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Messages - Victor Ivrii

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Final Exam / Re: FE-P5
« on: April 13, 2018, 12:20:38 PM »
General comment

This problem was included occasionally: I planned b.c.
 As a result the problem became really easy but also treacherous.

Typical errors:
* (The most widespread) Wrongly solve  problem for $X_n$: $X_n=\cos (2nx)$ $n=1,2,\ldots$ but forgetting $X_0=\frac{1}{2}$; as a result
u(x,y)={\color{blue}{\frac{A_0}{2}}} + \sum_{n=1}^\infty A_ne^{-2ny}\cos (2nx);
where highlighted term is missing. Then
{\color{blue}{-\frac{A_0}{2}}}- \sum _{n=1}^\infty (2n+1)\cos(2nx)=1.
This implies $A_n=0$ for $n=1,2,\ldots$ and $u(x,y)=0$ rather than correct $u(x,y)=-1$.

Remark: Basically one does not need to calculate $A_n$ here, since $g(x)=1$ is already a decomposition of $g(x)$ into $\cos$-Fourier series.

* Wrongly solved problem for $X_n$: $X_n=\sin (2nx)$...

APM346--Misc / Re: Cap on bonus from karma?
« on: April 13, 2018, 11:54:20 AM »
Yes, cap for Bonus Mark is 10 points. However there is a good reason to get more karma (when I am asked to write a recommendation letter to graduate school, I pay attention)

Do not worry about robbing. I put you and some others to "moderated" list. so your post, while submitted, would not appear without my approval, and I will wait with it ...

Final Exam / Re: FE-P4
« on: April 13, 2018, 06:40:51 AM »
Missing calculations for coefficients

You proved that $A_n=0$ for even $n$, but in the final formula you sum over all $n$. [/color]

 In polar coordinates $\{y=0\}$ is $\theta =0,\pi$.
 Separating variables we get
\frac{r^2 R''+rR'}{R}+\frac{P''}{P}=0\implies  &P''+\lambda P=0,\label{4-5}\\
 &r^2R''+rR'+\lambda R=0.\label{4-7}
This problem has solutions
$\lambda_n =n^2$, $X_n=\sin (n\theta)$, $n=1,2,\ldots$.

Then $r^2R''+rR'+n^2 R=0\implies R_n= A_n r^{n} + B_n r^{-n}$.  Then
&u= \sum_{n=1}^\infty \bigl(A_n r^{n} +B_{n}r^{-n}\bigr) \sin (n\theta)
and using (\ref{4-2}), (\ref{4-3})
&u|_{r=2}= \sum_{n=1}^\infty (2^{n}A_n +2^{-n}B_n)\sin(n\theta)=1,\\
&u|_{r=\frac{1}{2}}= \sum_{n=1}^\infty (2^{-n}A_n +2^{n}B_n)\sin(n\theta)=-1\\
which imply
&\sum_{n=1}^\infty (2^n+2^{-n})(A_n+B_n)\sin(n\theta)=0\implies (A_n+B_n)=0\\
&\sum_{n=1}^\infty (2^n-2^{-n})(A_n-B_n)\sin(n\theta)=2\implies
&\frac{8}{\pi (2m+1)}&& n=2m+1,\\
&0 &&n=2m\
which imply  $A_{2m}=B_{2m}=0$ and
&A_{2m+1}=-B_{2m+1}= \frac{4}{\pi (2m+1)(2^{2m+1}-2^{-2m-1})}
u= &\sum_{m=0}^\infty \frac{4}{\pi (2m+1)(2^{2m+1}-2^{-2m-1})}(r^{2m+1}-r^{-2m-1})
\sin ((2m+1)\theta).

General comments
Typical errors:
* Wrong bounds for $\theta$ and wrong b.c. for $P$ (usually periodic b.c., rather then $P(0)=P(\pi)=0$; then wrong eigenfunctions.
* Wrong bonds for $r$: $\frac{1}{4} \le r\le 4$ rather than $\frac{1}{2} \le r\le 2$.
* Throwing out $r^{-n}$ because it is unbounded as $r\to 0$ (but $r\ge 1/2$).
* Instead of plugging  (\ref{G}) into (\ref{4-2}), (\ref{4-3}) something like plugging with $r^n$ into (\ref{4-2}) and with $r^{-n}$ into (\ref{4-3})
* Wrong solutions to Euler's equation. $R_n(r)=e^{nr}$ and $R_n(r)=e^{-nr}$ (and variants) rather than $R_n(r)=r^n$ and $R_n(r)=r^{-n}$.
* Again, separate solutions for odd and even $n$.

