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Messages - XinYu Zheng

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FE / Re: FE6
« on: December 19, 2016, 11:14:28 PM »
$R(r)$ does not have a two-sided Dirichlet boundary condition. Such a boundary condition would be something like $v|_{r=0}=v|_{r=1}=0$. In particular, this must be true for all $t$, while in the problem you are only given that $v|_{r=1}=0$ at $t=0$.
I think the best way to solve this problem is to just ignore separation of variables and solve IVP for $v$ using the tools of chapter 2.

FE / Re: FE6
« on: December 18, 2016, 08:16:51 PM »
You cannot assume that the eigenvalues are integers. By writing the general solution as such a sum, you are assuming that $u=0$ at $r=\pi$, but this is not given. Indeed, there are no boundary conditions for $r>0$.

FE / Re: FE4
« on: December 14, 2016, 03:34:45 PM »
Note: The sector is not a half circle. There is an absolute value on $y$.

TT2 / Re: TT2-P5
« on: November 17, 2016, 11:13:51 AM »
Just a remark: if a function is different or undefined at a single point (or any finite number of points) it does not change the value of the integral. This is because a finite number of points has Jordan measure zero ( So there is no problem in not defining $\theta(0)$.

TT2 / Re: TT2-P4
« on: November 17, 2016, 08:08:36 AM »
Introduce $u=R(r)\Theta(\theta)$. Then after separating variables the angular equation will be $\Theta''+\lambda\Theta=0$ with Dirichlet B.C at 0 and $3\pi/2$. We know the solution to this problem:
With $\lambda_n=4n^2/9$, $n\geq 1$. Then the radial equation will have solution $R(r)=Ar^{2n/3}+Br^{-2n/3}$. We will have to drop the second term because they blow up at the origin. Thus our general solution is
$$u(r,\theta)=\sum_{n\geq 1} A_nr^{2n/3}\sin(2n\theta/3)$$
Now applying the B.C. at $r=8$ we have
$$1=\sum_{n\geq 1} A_n4^n\sin(2n\theta/3)$$
And the coefficients can be calculated the usual way:
$$A_n=\frac{1}{4^n}\frac{2}{\frac{3\pi}{2}}\int_0^{3\pi/2}\sin(2n\theta/3)\,d\theta=\frac{1}{4^n}\frac{4}{3\pi}\frac{3}{2n}\cos(2n\theta/3)|_{3\pi/2}^0=\frac{1}{4^{n-1}}\frac{1}{2n\pi}(1-(-1)^n)=\frac{1}{4^{n-1}}\frac{1}{n\pi}\,\,\,\,\,\,\text{n odd, 0 otherwise}$$
So our solution is
$$u(r,\theta)=\sum_{n\geq 1, n\,\,odd}\frac{1}{4^{n-1}}\frac{1}{n\pi}r^{2n/3}\sin(2n\theta/3)$$

TT2 / Re: TT2-P3
« on: November 17, 2016, 08:07:51 AM »
Introduce $u=X(x)Y(y)$ then applying separation of variables yields
Now we have two ODEs, namely $X''+\lambda_1X=0$ and $Y''+\lambda_2Y=0$ both with Neumann B.C. at $0,a$ and $0,b$ respectively. We know the solution to these problems:
\lambda_{1,0}=0& X_0=\frac{1}{2}\\
\lambda_{1,n}=\frac{n^2\pi^2}{a^2}& X_{n}=\cos(n\pi x/a)\\
\lambda_{2,0}=0& Y_0=\frac{1}{2}\\
\lambda_{2,n}=\frac{m^2\pi^2}{b^2}& Y_{m}=\cos(m\pi x/b)\\
Where $n,m\geq 1$. So the eigenfunctions are
$$u_{n,m}(x,y)=\cos(n\pi x/a)\cos(m\pi x/b)$$
With eigenvalues $\lambda_{m,n}=\lambda_{1,n}+\lambda_{2,n}=\pi^2(n^2/a^2+m^2/b^2)$. We can absorb the case where $m=0$ or $n=0$ by allowing us to plug $m,n=0$ in this equation.

