### Author Topic: Web bonus problem -- Week 2  (Read 2320 times)

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Re: Web bonus problem -- Week 2
« Reply #15 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
$$-t-1/2\leq x\leq t+1/2.$$
« Last Edit: January 15, 2018, 07:37:51 PM by Jingxuan Zhang »

#### Victor Ivrii ##### Re: Web bonus problem -- Week 2
« Reply #16 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.

#### Ziyuan Wang

• Newbie
• • Posts: 4
• Karma: 1 ##### Re: Web bonus problem -- Week 2
« Reply #17 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.

#### Ioana Nedelcu

• Full Member
•   • Posts: 29
• Karma: 3 ##### Re: Web bonus problem -- Week 2
« Reply #18 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!)