Toronto Math Forum
APM3462018S => APM346––Home Assignments => Web Bonus Problems => Topic started 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})

I just want to check if I did it right before writing out details:
u(t,x) = 3xt + xt^{2} for all (t,x)

This is my solution, not sure true or not. If it is correct, I will post more details. $u(x,t)=\sqrt2[(x+t)^\frac{3}{2}+(xt)^\frac{3}{2}]+\frac{\sqrt2}{3}[(x+t)^\frac{1}{2}(xt)^\frac{1}{2}]$

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.

May I ask how do we make a function of x+x^{2}/2 equals (x^{3}+x^{2})/2. I have been thinking for a long time and still can't figure it out.

Hint: What is the general solution of (\ref{A})?

Is it u(x,t)=C_{1}(x+t)+C_{2}(xt). Then I plug in t=x^{2}/2 for u and u_{t} and I get C_{1}(x+x^{2}/2)= (x^{3}+x^{2})/2 and C_{2}(xx^{2}/2) = (x^{3}x^{2})/2 which is the question i asked before. Can you tell me a bit more? I still don't get it.

Hint: What is the general solution of (\ref{A})?
For (1) it is $$ u(t,x)= \phi(x+t) + \psi(xt)$$
For the Cauchy Problem it is:
$$u(t,x)=\frac{1}{2}\bigl[g(x+t)+g(xt)\bigr]+
\frac{1}{2}\int_{xt}^{x+t} h(y)\,dy$$

Hint: What is the general solution of (\ref{A})?
For (1) it is $$ u(t,x)= \phi(x+t) + \psi(xt)$$
For the Cauchy Problem it is:
$$u(t,x)=\frac{1}{2}\bigl[g(x+t)+g(xt)\bigr]+
\frac{1}{2}\int_{xt}^{x+t} h(y)\,dy$$
I thought sol'n to Cauchy problem only works in IVP with t=0.

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

General solution for wave equation:
$$u(t,x)= \phi(x+t) + \psi(xt)$$
For the first initial condition: $u_{t=x^2/2}= x^3$
$$ x^3 = \phi(x+x^2/2) + \psi(xx^2/2)\tag{A}$$
Second condition: $u_t_{t=x^2/2}$
$$ 2x = \phi'(x+x^2/2)  \psi'(xx^2/2)\tag{B}$$
$$ \implies x^2 = \phi(x+x^2/2)  \psi(xx^2/2)\tag{C}$$
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

Second condition: $u_t_{t=x^2/2}$
$$ 2x = \phi'(x+x^2/2)  \psi'(xx^2/2) $$
$$ \implies x^2 = \phi(x+x^2/2)  \psi(xx^2/2) $$
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.

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 .....

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?

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..

Differentiating Ioana's $(A)$ and equating it with his(her?) $(B)$, we have the symbolic system
$$\left( \begin{array}{ccc}
1 & 1 & 2x\\
1+x& 1x&3x^{2}\\
\end{array} \right)
\implies
\left( \begin{array}{c}
\varphi'(X)\\
\psi'(Y)
\end{array} \right)
=
\left( \begin{array}{c}
X\\
Y
\end{array} \right)
\text{ where X, Y are the arguments of $\varphi, \psi$ resp. }
\implies
\left( \begin{array}{c}
\varphi(X)\\
\psi(Y)\end{array} \right)
=
\left( \begin{array}{c}
X^{2}/2+C_{1}\\
Y^{2}/2+C_{2}
\end{array} \right)
\implies
u=(x+t)^{2}/2  (xt)^{2}/2 + const.
$$
Fixed now. For uniqueness we impose that the characteristics intersect the initial data, that is, precisely when
$$x^{2}2x+2C, x^{2}+2x2C$$
both have solution. This happens whenever
$$t1/2\leq x\leq t+1/2.$$

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.

Differentiating Ioana's $(A)$ and equating it with his(her?) $(B)$, we have the symbolic system
$$\left( \begin{array}{ccc}
1 & 1 & 2x\\
1+x& 1x&3x^{2}\\
\end{array} \right)
\implies
\left( \begin{array}{c}
\varphi'(X)\\
\psi'(Y)
\end{array} \right)
=
\left( \begin{array}{c}
X\\
Y
\end{array} \right)
\text{ where X, Y are the arguments of $\varphi, \psi$ resp. }
\implies
\left( \begin{array}{c}
\varphi(X)\\
\psi(Y)\end{array} \right)
=
\left( \begin{array}{c}
X^{2}/2+C_{1}\\
Y^{2}/2+C_{2}
\end{array} \right)
\implies
u=(x+t)^{2}/2  (xt)^{2}/2 + Cx+D.
$$
The last two terms in the final step are empirical, and I urgently seek a theoretical account for it.
u=(x+t)^{2}/2(xt)^{2}/2, where u is define on t>=0 and x is any real number.

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!)