Mixing
Eigenvalues keep giving trivial solutions everytime.
$begingroup$
I am trying to find the eigenvalues of this Eigen BVP. $mu$ is the eigenvalue parameter
$$
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F = 0
$$ wit BC(s) $F(0)=0,frac{F''(0)}{F'(0)}=beta_h,frac{F''(1)}{F'(1)}=beta_h$
For $lambda_h=0.02$ and $beta_h=10$, I calculated the EVs using chebfun in MATLAB. There are infinite number of negative EVs. The general solution should be of the form
$$
F(x) = sum_k C_k e^{-delta_k(mu)x}
$$
Substituting the bc(s) yield three linear equations in $C_1,C_2,C_3$.
where $-delta_k(mu)$ are the three roots of the characteristic equation obtained after substituting one of the EVs.
But this keeps giving me trivial solutions i.e all $C_k=0$. I tried many EVs (ex. $-32.9463$)but to no avail.
Is there something horribly wrong in my understanding ? Is there some other approach i can apply to the problem ? I cannot figure out at all.
ATTEMPT
After @LutzL and @Christoph provided useful comments, i did some reading on null spaces of matrices and their use to solve the homogenous system of equations.
Using the EV mentioned above $-32.9463$ and $lambda_h = 0.02, beta_h = 10$, I arrive at three roots of the characteristic equation as:
$$
s_1 = 3.7421
$$
$$
s_2 = -11.8710+34.5722i
$$
$$
s_3 = -11.8710-34.5722i
$$
Now applying the three BC(s) i have the following matrix system $M(mu).C_k=0$ and i need to find $C_k$ which will be the non trivial solutions.
$$
begin{bmatrix}
1 & 1 & 1 \
{s_1}^2+beta_hs_1 & {s_2}^2+beta_hs_2 & {s_3}^2+beta_hs_3 \
e^{-s_1}({s_1}^2+beta_hs_1) & e^{-s_2}({s_2}^2+beta_hs_2) & e^{-s_3}({s_3}^2+beta_hs_3) \
end{bmatrix}
*
begin{bmatrix}
C_1 \
C_2 \
C_3 \
end{bmatrix}=0
$$
The above is a homogeneous system of equations which i solved using the following command in MATLAB
[U,S,V] = svd(A,'econ');
b = V(:,size(A,2));
Here $A$ is the coefficient matrix. The solver linsolve(A,B) kept giving me trivial solutions. svd means singular value decomposition. b is supposed to be the solution we are looking for. For the parameters i mentioned here it comes out to be
b =-0.4550 + 0.0000i
0.2278 + 0.5870i
0.2278 - 0.5870i
It is mentioned that
bwill be a solution if the corresponding smallest singular value is zero. If not, b will be the nearest you can come to a solution. If more than one singular value is zero, there are infinitely many solutions of which this will be one
So the values of b must be my corresponding $C_1,C_2,C_3$. But they are coming out to be complex. So should i just take the magnitudes of these complex numbers to form my solution ?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors linear-transformations boundary-value-problem
asked Jan 19 at 16:05
Indrasis MitraIndrasis Mitra
70111
$endgroup$
|
show 9 more comments
$begingroup$
I am trying to find the eigenvalues of this Eigen BVP. $mu$ is the eigenvalue parameter
$$
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F = 0
$$ wit BC(s) $F(0)=0,frac{F''(0)}{F'(0)}=beta_h,frac{F''(1)}{F'(1)}=beta_h$
For $lambda_h=0.02$ and $beta_h=10$, I calculated the EVs using chebfun in MATLAB. There are infinite number of negative EVs. The general solution should be of the form
$$
F(x) = sum_k C_k e^{-delta_k(mu)x}
$$
Substituting the bc(s) yield three linear equations in $C_1,C_2,C_3$.
where $-delta_k(mu)$ are the three roots of the characteristic equation obtained after substituting one of the EVs.
But this keeps giving me trivial solutions i.e all $C_k=0$. I tried many EVs (ex. $-32.9463$)but to no avail.
Is there something horribly wrong in my understanding ? Is there some other approach i can apply to the problem ? I cannot figure out at all.
ATTEMPT
After @LutzL and @Christoph provided useful comments, i did some reading on null spaces of matrices and their use to solve the homogenous system of equations.
