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Topics - Victor Ivrii

Pages: 1 ... 36 37 [38] 39 40 ... 47
556
HA1 / HA1 problem 2
« on: January 20, 2015, 06:47:46 AM »
Solutions to be posted as a "Reply" only after January 22, 21:00

a. Find the general solution of
\begin{equation}
xu_x+4 yu_y=0
\label{eq-HA1.2}
\end{equation}
in $ \{(x,y)\ne (0,0)\}$; when this solution is continuous at $(0,0)$?

b. Find the general solution of
\begin{equation}
xu_x-4yu_y=0
\label{eq-HA1.3}
\end{equation}
in $ \{(x,y)\ne (0,0)\}$; when this solution is continuous at $(0,0)$?

c. Explain the difference between (\ref{eq-HA1.2}) and (\ref{eq-HA1.3}).

557
HA1 / HA1 problem 1
« on: January 20, 2015, 06:45:17 AM »
Solutions to be posted as a "Reply" only after January 22, 21:00

a. Find general solution
\begin{equation}
4 u_x -3u_y=0;
\label{eq-HA1.1}
\end{equation}
b.  Solve IVP problem $u|_{x=0}=\sin (y)$ for equation (\ref{eq-HA1.1}) in $\mathbb{R}^2$;

c.  Consider equation (\ref{eq-HA1.1}) in $\{x>0, y>0\}$ with the initial condition $u|_{x=0}=y$ ($y>0$); where this solution defined? Is it defined everywhere in $\{x>0, y>0\}$ or do we need to impose condition at $y=0$?
In the latter case impose condition $u|_{y=0}=x$ ($x>0$) and solve this IVBP;

d. Consider equation (\ref{eq-HA1.1}) in $\{x<0, y>0\}$ with the  initial condition $u|_{x=0}=y$ ($y>0$); where this solution defined? Is it defined everywhere in $\{x<0, y>0\}$ or do we need to impose condition at $y=0$? In the latter case impose condition $u|_{y=0}=x$ ($x<0$) and solve this IVBP.

558
Web Bonus Problems / Web Bonus Problem 4
« on: January 17, 2015, 11:56:54 AM »
Find self-similar solution
\begin{equation}
u_t=u u_x\qquad -\infty <x <\infty,\ t>0
\label{eq-1}
\end{equation}
satisfying initial conditions
\begin{equation}
u|_{t=0}=\left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\label{eq-2}
\end{equation}

559
Web Bonus Problems / Web Bonus Problem 3
« on: January 17, 2015, 08:23:45 AM »
Find a self–similar solution $u$ of
\begin{equation}
u_t = (u u_x)_x  \qquad -\infty<x<\infty , t>0
\label{eq-1}
\end{equation}
with finite $\int_{-\infty}^\infty u\,dx$.

560
Web Bonus Problems / Web Bonus Problem 2
« on: January 14, 2015, 09:20:14 AM »
Solve
\begin{align}
&(t^2+1)u_{tt}+tu_t-u_{xx}=0,\label{eq-1}\\[3pt]
&u|_{t=0}=0, \qquad u_t|_{t=0}=1.\label{eq-2}
\end{align}
Hint: Make a change of variables $x=\frac{1}{2}(\xi+\eta)$, $t=\sinh (\frac{1}{2}(\xi-\eta))$ and calculate $u_\xi$, $u_\eta$, $u_{\xi\eta}$.

561
Web Bonus Problems / Web Bonus Problem 1
« on: January 08, 2015, 03:43:12 AM »
There are several subproblems and one can submit solution for any of them

Solve (for $t>0$)
\begin{align}
&u_t + u u_x=0,\label{eq-1}\\
&u|_{t=0}=f(x)\label{eq-2}
\end{align}
with one of the following initial data
\begin{align}
f(x)=&\left\{\begin{aligned}
-1& && x<-a,\\
x/a& && -a\le x \le a,\\
1& && x>a;
\end{aligned}\right.
\label{eq-3}\\
f(x)=&\left\{\begin{aligned}
1& && x<-a,\\
-x/a& && -a\le x \le a,\\
-1& && x>a;
\end{aligned}\right.
\label{eq-4}\\
f(x)=&\left\{\begin{aligned}
-1& && x<0,\\
1& && x> 0.\\
\end{aligned}\right.
\label{eq-5}
\end{align}
Here $a>0$ is a parameter. Plotting solutions for different $t>0$ would be appreciated

562
FE / Problems
« on: December 09, 2014, 03:39:57 PM »
I was advised by email

Quote
This morning I was looking through the questions you posted on the FE board, and I noticed that several of those questions were different from those that appeared on my final exam. Particularly, questions 3, 5, and 6.

