What exactly is a defective nodal source

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I attain repeated root eigenvalue of $-1$ with eigenvector $beginpmatrix2\1endpmatrix$

However, I'm uncertain if the general solution would be $c_1cdot e^-tbeginpmatrix2\1endpmatrix$ alone, or would it be $c_1cdot e^-tbeginpmatrix2\1endpmatrix + c_2cdot tcdot e^-tbeginpmatrix2\1endpmatrix$.

Likewise, when I plot the phase portriat in software, I receive an alert claiming that the phase protriat is a defective nodal source. What is that exactly?

differential-equations eigenvalues-eigenvectors homogeneous-equation

share|cite|improve this question

edited Jul 22 at 15:12

mrtaurho

740219

asked Jul 22 at 15:00

SleepApnea

426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your second attempt should be the right hence a general solution never consists only of one term. But I guess you need to rework the second eigenvector.
  – mrtaurho
  Jul 22 at 15:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so I got my general solution as e^t(c1*(2;1)+c2(t(2;1)+(-4,0)). However, how can I solve this IVP with just one initial conditions and two arbitrary constants?
  – SleepApnea
  Jul 22 at 15:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so I finalized my general solution to be e^t(c1(2,1)+c2(t(2,1)+(-4,0)) However, I do not understand how to solve the IVP with two constants unknown, and only one condition.
  – SleepApnea
  Jul 22 at 15:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you are working on a system of differential equations there is no problem in calculating the constants. Just set the general solution equal to the IVP and split it up in two equations : now you got a linear system of two equations and two variables.
  – mrtaurho
  Jul 22 at 15:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is my general solution correct to add the additional (-4,0) at the tail end of the problem?
  – SleepApnea
  Jul 22 at 15:35
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote

favorite

I attain repeated root eigenvalue of $-1$ with eigenvector $beginpmatrix2\1endpmatrix$

However, I'm uncertain if the general solution would be $c_1cdot e^-tbeginpmatrix2\1endpmatrix$ alone, or would it be $c_1cdot e^-tbeginpmatrix2\1endpmatrix + c_2cdot tcdot e^-tbeginpmatrix2\1endpmatrix$.

Likewise, when I plot the phase portriat in software, I receive an alert claiming that the phase protriat is a defective nodal source. What is that exactly?

differential-equations eigenvalues-eigenvectors homogeneous-equation

share|cite|improve this question

edited Jul 22 at 15:12

mrtaurho

740219

asked Jul 22 at 15:00

SleepApnea

426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your second attempt should be the right hence a general solution never consists only of one term. But I guess you need to rework the second eigenvector.
  – mrtaurho
  Jul 22 at 15:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so I got my general solution as e^t(c1*(2;1)+c2(t(2;1)+(-4,0)). However, how can I solve this IVP with just one initial conditions and two arbitrary constants?
  – SleepApnea
  Jul 22 at 15:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so I finalized my general solution to be e^t(c1(2,1)+c2(t(2,1)+(-4,0)) However, I do not understand how to solve the IVP with two constants unknown, and only one condition.
  – SleepApnea
  Jul 22 at 15:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you are working on a system of differential equations there is no problem in calculating the constants. Just set the general solution equal to the IVP and split it up in two equations : now you got a linear system of two equations and two variables.
  – mrtaurho
  Jul 22 at 15:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is my general solution correct to add the additional (-4,0) at the tail end of the problem?
  – SleepApnea
  Jul 22 at 15:35
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I attain repeated root eigenvalue of $-1$ with eigenvector $beginpmatrix2\1endpmatrix$

However, I'm uncertain if the general solution would be $c_1cdot e^-tbeginpmatrix2\1endpmatrix$ alone, or would it be $c_1cdot e^-tbeginpmatrix2\1endpmatrix + c_2cdot tcdot e^-tbeginpmatrix2\1endpmatrix$.

