Mixing
Eigen values of a Third Order Linear Homogenous ODE
$begingroup$
I have two third order linear ODE which have been arrived after applying separation of variables to a system of PDEs
begin{eqnarray}
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,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}
$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are
For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$
For $G$:
$$G(0)=0$$
$$frac{G''(0)}{G'(0)}=beta_c$$
$$frac{G''(1)}{G'(1)}=beta_c$$
$lambda_h$, $lambda_c$, $beta_h$ and $beta_c$ are all constants $>0$.
$mu$ is the separation constant .
I need to determine eigenvalues for each BVP involving $F$ and $G$. So for finding out eigenvalues i know i need to consider all the three cases
$mu>0$, $mu<0$ and $mu=0$ and then look for non-trivial solutions by applying the specific set of b.c. Although i am acquainted with the procedure to determine eigenvalues for a second order DE, the third order of the DE(s)is something i am not familiar with.
Any recommendations on how should i go about tackling this ?
Attempt
As per @Cesareo recommendations, I arrive at the following
linear equations
$$C_1+C_2+C_3=0$$
$$frac{F''(0)}{F'(0)}=frac{{C_1}{delta_1(mu)}^2+{C_2}{delta_2(mu)}^2+{C_3}{delta_3(mu)}^2}{-{C_1}{delta_1(mu)}-{C_2}{delta_2(mu)}-{C_3}{delta_3(mu)}}=beta_h$$
$$frac{F''(1)}{F'(1)}=frac{{C_1e^{-delta_1(mu)}}{delta_1(mu)}^2+{C_2e^{-delta_2(mu)}}{delta_2(mu)}^2+{C_3e^{-delta_3(mu)}}{delta_3(mu)}^2}{-{C_1e^{-delta_1(mu)}}{delta_1(mu)}-{C_2e^{-delta_2(mu)}}{delta_2(mu)}-{C_3e^{-delta_3(mu)}}{delta_3(mu)}}=beta_h$$
I reach the following form of $M(mu).C=0$
begin{vmatrix}
1 & 1 & 1 \
{delta_1(mu)}^2+beta_hdelta_1(mu) & {delta_2(mu)}^2+beta_hdelta_2(mu) & {delta_3(mu)}^2+beta_hdelta_3(mu) \
e^{-delta_1(mu)}({delta_1(mu)}^2+beta_hdelta_1(mu)) & e^{-delta_2(mu)}({delta_2(mu)}^2+beta_hdelta_2(mu)) & e^{-delta_3(mu)}({delta_3(mu)}^2+beta_hdelta_3(mu)) \
end{vmatrix}$=0$
Solving this determinant is supposed to give me the eigen values $mu_n$ and consequently the eigen functions. The determinant can be reduced to two $0$ in the first row by coloumn manipulation, but I do not find any way to handle the consequent equation that comes out of it which is something like this:
$$[(delta_1(mu)-delta_2(mu))(delta_1(mu)+delta_2(mu)+beta_h)[(e^{-delta_2(mu)}{delta_2(mu)}^2-e^{-delta_3(mu)}{delta_3(mu)}^2)+beta_h(e^{-delta_2(mu)}{delta_2(mu)}-e^{-delta_3(mu)}{delta_3(mu)})]]-[(delta_2(mu)-delta_3(mu))(delta_2(mu)+delta_3(mu)+beta_h)[(e^{-delta_1(mu)}{delta_1(mu)}^2-e^{-delta_2(mu)}{delta_2(mu)}^2)+beta_h(e^{-delta_1(mu)}{delta_1(mu)}-e^{-delta_2(mu)}{delta_2(mu)})]]=0$$
After this step i fail to proceed further to find the eigenvalues using this $mathbb{det}M=0$ equation
ordinary-differential-equations pde eigenvalues-eigenvectors eigenfunctions
asked Jan 11 at 12:06
Indrasis MitraIndrasis Mitra
2517
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add a comment |
$begingroup$
I have two third order linear ODE which have been arrived after applying separation of variables to a system of PDEs
begin{eqnarray}
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,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}
$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are
For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$
For $G$:
$$G(0)=0$$
$$frac{G''(0)}{G'(0)}=beta_c$$
$$frac{G''(1)}{G'(1)}=beta_c$$
$lambda_h$, $lambda_c$, $beta_h$ and $beta_c$ are all constants $>0$.