Using the EV mentioned above $-32.9463$ and $lambda_h = 0.02, beta_h = 10$, I arrive at three roots of the characteristic equation as:
$$
s_1 = 3.7421
$$
$$
s_2 = -11.8710+34.5722i
$$
$$
s_3 = -11.8710-34.5722i
$$
Now applying the three BC(s) i have the following matrix system $M(mu).C_k=0$ and i need to find $C_k$ which will be the non trivial solutions.
$$
begin{bmatrix}
1 & 1 & 1 \
{s_1}^2+beta_hs_1 & {s_2}^2+beta_hs_2 & {s_3}^2+beta_hs_3 \
e^{-s_1}({s_1}^2+beta_hs_1) & e^{-s_2}({s_2}^2+beta_hs_2) & e^{-s_3}({s_3}^2+beta_hs_3) \
end{bmatrix}
*
begin{bmatrix}
C_1 \
C_2 \
C_3 \
end{bmatrix}=0
$$
The above is a homogeneous system of equations which i solved using the following command in MATLAB
[U,S,V] = svd(A,'econ');
b = V(:,size(A,2));
Here $A$ is the coefficient matrix. The solver linsolve(A,B) kept giving me trivial solutions. svd means singular value decomposition. b is supposed to be the solution we are looking for. For the parameters i mentioned here it comes out to be
b =-0.4550 + 0.0000i
0.2278 + 0.5870i
0.2278 - 0.5870i
It is mentioned that
bwill be a solution if the corresponding smallest singular value is zero. If not, b will be the nearest you can come to a solution. If more than one singular value is zero, there are infinitely many solutions of which this will be one
So the values of b must be my corresponding $C_1,C_2,C_3$. But they are coming out to be complex. So should i just take the magnitudes of these complex numbers to form my solution ?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors linear-transformations boundary-value-problem
asked Jan 19 at 16:05
Indrasis MitraIndrasis Mitra
70111
$endgroup$
$begingroup$
Can't you get chebfun to compute the eigenfunctions as well? Obviously $F equiv 0$ is a solution to the BVP for any value of $mu$. But since this cannot be an eigenfunction, your linear system $AC=0$ for the coefficients $C_1, C_2, C_3$ must have non-trivial solutions if $mu$ is an eigenvalue. These are the solutions that you must find, for example by determining the null space of the linear map $C mapsto AC$.
$endgroup$
– Christoph
Jan 19 at 17:06
1
$begingroup$
Set $C_3=1$ and leave out the last equation, the first two should then give numerical values for $C_1,C_2$. It is not surprising that you get an invertible if ill-conditioned matrix instead of a singular one if evaluated in floating point arithmetic.
$endgroup$
– LutzL
Jan 19 at 17:15
$begingroup$
@Christoph . Thanks. Yes i know that the system $AX=0$ do have $X=0$ as a trivial solution always and i must look for the non trivial solutions. Also, $chebfun$ does give something called eigenmodes which are plots. Your last point , says "determine the null space of the linear map C to AC . I am sorry to say, but i have no idea how this is done and how i should move forward with it.
$endgroup$
– Indrasis Mitra
Jan 19 at 17:21
2
$begingroup$
Here you have complex conjugate $s_2 = overline{s_3}$ and $C_2 = overline{C_3}$. Therefore, the corresponding eigenfunction will be something like $F(x) = C_1 e^{-s_1 x} + 2 e^{-mathrm{Re}(s_2) x} left( mathrm{Re}(C_2) cos(mathrm{Im}(s_2) x) + mathrm{Im}(C_2) sin(mathrm{Im}(s_2) x) right)$. $C_1$ should probably be zero here.
$endgroup$
– Christoph
Jan 20 at 7:35
1
$begingroup$
Your characteristic roots have the wrong sign, if you set $y=ce^{-sx}$ for $0=a_3y'''+a_2y''+a_1y'+a_0y$, then the characteristic polynomial has to be $0=a_3s^3-a_2s^2+a_1s-a_0$.