Likewise, when I plot the phase portriat in software, I receive an alert claiming that the phase protriat is a defective nodal source. What is that exactly?

differential-equations eigenvalues-eigenvectors homogeneous-equation

share|cite|improve this question

edited Jul 22 at 15:12

mrtaurho

740219

asked Jul 22 at 15:00

SleepApnea

426

I attain repeated root eigenvalue of $-1$ with eigenvector $beginpmatrix2\1endpmatrix$

However, I'm uncertain if the general solution would be $c_1cdot e^-tbeginpmatrix2\1endpmatrix$ alone, or would it be $c_1cdot e^-tbeginpmatrix2\1endpmatrix + c_2cdot tcdot e^-tbeginpmatrix2\1endpmatrix$.

Likewise, when I plot the phase portriat in software, I receive an alert claiming that the phase protriat is a defective nodal source. What is that exactly?

differential-equations eigenvalues-eigenvectors homogeneous-equation

share|cite|improve this question

edited Jul 22 at 15:12

mrtaurho

740219

asked Jul 22 at 15:00

SleepApnea

426

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 15:12

mrtaurho

740219

edited Jul 22 at 15:12

mrtaurho

740219

edited Jul 22 at 15:12

mrtaurho

740219

740219

asked Jul 22 at 15:00

SleepApnea

426

asked Jul 22 at 15:00

SleepApnea

426

asked Jul 22 at 15:00

SleepApnea

426

426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your second attempt should be the right hence a general solution never consists only of one term. But I guess you need to rework the second eigenvector.
  – mrtaurho
  Jul 22 at 15:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so I got my general solution as e^t(c1*(2;1)+c2(t(2;1)+(-4,0)). However, how can I solve this IVP with just one initial conditions and two arbitrary constants?
  – SleepApnea
  Jul 22 at 15:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so I finalized my general solution to be e^t(c1(2,1)+c2(t(2,1)+(-4,0)) However, I do not understand how to solve the IVP with two constants unknown, and only one condition.
  – SleepApnea
  Jul 22 at 15:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you are working on a system of differential equations there is no problem in calculating the constants. Just set the general solution equal to the IVP and split it up in two equations : now you got a linear system of two equations and two variables.
  – mrtaurho
  Jul 22 at 15:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is my general solution correct to add the additional (-4,0) at the tail end of the problem?
  – SleepApnea
  Jul 22 at 15:35
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your second attempt should be the right hence a general solution never consists only of one term. But I guess you need to rework the second eigenvector.
  – mrtaurho
  Jul 22 at 15:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so I got my general solution as e^t(c1*(2;1)+c2(t(2;1)+(-4,0)). However, how can I solve this IVP with just one initial conditions and two arbitrary constants?
  – SleepApnea
  Jul 22 at 15:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so I finalized my general solution to be e^t(c1(2,1)+c2(t(2,1)+(-4,0)) However, I do not understand how to solve the IVP with two constants unknown, and only one condition.
  – SleepApnea
  Jul 22 at 15:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you are working on a system of differential equations there is no problem in calculating the constants. Just set the general solution equal to the IVP and split it up in two equations : now you got a linear system of two equations and two variables.
  – mrtaurho
  Jul 22 at 15:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is my general solution correct to add the additional (-4,0) at the tail end of the problem?
  – SleepApnea
  Jul 22 at 15:35
  

  
  
  
  

Your second attempt should be the right hence a general solution never consists only of one term. But I guess you need to rework the second eigenvector.
– mrtaurho
Jul 22 at 15:15

Your second attempt should be the right hence a general solution never consists only of one term. But I guess you need to rework the second eigenvector.
– mrtaurho
Jul 22 at 15:15

so I got my general solution as e^t(c1*(2;1)+c2(t(2;1)+(-4,0)). However, how can I solve this IVP with just one initial conditions and two arbitrary constants?
– SleepApnea
Jul 22 at 15:24

so I got my general solution as e^t(c1*(2;1)+c2(t(2;1)+(-4,0)). However, how can I solve this IVP with just one initial conditions and two arbitrary constants?
– SleepApnea
Jul 22 at 15:24

so I finalized my general solution to be e^t(c1(2,1)+c2(t(2,1)+(-4,0)) However, I do not understand how to solve the IVP with two constants unknown, and only one condition.
– SleepApnea
Jul 22 at 15:27

so I finalized my general solution to be e^t(c1(2,1)+c2(t(2,1)+(-4,0)) However, I do not understand how to solve the IVP with two constants unknown, and only one condition.
– SleepApnea
Jul 22 at 15:27

