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I'm studying for my PhD qualifying exam and have run into a problem that I cannot figure out how to solve. I'm sure there is a relatively simple trick that I'm forgetting. Can you help me solve this?

Problem Statement

Let $u(t)$ and $v(t)$ be two real functions that obey the differential relations:

$$ \begin{align} \dot{u} &= v - au\\ \dot{v} &= -u + \sin(bt) \end{align} $$

where $a$ and $b$ are positive constants and dot denotes the derivative with respect to $t$. Find the value of $b$ such that the largest value of $v$ is 5 at very large $t$ (i.e., as $t \to \infty$)

My thoughts so far:

It seems to me that transforming this system into the frequency domain will be the best way to find the answer. If we assume zero initial conditions $u(0) = v(0) = 0$, then we can solve for the Laplace transform of $v$:

$$ V(s) = \frac{b(s+a)}{(s^2 + b^2)(s^2 + as + 1)} $$

Attempt 1: I tried splitting this into pieces via partial fractions; things get very messy very quickly so I abandoned that path.

Attempt 2: Final value theorem (FVT) seemed to be the next answer, i.e.

$$ \underset{t \to \infty}{\lim} v(t) = \underset{s \to \infty}{\lim} s V(s) $$

However, FVT only applies if the roots of the denominator of $s V(s)$ all have negative real parts. Clearly, $V(s)$ has poles at $\pm bi$, so that condition is not satisfied...

Thoughts?

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    Eliminating $u$ gives the second order ODE $\ddot{v} + a\dot{v} + v = \sqrt{a^2+b^2}\sin(bt+\phi)$ which one can recognize as a [forced harmonic oscillator](https://en.wikipedia.org/wiki/Harmonic_oscillator#Sinusoidal_driving_force) with $\omega_0 = 1$, $\omega = b$, $\zeta = \frac{a}{2}$ and $F_0/m = \sqrt{a^2+b^2}$ for which the amplitude of oscillations is given by $A = \frac{\sqrt{a^2+b^2}}{\sqrt{a^2b^2 + (1-b^2)^2}}$. – 2017-05-28

2 Answers 2

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We have:

$$ \begin{cases} \text{u}'\left(t\right)=\text{v}\left(t\right)-\text{a}\cdot\text{u}\left(t\right)\\ \text{v}'\left(t\right)=\sin\left(\text{b}t\right)-\text{u}\left(t\right) \end{cases}\tag1 $$

Using, Laplace transform of both sides we get:

$$ \begin{cases} \text{s}\cdot\text{U}\left(\text{s}\right)-\text{u}\left(0\right)=\text{V}\left(\text{s}\right)-\text{a}\cdot\text{U}\left(\text{s}\right)\\ \\ \text{s}\cdot\text{V}\left(\text{s}\right)-\text{v}\left(0\right)=\frac{\text{b}}{\text{b}^2+\text{s}^2}-\text{U}\left(\text{s}\right) \end{cases}\tag2 $$

So, we get:

$$\text{s}\cdot\text{U}\left(\text{s}\right)-\text{u}\left(0\right)=\text{V}\left(\text{s}\right)-\text{a}\cdot\text{U}\left(\text{s}\right)\space\Longleftrightarrow\space\text{U}\left(\text{s}\right)=\frac{\text{V}\left(\text{s}\right)+\text{u}\left(0\right)}{\text{s}+\text{a}}\tag3$$

So:

$$\text{s}\cdot\text{V}\left(\text{s}\right)-\text{v}\left(0\right)=\frac{\text{b}}{\text{b}^2+\text{s}^2}-\frac{\text{V}\left(\text{s}\right)+\text{u}\left(0\right)}{\text{s}+\text{a}}\tag4$$

So, for $\text{V}\left(\text{s}\right)$ we get:

$$\text{V}\left(\text{s}\right)=\frac{\frac{\text{b}}{\text{b}^2+\text{s}^2}-\frac{\text{u}\left(0\right)}{\text{s}+\text{a}}+\text{v}\left(0\right)}{\text{s}+\frac{1}{\text{a}+\text{s}}}\tag5$$