TT2 / Re: TT2-P2
« on: November 17, 2016, 08:07:03 AM »
Applying fourier transform with respect to $x$ so that $u(x,y)\to \hat{u}(k,y)$ the PDE becomes
This PDE has general solution $\hat{u}=A(k)e^{-|k|y}+B(k)e^{|k|y}$. We drop the second term because it goes unbounded. Now applying the B.C. we see that $\hat{g}(k)=A(k)$. So we compute $\hat{g}(k)$:
$$\hat{g}(k)=\frac{1}{2\pi}\int_{-1}^1e^{-ikx}\,dx=\frac{1}{2\pi}\int_0^1 e^{-ikx}+e^{ikx}\,dx=\frac{1}{\pi}\int_0^1 \cos(kx)\,dx=\frac{\sin k}{k\pi}$$
Where in the middle we have split the integral in two parts and did a change of variables $x\to -x$ in the second one. Now we just need to apply IFT for the solution:
$$u(x,y)=\frac{1}{\pi}\int_{-\infty}^\infty \frac{\sin k}{k}e^{-|k|y+ikx}\,dk$$

TT2 / Re: TT2-P1
« on: November 17, 2016, 08:05:25 AM »
Introduce $u=X(x)T(t)$ and then separation of variables yield $X''+\lambda X=0$ and $T''+\lambda T=0$. Assume that $\lambda=\omega^2$ where $\omega\geq 0$. Then for $\omega\neq 0$, we have solutions $X_1=\sin \omega x$ and $X_2=\cos\omega x$. Applying the boundary conditions to $X_1$ we find
$$0=\sin \omega \pi$$
$$-1=\cos\omega \pi$$
Applying the boundary conditions to $X_2$ yields the same thing. The first equation suggests that $\omega=n$ where $n=1,2,...$. The second equation suggests that $\omega=2n+1$, $n=0,1,2,...$. To satisfy both, we must take the second one. So we have $\omega_n=2n+1$, $n=0,1,2,...$. Now for $\omega=0$ the solution is $Ax+B$. But the second B.C. requires $B=-B$ so $B=0$. Then applying the first B.C. we find $A=-A$ so $A=0$. So we do not have this eigenvalue. Thus our solutions for the spacial part is
$$X_{1,n}=\sin((2n+1)x), X_{2,n}=\cos((2n+1)x)$$
With $\lambda_n=(2n+1)^2$, $n=0,1,2...$. Then the equation for time can be solved immediately: $T=A\sin((2n+1)t)+B\cos((2n+1)t)$. Applying $u_t|_{t=0}=0$ we find $A=0$, and now we may write down the general solution:
$$u(x,t)=\sum_{n\geq 0}[A_n\sin((2n+1)x)+B_n\cos((2n+1)x)]\cos((2n+1)t)$$
Applying the condition $u_{t=0}=1$ we have
$$1=\sum_{n\geq 0}A_n\sin((2n+1)x)+B_n\cos((2n+1)x)$$
At this point the coefficients can be calculated via the standard method using orthogonality:
$$A_n=\frac{2}{\pi}\int_0^\pi \sin((2n+1)x)\,dx=\frac{2}{\pi(2n+1)}\cos((2n+1)x)|_\pi^0=\frac{4}{\pi (2n+1)}$$
$$B_n=\frac{2}{\pi}\int_0^\pi \cos((2n+1)x)\,dx=0$$
Where the second integral is zero because we will be evaluating sine functions at integer values of $\pi$. So we have our solution:
$$u(x,t)=\frac{4}{\pi}\sum_{n\geq 0}\frac{1}{2n+1}\sin((2n+1)x)\cos((2n+1)t)$$

Q6 / Q6
« on: November 10, 2016, 08:54:19 PM »
u_{xx}+u_{yy}=0& r<a\\
1 & 0<\theta<\pi\\
-1 & \pi<\theta <2\pi

(a) What is a necessary and sufficient condition on $f(\theta)$ for solution to exist? Is this condition satisfied here?
(b) Is the solution unique?

For Laplace's equation in the inner disk we know that the general solution takes the form
$$u(r,\theta)=\frac{A_0}{2}+\sum_{n\geq 1} r^n (A_n\sin(n\theta)+B_n\cos(n\theta))$$
Where we have dropped the logarithm and any terms with $r^{-n}$. Applying the boundary condition we have
$$f(\theta)=\sum_{n\geq 1}n a^{n-1}(A_n\sin(n\theta)+B_n\cos(n\theta))$$
At this point, the coefficients can be directly calculated:
$$A_n=\frac{a^{1-n}}{n\pi}\int_0^{2\pi}f(\theta)\sin(n\theta)\,d\theta=\frac{a^{1-n}}{n\pi}\left(\int_0^{\pi}\sin(n\theta)\,d\theta+\int_\pi^{2\pi}-\sin(n\theta)\,d\theta\right)=\frac{a^{1-n}}{n^2\pi}(\cos(n\theta)|_{\pi}^0+\cos(n\theta)|_\pi^{2\pi})=\frac{4a^{1-n}}{n^2\pi}\,\,\,\text{n odd, 0 otherwise}$$
Where the integrals are 0 because we will be evaluating sine functions at integer multiples of $\pi$. Thus we have our solution:
$$u(r,\theta)=\frac{A_0}{2}+\sum_{n\geq 1, n\,\,odd}\frac{4a^{1-n}}{n^2\pi}r^n\sin(n\theta)$$

