Is this Cauchy problem ill-posed? I have a problem that states :
Find the solution to the Cauchy problem consisting of the wave equation :
$$u_{xx}-u_{yy}=0$$ together with initial conditions: $$ u(x,0)=0,$$ $$u_{y}(x,0)=g(x)$$ for some known initial datum $g$.
Is the problem well-posed or ill-posed?
Why?
My effort is this:
By changing variables we have :
$$ u(x,y) = \xi(x-y)+\eta(x+y) $$
First , we note that the initial condition $u(x,0)=0$ gives us:
\begin{align*}
0 =& u(x,0) \\
=&\xi(x-0)+\eta(x+0) \\
=&\xi(x)+\eta(x)
\end{align*}
We then derived the first equation $$\xi(x)+\eta(x)=0.$$
Next,we determine the derivative $u(x,y)$ with respect to $y$:
\begin{align*}
u_{y}(x,y) =& \frac{\partial}{\partial y}(\xi(x-y)+\eta(x+y)) \\
=& -\xi'(x-y)+\eta'(x+y)
\end{align*}
Next the initial condition $$u_{y}(x,0)=g(x)$$ which implies:
\begin{align*}
g(x) =& u_{t}(x,0) \\
=& -\xi'(x-0)+\eta'(x+0) \\
=&-\xi'(x)+\eta'(x)
\end{align*}
We have then derived the second equation $-\xi'(x)-\eta'(x)=g(x)$
as well.
Solving the equation $$\xi(x)+\eta(x)=0$$ to $\eta(x)$ in the first equation we obtain: $$\xi(x)=-\eta(x).$$
Differentiate both sides of the previous equation with respect to $x$:
$$\xi'(x)=-\eta'(x)$$
Next , we replace $\eta'(x)$ by $-\xi'(x)$ in the equation $$-\xi'(x)-\eta'(x)=g(x)$$ we take:
\begin{align*}
g(x) =& -\xi'(x)+ \xi'(x)   \\
     =& -\xi'(x)- \xi'(x) 
\end{align*}
Combining terms :
$$-2\xi'(x) = g(x)$$
and dividing each side by $-2$
$$ \xi'(x) = -\frac{1}{2}g(x)$$
Integrating each side wrt $x$:
$$ \int_{x_{0}}^{x} \xi'(x)dx = \int_{x_{0}}^{x} -\frac{1}{2}g(x) dx $$
$$ \xi(x) -\xi(x_{0})  = -\frac{1}{2} \int_{x_{0}}^{x} g(\xi) d\xi  = $$
Add $\xi(x_{0})$ to each side:
$$\xi(x) =-\frac{1}{2} \int_{x_{0}}^{x} g(\xi) d\xi + \xi(x_{0})$$
We know that the general solutuon is $u(x,y) = \xi(x-y)+ \eta(x+y)$ with $\eta(x) = - \xi(x)$ and $\xi(x)-\frac{1}{2} \int_{x_{0}}^{x} g(\xi) d\xi + \xi(x_{0})$
\begin{align*}
u(x,y) =& \xi(x-y)+ \eta(x+y) \\
     =& \xi(x-y) -  \eta(x+y) \\
     =& -\frac{1}{2} \int_{x_{0}}^{x-y} g(\xi) d\xi + \xi(x_{0})+\frac{1}{2} \int_{x_{0}}^{x+y} g(\xi) d\xi - \xi(x_{0}) \\
     =& -\frac{1}{2} \int_{x_{0}}^{x-y} g(\xi) d\xi +\frac{1}{2} \int_{x_{0}}^{x} g(\xi) d\xi  \\
     =& \frac{1}{2} \left( \int_{x_{0}}^{x+y} g(\xi) d\xi -\frac{1}{2} \int_{x_{0}}^{x-y} g(\xi) d\xi  \right)\\
     =& \frac{1}{2} \int_{x_{0}-y}^{x+y} g(\xi) d\xi
\end{align*}
For the difference $$u(x,y)- \tilde{u}(x,y)$$ we will have :
\begin{align*}
u(x,y)- \tilde{u}(x,u) =& \frac{1}{2} \int_{x_{0}-y}^{x+y} g(\xi) d\xi -  \frac{1}{2} \int_{x_{0}-y}^{x+y} \tilde{g}(\xi)) d\xi \\ 
 =&  \frac{1}{2} G(\xi) + c -  \left( \frac{1}{2} \tilde{G}(\xi)+ k \right) 
\end{align*}
Knowning that $c+k=0$
then we have that the problem is ill-posed.
But I don't know if I am right.Can anybody check it or help with this one ?
 A: For a problem to be well-posed in the sense of Hadamard,$^{(a)}$ it must satisfy the following conditions:

*

*a solution exists;

*the solution is unique;

*the solution's behavior changes continuously with the initial conditions.

The derivation of the solution to the Cauchy problem presented in the question accounts for the first two conditions. In order to show that the third condition is also satisfied, let $\delta g(x)$ be a small perturbation in the initial condition on $u_y(x,0)$; by small, I mean
$$
\|\delta g\|_1=\int_{-\infty}^{\infty}|\delta g(x)|\,dx<\epsilon. \tag{1}
$$
Under such a perturbation, the new solution to the Cauchy problem is given by
$$
u(x,y)+\delta u(x,y)=\frac{1}{2}\int_{x-y}^{x+y}[g(\xi)+\delta g(\xi)]\,d\xi.
\tag{2}
$$
Therefore,
$$
|\delta u(x,y)|=\left|\frac{1}{2}\int_{x-y}^{x+y}\delta g(\xi)\,d\xi\,\right|
\leq\frac{1}{2}\int_{x-y}^{x+y}|\delta g(\xi)|\,d\xi
\leq\frac{1}{2}\int_{-\infty}^{\infty}|\delta g(\xi)|\,d\xi < \frac{\epsilon}{2}.
\tag{3}
$$
The above inequality shows that a small perturbation in $u_y(x,0)$ (in the sense defined in $(1)$) generates a small perturbation in $u(x,y)$ which is uniform in $x$ and $y$. The problem, therefore, is well-posed.$^{(b)}$

$^{(a)}$https://en.wikipedia.org/wiki/Well-posed_problem
$^{(b)}$I'm considering here only perturbations in $u_y(x,0)$; however, it's straightforward to show that a perturbation $\delta f(x)$ in $u(x,0)$ satisfying $|\delta f(x)|<\epsilon$ for all $x\in\Bbb{R}$ generates a perturbation $\delta u(x,y)$ in $u(x,y)$ such that $|\delta u(x,y)|<\epsilon$ for all $x$ and $y$.