Final Exam / Re: FE-P3
« on: April 13, 2018, 03:07:51 AM »
Missing calculations of the coefficient and the sign is wrong in the end.

Separating variables $u(x,t)=X(x)T(t)$ we get
&X''+\lambda X=0,\label{3-5}\\
&T''+(\lambda+4) T=0.\label{3-7}
Problem (\ref{3-5})--(\ref{3-6}) has solution
\lambda_n= n^2,\qquad X_n =\sin (nx),\qquad n=1,2,\ldots$$
and therefore
T_n = A_n \cos ((n^2+4)^{1/2}t)+B_n ((n^2+4)^{1/2}t),$$
and $$
u =\sum_{n=1}^\infty \Bigl[A_n \cos ((n^2+4)^{1/2}t)+B_n ((n^2+4)^{1/2}t)\Bigr]\sin (nx).
Plugging to (\ref{3-3})--(\ref{3-4}) we get
&\sum_{n=1}^\infty A_n \sin (nx)=0,\\
&\sum_{n=1}^\infty (n^2+4)^{1/2} B_n  \sin(nx)= x^2-\pi x.
and therefore $A_n=0$,
(n^2+4)^{1/2} B_n = \frac{2}{\pi} \int_0^\pi (x^2-\pi x) \sin (nx)\,dx=\\
-\frac{2}{\pi n}\int_0^\pi (x^2-\pi x)\,d\cos (nx) =\frac{2}{\pi n}\int_0^\pi (2x-\pi)\cos (nx)\,dx=
\frac{2}{\pi n^2}\int_0^\pi (2x-\pi)\,d\sin (nx)=-\frac{4}{\pi n^2}\int_0^\pi \sin (nx)\,dx =\\
\frac{4}{\pi n^3}\cos(nx)\bigl|_{x=0}^{x=\pi}=\left\{\begin{aligned}
&0 &&n=2m,\\
-&\frac{8}{\pi (2m+1)^3}  &&n=2m+1.
u(x,t)=-\sum_{m=0}^\infty \frac{8}{(2m+1)^3 \pi\sqrt{(2m+1)^2+4}} \sin((2m+1)x)\sin(\sqrt{(2m+1)^2+4}t).

General comment
Typical errors:
* $\sin(\sqrt{n^2{\color{red}-}4} t)$, $\sin(n t)$ and other errors in $T_n(t)$
* Errors in the calculation of $\int_0^\pi (x^2-\pi x) \sin(nx)\,dx$. It is calculated by integration by parts and it is easier not to break it into two, since function $g(x)$ vanishes at $0$ and $\pi$.
* In the end there is one solution, not two separate solutions for even and odd $n$.

Web Bonus Problems / Re: Exam Week
« on: April 13, 2018, 12:21:39 AM »
How do I modify (3) ?
Calculate $w|_{x^2+y^2=1}$ as $y>0$, if $u$ satisfies (\ref{3}) and $v=x^2-y^2$ you know.

Web Bonus Problems / Exam Week
« on: April 12, 2018, 03:38:18 PM »
Consider problem:
& \Delta u=0 &&\text{in   }x^2+y^2<1, \ \ y>0,
& u|_{y=0}=x^2 &&\text{as   }|x|<1,\label{2}\\
& u|_{x^2+y^2=1}=1,&& \text{as   } y>0.
We want to separate variable $r$ and $\theta$ but the conditions as $\theta=0,\pi$ are inhomogeneous.

So we want to make them homogeneous. Find $v$, so that $u:=v$ satisfies (\ref{1}) and (\ref{2}) but not necessarily (\ref{3}), so $v$ is not unique. Can you suggest a candidate?

Then $w=u-v$ will satisfy (\ref{1}), homogeneous condition (\ref{2}), modified (\ref{3}). Find $w$ by separation, and then $u=v+w$.

MAT244--Announcements / Grading FE
« on: April 12, 2018, 01:12:30 PM »
While other instructors and TAs are grading their parts of FE, I am busy with grading FE APM346, which was 5 hours earlier. I will deal with your class next week

Final Exam / Re: FE-P2
« on: April 12, 2018, 01:08:09 PM »
This looks familiar:)
Indeed, it looks familiar but in addition to misprints there are errors, leading to the errors in the answer.
Jingxuan, you are the second most prolific poster on this forum, you just made more than Emily, but this was a flood. Deleted.

\frac{1}{2t} (x+y )^2 + y \overset{?}{=}  \frac{1}{2t} (x+y +{\color{red}{2}}t)^2  - ...
Now it is fixed. I sketched $u(x,0)$ and $u(x,1)$.
General comments
Typical errors:
* Solving IVP rather than IBVP
* Improper square separation mentioned in (*)
* Forgetting to change the lower limit in $\int_0^\infty \ldots dy$ while changing variable $z= (x-y \pm c t)/\sqrt{2t}$.