(a) Note that in the fourier expansion of $f(\theta)$ we have no constant term. Thus for solution to exist we must demand
Which is fulfilled here.
(b) No. In Neumann problems the solution is defined up to a constant, in this case $A_0/2$.

Chapter 6 / Re: Small typos in Chapter 6
« on: November 09, 2016, 09:17:20 AM »
Some more typos in Section 6.5:

1. Equation (6) should contain $h(\theta ')$, not $g(\theta ')$.
2. In the derivation immediately following equation (6), we have a line that says $-\frac{a}{\pi}\mathrm{Re}\log (1-ra^{-1}e^{i(\theta-\theta')})=-\frac{a}{2\pi}\log(a^{-2}(1-ra^{-1}e^{i(\theta-\theta')})(1-ra^{-1}e^{-i(\theta-\theta')}))$. The $a^{-2}$ in the logarithm should not be there at this stage. It should be there in the line right after, so the final expression is correct.
3. At the very bottom of the page where it says "...where for sector $\{r<a, 0<\theta<\alpha\}$ we should set...", the $B_n$ is missing.

APM346--Misc / Interesting Problem: Ant on a rubber rope
« on: November 05, 2016, 06:17:23 PM »
Here is a problem with a counter intuitive result. Before reading the solution, I strongly encourage to try solving the problem yourself. It is rather fun to work out. This is not a PDE problem, but an ODE one. However, solving the ODE requires some cleverness.

Suppose we start with a rubber rope of rest length $L_0$ with one end attached to the wall. At $t=0$, an ant is placed on the rope at the loose end and begins crawling towards the wall with a speed of $u$ relative to the rope. Meanwhile, the free end of the rope is being pulled away from the wall at a constant speed $v\gg u$, causing the rope to stretch uniformly.

(a) Show that, no matter how large $v$ is compared to $u$, the ant will always reach the wall eventually. Find an explicit expression for the time it takes to do so in terms of $u$, $v$, and $L_0$.

(b) Suppose now that in addition to the rope being pulled away at speed $v$, it is also accelerating with acceleration $a>0$, such that $2aL_0<v^2$. Show that in this case, the ant may or may not reach the wall depending on what $u$ is. Show that there is a lower bound on $u$ in terms of $a$, $L_0$, and $v$ where if the ant crawls any slower it will never reach the wall.

(c) Examine the case where $2aL_0=v^2$. Why is this case interesting?

(d) Examine the case where $2aL_0>v^2$.

(e) It is sometimes said that because the universe is expanding, light from galaxies that are too distant may never reach us. In light of the results of this problem, comment on this statement.

(a) This is a one dimensional problem. Let us place our origin at the location of the wall. Then as a function of $t$, the location of the loose end of the rope is $L(t)=L_0+vt$. Let the location of the ant be $X(t)$. We know that the ant crawls towards the wall with speed $u$ relative to the rope. But at every moment it is also being pulled away from the wall by virtue of standing on the rope because the rope is stretching. Since the rope stretches uniformly, the speed at which the ant is being pulled away is $vX(t)/L(t)$ (ie. it is proportional to the fraction of the rope the ant occupies as measured from the wall). Then we can write the following differential equation describing the ant's motion:
Solving this equation can be rather tricky using standard techniques, so we make a change of variables $S(t)=X(t)/L(t)$. Then $X'=S'L+L'S$. Subbing this into the equation we get a major simplification:
$$S'L=-u\implies S'=-\frac{u}{L}=-\frac{u}{L_0+vt}$$
Now this can be solved easily:
Note that the initial condition at $t=0$ requires $S(0)=1$, thus $C=\frac{u}{v}\ln L_0+1$, and the final equation becomes
Requiring that the ant reaches the wall is equivalent to finding $T$ such that $S(T)=0$. Solving:
$$-\frac{u}{v}\ln\left(1+\frac{vt}{L_0}\right)+1=0\implies T=\frac{L_0}{v}(e^{v/u}-1)$$
So we can see that the time it takes the ant to reach the wall grows exponentially with the ratio $v/u$, but nevertheless always finite. Just as a comparison, suppose that $v=1km/s$ and $u=1m/s$. Then $T$ will be on the order of $e^{1000}\approx 1.97\times 10^{434}$ seconds, which is about $10^{417}$ times the age of the universe.

