Drawing a phase portrait given Eigenvectors
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I am a bit confused on how the author here drew the phase portraits in the following picture.
The second eigenvalue is larger than the first. For large and positive t’s this means that the solution for this eigenvalue will be smaller than the solution for the first eigenvalue. Therefore, as t increases the trajectory will move in towards the origin and do so parallel to . Likewise, since the second eigenvalue is larger than the first this solution will dominate for large and negative t’s. Therefore, as we decrease t the trajectory will move away from the origin and do so parallel to .
This is what really confuses me. How do I see in the picture that t is increasing/decreasing. The plot does have x1 and x2 on the axis which are dependent on t if I understand correctly. Also what confuses me also how the Eigenvectors can give us so much information on how to draw the phase portrait itself. Any explenation/clarification would be very much appreciated.
differential-equations
asked Aug 15 at 6:56
craft
132
add a comment |Â
up vote
0
down vote
favorite
I am a bit confused on how the author here drew the phase portraits in the following picture.
The second eigenvalue is larger than the first. For large and positive t’s this means that the solution for this eigenvalue will be smaller than the solution for the first eigenvalue. Therefore, as t increases the trajectory will move in towards the origin and do so parallel to . Likewise, since the second eigenvalue is larger than the first this solution will dominate for large and negative t’s. Therefore, as we decrease t the trajectory will move away from the origin and do so parallel to .
This is what really confuses me. How do I see in the picture that t is increasing/decreasing. The plot does have x1 and x2 on the axis which are dependent on t if I understand correctly. Also what confuses me also how the Eigenvectors can give us so much information on how to draw the phase portrait itself. Any explenation/clarification would be very much appreciated.
differential-equations
asked Aug 15 at 6:56
craft
132
You can also google for trace determinant plane. It does give a classification of all linear systems in the plane. The key fact is the trace and determinant of a $2times 2$ matrix $A$, when $x'(t)=Acdot x(t)$
– Fakemistake
Aug 15 at 9:23
Think also about the two functions $f(t)=e^-t$ and $g(t)=e^-6t$ for $tgeq 0$. For $t>0$ we have $f(t)>g(t)$ but both are converging to zero. Consequently $f$ is dominant. In the picture you see that for large $t$ the solutions are converging to the eigenspace of the eigenvalue $-1$. (Handwaving). For sure the solutions are converging to the origin.
– Fakemistake
Aug 15 at 9:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am a bit confused on how the author here drew the phase portraits in the following picture.
The second eigenvalue is larger than the first. For large and positive t’s this means that the solution for this eigenvalue will be smaller than the solution for the first eigenvalue. Therefore, as t increases the trajectory will move in towards the origin and do so parallel to . Likewise, since the second eigenvalue is larger than the first this solution will dominate for large and negative t’s. Therefore, as we decrease t the trajectory will move away from the origin and do so parallel to .
This is what really confuses me. How do I see in the picture that t is increasing/decreasing. The plot does have x1 and x2 on the axis which are dependent on t if I understand correctly. Also what confuses me also how the Eigenvectors can give us so much information on how to draw the phase portrait itself. Any explenation/clarification would be very much appreciated.
differential-equations
asked Aug 15 at 6:56
craft
132
I am a bit confused on how the author here drew the phase portraits in the following picture.
The second eigenvalue is larger than the first. For large and positive t’s this means that the solution for this eigenvalue will be smaller than the solution for the first eigenvalue. Therefore, as t increases the trajectory will move in towards the origin and do so parallel to . Likewise, since the second eigenvalue is larger than the first this solution will dominate for large and negative t’s. Therefore, as we decrease t the trajectory will move away from the origin and do so parallel to .
This is what really confuses me. How do I see in the picture that t is increasing/decreasing. The plot does have x1 and x2 on the axis which are dependent on t if I understand correctly. Also what confuses me also how the Eigenvectors can give us so much information on how to draw the phase portrait itself. Any explenation/clarification would be very much appreciated.
differential-equations
asked Aug 15 at 6:56
craft
132
asked Aug 15 at 6:56
craft
132
asked Aug 15 at 6:56
craft
132
asked Aug 15 at 6:56
craft
132
132
You can also google for trace determinant plane. It does give a classification of all linear systems in the plane. The key fact is the trace and determinant of a $2times 2$ matrix $A$, when $x'(t)=Acdot x(t)$
– Fakemistake
Aug 15 at 9:23
Think also about the two functions $f(t)=e^-t$ and $g(t)=e^-6t$ for $tgeq 0$. For $t>0$ we have $f(t)>g(t)$ but both are converging to zero. Consequently $f$ is dominant. In the picture you see that for large $t$ the solutions are converging to the eigenspace of the eigenvalue $-1$. (Handwaving). For sure the solutions are converging to the origin.