$endgroup$
– LutzL
Jan 20 at 9:56
|
show 9 more comments
$begingroup$
I am trying to find the eigenvalues of this Eigen BVP. $mu$ is the eigenvalue parameter
$$
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F = 0
$$ wit BC(s) $F(0)=0,frac{F''(0)}{F'(0)}=beta_h,frac{F''(1)}{F'(1)}=beta_h$
For $lambda_h=0.02$ and $beta_h=10$, I calculated the EVs using chebfun in MATLAB. There are infinite number of negative EVs. The general solution should be of the form
$$
F(x) = sum_k C_k e^{-delta_k(mu)x}
$$
Substituting the bc(s) yield three linear equations in $C_1,C_2,C_3$.
where $-delta_k(mu)$ are the three roots of the characteristic equation obtained after substituting one of the EVs.
But this keeps giving me trivial solutions i.e all $C_k=0$. I tried many EVs (ex. $-32.9463$)but to no avail.
Is there something horribly wrong in my understanding ? Is there some other approach i can apply to the problem ? I cannot figure out at all.
ATTEMPT
After @LutzL and @Christoph provided useful comments, i did some reading on null spaces of matrices and their use to solve the homogenous system of equations.
Using the EV mentioned above $-32.9463$ and $lambda_h = 0.02, beta_h = 10$, I arrive at three roots of the characteristic equation as:
$$
s_1 = 3.7421
$$
$$
s_2 = -11.8710+34.5722i
$$
$$
s_3 = -11.8710-34.5722i
$$
Now applying the three BC(s) i have the following matrix system $M(mu).C_k=0$ and i need to find $C_k$ which will be the non trivial solutions.
$$
begin{bmatrix}
1 & 1 & 1 \
{s_1}^2+beta_hs_1 & {s_2}^2+beta_hs_2 & {s_3}^2+beta_hs_3 \
e^{-s_1}({s_1}^2+beta_hs_1) & e^{-s_2}({s_2}^2+beta_hs_2) & e^{-s_3}({s_3}^2+beta_hs_3) \
end{bmatrix}
*
begin{bmatrix}
C_1 \
C_2 \
C_3 \
end{bmatrix}=0
$$
The above is a homogeneous system of equations which i solved using the following command in MATLAB
[U,S,V] = svd(A,'econ');
b = V(:,size(A,2));
Here $A$ is the coefficient matrix. The solver linsolve(A,B) kept giving me trivial solutions. svd means singular value decomposition. b is supposed to be the solution we are looking for. For the parameters i mentioned here it comes out to be
b =-0.4550 + 0.0000i
0.2278 + 0.5870i
0.2278 - 0.5870i
It is mentioned that
bwill be a solution if the corresponding smallest singular value is zero. If not, b will be the nearest you can come to a solution. If more than one singular value is zero, there are infinitely many solutions of which this will be one
So the values of b must be my corresponding $C_1,C_2,C_3$. But they are coming out to be complex. So should i just take the magnitudes of these complex numbers to form my solution ?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors linear-transformations boundary-value-problem
asked Jan 19 at 16:05
Indrasis MitraIndrasis Mitra
70111
$endgroup$
I am trying to find the eigenvalues of this Eigen BVP. $mu$ is the eigenvalue parameter
$$
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F = 0
$$ wit BC(s) $F(0)=0,frac{F''(0)}{F'(0)}=beta_h,frac{F''(1)}{F'(1)}=beta_h$
For $lambda_h=0.02$ and $beta_h=10$, I calculated the EVs using chebfun in MATLAB. There are infinite number of negative EVs. The general solution should be of the form
$$
F(x) = sum_k C_k e^{-delta_k(mu)x}
$$
Substituting the bc(s) yield three linear equations in $C_1,C_2,C_3$.
where $-delta_k(mu)$ are the three roots of the characteristic equation obtained after substituting one of the EVs.
But this keeps giving me trivial solutions i.e all $C_k=0$. I tried many EVs (ex. $-32.9463$)but to no avail.
Is there something horribly wrong in my understanding ? Is there some other approach i can apply to the problem ? I cannot figure out at all.
ATTEMPT
After @LutzL and @Christoph provided useful comments, i did some reading on null spaces of matrices and their use to solve the homogenous system of equations.