$mu$ is the separation constant .
I need to determine eigenvalues for each BVP involving $F$ and $G$. So for finding out eigenvalues i know i need to consider all the three cases
$mu>0$, $mu<0$ and $mu=0$ and then look for non-trivial solutions by applying the specific set of b.c. Although i am acquainted with the procedure to determine eigenvalues for a second order DE, the third order of the DE(s)is something i am not familiar with.
Any recommendations on how should i go about tackling this ?
Attempt
As per @Cesareo recommendations, I arrive at the following
linear equations
$$C_1+C_2+C_3=0$$
$$frac{F''(0)}{F'(0)}=frac{{C_1}{delta_1(mu)}^2+{C_2}{delta_2(mu)}^2+{C_3}{delta_3(mu)}^2}{-{C_1}{delta_1(mu)}-{C_2}{delta_2(mu)}-{C_3}{delta_3(mu)}}=beta_h$$
$$frac{F''(1)}{F'(1)}=frac{{C_1e^{-delta_1(mu)}}{delta_1(mu)}^2+{C_2e^{-delta_2(mu)}}{delta_2(mu)}^2+{C_3e^{-delta_3(mu)}}{delta_3(mu)}^2}{-{C_1e^{-delta_1(mu)}}{delta_1(mu)}-{C_2e^{-delta_2(mu)}}{delta_2(mu)}-{C_3e^{-delta_3(mu)}}{delta_3(mu)}}=beta_h$$
I reach the following form of $M(mu).C=0$
begin{vmatrix}
1 & 1 & 1 \
{delta_1(mu)}^2+beta_hdelta_1(mu) & {delta_2(mu)}^2+beta_hdelta_2(mu) & {delta_3(mu)}^2+beta_hdelta_3(mu) \
e^{-delta_1(mu)}({delta_1(mu)}^2+beta_hdelta_1(mu)) & e^{-delta_2(mu)}({delta_2(mu)}^2+beta_hdelta_2(mu)) & e^{-delta_3(mu)}({delta_3(mu)}^2+beta_hdelta_3(mu)) \
end{vmatrix}$=0$
Solving this determinant is supposed to give me the eigen values $mu_n$ and consequently the eigen functions. The determinant can be reduced to two $0$ in the first row by coloumn manipulation, but I do not find any way to handle the consequent equation that comes out of it which is something like this:
$$[(delta_1(mu)-delta_2(mu))(delta_1(mu)+delta_2(mu)+beta_h)[(e^{-delta_2(mu)}{delta_2(mu)}^2-e^{-delta_3(mu)}{delta_3(mu)}^2)+beta_h(e^{-delta_2(mu)}{delta_2(mu)}-e^{-delta_3(mu)}{delta_3(mu)})]]-[(delta_2(mu)-delta_3(mu))(delta_2(mu)+delta_3(mu)+beta_h)[(e^{-delta_1(mu)}{delta_1(mu)}^2-e^{-delta_2(mu)}{delta_2(mu)}^2)+beta_h(e^{-delta_1(mu)}{delta_1(mu)}-e^{-delta_2(mu)}{delta_2(mu)})]]=0$$
After this step i fail to proceed further to find the eigenvalues using this $mathbb{det}M=0$ equation
ordinary-differential-equations pde eigenvalues-eigenvectors eigenfunctions
asked Jan 11 at 12:06
Indrasis MitraIndrasis Mitra
2517
$endgroup$
add a comment |
$begingroup$
I have two third order linear ODE which have been arrived after applying separation of variables to a system of PDEs
begin{eqnarray}
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,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}
$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are
For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$
For $G$:
$$G(0)=0$$
$$frac{G''(0)}{G'(0)}=beta_c$$
$$frac{G''(1)}{G'(1)}=beta_c$$
$lambda_h$, $lambda_c$, $beta_h$ and $beta_c$ are all constants $>0$.