(b) This time, we have $L(t)=L_0+vt+at^2/2$. We make the same substitution $S=X/L$ and now the equation reduces to
Completing the square in the denominator yields
Where $k\equiv 1/\sqrt{v^2-2aL_0}$. The denominator can be factored as a difference of squares and then decomposed into partial fractions. The result after integration will be two logarithms (here combined into one):
Once again, we require $S(0)=1$. Then $C=ku\ln\left(\frac{1-kv}{1+kv}\right)+1$ and after some rearranging we obtain
Now let us try to solve $S(T)=0$ where $T>0$. Define $d=e^{-\frac{1}{ku}}$. Then we have
After some rearranging we arrive at
Note that since $d<1$ and $kv>1$, the numerator is positive. Thus for a positive solution of $T$ to exist we must demand that $1-kv+d(1+kv)>0$, which is equivalent to
Now subbing in for $d=e^{-\frac{1}{ku}}$ and solving for $u$ we obtain the lower bound
Thus in the case where there is non-zero acceleration satisfying $2aL_0<v^2$, there exists a lower bound on $u$. The ant can no longer make it to the wall all the time.

(c) In the borderline case where $2aL_0=v^2$, the ODE for $S(t)$ reduces to
Which can be solved right away:
Applying the initial condition $S(0)=1$ yields $C=1-2u/v$ and we have
Now, solving for $S(T)=0$ yields (after rearranging)
For positive $T$ solutions, we demand that $\frac{2uv}{2u-v}>v$, or $\frac{2u}{2u-v}>1$. For this to happen, we only need to demand that the denominator is positive, which yields the lower bound $u>v/2$. This case is particularly interesting because the lower bound only depends on $v$ and nothing else.

(d) Once again, the same story, except this time we define $k\equiv 1/\sqrt{2aL_0-v^2}$. Then the ODE for $S(t)$ becomes
The solution of this ODE is in the form of arctan:
Using initial condition $S(0)=1$ yields $C=2ku\arctan(kv)+1$ and thus we have
Note that this function is decreasing with increasing $t$ since $\arctan$ is strictly increasing. Thus to get a positive solution for $S(t)=0$, we can find the threshold by taking $t\to \infty$ and demand $\lim_{t\to\infty}S(t)=0$. Note that $\arctan(k(at+v))\to \pi/2$ in the limit, thus we have
$$0=-2ku(\pi/2-\arctan(kv))+1\implies u=\frac{1}{2k(\pi/2-\arctan(kv))}$$
This is the threshold velocity of crawling. Anything slower than this will cause $S(t)$ to never intersect the $x$ axis, thus the ant will never reach the wall.

(e) In light of the results of the problem, we can say that if the rate of expansion of the universe is constant and that the universe expands uniformly, then light from galaxies will always eventually reach us, regardless of how far away they are or how fast the galaxies are receding. This corresponds to the result of (a), where the light plays the role of the ant and the rubber rope is the fabric of spacetime. However, the results of Hubble tells us that the universe is accelerating in its expansion. From the conclusions of (b), (c), and (d) above, we can see that if this were the case, then it is possible that the light of some galaxies will never reach us. Their existence will never be known to us, nor to any future lifeforms that will ever inhabit our planet.

APM346--Misc / Re: Halloween Challenge 8
« on: November 04, 2016, 03:05:56 PM »
(a) Make fourier transform in $x$ to obtain $u(x,y)\to\hat{u}(k,y)$. Then the ODE becomes
\hat{u}|_{y=0}=\hat{f}(k), \hat{u}_y|_{y=0}=\hat{g}(k)
This has solution $\hat{u}(k,y)=A(k)\sinh(|k|y)+B(k)\cosh(|k|y)$. Plugging in first boundary condition we have get $B(k)=\hat{f}(k)$. The second condition gives $\hat{g}(k)=|k|A(k)\implies A(k)=\hat{g}(k)/|k|$. Note that at $|k|=0$, we have a problem since $A(k)$ could be singular. So we must require as a necessary condition that $\hat{g}(0)=0$, which means $\int_{-\infty}^\infty g(x)\,dx=0$. Now we obtain solution via IFT:
$$u(x,y)=\int_{-\infty}^\infty \left(\frac{\hat{g}(k)}{|k|}\sinh(|k|y)+\hat{f}(k)\cosh(|k|y)\right)e^{ikx}\,dk$$