To test if we can use the final value theorem, look if the poles of $\text{s}\cdot\text{V}\left(\text{s}\right)$ lie in the left half-plane:

$$\text{s}\cdot\text{V}\left(\text{s}\right)=\text{s}\cdot\frac{\frac{\text{b}}{\text{b}^2+\text{s}^2}-\frac{\text{u}\left(0\right)}{\text{s}+\text{a}}+\text{v}\left(0\right)}{\text{s}+\frac{1}{\text{a}+\text{s}}}\tag6$$

Using that $\text{U}\left(0\right)=\text{v}\left(0\right)=0$, we get:

$$\text{V}\left(\text{s}\right)=\frac{\frac{\text{b}}{\text{b}^2+\text{s}^2}}{\text{s}+\frac{1}{\text{a}+\text{s}}}\tag7$$

Using the convolution theorem:

$$\text{v}\left(t\right)=\mathcal{L}_\text{s}^{-1}\left[\frac{\text{b}}{\text{b}^2+\text{s}^2}\right]_{\left(t\right)}*\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}+\frac{1}{\text{a}+\text{s}}}\right]_{\left(t\right)}=$$ $$\sin\left(\text{b}t\right)\space*\space e^{-\frac{\text{a}t}{2}}\left\{\cosh\left(\frac{t\sqrt{\text{a}^2-4}}{2}\right)+\frac{\text{a}\sinh\left(\frac{t\sqrt{\text{a}^2-4}}{2}\right)}{\sqrt{\text{a}^2-4}}\right\}\tag8$$

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    First, the Laplace transform of $\sin(bt)$ is $b/(b^2 + s^2)$. Second, even when we include the initial conditions, the Laplace transform of $v(t)$ still has pure imaginary poles and the Final Value Theorem doesn't apply. How do we find the value of $b$ that limits the behavior of $v(t)$ to less than 5 as $t \to \infty$? – 2017-01-12
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    @AndrewCox First, if you react this way maybe you should not ask a question, it's better to be nicer to everyone – 2017-01-12
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    Sorry! I'm not trying to be mean, just concise; please don't read animosity into my comment. Thanks for the updates/edits - However, don't we need to expand and simplify Eq. 6 to find all the poles? If we work out a common denominator for the numerator, we find that $$ s V(s) = s \cdot \frac{ b(s + a) + v(0)(s+a)(s^2 + b^2) - u(0)(s^2 + b^2)}{(s^2+b^2)(s^2 + sa + 1)} $$ There are still two poles on the imaginary axis. – 2017-01-12
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    @AndrewCox It is oke, you're right. I thought about the convolution theorem maybe that will help :) – 2017-01-12
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    Hmmm - that's an interesting approach! – 2017-01-12
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    @AndrewCox I do not now for sure it helps, but let me know for more questions – 2017-01-12
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What about solving the equations exactly?

Let $x = \begin{bmatrix}u\\v\end{bmatrix}$, $A=\begin{bmatrix}-a&1\\0&-1\end{bmatrix}$, and $S = \begin{bmatrix}0\\\sin bt\end{bmatrix}$. Then your ODE is just $\dot{x} = Ax + S$.

Next, diagonalize: we have that $A = PDP^{-1}$ where $P = \begin{bmatrix}-1/(1-a)&1\\1&0\end{bmatrix}$ and $D=\mathrm{diag}(-1, a)$. Let $y = S^{-1}x$. Then the ODE becomes $\dot{y} = Dy + P^{-1}S$. Since the ODE is now separated, it can be solved with $y = y_0 + e^{Dt}\int_0^t e^{-Dt'}P^{-1}S(t')dt'$.

Then chug through some integrals, convert back to $u$, use some trig identities, and answer should appear...