– Fakemistake
Aug 15 at 9:28
add a comment |Â
You can also google for trace determinant plane. It does give a classification of all linear systems in the plane. The key fact is the trace and determinant of a $2times 2$ matrix $A$, when $x'(t)=Acdot x(t)$
– Fakemistake
Aug 15 at 9:23
Think also about the two functions $f(t)=e^-t$ and $g(t)=e^-6t$ for $tgeq 0$. For $t>0$ we have $f(t)>g(t)$ but both are converging to zero. Consequently $f$ is dominant. In the picture you see that for large $t$ the solutions are converging to the eigenspace of the eigenvalue $-1$. (Handwaving). For sure the solutions are converging to the origin.
– Fakemistake
Aug 15 at 9:28
You can also google for trace determinant plane. It does give a classification of all linear systems in the plane. The key fact is the trace and determinant of a $2times 2$ matrix $A$, when $x'(t)=Acdot x(t)$
– Fakemistake
Aug 15 at 9:23
You can also google for trace determinant plane. It does give a classification of all linear systems in the plane. The key fact is the trace and determinant of a $2times 2$ matrix $A$, when $x'(t)=Acdot x(t)$
– Fakemistake
Aug 15 at 9:23
Think also about the two functions $f(t)=e^-t$ and $g(t)=e^-6t$ for $tgeq 0$. For $t>0$ we have $f(t)>g(t)$ but both are converging to zero. Consequently $f$ is dominant. In the picture you see that for large $t$ the solutions are converging to the eigenspace of the eigenvalue $-1$. (Handwaving). For sure the solutions are converging to the origin.
– Fakemistake
Aug 15 at 9:28
Think also about the two functions $f(t)=e^-t$ and $g(t)=e^-6t$ for $tgeq 0$. For $t>0$ we have $f(t)>g(t)$ but both are converging to zero. Consequently $f$ is dominant. In the picture you see that for large $t$ the solutions are converging to the eigenspace of the eigenvalue $-1$. (Handwaving). For sure the solutions are converging to the origin.
– Fakemistake
Aug 15 at 9:28
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
As the $2$ by $2$ matrix has two real eigenvalues of multiplicity one, it can be diagonalized
$$
beginbmatrix
lambda_1 & 0
\
0 & lambda_2
endbmatrix.
$$
Look at diagonalization as a linear coordinate change. In the new coordinates, $(y_1, y_2)$, the ODE system has the form
$$
begincases
y'_1 = lambda_1 y_1
\
y'_2 = lambda_2 y_2,
endcases
$$
so its solutions are given by
$$
tag1
begincases
y_1(t) = C e^lambda_1 t
\
y_2(t) = D e^lambda_2 t,
endcases
$$
where $C$ and $D$ are real constants. $C = 0$ (or $D = 0$) correspond to the solutions whose first (or second) coordinate is constantly equal to zero.
Fix, for the moment, $C ne 0$ and $D ne 0$. By eliminating $t$ in $(1)$ we obtain
$$
leftlvert fracy_1C rightrvert^lambda_2 = leftlvert fracy_2D rightrvert^lambda_1,
$$
hence
$$
lvert y_1 rvert = E , lvert y_2 rvert^lambda_2/lambda_1,
$$
where $E = lvert D rvert lvert C rvert^-lambda_2 / lambda_1$. By symmetry, we can restrict ourselves to the first quadrant:
$$
y_2 = E , y_1^lambda_2/lambda_1, quad E > 0.
$$
As $lambda_2 < lambda_1 < 0$, $lambda_2/lambda_1 > 1$. We have thus obtained a family of "generalized parabolas", tangent at the origin to the $y_1$-axis.
Remember that we are still in the new coordinates. Returning to the old coordinates we do some rotating and stretching, and those operations do not change tangencies. In particular, the trajectories (except on the one-dimensional invariant subspaces spanned by eigenvectors) are curves tangent at the origin to an eigenvector corresponding to the larger eigenvalue, as in the picture.
