**Hint:** These *compatibility conditions* are on $f,g,h,s$ and may be their derivatives as $x=t=0$. Increasing smoothness by $1$ ads one condition. You do not need to solve problem, just plug and may be differentiate.

**1.** Consider problem:

\begin{align}

&u_{tt}-c^2u_{xx}=f(x,t),&&x>0, t>0,\\

&u|_{t=0}=g(x),\\

&u_t|_{t=0}=h(x),\\

&u|_{x=0}=r(t).

\end{align}

Assuming that all functions $g,h,r$ are very smooth, find conditions necessary for solution $u$ be

(a) $C$ in $\{x>0,t>0\}$ (was on the lecture);

(b) $C^1$ in $\{x>0,t>0\}$;

(c) $C^2$ in $\{x>0,t>0\}$;

(d) $C^3$ in $\{x>0,t>0\}$;

where $C^n$ is the class on $n$-times continuously differentiable functions.

**Solution**

(a) $g(0)=r(0)$ (since $u(x,0)=g(x)$, $u(0,t)=r(t)$).

(b) $h(0)=r'(0)$ (since $u_t(x,0)=h(x)$, $u_t(0,t)=r'(t)$)

(c) $f(0,0)=r''(0)-c^2g''(0)$ (from equation, since $u_{tt}(0,t)=r''(t)$, $u_{xx}(x,0)=g''(x)$).

(d) $f_t(0,0)=r'''(0)-c^2 h''(0)$ (differentiating equation by $t$, we get $u_{ttt}-c^2u_{xxt}=f_{t}$, and then like in (c)).

**2.** Consider problem:

\begin{align}

&u_{tt}-c^2u_{xx}=f(x,t),&&x>0, t>0,\\

&u|_{t=0}=g(x),\\

&u_t|_{t=0}=h(x),\\

&u_x|_{x=0}=s(t).

\end{align}

Assuming that all functions $g,h,s$ are very smooth, find conditions necessary for solution $u$ be

(a) $C$ in $\{x>0,t>0\}$ (automatically);

(b) $C^1$ in $\{x>0,t>0\}$;

(c) $C^2$ in $\{x>0,t>0\}$;

(d) $C^3$ in $\{x>0,t>0\}$.

**Solution.**

(a) Automatically

(b) $g'(0)=s(0)$

(c) $h'(0)=s'(0)$ (looking at $u_{xt}(x,t)$)

(d) $s''(0)-c^2 h'''(0)=f_x(0,0)$ (differentiating equation by $x$, we get $u_{ttx}-c^2u_{xxx}=f_{x}$, and then like in 1(c)).

**Remark** We checked the continuity etc ast $\{x\ge 0, t\ge 0\}$ rather than $\{x>0,t>0\}$ but for wave equation these are equivalent since singularities at $(0,0)$ would propagate in along $\{x=ct\}$. For other types of equations this would not be the case.