Diagonalization differential equation

Question

I have a problem in solving the diagonalization of this differential equation : $$\frac{d}{dt}\binom{x}{f} = \left(\begin{matrix} -\frac{1}{\tau} & 1 \\ 0 & -\frac{1}{\tau_{c}}\end{matrix}\right)\binom{x}{f}$$ Can anybody help me?

Answer 1

Making the change of variables

$$ p = (x,f)^{\dagger}\\ P = (X,F)^{\dagger} $$

such that

$$ p = T\cdot P $$

with $T$ invertible, we have

$$ \left(T\dot P\right) = A\cdot \left(T \cdot P\right)\Rightarrow \dot P = \left(T^{-1}\cdot A \cdot T\right)\cdot P $$

now choosing

$$ T = \left( \begin{array}{cc} \frac{\tau \tau_c}{\tau_c-\tau} & 1 \\ 1 & 0 \\ \end{array} \right) $$

which is the eigenvectors matrix for $A$ we have finally

$$ \dot P = \left( \begin{array}{cc} -\frac{1}{\tau_c} & 0 \\ 0 & -\frac{1}{\tau } \\ \end{array} \right)\cdot P $$

NOTE

Just in case you have not been introduced to the theory of eigenvalues, you can proceed with a generic invertible $T$ like

$$ T = \left( \begin{array}{cc} 1 & t_1 \\ 0 & t_2 \\ \end{array} \right) $$

and then solve the matrix equation

$$ T^{-1}\cdot A\cdot T = \Lambda $$

with

$$ \Lambda = \left( \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \\ \end{array} \right) $$

arriving to the conditions

$$ \left( \begin{array}{cc} -\frac{1}{\tau } & t_2 \left(\frac{t_1}{t_2 \tau_c}+1\right)-\frac{t_1}{\tau } \\ 0 & -\frac{1}{\tau_c} \\ \end{array} \right) = \left( \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \\ \end{array} \right) $$

now choosing conveniently $t_1,t_2$ to calcell

$$ t_2 \left(\frac{t_1}{t_2 \tau_c}+1\right)-\frac{t_1}{\tau } $$

such as

$$ t_1 = \frac{\tau\tau_c t_2}{\tau_c-\tau} $$

we solve the problem.