Question

Consider the IVP $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$with$\overrightarrow{\mathbf{x}}\left(0\right)\mathbf{=}{\overrightarrow{\mathbf{x}}}_{\mathbf{0}}$where A is an upper triangular$\mathit{n}\mathbf{\times }\mathit{n}$matrix with m distinct diagonal entries ${\mathbf{\lambda }}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{\dots }\mathbf{,}{\mathbf{\lambda }}_{\mathbf{m}}$. See the examples in Exercise 45 and 46.

(a) Show that this problem has a unique solution$\overrightarrow{\mathbf{x}}\left(t\right)$whose components${\mathit{x}}_{\mathbf{i}}\left(t\right)$are of the form

${\mathit{x}}_{\mathbf{i}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{P}}_{\mathbf{1}}\mathbf{\left(}\mathit{t}\mathbf{\right)}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{1}}\mathbf{t}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{P}}_{\mathbf{m}}\mathbf{\left(}\mathit{t}\mathbf{\right)}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{m}}\mathbf{t}}$,

for some polynomials ${\mathit{P}}_{\mathbf{j}}\left(t\right)$.Hint: Find first ${\mathit{x}}_{\mathbf{n}}\left(t\right)$, then ${\mathbf{x}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\left(t\right)$, and so on.

(b) Show that the zero state is a stable equilibrium solution of this system if (and only if) the real part of all the${\mathit{\lambda }}_{\mathbf{i}}$ is negative.

Short answer

(a) The solution is$x_i\left(t\right)=P_1\left(t\right)e^{{\lambda }_{1}t}+...+P_m\left(t\right)e^{{\lambda }_{m}t}$

(b)The system$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$is stable.

Step-by-step solution

Definition of first order linear differential equation.

Consider the differential equation $\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{af}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$where$\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}$is a smooth function and 'a'is a constant. Then the general solution will berole="math" localid="1660806568260" $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{at}}\mathbf{\int }{\mathbf{e}}^{\mathbf{-}\mathbf{at}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$.

Definition of characteristic polynomial

Consider the linear differential operator

$\mathrm{T}\left(\mathrm{f}\right)={\mathrm{f}}^{\left(\mathrm{n}\right)}+{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{f}}^{\left(\mathrm{n}-1\right)}+...+{\mathrm{a}}_1\mathrm{f}\text{'}+{\mathrm{a}}_0\mathrm{f}$.

The characteristic polynomial of T is defined as

${\mathrm{p}}_{\mathrm{\tau }}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^{\mathrm{n}}+{\mathrm{a}}_{\mathrm{n}-1}\mathrm{\lambda }+...+{\mathrm{a}}_1\mathrm{\lambda }+{\mathrm{a}}_0$.

(a) Explanation for the given first order linear differential equation

Consider the IVP $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$with $\overrightarrow{x}\left(0\right)={\overrightarrow{x}}_0$where A is an upper triangular$n\times n$ matrix with m distinct diagonal entries. ${\lambda }_1,\dots .,{\lambda }_m$

Hence by the above definitions we get the solution as follows

$x_i\left(t\right)=P_1\left(t\right)e^{{\lambda }_{1}t}+...+P_m\left(t\right)e^{{\lambda }_{m}t}$

Hence the solution.

(b) Explanation of the zero state stable equilibrium.

For a system$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$, here A is the matrix form.

The zero state is an asymptotically stable equilibrium solution if and only if the real parts of all eigen values of A are negative.

Here A has no real eigen value, which means the real part of all the eigen values are negative.

Thus, the system$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$is stable.