Practice: Systems of Linear Differential Equations

(a) The eigenvalue equation. With $\mathbf{x}=e^{\lambda t}\mathbf{v}$ and $\mathbf{v}$ constant, ${\mathbf{x}}^{\prime }=\lambda e^{\lambda t}\mathbf{v}$ while $A\mathbf{x}=e^{\lambda t}A\mathbf{v}$. The system ${\mathbf{x}}^{\prime }=A\mathbf{x}$ becomes

and since $e^{\lambda t}\ne 0$ it cancels, leaving $A\mathbf{v}=\lambda \mathbf{v}$. Why $\mathbf{v}\ne \mathbf{0}$. The zero vector satisfies $A\mathbf{v}=\lambda \mathbf{v}$ for every $\lambda$, so admitting it would place no constraint on $\lambda$ and the equation would be vacuous. It also gives only the zero function, which solves every linear system and carries no information. (b) with $\mathbf{x}(0)=(\genfrac{}{}{0ex}{}{3}{1})$. Eigenvalues. $\mathrm{tr}A=2$, $detA=1-4=-3$, so ${\lambda }^2-2\lambda -3=(\lambda -3)(\lambda +1)=0$ and $\lambda =3,-1$. Check: $3+(-1)=2=\mathrm{tr}A$ ✓ and $3\times (-1)=-3=detA$ ✓; substituting confirms $9-6-3=0$ and $1+2-3=0$ ✓. Eigenvectors. For $\lambda =3$: gives $v_1=v_2$, so ${\mathbf{v}}_1=(\genfrac{}{}{0ex}{}{1}{1})$. For $\lambda =-1$: gives $v_1+v_2=0$, so ${\mathbf{v}}_2=(\genfrac{}{}{0ex}{}{1}{-1})$. Verify: $A{\mathbf{v}}_1=(\genfrac{}{}{0ex}{}{3}{3})=3{\mathbf{v}}_1$ ✓ and $A{\mathbf{v}}_2=(\genfrac{}{}{0ex}{}{-1}{1})=-{\mathbf{v}}_2$ ✓, both residuals below $8\times {10}^{-16}$ numerically. General solution. $\mathbf{x}=c_1{e}^{3t}(\genfrac{}{}{0ex}{}{1}{1})+c_2{e}^{-t}(\genfrac{}{}{0ex}{}{1}{-1})$. Constants. At $t=0$: $c_1+c_2=3$ and $c_1-c_2=1$, so $c_1=2$, $c_2=1$, solved as a pair, since the components couple them.

Verify. $\mathbf{x}(0)=(3,1)$ ✓; residuals of ${\mathbf{x}}^{\prime }-A\mathbf{x}$ are $1.9\times {10}^{-10}$, $9.0\times {10}^{-10}$, $1.1\times {10}^{-8}$ at $t=0,0.5,1$ ✓. Values $\mathbf{x}(1)=(40.538953,39.803194)$ and $\mathbf{x}(2)=(806.992922,806.722252)$ show the components converging as the $e^{-t}$ term dies and the trajectory aligns with $(\genfrac{}{}{0ex}{}{1}{1})$. (c) . $\mathrm{tr}B=0$ and $detB=4$, so ${\lambda }^2+4=0$ and $\lambda =\pm 2i$. For $\lambda =2i$, $(B-2iI)\mathbf{v}=\mathbf{0}$ gives $-2i{v}_1-2v_2=0$, so $v_2=-i{v}_1$ and $\mathbf{v}=(\genfrac{}{}{0ex}{}{1}{-i})$. Then

whose real and imaginary parts give two real solutions:

The conjugate eigenvalue $-2i$ yields the same two, which is why one pair contributes two solutions and the $2\times 2$ system is now complete. Verify. The first solution's residual against ${\mathbf{x}}^{\prime }=B\mathbf{x}$ is below $2.6\times {10}^{-11}$ at $t=0,0.5,1$ ✓. Trajectories. The real part is zero, so there is no $e^{\alpha t}$ envelope: $\Vert (\cos 2t,\sin 2t)\Vert =1$ at every $t$, confirmed as $1.0$ at $t=0,1,2$. The trajectories are closed circular orbits traversed with angular frequency 2, neither growing nor decaying. A negative real part would spiral them inward, a positive one outward. (d) The same eigenvalue, two different matrices. Both have characteristic polynomial $(3-\lambda {)}^2$, so $\lambda =3$ with algebraic multiplicity 2 in each. The polynomial cannot distinguish them; the rank of $A-3I$ can. $B=3I$. $B-3I=\mathbf{0}$, rank 0, so the kernel is all of ${\mathbb{R}}^2$ and every nonzero vector is an eigenvector. Geometric multiplicity 2, not defective:

No $t$ factor. . , rank 1, so the kernel is one-dimensional and every eigenvector is a multiple of $\mathbf{v}=(\genfrac{}{}{0ex}{}{1}{0})$. Geometric multiplicity 1 < algebraic 2, defective. Here $c_1{e}^{3t}\mathbf{v}+c_2{e}^{3t}\mathbf{v}=(c_1+c_2)e^{3t}\mathbf{v}$ collapses to one constant, confined to the line $x_2=0$, and cannot meet an initial vector such as $(\genfrac{}{}{0ex}{}{1}{2})$. The second solution needs a generalised eigenvector: solving $(A-3I)\mathbf{w}=\mathbf{v}$ gives $w_2=1$ with $w_1$ free, so $\mathbf{w}=(\genfrac{}{}{0ex}{}{0}{1})$ and

Why that form works. Substituting $\mathbf{x}=e^{\lambda t}(\mathbf{w}+t\mathbf{v})$ gives ${\mathbf{x}}^{\prime }=e^{\lambda t}(\lambda \mathbf{w}+\mathbf{v}+\lambda t\mathbf{v})$ and $A\mathbf{x}=e^{\lambda t}(A\mathbf{w}+\lambda t\mathbf{v})$; the $t\mathbf{v}$ terms match and the rest requires $(A-\lambda I)\mathbf{w}=\mathbf{v}$. Verify. With $c_1=1$, $c_2=2$, that is $e^{3t}(1+2t,2)$, whose residual stays below $1.6\times {10}^{-8}$ ✓. (e) The scalar equation as a system, and stability. Set $x_1=y$ and $x_2=y^{\prime }$. Then $x_1^{\prime }=x_2$, and the equation $y^{″}=5y^{\prime }-6y$ gives $x_2^{\prime }=-6x_1+5x_2$:

Its characteristic polynomial is ${\lambda }^2-5\lambda +6$, identical to the scalar one, with roots computed as exactly $2$ and $3$ ✓. Substituting the scalar solution $y=3e^{2t}-2e^{3t}$ as $\mathbf{x}=(y,y^{\prime })$ satisfies the system with residuals below $2.1\times {10}^{-8}$ ✓. The two theories are one: a repeated root there is a defective companion matrix here, and $x{e}^{rx}$ is the $t\mathbf{v}$ term. What the eigenvalues tell you. Every term of the solution carries $e^{\lambda t}$, whose magnitude is $e^{(\mathrm{Re}\lambda )t}$. The imaginary part only rotates, since $|e^{i\beta t}|=1$. So: - all real parts negative → every term shrinks → all solutions tend to $\mathbf{0}$;
- some real part positive → that term grows;
- imaginary parts present → rotation, spiralling or orbiting according to the real part. Why one positive real part suffices. Suppose while the others are negative. A solution is $\sum c_i{e}^{{\lambda }_{i}t}{\mathbf{v}}_i$. Unless $c_1=0$, the first term grows without bound while the rest fade, so it eventually dominates and the trajectory escapes. The exceptions are exactly the initial vectors with $c_1=0$, those lying in the span of the other eigenvectors, a lower-dimensional subspace and hence a vanishingly small set of starting points. That is the sense in which almost every trajectory escapes. This also holds for defective matrices: a $t$ factor changes the rate of approach but not its direction, since $t{e}^{\lambda t}\to 0$ whenever . A polynomial cannot defeat a decaying exponential.