- #1
member 428835
Hi PF!
Here's an ODE (for now let's not worry about the solutions, as A LOT of preceding work went into reducing the PDEs and BCs to this BVP):
$$\lambda^2\phi-0.1 i\lambda\phi''-\phi'''=0$$ which admits analytic eigenvalues
$$\lambda =-2.47433 + 0.17337 I, 2.47433 + 0.17337 I, -10.5087 + 1.16033 I, 10.5087 + 1.16033 I, -22.0562 + 3.13425 I, 22.0562 + 3.13425 I...$$
There's an analytic technique to solve which is not terribly difficult (again, you'd need the entire system of equations to do this). However, another way to solve this ODE is to change the nature of the problem from an eigenvalue problem to a frequency response problem, where now the eigenvalues ##\lambda## are instead input parameters of the forcing frequency:
$$ \lambda^2\phi-0.1 i\lambda\phi''-\phi'''=f \lambda$$
By carefully selecting basis functions ##\phi_j = \sin(\pi(j-1/2)x)\cosh( \sqrt{ (\pi(j-1/2))^2 + 1} )## we can take inner products of this equation, resulting in a square matrix linear system of algebraic equations:
$$\lambda^2 M-0.1 i \lambda V - K = F : M_{ij} = \int_0^1 \int_0^1 \phi_i(x) \phi_j(y)\,dy\,dx,\\
M_{ij} = \int_0^1 \int_0^1 \phi_i(x) \phi_j(y)\,dy\,dx,\\
V_{ij} = \int_0^1 \int_0^1 \phi_i''(x) \phi_j(y)\,dy\,dx,\\
K_{ij} = \int_0^1 \int_0^1 \phi_i'''(x) \phi_j(y)\,dy\,dx,\\
F_{ij} = f\int_0^1 \phi_i(x) \,dx.$$
The vector solution for each ##\lambda## input is denoted ##C##. If we plot the magnitude of ##C## against several values of ##\lambda## we would arrive at a frequency response (see attached picture). The ##\lambda## values associated with the peaks are in fact the eigenvalues of the unforced problem.
Okay, now for my question: the ##\lambda## associated with each peak is real. However, I know the analytic solution (written above in decimal form) is complex. How would I separate the forced solutions of ##\lambda## to ascertain the real and imaginary components of the analytic eigenvalues?
EDIT: the preview does not display LaTeX form for me (google chrome, ubuntu 18.04). Also, the "\\" command no longer line breaks within latex.
Here's an ODE (for now let's not worry about the solutions, as A LOT of preceding work went into reducing the PDEs and BCs to this BVP):
$$\lambda^2\phi-0.1 i\lambda\phi''-\phi'''=0$$ which admits analytic eigenvalues
$$\lambda =-2.47433 + 0.17337 I, 2.47433 + 0.17337 I, -10.5087 + 1.16033 I, 10.5087 + 1.16033 I, -22.0562 + 3.13425 I, 22.0562 + 3.13425 I...$$
There's an analytic technique to solve which is not terribly difficult (again, you'd need the entire system of equations to do this). However, another way to solve this ODE is to change the nature of the problem from an eigenvalue problem to a frequency response problem, where now the eigenvalues ##\lambda## are instead input parameters of the forcing frequency:
$$ \lambda^2\phi-0.1 i\lambda\phi''-\phi'''=f \lambda$$
By carefully selecting basis functions ##\phi_j = \sin(\pi(j-1/2)x)\cosh( \sqrt{ (\pi(j-1/2))^2 + 1} )## we can take inner products of this equation, resulting in a square matrix linear system of algebraic equations:
$$\lambda^2 M-0.1 i \lambda V - K = F : M_{ij} = \int_0^1 \int_0^1 \phi_i(x) \phi_j(y)\,dy\,dx,\\
M_{ij} = \int_0^1 \int_0^1 \phi_i(x) \phi_j(y)\,dy\,dx,\\
V_{ij} = \int_0^1 \int_0^1 \phi_i''(x) \phi_j(y)\,dy\,dx,\\
K_{ij} = \int_0^1 \int_0^1 \phi_i'''(x) \phi_j(y)\,dy\,dx,\\
F_{ij} = f\int_0^1 \phi_i(x) \,dx.$$
The vector solution for each ##\lambda## input is denoted ##C##. If we plot the magnitude of ##C## against several values of ##\lambda## we would arrive at a frequency response (see attached picture). The ##\lambda## values associated with the peaks are in fact the eigenvalues of the unforced problem.
Okay, now for my question: the ##\lambda## associated with each peak is real. However, I know the analytic solution (written above in decimal form) is complex. How would I separate the forced solutions of ##\lambda## to ascertain the real and imaginary components of the analytic eigenvalues?
EDIT: the preview does not display LaTeX form for me (google chrome, ubuntu 18.04). Also, the "\\" command no longer line breaks within latex.