0
$\begingroup$

We have a linear system

$$\frac{dx_1}{dt}= -K_1x_1 +K_{21}x_2$$ $$\frac{dx_2}{dt} = K_{12}x_1-K_2x_2$$

So we have solutions of the form

$$x_1= a_1e^{-\lambda_1 t} + a_2e^{- \lambda_2 t}$$ $$x_2= b_1e^{-\lambda_1 t} + b_2e^{- \lambda_2 t}$$

What I don't understand is how the notes have solutions of the form

$$- \lambda_1 a_1= -K_{1}a_1+K_{21}b_1$$ $$- \lambda_1 b_1= -K_{12}a_1-K_{2}b_1$$ $$- \lambda_2 a_2= -K_{1}a_2+K_{21}b_2$$ $$- \lambda_1 b_2= -K_{12}a_2+K_{2}b_2$$

$\endgroup$
1
$\begingroup$

Since $$x_1= a_1e^{-\lambda_1 t} + a_2e^{- \lambda_2 t},$$ after differentiating we have $$\frac{dx_1}{dt}=-\lambda_1 a_1 e^{-\lambda_1 t} - \lambda_2 a_2e^{- \lambda_2 t}.$$ On the other hand, $$\frac{dx_1}{dt}= -K_1x_1 +K_{21}x_2=-K_1(a_1e^{-\lambda_1 t} + a_2e^{- \lambda_2 t})+K_{21}(b_1e^{-\lambda_1 t} + b_2e^{- \lambda_2 t})=(-K_1a_1+K_{21}b_1)e^{-\lambda_1 t}+(-K_1a_2+K_{21}b_2)e^{- \lambda_2 t}.$$

Since coefficients by $e^{\lambda_i t}$ are the same, we have $$- \lambda_1 a_1= -K_{1}a_1+K_{21}b_1,$$ $$- \lambda_2 a_2=-K_1a_2+K_{21}b_2.$$ Similarly we get other two relations after differentiating $x_2$.

$\endgroup$
0
$\begingroup$

Here is an "all-matricial" explanation:

$$K:=\pmatrix{-K_1&K_{21}\\K_{12}&-K_2}, \ \ P:=\pmatrix{a_1&a_2\\b_1&b_2}, \ \ D:=\pmatrix{\lambda_1 & 0 \\ 0 & \lambda_2}.$$

The set of 4 equations you have given can be expressed into a single matrix equation, up to a sign change:

$$\pmatrix{a_1&a_2\\b_1&b_2}\pmatrix{\lambda_1 & 0 \\ 0 & \lambda_2}=\pmatrix{-K_1&K_{21}\\K_{12}&-K_2}\pmatrix{a_1&a_2\\b_1&b_2}.$$

In other words:

$$\tag{1}PD=KP \ \ \iff \ \ D=P^{-1}KP,$$

in which we recognize the classical diagonalization relationship that is currently used in the solution of linear differential systems with constant coefficients.

Recall: Let us write the given differential system under the form $X'=KX$. Using change of variable $X=PY$, it becomes:

$$PY'=KPY \iff Y'= \underbrace{P^{-1}KP}_{D} Y$$

If we have chosen $P$ such that $P^{-1}KP=D$, (see equ. $(1)$) where $D$ is a diagonal matrix, we get two "decorellated" very simple differential equations $x'(t)=\lambda_k x(t) \ \ \Rightarrow \ \ x(t)=C_k e^{\lambda_k t}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.