$mu$ is the separation constant .
I need to determine eigenvalues for each BVP involving $F$ and $G$. So for finding out eigenvalues i know i need to consider all the three cases
$mu>0$, $mu<0$ and $mu=0$ and then look for non-trivial solutions by applying the specific set of b.c. Although i am acquainted with the procedure to determine eigenvalues for a second order DE, the third order of the DE(s)is something i am not familiar with.
Any recommendations on how should i go about tackling this ?
Attempt
As per @Cesareo recommendations, I arrive at the following
linear equations
$$C_1+C_2+C_3=0$$
$$frac{F''(0)}{F'(0)}=frac{{C_1}{delta_1(mu)}^2+{C_2}{delta_2(mu)}^2+{C_3}{delta_3(mu)}^2}{-{C_1}{delta_1(mu)}-{C_2}{delta_2(mu)}-{C_3}{delta_3(mu)}}=beta_h$$
$$frac{F''(1)}{F'(1)}=frac{{C_1e^{-delta_1(mu)}}{delta_1(mu)}^2+{C_2e^{-delta_2(mu)}}{delta_2(mu)}^2+{C_3e^{-delta_3(mu)}}{delta_3(mu)}^2}{-{C_1e^{-delta_1(mu)}}{delta_1(mu)}-{C_2e^{-delta_2(mu)}}{delta_2(mu)}-{C_3e^{-delta_3(mu)}}{delta_3(mu)}}=beta_h$$
I reach the following form of $M(mu).C=0$
begin{vmatrix}
1 & 1 & 1 \
{delta_1(mu)}^2+beta_hdelta_1(mu) & {delta_2(mu)}^2+beta_hdelta_2(mu) & {delta_3(mu)}^2+beta_hdelta_3(mu) \
e^{-delta_1(mu)}({delta_1(mu)}^2+beta_hdelta_1(mu)) & e^{-delta_2(mu)}({delta_2(mu)}^2+beta_hdelta_2(mu)) & e^{-delta_3(mu)}({delta_3(mu)}^2+beta_hdelta_3(mu)) \
end{vmatrix}$=0$
Solving this determinant is supposed to give me the eigen values $mu_n$ and consequently the eigen functions. The determinant can be reduced to two $0$ in the first row by coloumn manipulation, but I do not find any way to handle the consequent equation that comes out of it which is something like this:
$$[(delta_1(mu)-delta_2(mu))(delta_1(mu)+delta_2(mu)+beta_h)[(e^{-delta_2(mu)}{delta_2(mu)}^2-e^{-delta_3(mu)}{delta_3(mu)}^2)+beta_h(e^{-delta_2(mu)}{delta_2(mu)}-e^{-delta_3(mu)}{delta_3(mu)})]]-[(delta_2(mu)-delta_3(mu))(delta_2(mu)+delta_3(mu)+beta_h)[(e^{-delta_1(mu)}{delta_1(mu)}^2-e^{-delta_2(mu)}{delta_2(mu)}^2)+beta_h(e^{-delta_1(mu)}{delta_1(mu)}-e^{-delta_2(mu)}{delta_2(mu)})]]=0$$
After this step i fail to proceed further to find the eigenvalues using this $mathbb{det}M=0$ equation
ordinary-differential-equations pde eigenvalues-eigenvectors eigenfunctions
asked Jan 11 at 12:06
Indrasis MitraIndrasis Mitra
2517
$endgroup$
I have two third order linear ODE which have been arrived after applying separation of variables to a system of PDEs
begin{eqnarray}
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,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}
$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are
For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$
For $G$:
$$G(0)=0$$
$$frac{G''(0)}{G'(0)}=beta_c$$
$$frac{G''(1)}{G'(1)}=beta_c$$
$lambda_h$, $lambda_c$, $beta_h$ and $beta_c$ are all constants $>0$.