Since you are working on a system of differential equations there is no problem in calculating the constants. Just set the general solution equal to the IVP and split it up in two equations : now you got a linear system of two equations and two variables.
– mrtaurho
Jul 22 at 15:33

Since you are working on a system of differential equations there is no problem in calculating the constants. Just set the general solution equal to the IVP and split it up in two equations : now you got a linear system of two equations and two variables.
– mrtaurho
Jul 22 at 15:33

is my general solution correct to add the additional (-4,0) at the tail end of the problem?
– SleepApnea
Jul 22 at 15:35

is my general solution correct to add the additional (-4,0) at the tail end of the problem?
– SleepApnea
Jul 22 at 15:35

 | 
show 2 more comments

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

In the case of repeated real roots you need to calculate a different second eigenvector which satisfies the equation

$$beginpmatrixa-lambda & b\c & d-lambdaendpmatrixbeginpmatrixrho_1\rho_2endpmatrix~=~beginpmatrixeta_1\eta_2endpmatrix$$

where $beginpmatrixeta_1\eta_2endpmatrix$ is the first eigenvector and $lambda$ is the eigenvalue of the matrix. From there on you can conclude to the general solution of the form

$$vecx~=~c_1cdot e^lambda tveceta+c_2(tcdot e^lambda tveceta+e^lambda tvecrho)$$

---

The eigenvalues of the given systems are the repeated real roots $lambda_1,2=-1$. From there one you get the first eigenvector $beginpmatrix2\1endpmatrix$ as you already did. Now just calculate the second eigenvector and you are done.

share|cite|improve this answer

edited Jul 22 at 15:49

answered Jul 22 at 15:30

mrtaurho

740219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  based on my general solution: e^t(c1(2,1)+c2(t(2,1)+(-4,0)). Then if we use the IVP and plug in t = 0, I get (1,1) = c1(2,1) + c2(-4,0). I don't really understand how to isolate C1 and C2.
  – SleepApnea
  Jul 22 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just cosider it as a system of the two equations $1=c_1-4c_2$ and $1=-4c_1$
  – mrtaurho
  Jul 22 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Based on the phase portrait of this particular case, can I argue that when t either moves to infinity or negative infinity, the phase portrait is moving tangentially inwards towards the origin.
  – SleepApnea
  Jul 22 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you maybe link the phase potrait somewhere?
  – mrtaurho
  Jul 22 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ibb.co/cDCyby This is the phase portrait.
  – SleepApnea
  Jul 22 at 17:30
  
  

  
  
  
  

 | 
show 2 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

In the case of repeated real roots you need to calculate a different second eigenvector which satisfies the equation

$$beginpmatrixa-lambda & b\c & d-lambdaendpmatrixbeginpmatrixrho_1\rho_2endpmatrix~=~beginpmatrixeta_1\eta_2endpmatrix$$

where $beginpmatrixeta_1\eta_2endpmatrix$ is the first eigenvector and $lambda$ is the eigenvalue of the matrix. From there on you can conclude to the general solution of the form

$$vecx~=~c_1cdot e^lambda tveceta+c_2(tcdot e^lambda tveceta+e^lambda tvecrho)$$

---

The eigenvalues of the given systems are the repeated real roots $lambda_1,2=-1$. From there one you get the first eigenvector $beginpmatrix2\1endpmatrix$ as you already did. Now just calculate the second eigenvector and you are done.