Using the EV mentioned above $-32.9463$ and $lambda_h = 0.02, beta_h = 10$, I arrive at three roots of the characteristic equation as:
$$
s_1 = 3.7421
$$
$$
s_2 = -11.8710+34.5722i
$$
$$
s_3 = -11.8710-34.5722i
$$
Now applying the three BC(s) i have the following matrix system $M(mu).C_k=0$ and i need to find $C_k$ which will be the non trivial solutions.
$$
begin{bmatrix}
1 & 1 & 1 \
{s_1}^2+beta_hs_1 & {s_2}^2+beta_hs_2 & {s_3}^2+beta_hs_3 \
e^{-s_1}({s_1}^2+beta_hs_1) & e^{-s_2}({s_2}^2+beta_hs_2) & e^{-s_3}({s_3}^2+beta_hs_3) \
end{bmatrix}
*
begin{bmatrix}
C_1 \
C_2 \
C_3 \
end{bmatrix}=0
$$
The above is a homogeneous system of equations which i solved using the following command in MATLAB
[U,S,V] = svd(A,'econ');
b = V(:,size(A,2));
Here $A$ is the coefficient matrix. The solver linsolve(A,B) kept giving me trivial solutions. svd means singular value decomposition. b is supposed to be the solution we are looking for. For the parameters i mentioned here it comes out to be
b =-0.4550 + 0.0000i
0.2278 + 0.5870i
0.2278 - 0.5870i
It is mentioned that
bwill be a solution if the corresponding smallest singular value is zero. If not, b will be the nearest you can come to a solution. If more than one singular value is zero, there are infinitely many solutions of which this will be one
So the values of b must be my corresponding $C_1,C_2,C_3$. But they are coming out to be complex. So should i just take the magnitudes of these complex numbers to form my solution ?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors linear-transformations boundary-value-problem
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors linear-transformations boundary-value-problem
asked Jan 19 at 16:05
Indrasis MitraIndrasis Mitra
70111
asked Jan 19 at 16:05
Indrasis MitraIndrasis Mitra
70111
edited Jan 20 at 11:24
Indrasis Mitra
asked Jan 19 at 16:05
Indrasis MitraIndrasis Mitra
70111
asked Jan 19 at 16:05
Indrasis MitraIndrasis Mitra
70111
asked Jan 19 at 16:05
Indrasis MitraIndrasis Mitra
70111
70111
$begingroup$
Can't you get chebfun to compute the eigenfunctions as well? Obviously $F equiv 0$ is a solution to the BVP for any value of $mu$. But since this cannot be an eigenfunction, your linear system $AC=0$ for the coefficients $C_1, C_2, C_3$ must have non-trivial solutions if $mu$ is an eigenvalue. These are the solutions that you must find, for example by determining the null space of the linear map $C mapsto AC$.
$endgroup$
– Christoph
Jan 19 at 17:06
1
$begingroup$
Set $C_3=1$ and leave out the last equation, the first two should then give numerical values for $C_1,C_2$. It is not surprising that you get an invertible if ill-conditioned matrix instead of a singular one if evaluated in floating point arithmetic.
$endgroup$
– LutzL
Jan 19 at 17:15
$begingroup$
@Christoph . Thanks. Yes i know that the system $AX=0$ do have $X=0$ as a trivial solution always and i must look for the non trivial solutions. Also, $chebfun$ does give something called eigenmodes which are plots. Your last point , says "determine the null space of the linear map C to AC . I am sorry to say, but i have no idea how this is done and how i should move forward with it.
$endgroup$
– Indrasis Mitra
Jan 19 at 17:21
2
$begingroup$
Here you have complex conjugate $s_2 = overline{s_3}$ and $C_2 = overline{C_3}$. Therefore, the corresponding eigenfunction will be something like $F(x) = C_1 e^{-s_1 x} + 2 e^{-mathrm{Re}(s_2) x} left( mathrm{Re}(C_2) cos(mathrm{Im}(s_2) x) + mathrm{Im}(C_2) sin(mathrm{Im}(s_2) x) right)$. $C_1$ should probably be zero here.
$endgroup$
– Christoph
Jan 20 at 7:35
1
$begingroup$
Your characteristic roots have the wrong sign, if you set $y=ce^{-sx}$ for $0=a_3y'''+a_2y''+a_1y'+a_0y$, then the characteristic polynomial has to be $0=a_3s^3-a_2s^2+a_1s-a_0$.