$mu$ is the separation constant .
I need to determine eigenvalues for each BVP involving $F$ and $G$. So for finding out eigenvalues i know i need to consider all the three cases
$mu>0$, $mu<0$ and $mu=0$ and then look for non-trivial solutions by applying the specific set of b.c. Although i am acquainted with the procedure to determine eigenvalues for a second order DE, the third order of the DE(s)is something i am not familiar with.
Any recommendations on how should i go about tackling this ?
Attempt
As per @Cesareo recommendations, I arrive at the following
linear equations
$$C_1+C_2+C_3=0$$
$$frac{F''(0)}{F'(0)}=frac{{C_1}{delta_1(mu)}^2+{C_2}{delta_2(mu)}^2+{C_3}{delta_3(mu)}^2}{-{C_1}{delta_1(mu)}-{C_2}{delta_2(mu)}-{C_3}{delta_3(mu)}}=beta_h$$
$$frac{F''(1)}{F'(1)}=frac{{C_1e^{-delta_1(mu)}}{delta_1(mu)}^2+{C_2e^{-delta_2(mu)}}{delta_2(mu)}^2+{C_3e^{-delta_3(mu)}}{delta_3(mu)}^2}{-{C_1e^{-delta_1(mu)}}{delta_1(mu)}-{C_2e^{-delta_2(mu)}}{delta_2(mu)}-{C_3e^{-delta_3(mu)}}{delta_3(mu)}}=beta_h$$
I reach the following form of $M(mu).C=0$
begin{vmatrix}
1 & 1 & 1 \
{delta_1(mu)}^2+beta_hdelta_1(mu) & {delta_2(mu)}^2+beta_hdelta_2(mu) & {delta_3(mu)}^2+beta_hdelta_3(mu) \
e^{-delta_1(mu)}({delta_1(mu)}^2+beta_hdelta_1(mu)) & e^{-delta_2(mu)}({delta_2(mu)}^2+beta_hdelta_2(mu)) & e^{-delta_3(mu)}({delta_3(mu)}^2+beta_hdelta_3(mu)) \
end{vmatrix}$=0$
Solving this determinant is supposed to give me the eigen values $mu_n$ and consequently the eigen functions. The determinant can be reduced to two $0$ in the first row by coloumn manipulation, but I do not find any way to handle the consequent equation that comes out of it which is something like this:
$$[(delta_1(mu)-delta_2(mu))(delta_1(mu)+delta_2(mu)+beta_h)[(e^{-delta_2(mu)}{delta_2(mu)}^2-e^{-delta_3(mu)}{delta_3(mu)}^2)+beta_h(e^{-delta_2(mu)}{delta_2(mu)}-e^{-delta_3(mu)}{delta_3(mu)})]]-[(delta_2(mu)-delta_3(mu))(delta_2(mu)+delta_3(mu)+beta_h)[(e^{-delta_1(mu)}{delta_1(mu)}^2-e^{-delta_2(mu)}{delta_2(mu)}^2)+beta_h(e^{-delta_1(mu)}{delta_1(mu)}-e^{-delta_2(mu)}{delta_2(mu)})]]=0$$
After this step i fail to proceed further to find the eigenvalues using this $mathbb{det}M=0$ equation
ordinary-differential-equations pde eigenvalues-eigenvectors eigenfunctions
ordinary-differential-equations pde eigenvalues-eigenvectors eigenfunctions
asked Jan 11 at 12:06
Indrasis MitraIndrasis Mitra
2517
asked Jan 11 at 12:06
Indrasis MitraIndrasis Mitra
2517
edited Jan 13 at 3:42
Indrasis Mitra
asked Jan 11 at 12:06
Indrasis MitraIndrasis Mitra
2517
asked Jan 11 at 12:06
Indrasis MitraIndrasis Mitra
2517
asked Jan 11 at 12:06
Indrasis MitraIndrasis Mitra
2517
2517
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
Regarding the first DE the linear differential operator
$$
lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
$$
and the three roots $delta_i(mu), i = 1,2,3$ we have that
$$
F(t) = sum_k C_k e^{-delta_k(mu)t}
$$
using now the boundary conditions
$$
F(0) = sum_k C_k = 0 longrightarrow (1)
$$
and with
$$
F'(0) = -sum_k C_k delta_k(mu)\
F''(0) = sum_k C_k delta_k(mu)^2\
$$
giving
$$
-frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
$$
and similarly
$$
-frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
$$
then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as
$$
M(mu)cdot C = 0, C = (C_k)
$$
This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$
The procedure for $G$ is quite similar.
