# Toronto Math Forum

## APM346-2018S => APM346––Home Assignments => Web Bonus Problems => Topic started by: Victor Ivrii on January 12, 2018, 09:03:13 AM

Title: Web bonus problem -- Week 2
Post by: Victor Ivrii on January 12, 2018, 09:03:13 AM
Find solution $u=u(x,t)$ and describe domain, where it is uniquely defined
\begin{align}
&u_{tt}-u_{xx}=0,
\label{A}\\[2pt]
&u|_{t=x^2/2}= x^3,
\label{B}\\[2pt]
&u_t|_{t=x^2/2}= {\color{blue}{2}}x.
\label{C}
\end{align}
Correction: I replace $x$ by $2x$ in (\ref{C})
Title: Re: Web bonus problem -- Week 2
Post by: Jaisen Kuhle on January 12, 2018, 02:52:56 PM
I just want to check if I did it right before writing out details:

u(t,x) = 3xt + xt2 for all (t,x)
Title: Re: Web bonus problem -- Week 2
Post by: Sheng Gao on January 12, 2018, 07:47:21 PM
This is my solution, not sure true or not. If it is correct, I will post more details. $u(x,t)=\sqrt2[(x+t)^\frac{3}{2}+(x-t)^\frac{3}{2}]+\frac{\sqrt2}{3}[(x+t)^\frac{1}{2}-(x-t)^\frac{1}{2}]$
Title: Re: Web bonus problem -- Week 2
Post by: Victor Ivrii on January 12, 2018, 08:33:40 PM
Jaisen: definitely does not satisfy even equation.

Sheng: satisfies equation but not conditions (I checked solution against old conditions).

I made an error. Please ignore old initial conditions. You need to post calculations, not just answers.
Title: Re: Web bonus problem -- Week 2
Post by: Ziyuan Wang on January 14, 2018, 01:56:43 PM
May I ask how do we make a function of x+x2/2 equals (x3+x2)/2. I have been thinking for a long time and still can't figure it out.
Title: Re: Web bonus problem -- Week 2
Post by: Victor Ivrii on January 14, 2018, 02:00:06 PM
Hint: What is the general solution of (\ref{A})?
Title: Re: Web bonus problem -- Week 2
Post by: Ziyuan Wang on January 14, 2018, 05:00:30 PM
Is it u(x,t)=C1(x+t)+C2(x-t). Then I plug in t=x2/2 for u and ut and I get C1(x+x2/2)= (x3+x2)/2 and C2(x-x2/2) = (x3-x2)/2 which is the question i asked before. Can you tell me a bit more? I still don't get it.
Title: Re: Web bonus problem -- Week 2
Post by: Jaisen Kuhle on January 14, 2018, 06:08:52 PM
Hint: What is the general solution of (\ref{A})?

For (1) it is $$u(t,x)= \phi(x+t) + \psi(x-t)$$

For the Cauchy Problem it is:

$$u(t,x)=\frac{1}{2}\bigl[g(x+t)+g(x-t)\bigr]+ \frac{1}{2}\int_{x-t}^{x+t} h(y)\,dy$$
Title: Re: Web bonus problem -- Week 2
Post by: Ziyuan Wang on January 14, 2018, 06:28:20 PM
Hint: What is the general solution of (\ref{A})?

For (1) it is $$u(t,x)= \phi(x+t) + \psi(x-t)$$

For the Cauchy Problem it is:

$$u(t,x)=\frac{1}{2}\bigl[g(x+t)+g(x-t)\bigr]+ \frac{1}{2}\int_{x-t}^{x+t} h(y)\,dy$$

I thought sol'n to Cauchy problem only works in IVP with t=0.
Title: Re: Web bonus problem -- Week 2
Post by: Victor Ivrii on January 14, 2018, 07:24:10 PM
Ziyuan
What are $C_1$ and $C_2$ in your "solution"? Please, read Sect 2.3.

Jaisen
D'Alembert formula as you wrote works only in the case with the data at $t=0$ and could be modified for the case with the data at $t=t_0$ (which is constant).

You need to plug this general solution into initial conditions and then find $\phi$ and $\psi$ from there
Title: Re: Web bonus problem -- Week 2
Post by: Ioana Nedelcu on January 14, 2018, 09:20:00 PM
General solution for wave equation:
$$u(t,x)= \phi(x+t) + \psi(x-t)$$
For the first initial condition: $u|_{t=x^2/2}= x^3$
$$x^3 = \phi(x+x^2/2) + \psi(x-x^2/2)\tag{A}$$
Second condition: $u_t|_{t=x^2/2}$
$$2x = \phi'(x+x^2/2) - \psi'(x-x^2/2)\tag{B}$$
$$\implies x^2 = \phi(x+x^2/2) - \psi(x-x^2/2)\tag{C}$$
Combining the two equations,
$$\phi(x+x^2/2) = (x^3 + x^2) /2$$
$$\psi(x+x^2/2) = (-x^2 + x^3) /2$$