Was this discrepancy intentional? Did I perhaps choose the one desk with a different draft of the final (if such a draft exists)? If you could clarify the situation it would be greatly appreciated.


No, it was not. Probably because draft was rewritten so many times that I lost the track before Prof Milman submitted to Dean's office. I will correct what is posted when I get papers  for grading (I will be doing this after Profs Song and Milman since I am also dealing with the automatic calculations of the Final Grade

Since there is no bonus for these posts my error will have no consequences—and please post corrected version

563
FE / FE6
« on: December 08, 2014, 04:20:16 PM »
For the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t =4x^2y-2x^2-4xy+2y, \\
&y'_t =-4xy^2+2y^2+4xy-2x
\end{aligned}\right.
\end{equation*}

(a)  linearize the system at $\ x_0 = 1\, , \, y_0 = 1\ $ 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) describe the type of the critical point $\ x_0 = 12 \, , \, y_0 = 10\ $ of the nonlinear system.

Solution

(a) Let $f= 4x^2y -2x^2-4xy +2y$, $g= -4xy^2 + 2y^2+4xy -2x$; then $f_x(1,1)=0$, $f_y(1,1)= 2$,  $g_x(1,1)=-2$, $f_y(1,1)= 0$ and the linearized system is
\begin{equation*}\left\{\begin{aligned}
&X'_t =2Y, \\
&Y'_t = -2X
\end{aligned}\right.
\end{equation*}
with phase portrait consisting of clock-wise circles.

(b) Rewriting system as $fdx-gdy=0$ we get
$$
 (4xy^2-2y^2 -4xy+2x)\, dx+  (4x^2 y-2x^2 -4xy+2y)\,dy =0
$$
which is exact; then
$$
 H_x= 4xy^2-2y^2 -4xy+2x , \quad H_y=  4x^2 y-2x^2 -4xy+2y)$$
and the first equation implies that
$$
 H= 2x^2 y^2 -2xy^2 -2x^2y +x ^2 + \phi(y)$$
and the second equation implies that $\phi'=2y$ and $y=y^2$ and then
$$
 H=2x^2 y^2 -2xy^2 -2x^2y +x ^2 + y^2 = x^2(y-1)^2 + y^2(x-1)^2.
$$
 
(c) Since linearized system has a center and original system has a solution $H(x,y)=C$ the type of the stationary point is a center.


Remark
In fact the system has also critical point  $(0,0)$ of the type center, and critical point  $(\frac{1}{2},\frac{1}{2})$ of the type saddle






564
FE / FE5
« on: December 08, 2014, 04:16:29 PM »
For the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t = x (5-2x-3y)\, , \\
&y'_t = y (5-3x-2y)
\end{aligned}\right.
\end{equation*}

(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 phase portrait of this system of ODEs.



Solution

(a) Solving $x (5-2x-3y=0$, $y (5-3x-2y )$ we have 4 cases $x=y=0$, $x=5-3x-2y=0$, $y=5-2x-3y=0$ and $5-3x-2y =5-2x-3=0$ giving us 4 points $(0,0)$, $(0,\frac{5}{2})$, $(\frac{5}{2},0)$ and $(1,1)$.

(b) Let  $f= x (5-2x-3y)=5x-2x^2-3xy$, $g=y (5-3x-2y)=5y-2y^2-3xy$. Then $f_x=5-4x-3y$, $f_y=-3x$, $g_x=-3y$, $g_y=5-4y-3x$.

  • $(0,0)$; matrix $\begin{pmatrix} f_x & f_y \\ g_x &g_y\end{pmatrix}$ at this point equals $\begin{pmatrix} 5 & 0 \\ 0 &5\end{pmatrix}$ with eigenvalues $r_1=r_2=5$;  and eigenvectors $(1, 0)^T$ and $(0,1)^T$; unstable node;
  • $(0,\frac{5}{2})$; matrix $\begin{pmatrix} f_x & f_y \\ g_x &g_y\end{pmatrix}$ at this point equals  $\begin{pmatrix} -\frac{5}{2} & 0 \\ -\frac{15}{2} &-5\end{pmatrix}$ with eigenvalues $r_1=-\frac{5}{2},\ r_2=-5$ and eigenvectors $(1,-3)^T$ and $(1,0)^T$ respectively;  stable node;
  • $(\frac{5}{2},0)$; the same as in (2);
  • $(1,1)$; matrix $\begin{pmatrix} f_x & f_y \\ g_x &g_y\end{pmatrix}$ at this point equals  $\begin{pmatrix} -2 & -3 \\ -3 &-2\end{pmatrix}$ with eigenvalues $r_1=-5$ and $r_2=1$ and eigenvectors $(1,1)^T$ and $(1,-1)^T$ respectively; saddle.


