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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.

TT1 / Re: TT1-P1
« on: October 20, 2016, 12:09:07 AM »
Just to add onto Roro Sihui Yap's solution for (c), the solution is fully determined by condition at $t=0$ because all characteristics intersect the $x$ axis and do not intersect with each other.

TT1 / Re: TT1-P2
« on: October 19, 2016, 11:11:10 PM »
Before we begin, we need one result: $\int x^2e^{-x^2/2}\,dx$. Changing variables, this can be written as
$$-\int x\,d(e^{-x^2/2})$$
Now integrate by parts, set $u=x$, $du=dx$, $dv=d(e^{-x^2/2})$, $v=e^{-x^2/2}$, we obtain
$$-xe^{-x^2/2}+\int e^{-x^2/2}\,dx$$
Now we begin the problem.
By D'Alembert's formula and Duhamel integral, the solution is
$$u(x,t)=\frac{1}{2}\left(-e^{(x-t)^2/2}-e^{(x+t)^2/2}\right)+\frac{1}{2}\int_0^t\int_{x-(t-t')}^{x+(t-t')} x'^2e^{-x'^2/2}-e^{-x'^2/2}\,dx'dt'$$
Now consider the inside integral
$$\int_{x-(t-t')}^{x+(t-t')} x'^2e^{-x'^2/2}-e^{-x'^2/2}\,dx'\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(*)$$
Let $a=x-(t-t')$ and $b=x+(t-t')$ for notational simplicity.
Using the result above, the first term can be integrated:
$$\int_{a}^{b} x'^2e^{-x'^2/2}\,dx'=ae^{-a^2/2}-be^{-b^2/2}+\int_a^b e^{-x'^2/2}\,dx'$$
But note that the second integral here cancels with the integral of the second term in $(*)$. So we just have
$$\int_a^b x'^2e^{-x'^2/2}-e^{-x'^2/2}\,dx'=ae^{-a^2/2}-be^{-b^2/2}$$
Now we need to integrate this with respect to $t'$, from $0$ to $t$. Consider the first term first. Note that $da=dt'$. When $t'=0$, $a=x-t$, When $t'=t$, $a=x$. Thus the first term integrates
$$\int_{x-t}^x ae^{-a^2/2}\,da=e^{-(x-t)^2/2}-e^{-x^2/2}$$
Now for the second term, note that $db=-dt'$. When $t'=0$, $b=x+t$. When $t'=t$, $b=x$. Thus we have
$$-\int_{x+t}^x be^{-b^2/2}\,db=e^{-x^2/2}-e^{-(x+t)^2/2}$$
Now putting together all the results:
$$u(x,t)=\frac{1}{2}\left(-e^{(x-t)^2/2}-e^{(x+t)^2/2}\right)+\frac{1}{2}\left(e^{-(x-t)^2/2}-e^{-x^2/2}-e^{-x^2/2}+e^{-(x+t)^2/2}\right) =-e^{-x^2/2}$$
Since this does not depend on $t$, we have $\lim_{t\to\infty} u(x,t)=-e^{-x^2/2}$.

TT1 / Re: TT1-P3
« on: October 19, 2016, 10:47:13 PM »
Let $u(x,t)=\phi(x+3t)+\psi(x-3t)$. Applying D'Alembert's formula, for $x>0$ we have
$$\phi(x)=\frac{1}{2}\sin(x)+\frac{1}{6}\int_0^x 3\cos x'\,dx'=\sin(x)\\
\psi(x)=\frac{1}{2}\sin(x)-\frac{1}{6}\int_0^x 3\cos x'\,dx'=0$$
We need to find $\psi(x)$ for $x<0$. To do this, we apply boundary condition:
Therefore, we have
Thus we have the solution
which is valid for $0<x<3t$. But the original equation is defined on a domain that is a subset of this (since $x<t\implies x<3t$), so this is the solution to the original problem.

TT1 / Re: TT1-P5
« on: October 19, 2016, 10:36:22 PM »
Before we begin, we need one result:
$$\int z^2 e^{-z^2}\, dz$$
Changing variables, this can be written as
$$-\frac{1}{2}\int z\,d(e^{-z^2})$$
Integrate by parts, set $s=z$, $ds=dz$, $dv=d(e^{-z^2})$, $v=e^{-z^2}$ we have
$$-\frac{1}{2}ze^{-z^2}+\frac{1}{2}\int e^{-z^2}\,dz=\frac{\sqrt{\pi}}{4}erf(z)-\frac{1}{2} ze^{-z^2}$$

Now we begin the problem. The solution is
$$u(x,t)=\frac{1}{\sqrt{4\pi t}} \int_{-1}^1 e^{-\frac{(x-y)^2}{4t}}(1-y^2)\,dy=\frac{1}{\sqrt{4\pi t}} \left( \int_{-1}^1 e^{-\frac{(x-y)^2}{4t}}\,dy- \int_{-1}^1 y^2e^{-\frac{(x-y)^2}{4t}}\,dy\right)$$
From now on, define $a=(x+1)/\sqrt{4t}$ and $b=(x-1)/\sqrt{4t}$ for notational simplicity.
In the first integral, substitute $z=(x-y)/\sqrt{4t}$ so that $dz=-dy/\sqrt{4t}$, so that
$$\int_{-1}^1 e^{-\frac{(x-y)^2}{4t}}\,dy=-\sqrt{4t}\int_a^b e^{-z^2}\,dz=\frac{\sqrt{4\pi t}}{2}\left(erf(a)-erf(b)\right)$$
In the second integral, substitute the same thing. Note that $y^2=x^2-2xz\sqrt{4t}+4tz^2$. So we have
$$ \int_{-1}^1 y^2e^{-\frac{(x-y)^2}{4t}}\,dy=-\sqrt{4t}\int_a^b x^2e^{-z^2}-2x\sqrt{4t} ze^{-z^2}+4tz^2 e^{-z^2}\,dz$$
Now integrate the three terms. The first one can be done using error function, the second one can be integrated directly, the third one using the result we obtained above:
$$ \int_{-1}^1 y^2e^{-\frac{(x-y)^2}{4t}}\,dy=-\sqrt{4t}\left[ \frac{x^2\sqrt{\pi}}{2}\left(erf(b)-erf(a)\right)+x\sqrt{4t}\left(e^{-b^2}-e^{-a^2}\right)+4t\left(\frac{\sqrt{\pi}}{4}erf(b)-\frac{\sqrt{\pi}}{4}erf(a)-\frac{1}{2}\frac{x-1}{\sqrt{4t}}e^{-b^2}+\frac{1}{2}\frac{x+1}{\sqrt{4t}}e^{-a^2}\right)\right]$$
Now all that is left is to combine the two and simplify. In the end, one obtains

Chapter 3 / Re: A tentative solution to Assignment 5, Problem3(g)
« on: October 16, 2016, 04:21:44 PM »
Your function $w(x,t)$ is not continuous, so if you solve for $u$ in terms of $w$, it wouldn't be continuous either. There is a section in the textbook ( regarding inhomogeneous boundary conditions, but this has neither been covered in lectures nor tutorials.

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