R(x_0(t),y_0(t),z_0(t))\\ \frac{dx_0}{dt} & \frac{dz_0}{dt} \\ \end{vmatrix} \end{align} Then we have

1. unique solution iff $\Delta\neq 0$
2. infinitely many solutions when$\Delta=\Delta_1=\Delta_2=0$
3. no solution when $\Delta=0$ but $\Delta_1≠0$ or $\Delta_2≠0$

I found there is no case of finitely many solutions.

Now there is a question:

Question: let $u(x,t)$ be a function which satisfies PDE $$u_t+uu_x=1,\quad x\in \mathbb{R},t>0$$ and the initial condition $u\big(\frac{t^2}{4},t\big)=\frac{t}{2}$. Then how many solutions does this equation have?

I solved it through Lagrange's auxiliary equation and using Initial condition I get exactly two solutions $u=\frac{t}{2}±\sqrt{x-\frac{t^2}{4}}$ (I don't want to write whole solution here, since it will take a lot time in writing it). But we can parametrize the initial condition and get initial curve $\Gamma:\left(\frac{t^2}{4},t,\frac{t}{2}\right)$ and now I found that $\Delta=0$ and $\Delta_1,\Delta_2≠0$. Then according to the theorem stated the given equation doesn't have any solution, a contradiction. Can anybody see where I have gone wrong? Is there any mistake in the statement of theorem? Any help will be appreciable

Answer 1

The "solution" $u = \frac{t}2 \pm \sqrt{x-\frac14 t^2}$ with the bottom (minus) sign does not solve the PDE $u_t + uu_x=1$. The solution with the top (plus) sign satisfies both the PDE and the boundary condition. It is therefore likely to be the unique solution, which is defined real for $x\geq t^2/4$. Note that upon renaming the variables $(x,t)$ as $(x,y)$, the PDE reads $ zp + q = 1 $, so that $\Delta = 0$, $\Delta_1 = \frac12$, $\Delta_2 = -\frac14 t$ would yield no solution. There may be some missing validity conditions in the theorem's formulation to use it as done in OP. In fact, the theorem tells nothing about the solutions' range, while the solution $u = \frac{t}2 + \sqrt{x-\frac14 t^2}$ obtained via characteristics is only valid on a given range.