Notice that, as $t to infty$, both coordinates of any solution converge to zero, so the arrows on the curves must be directed "toward" the origin.
answered Aug 15 at 9:21
user539887
1,5491313
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
As the $2$ by $2$ matrix has two real eigenvalues of multiplicity one, it can be diagonalized
$$
beginbmatrix
lambda_1 & 0
\
0 & lambda_2
endbmatrix.
$$
Look at diagonalization as a linear coordinate change. In the new coordinates, $(y_1, y_2)$, the ODE system has the form
$$
begincases
y'_1 = lambda_1 y_1
\
y'_2 = lambda_2 y_2,
endcases
$$
so its solutions are given by
$$
tag1
begincases
y_1(t) = C e^lambda_1 t
\
y_2(t) = D e^lambda_2 t,
endcases
$$
where $C$ and $D$ are real constants. $C = 0$ (or $D = 0$) correspond to the solutions whose first (or second) coordinate is constantly equal to zero.
Fix, for the moment, $C ne 0$ and $D ne 0$. By eliminating $t$ in $(1)$ we obtain
$$
leftlvert fracy_1C rightrvert^lambda_2 = leftlvert fracy_2D rightrvert^lambda_1,
$$
hence
$$
lvert y_1 rvert = E , lvert y_2 rvert^lambda_2/lambda_1,
$$
where $E = lvert D rvert lvert C rvert^-lambda_2 / lambda_1$. By symmetry, we can restrict ourselves to the first quadrant:
$$
y_2 = E , y_1^lambda_2/lambda_1, quad E > 0.
$$
As $lambda_2 < lambda_1 < 0$, $lambda_2/lambda_1 > 1$. We have thus obtained a family of "generalized parabolas", tangent at the origin to the $y_1$-axis.
Remember that we are still in the new coordinates. Returning to the old coordinates we do some rotating and stretching, and those operations do not change tangencies. In particular, the trajectories (except on the one-dimensional invariant subspaces spanned by eigenvectors) are curves tangent at the origin to an eigenvector corresponding to the larger eigenvalue, as in the picture.
Notice that, as $t to infty$, both coordinates of any solution converge to zero, so the arrows on the curves must be directed "toward" the origin.
answered Aug 15 at 9:21
user539887
1,5491313
add a comment |Â
up vote
0
down vote
As the $2$ by $2$ matrix has two real eigenvalues of multiplicity one, it can be diagonalized
$$
beginbmatrix
lambda_1 & 0
\
0 & lambda_2
endbmatrix.
$$
Look at diagonalization as a linear coordinate change. In the new coordinates, $(y_1, y_2)$, the ODE system has the form
$$
begincases
y'_1 = lambda_1 y_1
\
y'_2 = lambda_2 y_2,
endcases
$$
so its solutions are given by
$$
tag1
begincases
y_1(t) = C e^lambda_1 t
\
y_2(t) = D e^lambda_2 t,
endcases
$$
where $C$ and $D$ are real constants. $C = 0$ (or $D = 0$) correspond to the solutions whose first (or second) coordinate is constantly equal to zero.
Fix, for the moment, $C ne 0$ and $D ne 0$. By eliminating $t$ in $(1)$ we obtain
$$
leftlvert fracy_1C rightrvert^lambda_2 = leftlvert fracy_2D rightrvert^lambda_1,
$$
hence
$$
lvert y_1 rvert = E , lvert y_2 rvert^lambda_2/lambda_1,
$$
where $E = lvert D rvert lvert C rvert^-lambda_2 / lambda_1$. By symmetry, we can restrict ourselves to the first quadrant:
$$
y_2 = E , y_1^lambda_2/lambda_1, quad E > 0.
$$
As $lambda_2 < lambda_1 < 0$, $lambda_2/lambda_1 > 1$. We have thus obtained a family of "generalized parabolas", tangent at the origin to the $y_1$-axis.
Remember that we are still in the new coordinates. Returning to the old coordinates we do some rotating and stretching, and those operations do not change tangencies. In particular, the trajectories (except on the one-dimensional invariant subspaces spanned by eigenvectors) are curves tangent at the origin to an eigenvector corresponding to the larger eigenvalue, as in the picture.
Notice that, as $t to infty$, both coordinates of any solution converge to zero, so the arrows on the curves must be directed "toward" the origin.
answered Aug 15 at 9:21
user539887
1,5491313
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As the $2$ by $2$ matrix has two real eigenvalues of multiplicity one, it can be diagonalized
$$
beginbmatrix
lambda_1 & 0
\
0 & lambda_2
endbmatrix.
$$
Look at diagonalization as a linear coordinate change. In the new coordinates, $(y_1, y_2)$, the ODE system has the form
$$
begincases
y'_1 = lambda_1 y_1
\
y'_2 = lambda_2 y_2,
endcases
$$
so its solutions are given by
$$
tag1
begincases
y_1(t) = C e^lambda_1 t
\
y_2(t) = D e^lambda_2 t,
endcases
$$
where $C$ and $D$ are real constants. $C = 0$ (or $D = 0$) correspond to the solutions whose first (or second) coordinate is constantly equal to zero.