Final Exam / Re: FE-P1
« on: April 12, 2018, 05:04:43 AM »
Solution is correct but should be written in the form
& ...... && \text{case 1},\\
& ...... && \text{case 2}
where "case 1" and "case 2" (or may be more cases?) should be described.

So that anyone could easily compare what he or she got with yours. I actually highlighted cases in your posts.

From  (\ref{1-1}):
u(x,t)=\phi (x+4t)+\psi (x-4t).
Plugging to (\ref{1-2})--(\ref{1-3}),   we conclude that for $x>0$  $\phi(x) +\psi(x)=4e^{-2x}$, $4 \phi'(x)-4\psi'(x)=16e^{-2x}$; integrating the second equation we get  $\phi (x)-\psi(x)=-2e^{-2x}$ (since we can select constant equal to $0$ here) and finally
\phi(x)=e^{-2x}, \quad \psi(x)=3e^{-2x}\qquad \text{for  } x>0.
We need to find $\psi(x)$ for $x<0$. Plugging (\ref{1-5}) into (\ref{1-4}) we conclude that
&&&\phi'(4t)-\phi(4t)+\psi'(-4t)-\psi(-4t)= e^{-8t}\qquad\text{for }t>0,\iff\\
&\text{and plugging $x=-4t$ we see that}\\
&&&\psi'(x)-\psi(x)=4e^{2x}\qquad\text{for  }x<0,\\
&\text{and solving this ODE}\\
&&&(\psi e^{-x})'= 4e^{x}\implies \psi(x) e^{-x}=4e^{x}+C
and therefore $\psi(x)=4e^{2x}+Ce^{x}$ for $x<0$. Since $\psi(+0)=3$, $\psi(-0)=4+C$ we need for continuity $C=-1$. So
 \psi(x)=4e^{2x}-e^{x}\qquad \text{for } x>0.
u(x,t)= \left\{\begin{aligned}
&e^{-2x-8t}+3e^{-2x+8t} \qquad &&x>4t>0,\\
&e^{-2x-8t} + 4e^{2x-8t}-e^{x-4t}  &&0<x< 4t.

Almost everybody got correctly $\phi(x),\psi(x)$ for $x>0$ and $u(x,t)$ for $x>4t$.
Most correctly wrote equation $\psi' -\psi =4e^{2x}$ for $x<0$.
But then correctly finding one solution $\psi= 4e^{2x}$ (mostly by the method of undetermined coefficient), they wrote $\psi=4e^{2x}+C$ rather than $\psi=4e^{2x}+Ce^{x}$ (WTH). Which in turn yielded term $-1$ instead of correct one $-e^{x-4t}$ in the very end for $0<x<4t$.

Final Exam / FE-P6
« on: April 11, 2018, 08:48:39 PM »
For the system of ODEs
&x'_t = -2xy\, , \\
&y'_t = x^2+y^2-1

a. Linearize the system at
stationary points and sketch the phase portrait of this linear system.

b.  Find the equation of the form $H(x,y) = C$, satisfied by the trajectories of the nonlinear system.

c. Sketch the full phase portrait.

Final Exam / FE-P5
« on: April 11, 2018, 08:47:26 PM »
For the system of ODEs
&x'_t = x (x -y+1)\, , \\
&y'_t = y (x - 2)\,.

a.  Describe the locations of all critical points.

b. Classify their types (including whatever relevant: stability, orientation, etc.).

c. Sketch the phase portraits near the critical points.

d.  Sketch the full phase portrait of this system of ODEs.

Final Exam / FE-P4
« on: April 11, 2018, 08:44:51 PM »
Find the general solution of the system of ODEs
&x'_t = \hphantom{-}x +   y +\, \frac{e^{t}}{\cos(t)}\, ,\\
&y'_t = - x + y +\, \frac{e^{t}}{\sin(t)}\,.

Final Exam / FE-P3
« on: April 11, 2018, 08:42:26 PM »
Find the general solution of
y''' -6y'' +11y'- 6y=2\frac{e^{3x}}{e^x+1} .

Final Exam / FE-P2
« on: April 11, 2018, 08:40:12 PM »
Find the general solution by method of the undetermined coefficients:
y'''-2y''+4y'-8y= 16 e^{2t} + 30\cos(t);

Final Exam / FE-P1
« on: April 11, 2018, 08:39:30 PM »
Find the general solution
\bigl(2xy \cos(y)-y^2\cos(x)\bigr)\,dx  +

Hint. Use the integrating factor.

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