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, $$G(0) = 0, \frac{G''(0)}{G'(0)}=\beta_c, \frac{G''(1)}{G'(1)}=\beta_c$$

Everything below is for $\lambda_h=\lambda_c=0.02$, $\beta_h=\beta_c=10$ 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 are:

  1. What should be my next step, substitute the same EV in the $F$ equation to find its constants?
  2. Or, the EVs of $F$ are to be found separately (which i already tried, they are of the same magnitude but opposite in sign) /
  3. How many EVs should i need to consider to build my solution (I tried the above mentioned process for one EV and my final $\theta_w$ function gives results of order $O(9)$ where it should not go above 1)

  4. In general what should be my path forward? I have no clue. Most of the texts mention second order eigenvalue problems which are not helping. Have been pestering a lot over this. Any advice will help.

NOTE When we solve the $$u_{xx}+u_{yy}=0$$ on $0<x<1$ and $0<y<1$ with bc(s) as $u(0,y)=0,u(1,y)=0,u(x,0)=0,u(x,1)=g_2(x)$ they assume a solution of the form $$u(x,y)=X(x)Y(y)$$ which results in $$X''+\lambda X=0$$ and $$Y''-\lambda Y=0$$ with the separated bc as $X(0)=0,X(1)=0,Y(0)=0,Y(1)=?$.

Solving the first eigenvalue problem gives EVs as

9.8696044010876 39.4784176043502 88.8264396097948 157.9136704174691 246.7401100272539 355.3057584392049 which is $\lambda_n={{n\pi}^2}$ where $n$ attains positive integer values. Then we substitute this $\lambda_n$ expression into the $Y$ problem to get a $Y_n$ expression.

This step makes it sure that $X$ and $Y$ are using the same EV values.

[From PDE:Methods,Applications and Theories; Harumi, Hattori]

When i solve the $Y$ expression separately for EVs with an assumed $Y(1)=0$, bc it gives me results as $\lambda_n=-{(n\pi)}^2$ viz. same magnitude and opposite sign which is something i observe in my problem too.Then since there are no intersecting $\lambda$ , this problem too should not have had a solution.

Hence i was contemplating of calculating the $F$ EVs and then substituting in $G$.

  • $\begingroup$ I think the first issue is that you need to find eigenpairs $(\mu,F,G)$ for $F$ and $G$ simultaneously. It is possible and not uncommon for boundary value problems to have no solution for certain combinations of parameters. $\endgroup$ – Christoph Jan 27 at 8:52
  • $\begingroup$ @Christoph If i understand you correctly , you mean that i need to look for values of $\mu$ that satisfy both the $F$ and $G$ equation together ? I would emphasize here that when i solved $F$ and $G$ as separate Eigenvalue problems using chebfun in MATLAB, for $\lambda_h$ and $\beta_h$ values as mentioned in my post, they had no intersecting EVs and the found EVs had same magnitude but were opposite in sign. $\endgroup$ – Indrasis Mitra Jan 27 at 10:22
  • $\begingroup$ Yes, there must be values of $\mu$ for which both problems have a solution. If there is no such $\mu$, then I believe your original BVP (with the PDE) has no solution for this combination of parameters. $\endgroup$ – Christoph Jan 27 at 10:26
  • $\begingroup$ @Christoph have added a note to my OP keeping in mind your points.It is something similar i find. Have a look. $\endgroup$ – Indrasis Mitra Jan 27 at 12:24
  • 1
    $\begingroup$ But for the Laplace equation example, we have $X_n(x) = \sin(n \pi x)$ and $Y_n(y) = C_n \sinh(n \pi y)$. These functions satisfy $X_n'' + \lambda_n X_n = 0$ and $Y_n'' - \lambda_n Y_n = 0$ with the same values $\lambda_n = (n\pi)^2$, $n \in \mathbb{N}$. $\endgroup$ – Christoph Jan 28 at 7:57

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