Exponential response formula

Example

Let's find the particular solution to 2nd order linear non-homogeneous ODE;

${\displaystyle 2x^{″}+x^{\prime }+x=1+2e^t+e^{-t}\cos (t).}$

The characteristic polynomial is ${\displaystyle P(s)=2s^2+s+1}$. Also, the non-homogeneous term, ${\displaystyle f(t)=1+2e^t+e^{-t}\cos (t)}$ can be written as follows

${\displaystyle f(t)=f_1(t)+f_2(t)+f_3(t),f_1(t)=1,f_2(t)=2e^t,f_3(t)=e^{-t}\cos (t).}$

Then, the particular solutions corresponding to ${\displaystyle f_1(t),f_2(t)}$ and ${\displaystyle f_3(t)}$, are found, respectively.

First, considering non-homogeneous term, ${\displaystyle f_1(t)=1}$. In this case, since ${\displaystyle f_1(t)=1=e^{0\cdot t},\gamma =0}$ and ${\displaystyle P(\gamma )=P(0)=1\ne 0}$.

from the ERF, a particular solution corresponding to ${\displaystyle f_1(t)}$ can be found.

${\displaystyle x_{1p}=\frac{f_1(t)}{P(0)}=\frac{1}{1}=1}$.

Similarly, a particular solution can be found corresponding to ${\displaystyle f_2(t)}$.

${\displaystyle x_{2p}=\frac{f_2(t)}{P(1)}=\frac{2e^t}{4}=\frac{e^t}{2}.}$

Let's find a particular solution to DE corresponding to 3rd term;

${\displaystyle 2x^{″}+x^{\prime }+x=e^{-t}\cos (t).}$

In order to do this, equation must be replaced by complex-valued equation, of which it is the real part:

${\displaystyle 2z^{″}+z^{\prime }+z=e^{(-1+i)t}.}$

Applying the exponential response formula (ERF), produces

${\displaystyle \begin{array}{cc}{z}_p& =\frac{e^{(-1+i)t}}{P(-1+i)}\\ & =\frac{i{e}^{(-1+i)t}}{3}& & P(-1+i)=2(-1+i{)}^2+(-1+i)+1=-3i\end{array}}$

and the real part is

${\displaystyle x_{3p}=-\frac{1}{3}{e}^{-t}\sin (t).}$

Therefore, the particular solution of given equation, ${\displaystyle x_p}$ is

${\displaystyle x_p=x_{1p}+x_{2p}+x_{3p}=1+\frac{e^t}{2}-\frac{1}{3}{e}^{-t}\sin (t).}$

Comparison with method of undetermined coefficients

The undetermined coefficients method is a method of appropriately selecting a solution type according to the form of the non-homogeneous term and determining the undetermined constant, so that it satisfies the non-homogeneous equation. On the other hand, the ERF method obtains a special solution based on differential operator. Similarity for both methods is that special solutions of non-homogeneous linear differential equations with constant coefficients are obtained, while form of the equation in consideration is the same in both methods.

For example, finding a particular solution of ${\displaystyle y^{″}+y=e^t}$ with the method of undetermined coefficients requires solving the characteristic equation ${\displaystyle {\lambda }^2+1=0,\lambda =\pm i}$. The non-homogeneous term ${\displaystyle f(t)=B{e}^{\gamma t},B=1,\gamma =1}$ is then considered and since ${\displaystyle \gamma =1}$ is not a characteristic root, it puts a particular solution in form of ${\displaystyle y_p(t)=A{e}^{\gamma t}}$, where ${\displaystyle A}$ is undetermined constant. Substituting into the equation to determine the tentative constant yields

${\displaystyle {\lambda }^{2}A{e}^{\lambda t}+A{e}^{\lambda t}=e^{\lambda t}}$

therefore

${\displaystyle A=\frac{1}{{\lambda }^2+1}.}$

The particular solution can be found in form:

${\displaystyle y_p(t)=A{e}^{\lambda t}=\frac{e^{\lambda t}}{{\lambda }^2+1}.}$

On the other hand, the exponential response formula method requires characteristic polynomial ${\displaystyle P(s)=s^2+1}$ to be found, after which the non-homogeneous terms ${\displaystyle f(t)=B{e}^{\gamma t},B=1,\gamma =1}$ is complex replaced. The particular solution is then found using formula

${\displaystyle y_p(t)=\frac{e^{\lambda t}}{P(\lambda )}=\frac{e^{\lambda t}}{{\lambda }^2+1}.}$

Generalized exponential response formula

The exponential response formula method was discussed in case of ${\displaystyle P(\gamma )\ne 0}$. In the case of ${\displaystyle P(\gamma )=0,P^{\prime }(\gamma )\ne 0}$, the resonant response formula is also considered.