NOTE
Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial
$$
s^3+20s^2+(110-mu)s+100 = 0
$$
with roots $delta_1(mu),delta_2(mu),delta_3(mu)$
The determinant after simplifications reads
$$
det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
_h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
_h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
_3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
_3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
left(delta _1+delta _2+delta _3right)right)
$$
discarding $cosh (delta_1+delta_2+delta_3)-sinh
(delta_1+delta_2+delta_3)=0$ we follow with
$$
Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
_1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
_1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
$$
and then after plotting we have
In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$
Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$
parms = {lh -> 1/4, bh -> -10};
F[t_, n_] := Sum[
!(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
!(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
!(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
sols = Solve[
lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
parms // FullSimplify
roots = s /. sols;
M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
{E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
det = -Det[M] // FullSimplify
subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
subdet2 = det/subdet1 // FullSimplify
subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
PlotRange -> {-10, 10}]
solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
Subscript[mu, 1] = mu /. solmu1;
Subscript[mu, 2] = mu /. solmu2;
Subscript[mu, 3] = mu /. solmu3;
Subscript[mu, 4] = mu /. solmu4;
Subscript[mu, 5] = mu /. solmu5;
Subscript[mu, 6] = mu /. solmu6;
Subscript[mu, 7] = mu /. solmu7;
Subscript[mu, 8] = mu /. solmu8;
Subscript[mu, 9] = mu /. solmu9;
exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
F[t, 9]
answered Jan 11 at 21:58
CesareoCesareo
8,6793516
$endgroup$
$begingroup$
Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
$endgroup$
– Indrasis Mitra
Jan 12 at 3:10
$begingroup$
Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
$endgroup$
– Indrasis Mitra
Jan 12 at 4:26
$begingroup$
my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
$endgroup$
– Indrasis Mitra
Jan 12 at 6:40
$begingroup$
I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
$endgroup$
– Indrasis Mitra
Jan 13 at 3:45
$begingroup$
@IndrasisMitra See attached note.
$endgroup$
– Cesareo
Jan 13 at 9:19
|
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1 Answer
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1 Answer
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$begingroup$
Regarding the first DE the linear differential operator
$$
lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
$$
and the three roots $delta_i(mu), i = 1,2,3$ we have that
$$
F(t) = sum_k C_k e^{-delta_k(mu)t}
$$
using now the boundary conditions
$$
F(0) = sum_k C_k = 0 longrightarrow (1)
$$
and with
$$
F'(0) = -sum_k C_k delta_k(mu)\
F''(0) = sum_k C_k delta_k(mu)^2\
$$
giving
$$
-frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
$$
and similarly
$$
-frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
$$
then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as
$$
M(mu)cdot C = 0, C = (C_k)
$$
This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$
The procedure for $G$ is quite similar.
