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Web bonus problem -- Week 4

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Victor Ivrii:
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.

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 $\{x>0,t>0\}$.

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.

Jingxuan Zhang:
Typo found: the boundary condition has the wrong argument. Indeed--fixed, V.I.
1. On the region $x>ct$, $u$ is automatically $C^{n}$ since the conditions are good. On $0<x<ct$:
a. intersection at origin requires $$r(0)=g(0)$$
b. plugging in the formula and match to $(3)$ gives $$r'(t)=\frac{1}{2}(h(ct)+h(-ct))+\frac{c}{2}(g'(ct)-g'(-ct))$$
c. d. I think the idea is to equate mixed partials but I am still trying to figure out what exactly is needed to infer the condition. Save for morrow.

Edit:
b. Was wrong. The approach was wrong, I should not plug in the general solution formula but rather just examine the initial data. To be $C^{1}$ it only needs
the various directional derivative agree to each other when close up to origin, viz.,
$$\lim_{x\to 0}u_{x}(x,0)=\lim_{t\to 0} u_{t}(0,t) \implies g'(0)=r'(0). $$

Wrong, since you equalize limits $u_t$ and $u_x$. Hint: equalize limits $u_t$ and $u_t$

c. This I think my idea was right. To have the mixed partial agree in particular we must have
$$\lim_{x\to 0} u_{tx}(x,0)=\lim_{x\to 0}u_{xt}(x,0) \implies h'(0)=(\frac{d}{dt}g')(0)=0. $$
Completely wrong. Use equation
d. Similarly
$$\lim_{x\to 0} u_{txx}(x,0)= \lim_{x\to 0}u_{xxt}(x,0)\implies h''(0)=(\frac{d}{dt}g'')(0)=0. $$
There are other equation arised by permuting partials but those are fulfilled automatically. Please correct me but anyway someone else please attempt the other part.
The same

Victor Ivrii:
Well, $r(0)=g(0)$ is obvious. And this is the only condition needed for the first problem, (a); no for the second (a).

You do not need to find solution to get compatibility conditions

Ioana Nedelcu:
A $C^1$ function has continuous derivative so maybe a condition should be that $$ u_{t} = h(x) = g'(x)$$ Otherwise if $ h(x) \neq g'(x) \forall x $ then there could be points of discontuity

Victor Ivrii:
Ioana, good attempt albeit not finished. And you compare wrong functions.

Hint: one of $g,h$ and $s$ (or $r$) should be involved

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