# Eigenvalues keep giving trivial solutions everytime.

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 ?

• 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 $C \mapsto AC$. – Christoph Jan 19 at 17:06
• 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. – LutzL Jan 19 at 17:15
• @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. – Indrasis Mitra Jan 19 at 17:21
• 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. – Christoph Jan 20 at 7:35
• 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 $0=a_3s^3-a_2s^2+a_1s-a_0$. – LutzL Jan 20 at 9:56