NOTE
Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial
$$
s^3+20s^2+(110-mu)s+100 = 0
$$
with roots $delta_1(mu),delta_2(mu),delta_3(mu)$
The determinant after simplifications reads
$$
det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
_h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
_h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
_3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
_3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
left(delta _1+delta _2+delta _3right)right)
$$
discarding $cosh (delta_1+delta_2+delta_3)-sinh
(delta_1+delta_2+delta_3)=0$ we follow with
$$
Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
_1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
_1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
$$
and then after plotting we have
In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$
Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$
parms = {lh -> 1/4, bh -> -10};
F[t_, n_] := Sum[
!(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
!(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
!(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
sols = Solve[
lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
parms // FullSimplify
roots = s /. sols;
M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
{E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
det = -Det[M] // FullSimplify
subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
subdet2 = det/subdet1 // FullSimplify
subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
PlotRange -> {-10, 10}]
solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
Subscript[mu, 1] = mu /. solmu1;
Subscript[mu, 2] = mu /. solmu2;
Subscript[mu, 3] = mu /. solmu3;
Subscript[mu, 4] = mu /. solmu4;
Subscript[mu, 5] = mu /. solmu5;
Subscript[mu, 6] = mu /. solmu6;
Subscript[mu, 7] = mu /. solmu7;
Subscript[mu, 8] = mu /. solmu8;
Subscript[mu, 9] = mu /. solmu9;
exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
F[t, 9]
answered Jan 11 at 21:58
CesareoCesareo
8,6793516
$endgroup$
$begingroup$
Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
$endgroup$
– Indrasis Mitra
Jan 12 at 3:10
$begingroup$
Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
$endgroup$
– Indrasis Mitra
Jan 12 at 4:26
$begingroup$
my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
$endgroup$
– Indrasis Mitra
Jan 12 at 6:40
$begingroup$
I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
$endgroup$
– Indrasis Mitra
Jan 13 at 3:45
$begingroup$
@IndrasisMitra See attached note.
$endgroup$
– Cesareo
Jan 13 at 9:19
|
show 11 more comments
$begingroup$
Regarding the first DE the linear differential operator
$$
lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
$$
and the three roots $delta_i(mu), i = 1,2,3$ we have that
$$
F(t) = sum_k C_k e^{-delta_k(mu)t}
$$
using now the boundary conditions
$$
F(0) = sum_k C_k = 0 longrightarrow (1)
$$
and with
$$
F'(0) = -sum_k C_k delta_k(mu)\
F''(0) = sum_k C_k delta_k(mu)^2\
$$
giving
$$
-frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
$$
and similarly
$$
-frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
$$
then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as
$$
M(mu)cdot C = 0, C = (C_k)
$$
This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$
The procedure for $G$ is quite similar.
NOTE
Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial
$$
s^3+20s^2+(110-mu)s+100 = 0
$$
with roots $delta_1(mu),delta_2(mu),delta_3(mu)$
The determinant after simplifications reads
$$
det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
_h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
_h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
_3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
_3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
left(delta _1+delta _2+delta _3right)right)
$$
discarding $cosh (delta_1+delta_2+delta_3)-sinh
(delta_1+delta_2+delta_3)=0$ we follow with
$$
Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
_1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
_1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
$$
and then after plotting we have
In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$
Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$
parms = {lh -> 1/4, bh -> -10};
F[t_, n_] := Sum[
!(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
!(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
!(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
sols = Solve[
lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
parms // FullSimplify
roots = s /. sols;
M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
{E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
det = -Det[M] // FullSimplify
subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
subdet2 = det/subdet1 // FullSimplify
subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
PlotRange -> {-10, 10}]
solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
Subscript[mu, 1] = mu /. solmu1;
Subscript[mu, 2] = mu /. solmu2;
Subscript[mu, 3] = mu /. solmu3;
Subscript[mu, 4] = mu /. solmu4;
Subscript[mu, 5] = mu /. solmu5;
Subscript[mu, 6] = mu /. solmu6;
Subscript[mu, 7] = mu /. solmu7;
Subscript[mu, 8] = mu /. solmu8;
Subscript[mu, 9] = mu /. solmu9;
exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
F[t, 9]
answered Jan 11 at 21:58
CesareoCesareo
8,6793516
$endgroup$
$begingroup$
Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
$endgroup$
– Indrasis Mitra
Jan 12 at 3:10
$begingroup$
Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
$endgroup$
– Indrasis Mitra
Jan 12 at 4:26
$begingroup$
my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
$endgroup$
– Indrasis Mitra
Jan 12 at 6:40
$begingroup$
I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
$endgroup$
– Indrasis Mitra
Jan 13 at 3:45
$begingroup$
@IndrasisMitra See attached note.
