Convolution Integrals: Question 4-9.10 & Solution

Question 10

All functions are causal (zero for $t<0$ ). Use the one-sided Laplace transform and $(f*g)(t)=\int_0^t f(t-u)g(u)\,du$ . Write $H_a(t)=0$ for $t<a$ and $1$ for $t\ge a$ . Values at isolated endpoints do not affect an ordinary integral.

Let $f\in C^1[0,\infty)$ satisfy $|f'(t)|\le L$ , with $L\ge 0$ . For $\varepsilon>0$ define $k_\varepsilon(t)=\frac 1\varepsilon e^{-t/\varepsilon},\qquad y_\varepsilon=k_\varepsilon*f.$ This causal averaging kernel becomes concentrated near zero as $\varepsilon$ decreases.

Tasks

Derive an IVP for $y_\varepsilon$ and use integration by parts to express $y_\varepsilon-f$ in terms of $f(0)$ and $f'$ .
Prove an explicit error bound containing both the initial transient and a term at most $L\varepsilon$ .
Distinguish uniform convergence on $[0,\infty)$ from uniform convergence on $[t_0,\infty)$ for $t_0>0$ . Show the $L\varepsilon$ bound is sharp when $f(0)=0$ .
For $f(t)=1+t$ , decide separately whether $\varepsilon=1/20$ and $\varepsilon=1/40$ ensure $|y_\varepsilon(t)-f(t)|\le 1/20$ for every $t\ge 1$ . Justify the strict inequalities involved.

Original worksheet page 1: question and worked solution for 4-9-010

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Question 10 – Solution

Strategy. Separate the error due to starting from rest from the error due to variation of the input.

Step 1: Obtain the equation and exact error identity. Differentiation gives $\varepsilon y_\varepsilon'+y_\varepsilon=f$ , $y_\varepsilon(0)=0$ . Integrating the convolution by parts in $u$ gives $y_\varepsilon(t)-f(t)=-e^{-t/\varepsilon}f(0) -\int_0^t e^{-(t-u)/\varepsilon}f'(u)\,du.$

Step 2: Bound the two contributions. Taking absolute values yields $\boxed{|y_\varepsilon-f|\le |f(0)|e^{-t/\varepsilon} +L\varepsilon(1-e^{-t/\varepsilon}).}$ This accounts for the kernel mass missing before time zero as well as the variation over its recent history.

Step 3: Distinguish the convergence statements. If $f(0)=0$ , the error is uniformly at most $L\varepsilon$ on the half-line. For the ramp $f=Lt$ it equals $L\varepsilon(1-e^{-t/\varepsilon})$ in magnitude, whose supremum is $L\varepsilon$ (approached as $t\to\infty$ when $L>0$ ). Thus the constant is sharp. If $f(0)\ne 0$ , the error at zero is always $|f(0)|$ , preventing uniform convergence on $[0,\infty)$ . On $[t_0,\infty)$ it is bounded by $|f(0)|e^{-t_0/\varepsilon}+L\varepsilon\to 0$ for every fixed $t_0>0$ .

Step 4: Test the two proposed averaging scales. For $f=1+t$ , $y_\varepsilon=t+1-\varepsilon-(1-\varepsilon)e^{-t/\varepsilon}$ . For $0<\varepsilon<1$ the error magnitude is $E_\varepsilon=\varepsilon+(1-\varepsilon)e^{-t/\varepsilon}$ . At $\varepsilon=1/20$ this is strictly greater than $1/20$ at every finite time, so that choice fails. At $\varepsilon=1/40$ and $t\ge 1$ , $E_\varepsilon\le\frac 1{40}+\frac{39}{40}e^{-40}<\frac 1{20},$ since $e^{40}>1+40=41>39$ . Thus the smaller choice succeeds. The figure shows the initial error layer and the $1/20$ tolerance.

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 4-9-010

Convolution Integrals — Question 10

Question 10

Question 10 – Solution