Fix, for the moment, $C ne 0$ and $D ne 0$. By eliminating $t$ in $(1)$ we obtain
$$
leftlvert fracy_1C rightrvert^lambda_2 = leftlvert fracy_2D rightrvert^lambda_1,
$$
hence
$$
lvert y_1 rvert = E , lvert y_2 rvert^lambda_2/lambda_1,
$$
where $E = lvert D rvert lvert C rvert^-lambda_2 / lambda_1$. By symmetry, we can restrict ourselves to the first quadrant:
$$
y_2 = E , y_1^lambda_2/lambda_1, quad E > 0.
$$
As $lambda_2 < lambda_1 < 0$, $lambda_2/lambda_1 > 1$. We have thus obtained a family of "generalized parabolas", tangent at the origin to the $y_1$-axis.
Remember that we are still in the new coordinates. Returning to the old coordinates we do some rotating and stretching, and those operations do not change tangencies. In particular, the trajectories (except on the one-dimensional invariant subspaces spanned by eigenvectors) are curves tangent at the origin to an eigenvector corresponding to the larger eigenvalue, as in the picture.
Notice that, as $t to infty$, both coordinates of any solution converge to zero, so the arrows on the curves must be directed "toward" the origin.
answered Aug 15 at 9:21
user539887
1,5491313
As the $2$ by $2$ matrix has two real eigenvalues of multiplicity one, it can be diagonalized
$$
beginbmatrix
lambda_1 & 0
\
0 & lambda_2
endbmatrix.
$$
Look at diagonalization as a linear coordinate change. In the new coordinates, $(y_1, y_2)$, the ODE system has the form
$$
begincases
y'_1 = lambda_1 y_1
\
y'_2 = lambda_2 y_2,
endcases
$$
so its solutions are given by
$$
tag1
begincases
y_1(t) = C e^lambda_1 t
\
y_2(t) = D e^lambda_2 t,
endcases
$$
where $C$ and $D$ are real constants. $C = 0$ (or $D = 0$) correspond to the solutions whose first (or second) coordinate is constantly equal to zero.
Fix, for the moment, $C ne 0$ and $D ne 0$. By eliminating $t$ in $(1)$ we obtain
$$
leftlvert fracy_1C rightrvert^lambda_2 = leftlvert fracy_2D rightrvert^lambda_1,
$$
hence
$$
lvert y_1 rvert = E , lvert y_2 rvert^lambda_2/lambda_1,
$$
where $E = lvert D rvert lvert C rvert^-lambda_2 / lambda_1$. By symmetry, we can restrict ourselves to the first quadrant:
$$
y_2 = E , y_1^lambda_2/lambda_1, quad E > 0.
$$
As $lambda_2 < lambda_1 < 0$, $lambda_2/lambda_1 > 1$. We have thus obtained a family of "generalized parabolas", tangent at the origin to the $y_1$-axis.
Remember that we are still in the new coordinates. Returning to the old coordinates we do some rotating and stretching, and those operations do not change tangencies. In particular, the trajectories (except on the one-dimensional invariant subspaces spanned by eigenvectors) are curves tangent at the origin to an eigenvector corresponding to the larger eigenvalue, as in the picture.
Notice that, as $t to infty$, both coordinates of any solution converge to zero, so the arrows on the curves must be directed "toward" the origin.
answered Aug 15 at 9:21
user539887
1,5491313
answered Aug 15 at 9:21
user539887
1,5491313
answered Aug 15 at 9:21
user539887
1,5491313
answered Aug 15 at 9:21
user539887
1,5491313
1,5491313
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2883271%2fdrawing-a-phase-portrait-given-eigenvectors%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
You can also google for trace determinant plane. It does give a classification of all linear systems in the plane. The key fact is the trace and determinant of a $2times 2$ matrix $A$, when $x'(t)=Acdot x(t)$
– Fakemistake
Aug 15 at 9:23
Think also about the two functions $f(t)=e^-t$ and $g(t)=e^-6t$ for $tgeq 0$. For $t>0$ we have $f(t)>g(t)$ but both are converging to zero. Consequently $f$ is dominant. In the picture you see that for large $t$ the solutions are converging to the eigenspace of the eigenvalue $-1$. (Handwaving). For sure the solutions are converging to the origin.
– Fakemistake
Aug 15 at 9:28