But now I'm not sure how to change the functions to get their argument to $x + t$ and $x - t$ to find $u$ in terms of x and t
Title: Re: Web bonus problem -- Week 2
Post by: Jaisen Kuhle on January 14, 2018, 10:39:07 PM
Second condition: $u_t|_{t=x^2/2}$

$$2x = \phi'(x+x^2/2) - \psi'(x-x^2/2)$$
$$\implies x^2 = \phi(x+x^2/2) - \psi(x-x^2/2)$$

I'm not following your implication here. It seems we have:

$$u_{x^2/2} = \phi_{x^2/2} - \psy_{x^2/2) = 2x$$

So don't we need to integrate with respect to $x^2/2$, in which case:

$$\phi(x+x^2/2) - \psi(x+x^2/2) = x^3 + C$$

Also, is this C truly constant?

Edit: Not sure what's wrong with my script for it not to display.
Title: Re: Web bonus problem -- Week 2
Post by: Victor Ivrii on January 14, 2018, 10:42:20 PM
Ioana
Equation (A), (B) (I put tags here) are correct, but (C) does not follow from (B). Explain why.

Hint. We cannot integrate (B) but .....
Title: Re: Web bonus problem -- Week 2
Post by: Ioana Nedelcu on January 15, 2018, 12:06:43 AM
Thank you! I'm pretty sure my integration was wrong to begin with - I was confused with the functions' arguments and what the constants of integration should be (like Jaisen said)

If we can't integrate B, maybe we can differentiate A and solve for the functions this way?
Title: Re: Web bonus problem -- Week 2
Post by: Victor Ivrii on January 15, 2018, 12:32:21 AM
I'm pretty sure my integration was wrong to begin with - I was confused with the functions' arguments...
If we can't integrate (B), maybe we can differentiate (A)?
Yes, indeed. $\phi '(...)$ is a derivative with respect to ... , not $x$. And if you have ideas try them, rather than waiting my approval..
Title: Re: Web bonus problem -- Week 2
Post by: Jingxuan Zhang on January 15, 2018, 06:50:44 AM
Differentiating Ioana's $(A)$ and equating it with his(her?) $(B)$, we have the symbolic system
$$\left( \begin{array}{cc|c} 1 & -1 & 2x\\ 1+x& 1-x&3x^{2}\\ \end{array} \right) \implies \left( \begin{array}{c} \varphi'(X)\\ \psi'(Y) \end{array} \right) = \left( \begin{array}{c} X\\ -Y \end{array} \right) \text{ where X, Y are the arguments of \varphi, \psi resp. } \implies \left( \begin{array}{c} \varphi(X)\\ \psi(Y)\end{array} \right) = \left( \begin{array}{c} X^{2}/2+C_{1}\\ -Y^{2}/2+C_{2} \end{array} \right) \implies u=(x+t)^{2}/2 - (x-t)^{2}/2 + const.$$

Fixed now. For uniqueness we impose that the characteristics intersect the initial data, that is, precisely when
$$x^{2}-2x+2C, x^{2}+2x-2C$$
both have solution. This happens whenever
$$-t-1/2\leq x\leq t+1/2.$$
Title: Re: Web bonus problem -- Week 2
Post by: Victor Ivrii on January 15, 2018, 07:00:58 AM
Jingxuan
Misprint at the very end, correct it. However $\phi'(X)=X$ implies $\phi(x)=X^2/2+c$, etc. So you need to take it into account and find it from (B).

Just for fun, simplify

We need also answer the question, where this solution is uniquely determined.
Title: Re: Web bonus problem -- Week 2
Post by: Ziyuan Wang on January 15, 2018, 02:38:01 PM
Differentiating Ioana's $(A)$ and equating it with his(her?) $(B)$, we have the symbolic system
$$\left( \begin{array}{cc|c} 1 & -1 & 2x\\ 1+x& 1-x&3x^{2}\\ \end{array} \right) \implies \left( \begin{array}{c} \varphi'(X)\\ \psi'(Y) \end{array} \right) = \left( \begin{array}{c} X\\ -Y \end{array} \right) \text{ where X, Y are the arguments of \varphi, \psi resp. } \implies \left( \begin{array}{c} \varphi(X)\\ \psi(Y)\end{array} \right) = \left( \begin{array}{c} X^{2}/2+C_{1}\\ -Y^{2}/2+C_{2} \end{array} \right) \implies u=(x+t)^{2}/2 - (x-t)^{2}/2 + Cx+D.$$

The last two terms in the final step are empirical, and I urgently seek a theoretical account for it.
u=(x+t)2/2-(x-t)2/2, where u is define on t>=0 and x is any real number.
Title: Re: Web bonus problem -- Week 2
Post by: Ioana Nedelcu on January 15, 2018, 11:32:33 PM
Simplifying further, we get $$u = 2xt + C$$ but using the second initial condition implies C = 0 so the solution is $u(x,t) = 2xt$ (thanks for going over this in class!)