share|cite|improve this answer

edited Jul 22 at 15:49

answered Jul 22 at 15:30

mrtaurho

740219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  based on my general solution: e^t(c1(2,1)+c2(t(2,1)+(-4,0)). Then if we use the IVP and plug in t = 0, I get (1,1) = c1(2,1) + c2(-4,0). I don't really understand how to isolate C1 and C2.
  – SleepApnea
  Jul 22 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just cosider it as a system of the two equations $1=c_1-4c_2$ and $1=-4c_1$
  – mrtaurho
  Jul 22 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Based on the phase portrait of this particular case, can I argue that when t either moves to infinity or negative infinity, the phase portrait is moving tangentially inwards towards the origin.
  – SleepApnea
  Jul 22 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you maybe link the phase potrait somewhere?
  – mrtaurho
  Jul 22 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ibb.co/cDCyby This is the phase portrait.
  – SleepApnea
  Jul 22 at 17:30
  
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote



accepted

In the case of repeated real roots you need to calculate a different second eigenvector which satisfies the equation

$$beginpmatrixa-lambda & b\c & d-lambdaendpmatrixbeginpmatrixrho_1\rho_2endpmatrix~=~beginpmatrixeta_1\eta_2endpmatrix$$

where $beginpmatrixeta_1\eta_2endpmatrix$ is the first eigenvector and $lambda$ is the eigenvalue of the matrix. From there on you can conclude to the general solution of the form

$$vecx~=~c_1cdot e^lambda tveceta+c_2(tcdot e^lambda tveceta+e^lambda tvecrho)$$

---

The eigenvalues of the given systems are the repeated real roots $lambda_1,2=-1$. From there one you get the first eigenvector $beginpmatrix2\1endpmatrix$ as you already did. Now just calculate the second eigenvector and you are done.

share|cite|improve this answer

edited Jul 22 at 15:49

answered Jul 22 at 15:30

mrtaurho

740219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  based on my general solution: e^t(c1(2,1)+c2(t(2,1)+(-4,0)). Then if we use the IVP and plug in t = 0, I get (1,1) = c1(2,1) + c2(-4,0). I don't really understand how to isolate C1 and C2.
  – SleepApnea
  Jul 22 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just cosider it as a system of the two equations $1=c_1-4c_2$ and $1=-4c_1$
  – mrtaurho
  Jul 22 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Based on the phase portrait of this particular case, can I argue that when t either moves to infinity or negative infinity, the phase portrait is moving tangentially inwards towards the origin.
  – SleepApnea
  Jul 22 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you maybe link the phase potrait somewhere?
  – mrtaurho
  Jul 22 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ibb.co/cDCyby This is the phase portrait.
  – SleepApnea
  Jul 22 at 17:30
  
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote



accepted

up vote
0
down vote



accepted

In the case of repeated real roots you need to calculate a different second eigenvector which satisfies the equation

$$beginpmatrixa-lambda & b\c & d-lambdaendpmatrixbeginpmatrixrho_1\rho_2endpmatrix~=~beginpmatrixeta_1\eta_2endpmatrix$$

where $beginpmatrixeta_1\eta_2endpmatrix$ is the first eigenvector and $lambda$ is the eigenvalue of the matrix. From there on you can conclude to the general solution of the form

$$vecx~=~c_1cdot e^lambda tveceta+c_2(tcdot e^lambda tveceta+e^lambda tvecrho)$$

---

The eigenvalues of the given systems are the repeated real roots $lambda_1,2=-1$. From there one you get the first eigenvector $beginpmatrix2\1endpmatrix$ as you already did. Now just calculate the second eigenvector and you are done.

share|cite|improve this answer

edited Jul 22 at 15:49

answered Jul 22 at 15:30

mrtaurho

740219

In the case of repeated real roots you need to calculate a different second eigenvector which satisfies the equation

$$beginpmatrixa-lambda & b\c & d-lambdaendpmatrixbeginpmatrixrho_1\rho_2endpmatrix~=~beginpmatrixeta_1\eta_2endpmatrix$$

where $beginpmatrixeta_1\eta_2endpmatrix$ is the first eigenvector and $lambda$ is the eigenvalue of the matrix. From there on you can conclude to the general solution of the form

$$vecx~=~c_1cdot e^lambda tveceta+c_2(tcdot e^lambda tveceta+e^lambda tvecrho)$$

---

The eigenvalues of the given systems are the repeated real roots $lambda_1,2=-1$. From there one you get the first eigenvector $beginpmatrix2\1endpmatrix$ as you already did. Now just calculate the second eigenvector and you are done.

