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For the differential equation (physical friction)

$\ddot x=-a\cdot \dot x$

The solution can be easily found using exponential ansatz and is

$x(t)=c_1+c_2 \exp(-a\cdot t)$

Or expressing this using initial conditions $x_0:= x(0)=c_1+c_2 $, $v_0:=\dot x(0)=-a c_2$ it can be written as $x(t)=1\cdot x_0+\left(\frac 1 a - \frac {\exp(-at)} a \right) {v_0 } $, $v(t)={\exp(-at)}\cdot {v_0 } $ .

Putting into matrix form it is:

$ \begin{pmatrix} x(t) \\ \dot x(t) \end{pmatrix} = \begin{pmatrix} 1 & \frac {1-\exp(-at)} a \\ 0 & \exp(-at) \end{pmatrix} \begin{pmatrix} x(0) \\ \dot x(0) \end{pmatrix} $

Now however when we use matrix exponential the solution differs! The upper solution seems more reasonable since for vanishing speed we get the position back.

$\mathbf{\dot x} = \begin{pmatrix} \dot x \\ \ddot x \end{pmatrix} = \begin{pmatrix} 0& 1\\ 0 & -a \end{pmatrix} \begin{pmatrix} x \\ \dot x \end{pmatrix} =A \mathbf{x} $

It should be $ \mathbf{x}=\exp(At) \mathbf{x_0}$, so that both solutions could be compared using $ \exp(At) = \begin{bmatrix} 1 & \exp(t)\\ 1 & \exp(-at) \end{bmatrix} $ but they do not coincide.

Matrix exponential calculation

What/where is the issue/error/misunderstanding?

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1 Answer 1

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The exponential matrix is incorrect. Observe that

$$(s I - A)^{-1} = \begin{bmatrix} s & -1 \\ 0 & s + a\end{bmatrix}^{-1} = \frac{1}{s(s+a)} \begin{bmatrix} s + a & 1\\ 0 & s\end{bmatrix}.$$

Simplifying,

$$(s I - A)^{-1} = \begin{bmatrix} \frac{1}{s} & \frac{1/a}{s} + \frac{-1/a}{s+a} \\ 0 & \frac{1}{s + a}\end{bmatrix}.$$

Computing the inverse Laplace we find,

$$\exp(A t) = \mathcal{L}^{-1}\{ (s I - A)^{-1} \} = \begin{bmatrix} 1 & \frac{1}{a} - \frac{1}{a}\exp(-a t) \\ 0 & \exp(-a t) \end{bmatrix}$$

which is consistent with your ansatz solution. Check how you computed your exponential matrix. It looks like you did a term-by-term exponential, which is almost never correct (only works for diagonal matrices).

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  • $\begingroup$ Did you use Neuman inversion of geometric series or how does that work with Laplacian, could you elaborate? L(exp(At))=[sI-A]^(-1) $\endgroup$
    – camel
    Commented Mar 5, 2022 at 16:56
  • $\begingroup$ That's the Laplace transform (and its associated inverse operation). The term by term Laplace transform of $\exp(A t)$ is $(s I - A)^{-1}.$ $\endgroup$ Commented Mar 5, 2022 at 17:33
  • $\begingroup$ Is it a more robust or easier method to calculate matrix exponential? Btw I used WolframAlpha as linked in the question, is my input wrong? $\endgroup$
    – camel
    Commented Mar 5, 2022 at 22:28
  • $\begingroup$ You'll want to use this command instead; it is (expm) not (exp). $\endgroup$ Commented Mar 5, 2022 at 22:31
  • $\begingroup$ Good to know, it was only the wrong input. One has to watch out with the commands, also for MatrixLog. $\endgroup$
    – camel
    Commented Mar 6, 2022 at 10:17

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