Jtdylktuy

Exercise :

Consider the equation
$$z z_x + y z_y = x$$
and the initial curve
$$C : x=t, y=t ; ; ; t >0$$
Decide whether there is a unique solution, no solution or infinitely many solutions in a neighborhood of the point $(1,1)$, for the initial data $z=2t$.

Attempt :

We yield the Lagrange equations :

$$frac{mathrm{d}x}{z} = frac{mathrm{d}y}{y} = frac{mathrm{d}z}{x}$$

and, easily, we can calculate two linearly independent integral curves :

$$frac{mathrm{d}x}{z} = frac{mathrm{d}y}{y} implies z_1 = y(x-z) $$

$$frac{mathrm{d}x}{z} = frac{mathrm{d}z}{x} implies z_2 = frac{1}{2}(z^2-x^2)$$

Thus, the general solution is given by an expression of a $C^1$ differentiable function $F$, such that :

$$F(z_1,z_2) = 0 Rightarrow Fbigg(y(x-z), frac{1}{2}(z^2-x^2)bigg) = 0 $$

Taking into account the parameters of the initial curve $C$, we yield :

$$C : mathbf{r}_C(t) = (t,t,2t) ; ; ; t > 0$$
$$frac{mathrm{d}mathbf{r}_C(t)}{mathrm{d}t} = (1,1,2) ; ; ; t > 0$$

Since :

$$frac{mathrm{d}mathbf{r}_C(t)}{mathrm{d}t} = frac{1}{t} mathbf{r}_C(t)$$

there exists a proportionality function-constant $μ(t) = frac{1}{t}$ such that $frac{mathrm{d}mathbf{r}_C(t)}{mathrm{d}t} = μ(t) mathbf{r}_C(t)$, which means that there exist infinitely many solutions.

Alternativelly, substituting the parameters into $F$, we get :

$$Fbigg(-t^2, frac{3}{2}t^2bigg) = z(t,t) = 2t = 0$$

But that would only hold for $t=0$, thus for $F(0,0) = 0$ and there can be infinitely many $C^1$ differentiable functions $F$ such that this case holds, thus we have infinitely many solutions again.

Questions : Is my approach correct ? Is there a more standard way of approaching the existence and uniqueness of a PDE ? Maybe more strictly and formally ? Am I missing something ?

Let $c_1=y(z-x)$ and $c_2=z^2-x^2$ be the constants of the characteristic curves. None are constant on the initial data, indeed there $c_1=t^2$ and $c_2=3t^2$ so that $c_2=3c_1$ and inserting that relation backwards
$$
z^2-x^2=3y(z-x)iff z=xlor z=3y-x
$$
Both are solutions of the PDE, $z(x,y)=x$ and $z(x,y)=3y-x$, but the first does not satisfy the initial conditions except at $t=0$, which is not enough.