(c-d) Plotting


Remark This is ``two competing species'' system.

565
FE / FE4
« on: December 08, 2014, 04:14:27 PM »
Find the general solution of the ODE
\begin{equation*}
x y' = y - x e^{\frac{y}{x}}
\end{equation*}
and solve the initial value problem $\ y(1) = -2\ $.

Solution
Since it is homogeneous equation we plug $y=ux$ and then
\begin{equation*}
u'x^2+ux=ux -xe^{u}\implies u'=-e^{u}\implies x^{-1}dx=-e^{-u}du\implies
\ln x = e^{-u}+\ln C\implies u =-\ln \ln (Cx)\implies y=-x\ln \ln (Cx).
\end{equation*}
As $x=1$, $y=-2$, $u=-2$ we get $\ln \ln C=2$, and $y= -x \ln (e^2+\ln x)$.

566
FE / FE3
« on: December 08, 2014, 04:13:30 PM »
Find the general solution of the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t = -\frac{5}{4} x + \, \frac{3}{4} y + \frac{2}{1+e^t} , \\
&y'_t =\,\hphantom{-}\frac{3}{4} x - \frac{5}{4} y\ .
\end{aligned}\right.
\end{equation*}

Solution
Characteristic equation is
\begin{equation*}
\left| \begin{matrix} -\frac{5}{4}-r & \frac{3}{4}\\ \frac{3}{4} & \frac{5}{4}-r\end{matrix}\right| =( r+\frac{5}{4})^2-\frac{9}{16} =0
\end{equation*} with characteristic roots $r_{1,2}=-\frac{5}{4}\pm \frac{3}{4} $, $r_1=-\frac{1}{2}$, $r_2=-2$.

Finding corresponding eigenvectors: (a) $r_1=1$,
\begin{equation*}
\begin{pmatrix} -\frac{3}{4} & \frac{3}{4}\\ \frac{3}{4} & -\frac{3}{4}\end{pmatrix}\begin{pmatrix}\alpha \\ \beta\end{pmatrix}=0
\end{equation*}
and then $\alpha=\beta=1$ and eigenvector is $\mathbf{e}_1=\begin{pmatrix} 1 \\ 1\end{pmatrix}$.


$r_2=2$ and $\mathbf{e}_2=\begin{pmatrix} 1 \\ -1\end{pmatrix}$ (since matrix is symmetric eigenvectors are orthogonal).

Therefore the general solution of the homogeneous system is
\begin{equation}
\begin{pmatrix} x^* \\ y^*\end{pmatrix}=C_1 \begin{pmatrix} 1 \\ 1\end{pmatrix} e^{-\frac{1}{2}t} + C_2\begin{pmatrix} 1 \\ -1\end{pmatrix}e^{-2t}.
\label{eq-3-1}
\end{equation}


To solve inhomogeneous system we use method of variation of parameters leading to
\begin{equation*}
\begin{pmatrix} e^{-\frac{t}{2}} & e^{-2t}\\  e^{-\frac{t}{2}} & -e^{-2t}\end{pmatrix}\begin{pmatrix} C'_1 \\  C'_2 \end{pmatrix}=
\begin{pmatrix} \frac{2}{1+e^{t} }\\ 0\end{pmatrix}\implies\\
\begin{aligned}
&C'_1=\frac{ e^{\frac{t}{2}}}  { 1+e^t }\implies C_1= \int \frac{ e^{\frac{t}{2}}}  { 1+e^t } \,dt = 2\arctan (e^{\frac{t}{2}})+ c_1, \\
&C'_2= \frac{e^{2t}}{1+e^{2t}} \implies C_2= \int \frac{ e^{2t}}  { 1+e^t } \,dt = \int \Bigl(e^t -  \frac{ e^{t}}  { 1+e^t } \bigr)\,dt=
e^t - \ln (1+e^t)+c_2\end{aligned}
\end{equation*}
where the first integral is taken by substitution $u=e^{\frac{t}{2}}$ and the second by substitution $u=1+e^{t}$.

