The energy is
$$
E(t)=c_1^2e^{-2\lambda_1t}
+c_2^2e^{-2\lambda_2t}
+2\alpha c_1c_2e^{-(\lambda_1+\lambda_2)t}.
$$
Thus
$$
E'(0)=-2[\lambda_1c_1^2+\lambda_2c_2^2
+(\lambda_1+\lambda_2)\alpha c_1c_2].
$$
This quadratic form can be negative for positive $c_1,c_2$ precisely when
$$
\boxed{\alpha<-\frac{2\sqrt{\lambda_1\lambda_2}}
{\lambda_1+\lambda_2}}.
$$
The decaying vectors must therefore be sufficiently nonorthogonal and oppositely directed. This is <transient growth from non-normal modes>.

With $c_2=1$ and $\mu=c_1/c_2$, maximizing $E'(0)$ gives
$$
2\lambda_1\mu+(\lambda_1+\lambda_2)\alpha=0,
$$
so
$$
\boxed{\mu_{\rm opt}
=-\frac{(\lambda_1+\lambda_2)\alpha}{2\lambda_1}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 331 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-331/2/a/solution