$endgroup$
– LutzL
Jan 20 at 9:56
|
show 9 more comments
$begingroup$
Can't you get chebfun to compute the eigenfunctions as well? Obviously $F equiv 0$ is a solution to the BVP for any value of $mu$. But since this cannot be an eigenfunction, your linear system $AC=0$ for the coefficients $C_1, C_2, C_3$ must have non-trivial solutions if $mu$ is an eigenvalue. These are the solutions that you must find, for example by determining the null space of the linear map $C mapsto AC$.
$endgroup$
– Christoph
Jan 19 at 17:06
1
$begingroup$
Set $C_3=1$ and leave out the last equation, the first two should then give numerical values for $C_1,C_2$. It is not surprising that you get an invertible if ill-conditioned matrix instead of a singular one if evaluated in floating point arithmetic.
$endgroup$
– LutzL
Jan 19 at 17:15
$begingroup$
@Christoph . Thanks. Yes i know that the system $AX=0$ do have $X=0$ as a trivial solution always and i must look for the non trivial solutions. Also, $chebfun$ does give something called eigenmodes which are plots. Your last point , says "determine the null space of the linear map C to AC . I am sorry to say, but i have no idea how this is done and how i should move forward with it.
$endgroup$
– Indrasis Mitra
Jan 19 at 17:21
2
$begingroup$
Here you have complex conjugate $s_2 = overline{s_3}$ and $C_2 = overline{C_3}$. Therefore, the corresponding eigenfunction will be something like $F(x) = C_1 e^{-s_1 x} + 2 e^{-mathrm{Re}(s_2) x} left( mathrm{Re}(C_2) cos(mathrm{Im}(s_2) x) + mathrm{Im}(C_2) sin(mathrm{Im}(s_2) x) right)$. $C_1$ should probably be zero here.
$endgroup$
– Christoph
Jan 20 at 7:35
1
$begingroup$
Your characteristic roots have the wrong sign, if you set $y=ce^{-sx}$ for $0=a_3y'''+a_2y''+a_1y'+a_0y$, then the characteristic polynomial has to be $0=a_3s^3-a_2s^2+a_1s-a_0$.
$endgroup$
– LutzL
Jan 20 at 9:56
$begingroup$
Can't you get chebfun to compute the eigenfunctions as well? Obviously $F equiv 0$ is a solution to the BVP for any value of $mu$. But since this cannot be an eigenfunction, your linear system $AC=0$ for the coefficients $C_1, C_2, C_3$ must have non-trivial solutions if $mu$ is an eigenvalue. These are the solutions that you must find, for example by determining the null space of the linear map $C mapsto AC$.
$endgroup$
– Christoph
Jan 19 at 17:06
$begingroup$
Can't you get chebfun to compute the eigenfunctions as well? Obviously $F equiv 0$ is a solution to the BVP for any value of $mu$. But since this cannot be an eigenfunction, your linear system $AC=0$ for the coefficients $C_1, C_2, C_3$ must have non-trivial solutions if $mu$ is an eigenvalue. These are the solutions that you must find, for example by determining the null space of the linear map $C mapsto AC$.
$endgroup$
– Christoph
Jan 19 at 17:06
1
1
$begingroup$
Set $C_3=1$ and leave out the last equation, the first two should then give numerical values for $C_1,C_2$. It is not surprising that you get an invertible if ill-conditioned matrix instead of a singular one if evaluated in floating point arithmetic.
$endgroup$
– LutzL
Jan 19 at 17:15
$begingroup$
Set $C_3=1$ and leave out the last equation, the first two should then give numerical values for $C_1,C_2$. It is not surprising that you get an invertible if ill-conditioned matrix instead of a singular one if evaluated in floating point arithmetic.
$endgroup$
– LutzL
Jan 19 at 17:15
$begingroup$
@Christoph . Thanks. Yes i know that the system $AX=0$ do have $X=0$ as a trivial solution always and i must look for the non trivial solutions. Also, $chebfun$ does give something called eigenmodes which are plots. Your last point , says "determine the null space of the linear map C to AC . I am sorry to say, but i have no idea how this is done and how i should move forward with it.
