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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

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1 Answer 1

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

The determinant after simplifications reads

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

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

enter image description here

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]
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  • $\begingroup$ Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$ $\endgroup$
    – Avrana
    Jan 12, 2019 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 ? $\endgroup$
    – Avrana
    Jan 12, 2019 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 ? $\endgroup$
    – Avrana
    Jan 12, 2019 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 ? $\endgroup$
    – Avrana
    Jan 13, 2019 at 3:45
  • $\begingroup$ @IndrasisMitra See attached note. $\endgroup$
    – Cesareo
    Jan 13, 2019 at 9:19

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