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[SOLVED] Solvability condition

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
A_{\lambda}=
\begin{pmatrix}
-\mu\lambda k^2 - k^2 - s & i\tau k & i\tau k - i\beta k^3\\
i\lambda k & \lambda + Dk^2 & -\alpha k^2\\
i\lambda k & 0 & \lambda
\end{pmatrix}
$$

where $\lambda = \lambda(k^2)$ (this is confusing) is the growth rate and k is the wave number, and tau, mu, D, alpha, s, and beta are biological parameters. The solvability condition for a matrix $A_{\lambda}$ requires that the matrix is singular.

So the I found the determinant to be $-Dk^4\lambda -Dik^2\lambda -Dk^2s\lambda +Di^2k^6\beta\lambda -k^2\lambda^2-ik^2\lambda^2-s\lambda^2+i^2k^4\beta\lambda^2-Dk^4\lambda^2\mu-k^2\lambda^3\mu-Di^2k^3\lambda\tau-i^2k^4\alpha\lambda\tau-i^2k\lambda^2\tau-i^2k^2\lambda^2\tau=0$
for it to be singular.

Then it says show that the polynomial associated with the solvability condition is given by
$$
a(k^2)\lambda^2 + b(k^2)\lambda +c(k^2) =0
$$
where
\begin{align}
a(k^2)=& \mu k^2\\
b(k^2)=& (\beta+\mu D)k^4 - (2\tau-1)k^2 s\\
c(k^2)=& \beta Dk^6-\{(\alpha+D)\tau-D\}k^4+sDk^2
\end{align}

From what I have, I can obtain that. I am not sure if i is the imaginary constant or just some constant i.
Even if i is the imaginary constant, we would still have some i's.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
A_{\lambda}=
\begin{pmatrix}
-\mu\lambda k^2 - k^2 - s & i\tau k & i\tau k - i\beta k^3\\
i\lambda k & \lambda + Dk^2 & -\alpha k^2\\
i\lambda k & 0 & \lambda
\end{pmatrix}
$$

where $\lambda = \lambda(k^2)$ (this is confusing because that is just 1 = k^2) is the growth rate and k is the wave number, and tau, mu, D, alpha, s, and beta are biological parameters. The solvability condition for a matrix $A_{\lambda}$ requires that the matrix is singular.

So the I found the determinant to be
$-Dk^4\lambda -Dik^2\lambda -Dk^2s\lambda +Di^2k^6\beta\lambda -k^2\lambda^2$
$-ik^2\lambda^2-s\lambda^2+i^2k^4\beta\lambda^2-Dk^4\lambda^2\mu-k^2\lambda^3\mu-Di^2k^3\lambda\tau-i^2k^4\alpha\lambda\tau-i^2k\lambda^2\tau-i^2k^2\lambda^2\tau=0$
for it to be singular.

Then it says show that the polynomial associated with the solvability condition is given by
$$
a(k^2)\lambda^2 + b(k^2)\lambda +c(k^2) =0
$$
where
\begin{align}
a(k^2)=& \mu k^2\\
b(k^2)=& (\beta+\mu D)k^4 - (2\tau-1)k^2 s\\
c(k^2)=& \beta Dk^6-\{(\alpha+D)\tau-D\}k^4+sDk^2
\end{align}

From what I have, I can't obtain that. I am not sure if i is the imaginary constant or just some constant i.
Even if i is the imaginary constant, we would still have some i's left after simplifying.
Typo in the original question. Also, fixed the equation from going of the screen.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
​Solved
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197

dwsmith

Well-known member
Feb 1, 2012
1,673
Do you have the ability to mark threads as solved now, under Thread Tools? It should be there...
I don't see it as an option.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
I don't see it as an option.
Oh, silly me. I already marked your thread as Solved. Do you have the "Mark this thread as unsolved" option in your Thread Tools?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Oh, silly me. I already marked your thread as Solved. Do you have the "Mark this thread as unsolved" option in your Thread Tools?
It doesn't appear for any threads I have. Even if it is marked solved, when the option is there, it then says to mark as unsolved--which is nonexistence to me.

View attachment 89
 
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