share|cite|improve this answer

edited Jul 22 at 15:49

answered Jul 22 at 15:30

mrtaurho

740219

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 15:49

edited Jul 22 at 15:49

edited Jul 22 at 15:49

answered Jul 22 at 15:30

mrtaurho

740219

answered Jul 22 at 15:30

mrtaurho

740219

answered Jul 22 at 15:30

mrtaurho

740219

740219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  based on my general solution: e^t(c1(2,1)+c2(t(2,1)+(-4,0)). Then if we use the IVP and plug in t = 0, I get (1,1) = c1(2,1) + c2(-4,0). I don't really understand how to isolate C1 and C2.
  – SleepApnea
  Jul 22 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just cosider it as a system of the two equations $1=c_1-4c_2$ and $1=-4c_1$
  – mrtaurho
  Jul 22 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Based on the phase portrait of this particular case, can I argue that when t either moves to infinity or negative infinity, the phase portrait is moving tangentially inwards towards the origin.
  – SleepApnea
  Jul 22 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you maybe link the phase potrait somewhere?
  – mrtaurho
  Jul 22 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ibb.co/cDCyby This is the phase portrait.
  – SleepApnea
  Jul 22 at 17:30
  
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  based on my general solution: e^t(c1(2,1)+c2(t(2,1)+(-4,0)). Then if we use the IVP and plug in t = 0, I get (1,1) = c1(2,1) + c2(-4,0). I don't really understand how to isolate C1 and C2.
  – SleepApnea
  Jul 22 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just cosider it as a system of the two equations $1=c_1-4c_2$ and $1=-4c_1$
  – mrtaurho
  Jul 22 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Based on the phase portrait of this particular case, can I argue that when t either moves to infinity or negative infinity, the phase portrait is moving tangentially inwards towards the origin.
  – SleepApnea
  Jul 22 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you maybe link the phase potrait somewhere?
  – mrtaurho
  Jul 22 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ibb.co/cDCyby This is the phase portrait.
  – SleepApnea
  Jul 22 at 17:30
  
  

  
  
  
  

based on my general solution: e^t(c1(2,1)+c2(t(2,1)+(-4,0)). Then if we use the IVP and plug in t = 0, I get (1,1) = c1(2,1) + c2(-4,0). I don't really understand how to isolate C1 and C2.
– SleepApnea
Jul 22 at 15:36

based on my general solution: e^t(c1(2,1)+c2(t(2,1)+(-4,0)). Then if we use the IVP and plug in t = 0, I get (1,1) = c1(2,1) + c2(-4,0). I don't really understand how to isolate C1 and C2.
– SleepApnea
Jul 22 at 15:36

Just cosider it as a system of the two equations $1=c_1-4c_2$ and $1=-4c_1$
– mrtaurho
Jul 22 at 15:44

Just cosider it as a system of the two equations $1=c_1-4c_2$ and $1=-4c_1$
– mrtaurho
Jul 22 at 15:44

Based on the phase portrait of this particular case, can I argue that when t either moves to infinity or negative infinity, the phase portrait is moving tangentially inwards towards the origin.
– SleepApnea
Jul 22 at 16:54

Based on the phase portrait of this particular case, can I argue that when t either moves to infinity or negative infinity, the phase portrait is moving tangentially inwards towards the origin.
– SleepApnea
Jul 22 at 16:54

Could you maybe link the phase potrait somewhere?
– mrtaurho
Jul 22 at 16:58

Could you maybe link the phase potrait somewhere?
– mrtaurho
Jul 22 at 16:58

ibb.co/cDCyby This is the phase portrait.
– SleepApnea
Jul 22 at 17:30

ibb.co/cDCyby This is the phase portrait.
– SleepApnea
Jul 22 at 17:30

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