# Jordan Chains to solve x'=Ax, complex-valued.

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1. Feb 8, 2016

### kostoglotov

This isn't a homework help question, I am asking about a general case as the text hasn't fleshed out it's explanation enough for me.

So, I've just seen how Jordan Chains can be applied to solve some applied problems involving solving x'=Ax.

My question is about what they specifically mean in this copy-pasted paragraph above.

Would they mean that, I should find a Jordan Chain of length k for lambda = alpha and then another Jordan Chain of length k for lambda = beta? And if so, what would the general form of the solution look like?

Take a three by three case with complex eigenvalue of defect 2. Would the general solution look like:

$$\lambda = \alpha, \left \{ \vec{u_1},\vec{u_2},\vec{u_3} \right \}$$
$$\lambda = \beta, \left \{ \vec{v_1},\vec{v_2},\vec{v_3} \right \}$$
$$\vec{x_1} = \vec{u_1}e^{\alpha t} \\ \vec{x_2} = (t\vec{u_1} + \vec{u_2})e^{\alpha t} \\ \vec{x_3} = \left(\frac{1}{2}t^2\vec{u_1} + t\vec{u_2}+\vec{u_3}\right)e^{\alpha t} \\ \\ \vec{x_4} = \vec{v_1}(\cos{\beta t} + \sin{\beta t}) \\ \vec{x_5} = (t\vec{v_1} + \vec{v_2})(\cos{\beta t} + \sin{\beta t}) \\ \vec{x_6} = \left(\frac{1}{2}t^2\vec{v_1} + t\vec{v_2}+\vec{v_3}\right)(\cos{\beta t} + \sin{\beta t})$$

Or

$$\vec{x_1} = \vec{u_1}e^{\alpha t}(\cos{\beta t} + \sin{\beta t}) \\ \vec{x_2} = (t\vec{u_1} + \vec{u_2})e^{\alpha t}(\cos{\beta t} + \sin{\beta t}) \\ \vec{x_3} = \left(\frac{1}{2}t^2\vec{u_1} + t\vec{u_2}+\vec{u_3}\right)e^{\alpha t}(\cos{\beta t} + \sin{\beta t}) \\ \\ \vec{x_4} = \vec{v_1}e^{\alpha t}(\cos{\beta t} + \sin{\beta t}) \\ \vec{x_5} = (t\vec{v_1} + \vec{v_2})e^{\alpha t}(\cos{\beta t} + \sin{\beta t}) \\ \vec{x_6} = \left(\frac{1}{2}t^2\vec{v_1} + t\vec{v_2}+\vec{v_3}\right)e^{\alpha t}(\cos{\beta t} + \sin{\beta t})$$

or something else?

2. Feb 13, 2016