14.10.2: Second-order Differential Equations

Second-order differential equations

We consider the general Second-order differential equation:

\[\tau^{2} \frac{d^{2} Y(t)}{d t^{2}}+2 \zeta \tau \frac{d Y(t)}{d t} + Y(t)=X(t)\]

If you expand the previous Second-order differential equation:

\[ \begin{align} \tau_{1} \tau_{2} \frac{d^{2} Y(t)}{d t^{2}}+\left(\tau_{1}+\tau_{2}\right) \frac{d Y(t)}{d t} + Y(t) &=X(t) \\[4pt] \left(\tau_{1} \frac{d}{d t}+1\right)\left(\tau_{2} \frac{d}{d t}+1\right) (Y(t) &=X(t) \end{align}\]

where:

\[\tau=\sqrt{\tau_{1} \tau_{2}}\]

\[\zeta=\frac{\tau_{1}+\tau_{2}}{2 \sqrt{\tau_{1} \tau_{2}}}\]

Expansion of the differential equation allows you to guess what the shape of the solution (Y(t)) will look like when X(t)=1.

The following rules apply when τ₁ = Re(τ₁)+ i*Im(τ₁) and τ₂ = Re(τ₂)+ i*Im(τ₂):

• If Re(τ₁) and Re(τ₂) are both positive, and there are NO imaginary parts, \(Y(t)\) will exponentially decay (overdamped). Yellow in graph.
• If Re(τ₁) and Re(τ₂) are both positive, and there ARE imaginary parts, \(Y(t)\) will oscillate until it reaches steady state (underdamped). Green in graph.
• If Re(τ₁) and Re(τ₂) are both negative, and there are NO imaginary parts, \(Y(t)\) will exponentially grow (unstable). Pink in graph.
• If Re(τ₁) and Re(τ₂) are both negative, and there ARE imaginary parts, \(Y(t)\) will oscillate and grow exponentially (unstable). Blue in graph.
• If Re(τ₁) and Re(τ₂) are both zero, and there ARE imaginary parts, \(Y(t)\) will oscillate and neither grow nor decay. Purple in graph.
• If τ₁ and τ₂ are both zero, \(Y(t)\) is equal to X(t). Red in graph.

Solution of the General Second-Order System (When X(t)= θ(t))

The solution for the output of the system, \(Y(t)\), can be found in the following section, if we assume that the input, \(X(t)\), is a step function \(θ(t)\). The solution will depend on the value of ζ. If ζ is less than one, \(Y(t)\) will be underdamped. This means that the output will overshoot and oscillate. If ζ is equal to one, \(Y(t)\) will be critically damped. This means that the output will reach the steady state value quickly, without overshoot or oscillation. If ζ is greater than one, \(Y(t)\) will be overdamped. This means that the output will not reach the steady state value as quickly as a critically damped system, but there will be no overshoot or oscillation¹.

Underdamped (ζ<1)

If ζ < 1, the solution is:

\[Y(t) = 1 - \dfrac{1}{\sqrt{1 - \zeta^2}} e^{-\zeta t / \tau} \sin \left( \sqrt{\left(1 - \zeta^2\right)}\dfrac{t}{\tau} + \phi\right)\]

where:

\[\phi=-\tan ^{-1}\left(\frac{\sqrt{1-\zeta^{2}}}{\zeta}\right)\]

The decay ratio (C/A) can be calculated using the following equation:

\[\dfrac{C}{A} = e^{-2 \pi \zeta / \sqrt{1-\zeta^2}}\]

The overshoot (A/B) can be calculated using the following equation:

\[\dfrac{A}{B} = e^{-\pi \zeta / \sqrt{1-\zeta^2}}\]

The period (T) and the frequency (ω) are the following:

\[T=t_{2}-t_{1}=\frac{2 \pi \tau}{\sqrt{1-\zeta^{2}}}\]

\[\omega=\frac{2 \pi}{T}=\frac{\sqrt{1-\zeta^{2}}}{\tau}\]

Critically Damped (ζ=1)

If ζ = 1, the solution is:

\[Y(t)=1-\left(1+\frac{t}{\tau}\right) e^{-t / \tau}\]

Overdamped (ζ>1)

If \(ζ > 1\), the solution is:

\[Y(t) = 1 - \dfrac{1}{\sqrt{1 - \zeta^2}} e^{-\zeta t / \tau} \sinh \left( \sqrt{\left(\zeta^2 - 1\right)}\dfrac{t}{\tau} + \phi\right)\]

where:

\[\phi=-\tanh ^{-1}\left(\frac{\sqrt{\zeta^{2}-1}}{\zeta}\right)\]

Example \(\PageIndex{1}\)

Given:

• A₁ = 1 m²
• A₂ = 1.5 m²
• R₁ = 0.25 s/m²
• R₂ = 0.75 s/m²

where:

