I have the following eigenvalue BVP $$ \lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2 F = 0 $$ wit BC(s) $F(0)=0,\frac{F''(0)}{F'(0)}=\beta_h,\frac{F''(1)}{F'(1)}=\beta_h$

$\mu$ is the eigenvalue parameter. For $\lambda_h=0.02,\beta_h=10$, I solved this eigen value problem numerically using the chebfun method in MATLAB to attain the follwoing 6 EVs


I took one of the eigenvalues $\mu=-32.9463$ and made a charcteristic equation from the BVP as:

$$0.02x^3-0.4x^2+24.9463x+100=0$$ which gave me three roots. $-3.74207, (11.871-34.5722i), (11.871+34.5722i)$

How should i proceed further to reach a solution form ?

I know that probably i need to guess a solution structure considering a real root and two complex and then proceed further to find the constants which will be three in this case because of the cubic nature of the ODE.

How many of EVs i need to consider ?

Also there will be three linear equations because of the three BC(s).

But i doubt won't this all just give me a trivial solution ?


I tried the follwing:

the three roots $\delta_k(\mu),\ \ k = 1,2,3$ we have the general solution form as

$$ F(x) = \sum_k C_k e^{-\delta_k(\mu)x} $$

using now the boundary conditions

$$ F(0) = \sum_k C_k = 0 \longrightarrow (1) $$

and with

$$ F'(0) = -\sum_k C_k \delta_k(\mu)\\ F''(0) = \sum_k C_k \delta_k(\mu)^2\\ $$


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

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

$$ M(\mu)\cdot C = 0,\ \ C = (C_k) $$

I arrive at the following linear equations




This could be rearranged in the form of $M(\mu).C_k=0$, which becomes a system of linear equations with $C_1$,$C_2$,$C_3$ as unknowns.

Here $\delta_1(\mu)$, $\delta_2(\mu)$ and $\delta_3(\mu)$ are the three roots i mentioned above which were found after substituting the first eigen value $-32.9463$

But when i try to solve this system of equations I get $C_1=0$, $C_2=0$ and $C_3=0$. (I solved it using matlab $AX=B$ solver)

Is there something really silly i am doing wrong here ? or is my assumption of the solution form misplaced ?

I must mention that there are infinite number of negative EVs and chebfun returns these 6 calling it the smoothest eigenmodes

  • $\begingroup$ Using the two conditions at endpoint $0$, you should get a one-parameter family of solutions. If your eigenvalue is really correct, these should all satisfy the condition at endpoint $1$. $\endgroup$ – Robert Israel Jan 18 at 17:43
  • $\begingroup$ @RobertIsrael from one parameter do you mean with "one undetermined constant" ? Can you elaborate a bit more ? $\endgroup$ – Indrasis Mitra Jan 19 at 2:12
  • $\begingroup$ @RobertIsrael I made an attempt to solve the problem. Have edited the original question to reflect it. $\endgroup$ – Indrasis Mitra Jan 19 at 3:39
  • $\begingroup$ @RobertIsrael Update: I also tried the Evans Circle method for finding eigenvalues using a Mathematica function and get the same Eigenvalues. They still give me a trivial solution. I cannot figure out at all what i am doing wrong. $\endgroup$ – Indrasis Mitra Jan 19 at 7:27

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