Thus
\begin{equation*}
\begin{pmatrix} x \\ y\end{pmatrix}=\bigl(2\arctan (e^{\frac{t}{2}})+ c_1\bigr) \begin{pmatrix} 1 \\ 1\end{pmatrix} e^{-\frac{1}{2}t} +
\bigl(e^t - \ln (1+e^t)+c_2\bigr)\begin{pmatrix} 1 \\ -1\end{pmatrix}e^{-2t}.
\end{equation*}


567
FE / FE2
« on: December 08, 2014, 04:13:04 PM »
Find the general solution of
\begin{equation*}
x^4 y^{(4)} + 6 x^3 y^{(3)} + 7 x^2 y^{(2)} + x y' - y = 3 \ln x + \cos (\ln x) \, .
\end{equation*}



Solution
It is Euler's equation. Its characteristic polynomial is
\begin{equation*}
r(r-1)(r-2)(r-3)+6r(r-1)(r-2) +7r(r-1)+r -1=\\
r^4 - 6r^3 +11r^2 -6r + 6r^3 -18r^2 +12r +7r^2-7r+r-1=
r^4-1
\end{equation*}
with characteristic roots $r_{1,2}=\pm 1$, $r_{3,4}=\pm i$ and plugging $t=\ln x$ we arrive to
\begin{equation}
y^{(4)}_t-y= 3t + \cos (t).
\label{eq-2-1}
\end{equation}
Solution to homogeneous equation is
\begin{equation}
z= C_1 e^t +C_2 e^{-t}+C_3 \cos (t)+ C_4\sin (t)=\\
 C _1 x +C_2 x^{-1}+ C_3\cos (\ln x) + C_4\sin(\ln x)
\label{eq-2-2}
\end{equation}
and the particular solution to inhomogeneous equation is $y_p = y_{p1}+y_{p2}$ with $y_{p1}= at +b$ and $y_{p2}=  (c\cos (t) +d \sin(t))t$ solving equation with right hand expressions $f_1=3\ln x$ and $f_2= \cos(\ln x)$ respectively.

Plugging $y_{p1}$ we get
\begin{equation}
-at -b=3t \implies a=-3,b=0\implies y_{p1}=-3t=-3\ln x
\label{2-3}
\end{equation}
and plugging $y_{p2}$ we get
\begin{equation}
3(c\sin (t)-d\cos (t)) = \cos(t) \implies c=0,d=-\frac{1}{3}\implies \\
y_{p2}=-\frac{1}{3}\sin (t) t=-\frac{1}{3}\sin (\ln x)\ln x.
\label{eq-2-4}
\end{equation}
Adding (\ref{eq-2-2})--(\ref{eq-2-4}) we get
\begin{equation*}
y= C _1 x +C_2 x^{-1}+ C_3\cos (\ln x) + C_4\sin(\ln x)-3\ln x -\frac{1}{3}\sin (\ln x)\ln x.
\end{equation*}

568
FE / FE1
« on: December 08, 2014, 04:12:22 PM »
Solve the initial value problem
\begin{equation*}
\left(3x + 2y\right) dx + \left(x + \frac{6 y^2}{x}\right) dy = 0\, , \qquad  y(0) = 3\, .
\end{equation*}

Solution
As $M=3x+2y$, $N=x+ \frac{6 y^2}{x}$, $M_y-N_x= 2- \bigl(1-\frac{6y^2}{x^2}\bigr)$ and $(M_y-N_x )/N= 1/x$ is a function of $x$ only. So we can find integrating factor $\mu =\mu(x)$ from $\mu'/\mu = 1/x \implies \ln \mu =\ln x$ (modulo constant factor) and $\mu=x$. Therefore
$$3x^2 + 2xy) dx + (x^2 + 6 y^2) dy = 0.$$
Then
$$
U_x= 3x^2+2xy ,\qquad U_y=x^2+6y^2$$
where the first equation implies that
$$
U= x^3 + x^2y+ \phi(y)$$
and plugging to the second equation we see that
$$
\phi'=6y^2\implies y=2y^3.$$
Then
$$
U:=x^3+x^2y+2y^3 =C $$
is a general solution and finding $C=54$ from initial condition we arrive to
$$
U:=x^3+x^2y+2y^3 =54.
$$

569
Quiz 5 / Quiz 5
« on: November 29, 2014, 01:03:00 PM »
For system
\begin{equation*}
\left\{\begin{aligned}
&\frac{dx}{dt} = (1 - y)(2x- y), \\
&\frac{dy}{dt}  = (2 + x)(x - 2y)
\end{aligned}\right.
\end{equation*}


(a) Determine all critical points of the given system of equations and find the corresponding linear system near each critical point.

(b) Draw phase portraits of the linear systems of ODEs from the previous item and describe completely their types of the stationary points  indicating if it is stable or unstable and in the cases of the center or focus indicate orientation (clockwise or counter-clockwise).

(c) What conclusions can you then draw (and why) about the types of the critical points and phase portraits of the nonlinear system near each critical point ?

(d) Combine answers from the previous item into a (probable) phase portrait of the original non linear system.


570
TT2 / TT2 # 4
« on: November 19, 2014, 08:47:30 PM »
Find the general solution and determine the type of behavior near the origin of the the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t= -6x + 5y\ , \\
&y'_t= -5x + 4y .
\end{aligned}\right.
\end{equation*}

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