# Toronto Math Forum

## APM346-2018S => APM346––Home Assignments => Web Bonus Problems => Topic started by: Victor Ivrii on January 19, 2018, 05:53:52 AM

Title: Web bonus problem -- Week 4
Post by: Victor Ivrii on January 19, 2018, 05:53:52 AM
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.

Title: Re: Web bonus problem -- Week 4
Post by: Jingxuan Zhang on January 22, 2018, 09:30:08 PM
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
Title: Re: Web bonus problem -- Week 4
Post by: Victor Ivrii on January 23, 2018, 05:34:09 AM
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

Title: Re: Web bonus problem -- Week 4
Post by: Ioana Nedelcu on January 24, 2018, 01:22:42 AM
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
Title: Re: Web bonus problem -- Week 4
Post by: Victor Ivrii on January 24, 2018, 03:28:37 AM
Ioana, good attempt albeit not finished. And you compare wrong functions.

Hint: one of $g,h$ and $s$ (or $r$) should be involved
Title: Re: Web bonus problem -- Week 4
Post by: Jingxuan Zhang on January 24, 2018, 06:18:10 AM
Ioana would you explain how to you get this? I thought the condition should be only at a point, such as origin as in my edited post. For I don't see the connexion from $$u(x,0)=g(x)$$ to $$u_{t}(x,0)=g'(x)$$, and how this helps to ensure continuity.
Title: Re: Web bonus problem -- Week 4
Post by: Ioana Nedelcu on January 24, 2018, 05:54:17 PM
Jingxuan: why should there be only continuity at one point? The entire derivative function should be continuous on its domain.

My thinking in equating the two functions is that the derivative with respect to t is given by the two functions, so to ensure that they both give the same value at a point x they should be equal everywhere (since they are already smooth themselves).

The arguments of the functions threw me off a bit in the beginning. Now I'm thinking it should be $r'(0) = h(x) = u_{t}, t=0$
Title: Re: Web bonus problem -- Week 4
Post by: Victor Ivrii on January 25, 2018, 04:29:49 AM
1. All functions $f,g,h,r,s$ are assumed to be very good functions. "Bad things" can happen only at $(0,0)$ and propagate along $x=ct$.

a. The first, as we talked is continuity of $u$ itself. But $u|_{t=0}=g(x)$ and $u|_{x=0}=r(t)$ (in the case of this condition). What conflict could be? $u(0,0)=g(0)$, $u(0,0)=r(0)$ and they are not equal. So, we have one condition:
$$g(0)=r(0).$$

b. Now try $u_x(0,0)$ or $u_t(0,0)$. Which of them could be calculated in two ways?
Title: Re: Web bonus problem -- Week 4
Post by: Jingxuan Zhang on January 26, 2018, 05:50:25 AM
1. All functions $f,g,h,r,s$ are assumed to be very good functions. "Bad things" can happen only at $(0,0)$ and propagate along $x=ct$.

Hence I again advocate my answer in reply #2. Ioana this is what I meant, and especially that imposing the function to coincide at one point does not require it to be equal everywhere.
Title: Re: Web bonus problem -- Week 4
Post by: Victor Ivrii on January 28, 2018, 04:25:50 AM
Title: Re: Web bonus problem -- Week 4
Post by: Victor Ivrii on January 30, 2018, 07:15:01 AM
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.