Convolution Integrals — Question 5

Question 5 – Solution

Strategy. Feedback changes the effective growth rate. Verify the transformed candidate through an equivalent IVP so no growth assumption is needed for uniqueness.

Step 1: Solve the transformed equation. Initially for sufficiently large $s$, $$U=\frac 1s+\frac{\lambda U}{s+1},\qquad U=\frac{s+1}{s(s+1-\lambda)}.$$ Partial fractions give $$\boxed{u(t)=\frac{1-\lambda e^{(\lambda-1)t}}{1-\lambda} \quad(\lambda\ne 1),\qquad u(t)=1+t\quad(\lambda=1).}$$ In particular $\lambda=0$ gives $u=1$; the factor $s+1$ then cancels in $U$.

Step 2: Prove equivalence without assuming a transform. For continuous $u$, define $v=e^{-t}*u$. Differentiation gives $v'=u-v$, $v(0)=0$. The integral equation makes $u$ continuously differentiable and yields $$u'=(\lambda-1)u+1,\qquad u(0)=1.$$ Conversely, let $u$ solve this IVP and form $v$ as above. The residual $w=u-1-\lambda v$ satisfies $w'=-u+1+\lambda v=-w$ and $w(0)=0$. Hence $w=0$ and the integral equation holds, including at $\lambda=0$. The linear IVP has a unique global solution, which proves the asserted existence and uniqueness.

Step 3: Classify all parameters. For $\lambda<1$, the solution is bounded and tends to $1/(1-\lambda)>0$. For $\lambda=1$ it grows linearly; for $\lambda>1$ it grows exponentially with positive leading coefficient $\lambda/(\lambda-1)$. For $\lambda\ne 1$, $u'=\lambda e^{(\lambda-1)t}$: it decreases from $1$ to a positive limit when $\lambda<0$, is constant when $\lambda=0$, and increases when $\lambda>0$. Thus it is positive for every real $\lambda$ and every $t\ge 0$.

Step 4: Inspect the actual transform domain. For $\lambda<1$ the positive nonzero limiting constant forces $s>0$; for $\lambda=1$ the linear term also requires $s>0$. For $\lambda>1$ the nonzero exponential requires $s>\lambda-1$. Consequently $$\boxed{s>0\ (\lambda\le 1),\qquad s>\lambda-1\ (\lambda>1).}$$ Positive feedback can offset or exceed the kernel’s decay. The rate in the equivalent equation is $\lambda-1$, not simply the kernel’s rate $-1$.