Vector Fields — Question 8

Question 8 – Solution

Strategy. Solve the two component equations and separate the parameter value that makes them dependent.

Step 1: Solve generally The first component gives $x=y$. Substituting into the second gives $$\lambda x+y=(\lambda+1)x=0.$$ If $\lambda\ne-1$, then $x=0$ and $y=0$, so $$\boxed{\lambda\ne-1:\ \text{the origin is the only equilibrium}}.$$

Step 2: Exceptional value If $\lambda=-1$, the second component is $-x+y$, the negative of the first. Every point with $x=y$ is then a zero: $$\boxed{\lambda=-1:\ \{(x,y):y=x\}\text{ is the equilibrium set}}.$$

Step 3: Behavior off the line For $\lambda=-1$, set $d=x-y$. Then $$\mathbf F_{-1}=\langle d,-d\rangle=d\langle 1,-1\rangle.$$ This vector is normal to $y=x$. When $x>y$, it points southeast; when $x<y$, it points northwest. Both directions lead away from the equilibrium line.

Verification The coefficient determinant is $1(1)-(-1)\lambda=1+\lambda$, which vanishes exactly at $\lambda=-1$, confirming the case split.