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APM346-2015F => APM346--Home Assignments => HA4 => Topic started by: Victor Ivrii on October 10, 2015, 07:23:42 AM

Title: HA4-P2
Post by: Victor Ivrii on October 10, 2015, 07:23:42 AM
Problem 2: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.2 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.2)
Title: Re: HA4-P2
Post by: Yeming Wen on October 15, 2015, 04:20:46 PM
Suppose $u=\phi_1(x+c_1t)+\psi_1(x-c_1t)$ for $x>0$ and $u=\phi_2(x+c_2t)+\psi_2(x-c_2t)$ for $x<0$.
Also we can solve $\phi_1(x)=\phi(x)-\frac{1}{2}\phi(0)$, $\psi_1(x)=\frac{1}{2}\phi(0)$ for $x>0$.
And $\phi_2=\psi_2=0$ for $x<0$.
It follows that $u=\phi(x+c_1t)$ for $x>c_1t$ and $u=0$ for $x<-c_2t$.
Now we are interested in $\psi_1(x)$ when $x<0$ and $\phi_2(x)$ when $x>0$.
Notice from boundary condition, for $t>0$
$$\phi_1(c_1t)-\psi_1(-c_1t)=\frac{c_1\beta}{c_2}\phi_2(c_2t)$$
$$\phi_1(c_1t)+\psi_1(-c_1t)=a\phi_2(c_2t)$$
Solve for $\psi_1$ and $\phi_2$, we have
$$\psi_1(x)=\frac{c_1\beta-ac_2}{ac_2+c_1\beta}\phi_1(-x) \enspace \text{for } x<0$$
$$\phi_2(x)=\frac{2c_2\phi_1(\frac{c_1x}{c_2})}{ac_2+c_1\beta} \enspace \text{for } x>0$$
Thus, for $-c_2t<x<0$,
$$u=\phi_2(x+c_2t)=\frac{2c_2}{ac_2+c_1\beta}[\phi(\frac{c_1}{c_2}(x+c_2t))-\frac{1}{2}\phi(0)]$$
For $0<x<c_1t$,
$$u=\phi_1(x+c_1t)+\psi_1(x-c_1t)=\phi(x+c_1t)-\frac{1}{2}\phi(0)+\frac{c_1\beta-ac_2}{ac_2+c_1\beta}[\phi(c_1t-x)-\frac{1}{2}\phi(0)]$$
Title: Re: HA4-P2
Post by: Victor Ivrii on October 15, 2015, 04:50:34 PM

:D
(http://www.math.toronto.edu/courses/apm346h1/20159/Forum/P2.6.2a.svg)

(http://www.math.toronto.edu/courses/apm346h1/20159/Forum/P2.6.2b.svg)
Post by: Bruce Wu on October 15, 2015, 10:56:49 PM
Remember that it is just HA4 problem 2 part a): http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.P.2
With $c_{1}=1,c_{2}=2,\alpha=\beta=1$

For $x<-2t$: $$u(x,t)=0$$
For $-2t<x<0$: $$u(x,t)=\frac{4}{3}\phi\left(\frac{1}{2}x+t\right)-\frac{1}{3}\phi(0)$$
For $0<x<t$: $$u(x,t)=\phi(x+t)+\frac{1}{3}\phi(t-x)-\frac{1}{3}\phi(0)$$
For $x>t$: $$u(x,t)=\phi(x+t)$$

Visualization can be found here: http://forum.math.toronto.edu/index.php?topic=660.msg2418#new
Substitute the appropriate constants.
Post by: Yumeng Wang on October 16, 2015, 07:31:26 PM
Where does 1\3 Ï•(0) come from ?
When I plug numbers into graph which was posted by prof, I do not get this term.
Post by: Rong Wei on October 16, 2015, 11:53:51 PM
because plug x = 0 in U|t = 0 = Ï•(x), we will have
U|t = 0, x = 0 = Ï•(0)
(1)

and plug x = 0, t = 0 in u(x,t)=4/3 Ï•(12x+t), we will have

U(0,0) = 4/3 Ï•(0)
(2)

conclude (1) and (2), we will have the answer by Fei Fan Wu
Title: Re: HA4-P2
Post by: Rong Wei on October 17, 2015, 12:10:11 AM
the answer for âˆ’c2t<x<0, Yeming Wen add one constant - 1 / 2  Ï•(0) in the bracket [].
let k be the constant multiplying Ï•(c1/c2(x+c2t)),
plug x = 0 in U | t = 0 by the initial condition, then we will have U(0,0) = Ï•(0)
so this constant should be Ï•(0) (1 / k - 1) in the bracket [].

for 0<x<c1t, same steps used in calculate the constant in the brackets [].
In fact both answers are correct. This IBVP and many other have a solution up to a constant. The problem here is that all solutions save may be one have jumps and we need to explain (in terms of distributions , in the end of the course) those solutions. Which of them is a correct one? Without jumps, if it exists--and this is possible for $\phi(0)=0$ only.
Without jumps solution exists iff $\phi(0)=0$ and $\alpha=1$. There are other solutions anyway.