# Eigen values of a Third Order Linear Homogenous ODE

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 $$M(\mu).C=0$$

$$\begin{vmatrix} 1 & 1 & 1 \\ {\delta_1(\mu)}^2+\beta_h\delta_1(\mu) & {\delta_2(\mu)}^2+\beta_h\delta_2(\mu) & {\delta_3(\mu)}^2+\beta_h\delta_3(\mu) \\ e^{-\delta_1(\mu)}({\delta_1(\mu)}^2+\beta_h\delta_1(\mu)) & e^{-\delta_2(\mu)}({\delta_2(\mu)}^2+\beta_h\delta_2(\mu)) & e^{-\delta_3(\mu)}({\delta_3(\mu)}^2+\beta_h\delta_3(\mu)) \\ \end{vmatrix}$$$$=0$$

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

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

$$-\frac{\sum_k C_k \delta_k(\mu)^2}{\sum_k C_k \delta_k(\mu)}=\beta_h\longrightarrow (2)$$

and similarly

$$-\frac{\sum_k C_k \delta_k(\mu)^2e^{-\delta_k(\mu)}}{\sum_k C_k \delta_k(\mu)e^{-\delta_k(\mu)}}=\beta_h\longrightarrow (3)$$

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 $$\delta_1(\mu),\delta_2(\mu),\delta_3(\mu)$$

$$\det(M) = \left(e^{\delta _1+\delta _2} \left(\delta _1-\delta _2\right) \delta _3 \left(\beta _h+\delta _1+\delta _2\right) \left(\beta _h+\delta _3\right)-e^{\delta _1+\delta _3} \delta _2 \left(\delta _1-\delta _3\right) \left(\beta _h+\delta _2\right) \left(\beta _h+\delta _1+\delta _3\right)+e^{\delta _2+\delta _3} \delta _1 \left(\delta _2-\delta _3\right) \left(\beta _h+\delta _1\right) \left(\beta _h+\delta _2+\delta _3\right)\right) \left(\cosh \left(\delta _1+\delta _2+\delta _3\right)-\sinh \left(\delta _1+\delta _2+\delta _3\right)\right)$$

discarding $$\cosh (\delta_1+\delta_2+\delta_3)-\sinh (\delta_1+\delta_2+\delta_3)=0$$ we follow with

$$\Delta(\mu)=e^{\delta _1+\delta _2} \left(\delta _1-\delta _2\right) \delta _3 \left(\beta _h+\delta _1+\delta _2\right) \left(\beta _h+\delta _3\right)-e^{\delta _1+\delta _3} \delta _2 \left(\delta _1-\delta _3\right) \left(\beta _h+\delta _2\right) \left(\beta _h+\delta _1+\delta _3\right)+e^{\delta _2+\delta _3} \delta _1 \left(\delta _2-\delta _3\right) \left(\beta _h+\delta _1\right) \left(\beta _h+\delta _2+\delta _3\right)=0$$

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]

• Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$ – Indrasis Mitra Jan 12 at 3:10
• 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 ? – Indrasis Mitra Jan 12 at 4:26
• 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 ? – Indrasis Mitra Jan 12 at 6:40
• 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 ? – Indrasis Mitra Jan 13 at 3:45
• @IndrasisMitra See attached note. – Cesareo Jan 13 at 9:19