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I am trying to find the eigenvalues of this Eigen BVP. $\mu$ is the eigenvalue parameter

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

For $\lambda_h=0.02$ and $\beta_h=10$, I calculated the EVs using chebfun in MATLAB. There are infinite number of negative EVs. The general solution should be of the form

$$ F(x) = \sum_k C_k e^{-\delta_k(\mu)x} $$ Substituting the bc(s) yield three linear equations in $C_1,C_2,C_3$. where $-\delta_k(\mu)$ are the three roots of the characteristic equation obtained after substituting one of the EVs.

But this keeps giving me trivial solutions i.e all $C_k=0$. I tried many EVs (ex. $-32.9463$)but to no avail.

Is there something horribly wrong in my understanding ? Is there some other approach i can apply to the problem ? I cannot figure out at all.

ATTEMPT

After @LutzL and @Christoph provided useful comments, i did some reading on null spaces of matrices and their use to solve the homogenous system of equations.

Using the EV mentioned above $-32.9463$ and $\lambda_h = 0.02, \beta_h = 10$, I arrive at three roots of the characteristic equation as:

$$ s_1 = 3.7421 $$ $$ s_2 = -11.8710+34.5722i $$ $$ s_3 = -11.8710-34.5722i $$

Now applying the three BC(s) i have the following matrix system $M(\mu).C_k=0$ and i need to find $C_k$ which will be the non trivial solutions. $$ \begin{bmatrix} 1 & 1 & 1 \\ {s_1}^2+\beta_hs_1 & {s_2}^2+\beta_hs_2 & {s_3}^2+\beta_hs_3 \\ e^{-s_1}({s_1}^2+\beta_hs_1) & e^{-s_2}({s_2}^2+\beta_hs_2) & e^{-s_3}({s_3}^2+\beta_hs_3) \\ \end{bmatrix} * \begin{bmatrix} C_1 \\ C_2 \\ C_3 \\ \end{bmatrix}=0 $$ The above is a homogeneous system of equations which i solved using the following command in MATLAB

[U,S,V] = svd(A,'econ');
b = V(:,size(A,2));

Here $A$ is the coefficient matrix. The solver linsolve(A,B) kept giving me trivial solutions. svd means singular value decomposition. b is supposed to be the solution we are looking for. For the parameters i mentioned here it comes out to be 

b =-0.4550 + 0.0000i
    0.2278 + 0.5870i
    0.2278 - 0.5870i

It is mentioned that

b will be a solution if the corresponding smallest singular value is zero. If not, b will be the nearest you can come to a solution. If more than one singular value is zero, there are infinitely many solutions of which this will be one

So the values of b must be my corresponding $C_1,C_2,C_3$. But they are coming out to be complex. So should i just take the magnitudes of these complex numbers to form my solution ?

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  • $\begingroup$ Can't you get chebfun to compute the eigenfunctions as well? Obviously $F \equiv 0$ is a solution to the BVP for any value of $\mu$. But since this cannot be an eigenfunction, your linear system $AC=0$ for the coefficients $C_1, C_2, C_3$ must have non-trivial solutions if $\mu$ is an eigenvalue. These are the solutions that you must find, for example by determining the null space of the linear map $C \mapsto AC$. $\endgroup$
    – Christoph
    Jan 19, 2019 at 17:06
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    $\begingroup$ Set $C_3=1$ and leave out the last equation, the first two should then give numerical values for $C_1,C_2$. It is not surprising that you get an invertible if ill-conditioned matrix instead of a singular one if evaluated in floating point arithmetic. $\endgroup$
    – Lutz Lehmann
    Jan 19, 2019 at 17:15
  • $\begingroup$ @Christoph . Thanks. Yes i know that the system $AX=0$ do have $X=0$ as a trivial solution always and i must look for the non trivial solutions. Also, $chebfun$ does give something called eigenmodes which are plots. Your last point , says "determine the null space of the linear map C to AC . I am sorry to say, but i have no idea how this is done and how i should move forward with it. $\endgroup$
    – Avrana
    Jan 19, 2019 at 17:21
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    $\begingroup$ Here you have complex conjugate $s_2 = \overline{s_3}$ and $C_2 = \overline{C_3}$. Therefore, the corresponding eigenfunction will be something like $F(x) = C_1 e^{-s_1 x} + 2 e^{-\mathrm{Re}(s_2) x} \left( \mathrm{Re}(C_2) \cos(\mathrm{Im}(s_2) x) + \mathrm{Im}(C_2) \sin(\mathrm{Im}(s_2) x) \right)$. $C_1$ should probably be zero here. $\endgroup$
    – Christoph
    Jan 20, 2019 at 7:35
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    $\begingroup$ Your characteristic roots have the wrong sign, if you set $y=ce^{-sx}$ for $0=a_3y'''+a_2y''+a_1y'+a_0y$, then the characteristic polynomial has to be $0=a_3s^3-a_2s^2+a_1s-a_0$. $\endgroup$
    – Lutz Lehmann
    Jan 20, 2019 at 9:56

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