(b) Same story. We still have $A(k)=\hat{g}(k)/|k|$, but this time the first condition gives $|k|^2B(k)=\hat{f}(k)\implies B(k)=\hat{f}(k)/|k|^2$. Once again, at $|k|=0$ we have a problem, so we require as necessary conditions that $\hat{f}(0)=0$ and $\hat{g}(0)=0$, which is the same as $\int_{-\infty}^\infty g(x)\,dx=\int_{-\infty}^\infty f(x)\,dx=0$. Then the solution is given by the IFT:
$$u(x,y)=\int_{-\infty}^\infty \left(\frac{\hat{g}(k)}{|k|}\sinh(|k|y)+\frac{\hat{f}(k)}{|k|^2}\cosh(|k|y)\right)e^{ikx}\,dk$$

APM346--Misc / Re: Halloween Challenge 6
« on: November 02, 2016, 12:55:18 PM »
Ah I see. Then there is a typo in, where the sign on $ix^2/(4kt)$ in the exponent should be positive, and the sign on $i\pi/4$ should be $\mp$ not $\pm$.

Edit: Wait a second. Why does switching from $ik+0$ to $ik-0$ change the argument of the complex number? $ik+0$ and $ik-0$ have the same angle in the complex plane.

APM346--Misc / Re: Halloween Challenge 6
« on: November 01, 2016, 02:52:20 PM »
Suppose for a moment that $t>0$. We know that the heat equation $u_t=ku_{xx}$ has a self similar solution of $u(x,t)=(4\pi kt)^{-1/2}e^{-\frac{x^2}{4kt}}$. For this equation, we just make the change $k\to ik$ and obtain
$$u(x,t)=(4\pi i kt)^{-1/2}e^{\frac{i x^2}{4kt}}$$
Note that $i^{-1/2}=e^{-(\log i)/2}$. Taking the principle branch of the complex logarithm where the argument lies in $[-\pi, \pi)$, we have $i^{-1/2}=e^{-i \pi/4}$. Thus for $t>0$ we have the solution
$$u(x,t)=(4\pi kt)^{-1/2}e^{\frac{i x^2}{4kt}-\frac{i \pi}{4}}$$
Now, if $t<0$, note that $u(ix,-t)$ satisfies the original ODE. So we make the change $x\to ix$ and $t\to -t$ to obtain the solution for $t<0$:
$$u(x,t)=(-4\pi kt)^{-1/2}e^{\frac{i x^2}{4kt}-\frac{i \pi}{4}}$$
Putting this together we have
$$u(x,t)=(4\pi k|t|)^{-1/2}e^{\frac{i x^2}{4kt}-\frac{i \pi}{4}}$$

APM346--Misc / Re: Halloween Challenge 7
« on: October 30, 2016, 11:47:02 PM »
Making Fourier Transform $u(x,y)\to \hat{u}(k,y)$ we can rewrite the ODE as
This ODE has solution $\hat{u}(k,y)=A(k)e^{-|k|y}+B(k)e^{|k|y}$. We will discard the second term because it goes unbounded as $y\to\infty$. Now using the boundary condition we have
$$(\hat{u}_y+\alpha\hat{u})_{y=0}=\hat{f}(k)\implies (-|k|+\alpha)A(k)=\hat{f}(k)\implies A(k)=\frac{\hat{f}(k)}{\alpha-|k|}$$
Now, if $\alpha<0$, then the RHS function has no singularities and we are good. To write the solution as fourier integral, we want IFT of $e^{-|k|y}/(\alpha-|k|)$, which we cannot find in closed form. So for now we can just write
$$u(x,y)=\frac{1}{2\pi}\int_{-\infty}^\infty f(x-z) \left(\int_{-\infty}^\infty \frac{e^{-|k|y+ikz}}{\alpha-|k|}\,dk\right)\,dz$$
If $\alpha>0$, then we have a problem as $\frac{\hat{f}(k)}{\alpha-|k|}$ could be singular. In this case, we need to require $\hat{f}(\alpha)=0$, which means
$$\int_{-\infty}^\infty f(x)e^{-i\alpha x}\,dx=0$$
This is a necessary condition, but not sufficient. $\hat{f}(k)$ needs to approach 0 faster than $\alpha-|k|$ does as $|k|\to\alpha$. Otherwise, we may still have problems with convergence. Assuming $\hat{f}(k)$ vanishes quickly enough at $\alpha$, the solution is given by the same equation.

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