I wonder how can I find the largest delta in which if initial condition is inside circle (How do I specify radius of that circle) bounded by delta then the solution is always within the dashed circle in the figure? How do you approach this solely based on geometry of phase portrait since no equation is given?
On
Look at my diagram. You'll notice that any solution that start within the black circle will stay within epsilon for the remainder of the trip. Solutions that start outside of the black circle can extend out past epsilon. (Mainly ones that start at the bottom-left, and top-right of the black circle)
This should be enough to let you get the other 3 on your homework by yourself.
It appears you have a spiral. A spiral is either unstable or stable. The spiral is produced by two cases.
In order for stability, for $\varepsilon = 0.6$, then there exists a $\delta$ such that $\lvert f(t_0) - 0\rvert < \delta$ of the equilibrium $\lvert f(t) - 0\rvert < \varepsilon = 0.6$ for $t\geq t_0$. To be asymptotically stable, there exists a $\delta_0$ such that $\delta_0 > \lvert f(t_0) - 0\rvert$ whenever $f(t)\to 0 $ when $t\to\infty$.
Case 1:
In order to have a spiral, you must have complex eigenvalues. If an initial condition is going to stay within that circle, it must be on a trajectory to the fixed point; that is, you must have a stable spiral so $Re(\lambda) < 0$. If you had an unstable spiral all initial conditions will be thrown out of the circle no matter what some radius of this circle is. With this being the case, the real part of the eigenvalue must be negative in order to have a stable spiral.
Case 2:
One eigenvalue with multiplicity two. Let $\lambda_1\in\mathbb{R}$ be the eigenvalue. Then by the Cayley-Hamilton Theorem, $(A - \lambda_1\mathbb{I})^2 = 0$. Then either $A = \lambda_1\mathbb{I}$ or $A\neq \lambda_1\mathbb{I}$.
For $A =\lambda_1\mathbb{I}$, and $\lambda_1 < 0$, all solutions approach the origin asymptotically. For $\lambda_1 > 0$, all solutions go to $\infty$ along the radial lines. Finally, if $\lambda_1 = 0$, all points are fixed.
Now let's consider $A\neq \lambda_1\mathbb{I}$. Let $X = A - \lambda_1\mathbb{I}$. Then we have $X^2 = 0$ where $X\neq 0$. Then there exist a vector such that $v_1 = Xv_2\neq 0$ where $v_1$ nd $v_2$ are linearly independent. \begin{align} Av_1 &= \lambda_1v_1\\ Av_2 &= \lambda_1v_2 + v_1 \end{align} Let $T$ be the matrix of linear transformation, then $$ T = \begin{bmatrix} \lambda_1 & 1\\ 0 & \lambda_1 \end{bmatrix} $$ Exponentiating $T$, we have $$ e^{Tt} = \begin{bmatrix} e^{\lambda_1t} & te^{\lambda_1t}\\ 0 & e^{\lambda_1t} \end{bmatrix} $$ Then \begin{align} x(t) &= x_0e^{\lambda_1t} + y_0e^{\lambda_1t}\\ y(t) &= y_0e^{\lambda_1t} \end{align} Now if $\lambda < 0$, all solutions will approach the origin. If $y_0 = 0$, then the solutions approach the origin along the x axis. If $y_0 \neq 0$, then $y(t) - x(t)$ approaches the origin as $t\to\infty$ so trajectory is asymptotic to $y = x$. For $\lambda > 0$, all points go to $\infty$ except zero. If $\lambda_1 = 0$, then the points on the x axis are fixed and all others go to $\infty$.