# Cauchy problem ill-posed?

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 thoughts:

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.

Am I correct?

• If we interpret $y$ as time then we could use some physical arguments here. Suppose you have a system supporting 1D linear waves, e.g., acoustic waves in a string, and you give initial velocity to segments of the string according to your initial condition. This should be enough to determine the further time evolution - correct? – Maxim Umansky Feb 10 at 15:01
• What's our domain? If you have a finite domain, you also need boundary values in $x$ direction. – Wolfgang Bangerth Feb 10 at 17:39