• A is the area of the tank
• Q is the volumetric flow rate
• R is the resistance to the flow of the stream exiting the tank
• H is the height of liquid in the tank

develop an expression describing the response of H₂ to Q_in. Determine if the system is over, under or critically damped and determine what the graph of the expression would look like using the complex τ plane above. A diagram of the system is shown below:

Solution

Performing a mass balance on each tank:

\[A_{1} \frac{d H_{1}}{d t}=Q_{i n}-\frac{H_{1}}{R_{1}} \label{1}\]

\[A_{2} \frac{d H_{2}}{d t}=\frac{H_{1}}{R_{1}}-\frac{H_{2}}{R_{2}} \label{2}\]

where the left hand terms account for the accumulation in the tank and the right hand terms account for the flow in the entering and exiting streams

Let τ₁ = R₁A₁ and τ₂ = R₂A₂

Equations \ref{1} and \ref{2} now become

\[\tau_{1} \frac{d H_{1}}{d t}=R_{1} Q_{i n}-H_{1} \label{3}\]

\[\tau_{2} \frac{d H_{2}}{d t}=\frac{R_{2}}{R_{1}} H_{1}-H_{2} \label{4}\]

Put like terms on the same side and factor

\[\left(\tau_{1} \frac{d}{d t}+1\right) H_{1}=R_{1} Q_{i n} \label{5}\]

\[\left(\tau_{2} \frac{d}{d t}+1\right) H_{2}=\frac{R_{2}}{R_{1}} H_{1} \label{6}\]

Apply \(\left(\tau_1 \dfrac{d}{dt} + 1\right)\) operator from Equation \ref{5} to Equation \ref{6}

\[\left(\tau_{1} \frac{d}{d t}+1\right)\left(\tau_{2} \frac{d}{d t}+1\right) H_{2}=\left(\tau_{1} \frac{d}{d t}+1\right) \frac{R_{2}}{R_{1}} H_{1} \label{7}\]

The \(\left(\tau_1 \dfrac{d}{dt} + 1\right) H_1\) term from the left hand portion of Equation \ref{5} can be substituted into the right hand side of Equation \ref{7}

\[\left(\tau_{1} \frac{d}{d t}+1\right)\left(\tau_{2} \frac{d}{d t}+1\right) H_{2}=R_{1} Q_{i n} \frac{R_{2}}{R_{1}}\]

\[\left(\tau_{1} \frac{d}{d t}+1\right)\left(\tau_{2} \frac{d}{d t}+1\right) H_{2}=Q_{i n} R_{2}\]

This expression shows the response of H₂ to Q_in as a second order solution like those pictured above. Here Y(t)=H₂ and X(t)=R₂ Q_in

\[\zeta=\frac{\tau_{1}+\tau_{2}}{2 \sqrt{\tau_{1} \tau_{2}}}=\frac{(0.25 * 1)+(0.75 * 1.5)}{2 \sqrt{(0.25 * 1)(0.75 * 1.5)}}=1.296\]

Overdamped.

Both values of τ are positive real numbers, and the behavior of the graph of the equation can be found on the complex τ plane above.

Example \(\PageIndex{2}\): Analogy to Physics - Spring System

It might be helpful to use a spring system as an analogy for our second order systems.

From Newton's second law of motion,

\[F = ma\]

where:

• F is Force
• m is mass
• a is acceleration

For the spring system, this equation can be written as:

\[F_{\text {applied}}-F_{\text {friction}}-F_{\text {restoring}}=m x^{\prime \prime}\]

where x'' is the acceleration of the car in the x-direction

\[F_{\text {applied}}-f x-k x=m x^{\prime \prime}\]

where:

• k is the spring constant, which relates displacement of the object to the force applied
• f is the frequency of oscillation

\[\frac{m}{k} x^{\prime \prime}+\frac{f}{k} x^{\prime}+x=F_{a p p l i e d}\]

As you can see, this equation resembles the form of a second order equation². The equation can be then thought of as:

\[\mathrm{T}^{2} X^{\prime \prime}+2 \zeta \mathrm{T} X^{\prime}+X=F_{\text {applied }}\]

\[\tau=\sqrt{\frac{m}{k}}\]

\[\zeta = \dfrac{f}{2\sqrt{mk}}\]

Because of this, the spring exhibits behavior like a second order differential equations:

• If \(ζ > 1\) or \(f \gt 2\sqrt{mk}\) it is overdamped
• If \(ζ = 1\) or \(f=2\sqrt{mk}\) it is critically damped
• If \(ζ < 1\) or \(f \lt 2\sqrt{mk}\) it is underdamped

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¹Note that there is no "good" or "bad" associated with overdamped, critically damped, and underdamped. Any of these conditions could be desired (or not desired) depending on the applications.

²While this equation be though of as a mechanical or civil engineering equation, a similar equation exists in electrical engineering with capacitors, inductors, and resistors. See the signals and systems (control systems) section herein.