KindiakMath

Breaking the Rules

Problem 1. For any k \in \mathbb R, evaluate \displaystyle \frac 1{\mathcal D - k}(U(t)).

(Click for Solution)

Solution. We require f such that

(\mathcal D-k)(f(t)) = U(t).

If k = 0, then we leave it as an exercise to verify that f(t) := t U(t) works.

Suppose k \neq 0. Expanding the left-hand side,

\mathcal D(f(t)) - k f(t) = U(t).

We use the method of integrating factors, setting P(t) = -k and Q(t) = U(t). The integrating factor

I(t) = e^{\int P(t)\, \mathrm dt} = e^{\int -k\, \mathrm dt} = e^{-kt}

yields the general solution

\displaystyle \begin{aligned} f(t) \cdot I(t) &= \int I(t) \cdot Q(t)\, \mathrm dt \\ f(t) e^{-kt} &= \int e^{-kt} \cdot U(t)\, \mathrm dt. \end{aligned}

Plugging in the limits from 0 to x,

\begin{aligned} f(x) e^{-kx} &= f(0) + \int_0^x e^{-kt} \cdot U(t)\, \mathrm dt.\end{aligned}

Therefore, we must have

\displaystyle f(x) = f(0) e^{kx} + e^{kx} \int_0^x e^{-kt} \cdot U(t)\, \mathrm dt.

Setting f(0) = 0 and relabeling,

\displaystyle f(t) = e^{kt} \int_0^t e^{-k v} \cdot U(v)\, \mathrm dv.

To check that this solution works,

\displaystyle \begin{aligned}\mathcal D(f(t)) &= \mathcal D\left( e^{kt} \int_0^t e^{-k v} \cdot U(v)\, \mathrm dv \right) \\ &= ke^{kt} \int_0^t e^{-k v} \cdot U(v)\, \mathrm dv + e^{kt} \cdot e^{-k t} \cdot U(t) \\ &= k f(t) + U(t) \end{aligned}

Therefore,

(\mathcal D - k)(f(t)) = (k f(t) + U(t)) - kf(t) = U(t),

as required. Finally, evaluating the integral for t \neq 0,

\displaystyle \begin{aligned} \frac 1{\mathcal D - k}(U(t)) = f(t) &= U(t) \cdot e^{kt} \int_0^t e^{-k v}\, \mathrm dv \\ &= U(t) \cdot e^{kt} \cdot \frac 1k \cdot (1 - e^{-kt} ) \\ &= \frac {e^{kt} - 1}k \cdot U(t). \end{aligned}

Fixing t, since (e^{kt}-1)/k \to t as k \to 0, we will conveniently define

\displaystyle \frac 1{\mathcal D}(U(t)) := \lim_{k \to 0} \frac 1{\mathcal D - k}(U(t)),\quad t \neq 0.

Problem 2. For any \alpha, \beta \in \mathbb R, evaluate \displaystyle \frac 1{(\mathcal D - \alpha)(\mathcal D - \beta)} (U(t)).

(Click for Solution)

Solution. Assume \alpha, \beta \neq 0 and \alpha \neq \beta. Applying the result of Problem 1,

\displaystyle \begin{aligned}\frac 1{(\mathcal D - \alpha)(\mathcal D - \beta)} (U(t)) &= \frac 1{\mathcal D-\alpha} \left( \frac{e^{\beta t} - 1}{\beta} \cdot U(t) \right) \\ &= \frac 1{\beta} \cdot \frac 1{\mathcal D-\alpha} (e^{\beta t} \cdot U(t) - U(t)) \\ &= \frac 1{\beta} \cdot \left( \frac 1{\mathcal D-\alpha} (e^{\beta t} \cdot U(t) ) - \frac 1{\mathcal D-\alpha} ( U(t) ) \right) \\ &= \frac 1{\beta} \cdot \left( e^{\beta t} \cdot \frac 1{\mathcal D-(\alpha - \beta)} (U(t) ) - \frac 1{\mathcal D-\alpha} ( U(t) ) \right) \\ &= \frac 1{\beta} \cdot \left( e^{\beta t} \cdot \frac {e^{(\alpha - \beta)t} - 1}{\alpha - \beta} \cdot U(t) - \frac {e^{\alpha t} - 1}{\alpha } \cdot U(t) \right) \\ &= \frac 1{\beta} \cdot \left( \frac {e^{\alpha t} - e^{\beta t}}{\alpha - \beta} - \frac {e^{\alpha t} - 1}{\alpha } \right) \cdot U(t) \\ &= \frac {\alpha(e^{\alpha t} - e^{\beta t}) - (e^{\alpha t} - 1)(\alpha - \beta)}{\alpha \beta (\alpha - \beta)} \cdot U(t) \\ &= \frac { - \alpha e^{\beta t} + (\alpha - \beta + \beta e^{\alpha t} )}{\alpha \beta (\alpha - \beta)} \cdot U(t) \\ &= \left( \frac 1{\alpha \beta} + \frac {1}{\alpha(\alpha -\beta)} e^{\alpha t} - \frac 1{\beta(\alpha - \beta)} e^{\beta t} \right) \cdot U(t) \end{aligned}

The other cases are left as an exercise.

—Joel Kindiak, 22 Apr 25, 2206H

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