$endgroup$
– Cesareo
Jan 13 at 9:19
|
show 11 more comments
$begingroup$
Regarding the first DE the linear differential operator
$$
lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
$$
and the three roots $delta_i(mu), i = 1,2,3$ we have that
$$
F(t) = sum_k C_k e^{-delta_k(mu)t}
$$
using now the boundary conditions
$$
F(0) = sum_k C_k = 0 longrightarrow (1)
$$
and with
$$
F'(0) = -sum_k C_k delta_k(mu)\
F''(0) = sum_k C_k delta_k(mu)^2\
$$
giving
$$
-frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
$$
and similarly
$$
-frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
$$
then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as
$$
M(mu)cdot C = 0, C = (C_k)
$$
This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$
The procedure for $G$ is quite similar.
NOTE
Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial
$$
s^3+20s^2+(110-mu)s+100 = 0
$$
with roots $delta_1(mu),delta_2(mu),delta_3(mu)$
The determinant after simplifications reads
$$
det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
_h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
_h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
_3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
_3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
left(delta _1+delta _2+delta _3right)right)
$$
discarding $cosh (delta_1+delta_2+delta_3)-sinh
(delta_1+delta_2+delta_3)=0$ we follow with
$$
Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
_1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
_1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
$$
and then after plotting we have
In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$
Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$
parms = {lh -> 1/4, bh -> -10};
F[t_, n_] := Sum[
!(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
!(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
!(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
sols = Solve[
lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
parms // FullSimplify
roots = s /. sols;
M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
{E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
det = -Det[M] // FullSimplify
subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
subdet2 = det/subdet1 // FullSimplify
subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
PlotRange -> {-10, 10}]
solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
Subscript[mu, 1] = mu /. solmu1;
Subscript[mu, 2] = mu /. solmu2;
Subscript[mu, 3] = mu /. solmu3;
Subscript[mu, 4] = mu /. solmu4;
Subscript[mu, 5] = mu /. solmu5;
Subscript[mu, 6] = mu /. solmu6;
Subscript[mu, 7] = mu /. solmu7;
Subscript[mu, 8] = mu /. solmu8;
Subscript[mu, 9] = mu /. solmu9;
exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
F[t, 9]
answered Jan 11 at 21:58
CesareoCesareo
8,6793516
$endgroup$
Regarding the first DE the linear differential operator
$$
lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
$$
and the three roots $delta_i(mu), i = 1,2,3$ we have that
$$
F(t) = sum_k C_k e^{-delta_k(mu)t}
$$
using now the boundary conditions
$$
F(0) = sum_k C_k = 0 longrightarrow (1)
$$
and with
$$
F'(0) = -sum_k C_k delta_k(mu)\
F''(0) = sum_k C_k delta_k(mu)^2\
$$
giving
$$
-frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
$$
and similarly
$$
-frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
$$
then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as
$$
M(mu)cdot C = 0, C = (C_k)
$$
This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$
The procedure for $G$ is quite similar.
