I have a eigen boundary value problem $$ \lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2F=0 $$ $\mu$ is the separation variable or the eigenvalue variable with b.c. as $F(0)=0,\frac{F''(0)}{F'(0)}=\beta_h,\frac{F''(1)}{F'(1)}=\beta_h$

Typically the eigenvalue and eigen function relation is shown as

$$ op(\bar{f})=\mu \bar{f} $$ If the above relation is satisfied, then $\mu$ is called the eigenvalue of operator $op$ corresponding to eigen function $\bar{f}$

I am failing to express my EBVP in the standard form as my separation constant $\mu$ is attached to $F'$


$$ (\lambda_h\frac{d^2}{dx^2}-2\beta_h\lambda_h\frac{d}{dx}+{\beta_h}^2\int\mathbb{d}x)f=(-\lambda_h{\beta_h}^2+\beta_h+\mu)f $$

assuming $F:=\int f(x)$

but then i cannot form a characteristic equation to go about finding eigenvalues


I must make a note here that the solution $F$ or $f$ would be used in the following function

$$ \theta_w=e^{-\beta_hx}F'(x)e^{-\beta_cy}G'(y)$$ or $$\theta_w=e^{-\beta_hx}f(x)e^{-\beta_cy}g(y)$$

The initial EBVP gets complicated because it is third order and my described attempt makes it a Integro - Differential equation. Stuck at this for a while now.


Substituted $F=e^{kx}$ into the original EBVP yields

$$ e^{kx}(\lambda_h k^3- 2 \lambda_h \beta_h k^2 + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) k + \beta_h^2)=0 $$ So the characteristic polynomial is $$\lambda_h k^3- 2 \lambda_h \beta_h k^2 + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) k + \beta_h^2=0$$

which can now have three roots

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    $\begingroup$ Do you have a typo? $\beta_h^2$ is a free variable in the original EBVP, but it is a prefactor for an integral operator in your attempt $\endgroup$ – Aleksejs Fomins Jan 17 at 7:48
  • $\begingroup$ Also, what exactly is the problem? Substitute $F = e^{kx}$ into the original equation, and you will have a 3rd order polynomial equation involving $k$ and $m$. Why does that not pass for a characteristic equation? It can be solved to express $k = K(m)$ $\endgroup$ – Aleksejs Fomins Jan 17 at 7:58
  • $\begingroup$ @AleksejsFomins Yes,you were right. I had a typo in the original EBVP. Corrected it. $\endgroup$ – Indrasis Mitra Jan 17 at 9:21
  • $\begingroup$ @AleksejsFomins The third order polynomial which i was getting without any substitution , just changed the problem it into something like this $\lambda_h s^3 - 2 \lambda_h \beta_h s^2 + ((\lambda_h \beta_h - 1)\beta_h - \mu) s + {\beta_h}^2$. Solving for the three roots of it made it complicated $\endgroup$ – Indrasis Mitra Jan 17 at 9:27
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    $\begingroup$ I do not think that there is another way. If you want an analytical solution you need to be able to solve for the roots of the equation. If the analytical solution ends up being too messy, you might want to solve the equation numerically $\endgroup$ – Aleksejs Fomins Jan 17 at 10:00

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