$endgroup$
– Indrasis Mitra
Jan 19 at 17:21
$begingroup$
@Christoph . Thanks. Yes i know that the system $AX=0$ do have $X=0$ as a trivial solution always and i must look for the non trivial solutions. Also, $chebfun$ does give something called eigenmodes which are plots. Your last point , says "determine the null space of the linear map C to AC . I am sorry to say, but i have no idea how this is done and how i should move forward with it.
$endgroup$
– Indrasis Mitra
Jan 19 at 17:21
2
2
$begingroup$
Here you have complex conjugate $s_2 = overline{s_3}$ and $C_2 = overline{C_3}$. Therefore, the corresponding eigenfunction will be something like $F(x) = C_1 e^{-s_1 x} + 2 e^{-mathrm{Re}(s_2) x} left( mathrm{Re}(C_2) cos(mathrm{Im}(s_2) x) + mathrm{Im}(C_2) sin(mathrm{Im}(s_2) x) right)$. $C_1$ should probably be zero here.
$endgroup$
– Christoph
Jan 20 at 7:35
$begingroup$
Here you have complex conjugate $s_2 = overline{s_3}$ and $C_2 = overline{C_3}$. Therefore, the corresponding eigenfunction will be something like $F(x) = C_1 e^{-s_1 x} + 2 e^{-mathrm{Re}(s_2) x} left( mathrm{Re}(C_2) cos(mathrm{Im}(s_2) x) + mathrm{Im}(C_2) sin(mathrm{Im}(s_2) x) right)$. $C_1$ should probably be zero here.
$endgroup$
– Christoph
Jan 20 at 7:35
1
1
$begingroup$
Your characteristic roots have the wrong sign, if you set $y=ce^{-sx}$ for $0=a_3y'''+a_2y''+a_1y'+a_0y$, then the characteristic polynomial has to be $0=a_3s^3-a_2s^2+a_1s-a_0$.
$endgroup$
– LutzL
Jan 20 at 9:56
$begingroup$
Your characteristic roots have the wrong sign, if you set $y=ce^{-sx}$ for $0=a_3y'''+a_2y''+a_1y'+a_0y$, then the characteristic polynomial has to be $0=a_3s^3-a_2s^2+a_1s-a_0$.
$endgroup$
– LutzL
Jan 20 at 9:56
|
show 9 more comments
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Can't you get chebfun to compute the eigenfunctions as well? Obviously $F equiv 0$ is a solution to the BVP for any value of $mu$. But since this cannot be an eigenfunction, your linear system $AC=0$ for the coefficients $C_1, C_2, C_3$ must have non-trivial solutions if $mu$ is an eigenvalue. These are the solutions that you must find, for example by determining the null space of the linear map $C mapsto AC$.
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– Christoph
Jan 19 at 17:06
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Set $C_3=1$ and leave out the last equation, the first two should then give numerical values for $C_1,C_2$. It is not surprising that you get an invertible if ill-conditioned matrix instead of a singular one if evaluated in floating point arithmetic.
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– LutzL
Jan 19 at 17:15
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@Christoph . Thanks. Yes i know that the system $AX=0$ do have $X=0$ as a trivial solution always and i must look for the non trivial solutions. Also, $chebfun$ does give something called eigenmodes which are plots. Your last point , says "determine the null space of the linear map C to AC . I am sorry to say, but i have no idea how this is done and how i should move forward with it.
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– Indrasis Mitra
Jan 19 at 17:21
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Here you have complex conjugate $s_2 = overline{s_3}$ and $C_2 = overline{C_3}$. Therefore, the corresponding eigenfunction will be something like $F(x) = C_1 e^{-s_1 x} + 2 e^{-mathrm{Re}(s_2) x} left( mathrm{Re}(C_2) cos(mathrm{Im}(s_2) x) + mathrm{Im}(C_2) sin(mathrm{Im}(s_2) x) right)$. $C_1$ should probably be zero here.
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– Christoph
Jan 20 at 7:35
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Your characteristic roots have the wrong sign, if you set $y=ce^{-sx}$ for $0=a_3y'''+a_2y''+a_1y'+a_0y$, then the characteristic polynomial has to be $0=a_3s^3-a_2s^2+a_1s-a_0$.
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– LutzL
Jan 20 at 9:56