NOTE
Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial
$$
s^3+20s^2+(110-mu)s+100 = 0
$$
with roots $delta_1(mu),delta_2(mu),delta_3(mu)$
The determinant after simplifications reads
$$
det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
_h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
_h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
_3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
_3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
left(delta _1+delta _2+delta _3right)right)
$$
discarding $cosh (delta_1+delta_2+delta_3)-sinh
(delta_1+delta_2+delta_3)=0$ we follow with
$$
Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
_1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
_1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
$$
and then after plotting we have
In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$
Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$
parms = {lh -> 1/4, bh -> -10};
F[t_, n_] := Sum[
!(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
!(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
!(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
sols = Solve[
lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
parms // FullSimplify
roots = s /. sols;
M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
{E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
det = -Det[M] // FullSimplify
subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
subdet2 = det/subdet1 // FullSimplify
subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
PlotRange -> {-10, 10}]
solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
Subscript[mu, 1] = mu /. solmu1;
Subscript[mu, 2] = mu /. solmu2;
Subscript[mu, 3] = mu /. solmu3;
Subscript[mu, 4] = mu /. solmu4;
Subscript[mu, 5] = mu /. solmu5;
Subscript[mu, 6] = mu /. solmu6;
Subscript[mu, 7] = mu /. solmu7;
Subscript[mu, 8] = mu /. solmu8;
Subscript[mu, 9] = mu /. solmu9;
exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
F[t, 9]
answered Jan 11 at 21:58
CesareoCesareo
8,6793516
edited Jan 19 at 18:38
answered Jan 11 at 21:58
CesareoCesareo
8,6793516
answered Jan 11 at 21:58
CesareoCesareo
8,6793516
answered Jan 11 at 21:58
CesareoCesareo
8,6793516
8,6793516
$begingroup$
Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
$endgroup$
– Indrasis Mitra
Jan 12 at 3:10
$begingroup$
Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
$endgroup$
– Indrasis Mitra
Jan 12 at 4:26
$begingroup$
my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
$endgroup$
– Indrasis Mitra
Jan 12 at 6:40
$begingroup$
I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
$endgroup$
– Indrasis Mitra
Jan 13 at 3:45
$begingroup$
@IndrasisMitra See attached note.
$endgroup$
– Cesareo
Jan 13 at 9:19
|
show 11 more comments
$begingroup$
Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
$endgroup$
– Indrasis Mitra
Jan 12 at 3:10
$begingroup$
Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
$endgroup$
– Indrasis Mitra
Jan 12 at 4:26
$begingroup$
my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
$endgroup$
– Indrasis Mitra
Jan 12 at 6:40
$begingroup$
I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
$endgroup$
– Indrasis Mitra
Jan 13 at 3:45
$begingroup$
@IndrasisMitra See attached note.
$endgroup$
– Cesareo
Jan 13 at 9:19
$begingroup$
Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
$endgroup$
– Indrasis Mitra
Jan 12 at 3:10
$begingroup$
Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
$endgroup$
– Indrasis Mitra
Jan 12 at 3:10
$begingroup$
Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
$endgroup$
– Indrasis Mitra
Jan 12 at 4:26
$begingroup$
Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
$endgroup$
– Indrasis Mitra
Jan 12 at 4:26
$begingroup$
my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
$endgroup$
– Indrasis Mitra
Jan 12 at 6:40
$begingroup$
my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
$endgroup$
– Indrasis Mitra
Jan 12 at 6:40
$begingroup$
I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
$endgroup$
– Indrasis Mitra
Jan 13 at 3:45
$begingroup$
I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
$endgroup$
– Indrasis Mitra
Jan 13 at 3:45
$begingroup$
@IndrasisMitra See attached note.
$endgroup$
– Cesareo
Jan 13 at 9:19
$begingroup$
@IndrasisMitra See attached note.
$endgroup$
– Cesareo
Jan 13 at 9:19
|
show 11 more comments
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