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I (with help from a MSE user) used the following substitution to seperate variables in a second order linear PDE

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

The following two ODEs (Eigenvalue problems) are a result of applying variable seperation to a system of three coupled 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} with some separation constant $\mu \in \mathbb{R}$.

BC(s) $$F(0) = 0, \frac{F''(0)}{F'(0)}=\beta_h, \frac{F''(1)}{F'(1)}=\beta_h$$ Similarly, $$G(0) = 0, \frac{G''(0)}{G'(0)}=\beta_c, \frac{G''(1)}{G'(1)}=\beta_c$$

These two EV problem need to be solved .Let us take the parameter values as $\lambda_h=\lambda_c=0.02$, $\beta_h=\beta_c=10$ and $V=1$.

The next step i did was solve the Eigen BVP for $G$ using chebfun in MATLAB which gave me

14.364332916201686 17.484587457962977 20.888494184298537 24.587309921467451 28.600217347815317 32.946305486743015,.... and infinitely many towards positive as the eigenvalues. The general solution form for this ODE is:

$$ F(x) = \sum_k C_k e^{-\delta_k(\mu)x} $$ the three constants i can determine by the three linear equations in $C_1,C_2,C_3$ using the boundary conditions.

My questions:

  1. When i use one EV and substitute in the ODE of $G$ it gives me 3 roots and using the BC(s) on the general solution stated above I get a $MC_k=0$ form of homogeneous equation where $M$ is the coefficient matrix. Using svd in MATLAB i can find $C_k$. For how may EVs should i repeat this procedure ? OR Is there any way of extracting the Eigen functions from chebfun and building a solution ?

  2. For function $F$ should i calculate the eigenvalues again separately or use the same EVs from G (cause variable separation means both the separated Equations are equal to the same constant $\mu$). I did find the EVs of $F$ separately using chebfun and they are of the same magnitude albeit negative and infinite in number. This got me really confused.

  3. Is the whole approach i am undertaking plausible ? Or is there some other pathway i could take from the separated equations to solve the EBVPs which i am not thinking about or do not know ?

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  • $\begingroup$ I have problems identifying the eigenvalue problem. The eigenfunction-pair is $(F, G)$, but, what is the eigenvalue in your equation? $\endgroup$ – nicoguaro Jan 29 at 21:10
  • $\begingroup$ @nicoguaro $\mu$ is the eigenvalue in the equations. I mentioned it as a separation constant. $\endgroup$ – Indrasis Mitra Jan 30 at 2:41
  • $\begingroup$ I thought that an eigenvalue problem is of the form $\mathcal{L}u = \lambda u$. $\endgroup$ – nicoguaro Jan 30 at 2:43
  • $\begingroup$ @nicoguaro You are right. This is a generalized Eigen value problem. The same problem can be expressed as $\lambda_c Vg'' - 2 \lambda_c \beta_c Vg' + ( (\lambda_c \beta_c - 1) \beta_cV) g + \beta_c^2 \int g \mathrm{d}y = -\mu g $ where $G := \int g \mathrm{d}y$ . Then the problem can be expressed as $Au=\lambda u$ but then there will be an integral operator $\endgroup$ – Indrasis Mitra Jan 30 at 2:51

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