In the case of ${\displaystyle P(\gamma )=P^{\prime }(\gamma )=\cdots =P^{(k-1)}(\gamma )=0,P^k(\gamma )\ne 0}$, we will discuss how the ERF method will be described in this section.

Let ${\displaystyle P(D)}$ be a polynomial operator with constant coefficients, and ${\displaystyle P^{(m)}}$ its ${\displaystyle m}$-th derivative. Then ODE

${\displaystyle P(D)y=B{e}^{\gamma t}}$, where ${\displaystyle \gamma }$ is real or complex.

has the particular solution as following.

• ${\displaystyle P(\gamma )\ne 0}$. In this case, a particular solution will be given by ${\displaystyle y_p(t)=\frac{B{e}^{\gamma t}}{P(\gamma )}}$.(exponent response formula)
• ${\displaystyle P(\gamma )=0}$ but ${\displaystyle P^{\prime }(\gamma )\ne 0}$. In this case, a particular solution will be given by ${\displaystyle y_p(t)=\frac{Bt{e}^{\gamma t}}{P^{\prime }(\gamma )}}$.(resonant response formula)
• ${\displaystyle P(\gamma )=P^{\prime }(\gamma )=\cdots =P^{(k-1)}(\gamma )=0}$ but ${\displaystyle P^k(\gamma )\ne 0}$. In this case, a particular solution will be given by

${\displaystyle y_p(t)=\frac{B{t}^k{e}^{\gamma t}}{P^{(k)}(\gamma )},k=2,\dots ,m}$

Above equation is called generalized exponential response formula.

Example

To find a particular solution of the following ODE;

${\displaystyle y^{‴}-3y^{\prime }+2y=6e^t.}$

the characteristic polynomial is ${\displaystyle P(s)=s^3-3s+2}$.

By the calculating, we get the following:

${\displaystyle P(1)=0,P^{\prime }(1)=0,P^{″}(1)=6\ne 0.}$

Original exponential response formula is not applicable to this case due to division by zero. Therefore, using the generalized exponential response formula and calculated constants, particular solution is

${\displaystyle y_p(t)=\frac{6t^2{e}^t}{P^{″}(1)}=\frac{6t^2{e}^t}{6}=t^2{e}^t.}$

Application examples

Motion of object hanging from a spring

Object hanging from a spring with displacement ${\displaystyle d}$. The force acting is gravity, spring force, air resistance, and any other external forces.

From Hooke’s law, the motion equation of object is expressed as follows;

${\displaystyle m\frac{d^{2}x}{d{t}^2}+r\frac{dx}{dt}+kx=F(t),}$

where ${\displaystyle F(t)}$ is external force.

Now, assuming drag is neglected and ${\displaystyle F(t)=F_0\cos (\omega t)}$, where ${\displaystyle \omega =\sqrt{\frac{k}{m}}}$ (the external force frequency coincides with the natural frequency). Therefore, the harmonic oscillator with sinusoidal forcing term is expressed as following:

${\displaystyle m\frac{d^{2}x}{d{t}^2}+kx=F(t).}$

Then, a particular solution is

${\displaystyle x_p=\frac{F_0}{2\sqrt{km}}t\sin (\omega t).}$

Applying complex replacement and the ERF: if ${\displaystyle z_p}$ is a solution to the complex DE

${\displaystyle m\frac{d^{2}z}{d{t}^2}+kz=F_0{e}^{i\omega t},}$

then ${\displaystyle x_p=\mathrm{Re}(z_p)}$ will be a solution to the given DE.

The characteristic polynomial is ${\displaystyle P(s)=m{s}^2+k}$, and ${\displaystyle \gamma =i\omega }$, so that ${\displaystyle P(\gamma )=0}$. However, since ${\displaystyle P^{\prime }(s)=2ms}$, then ${\displaystyle P^{\prime }(\gamma )=P^{\prime }(i\omega )=2m\omega i\ne 0}$. Thus, the resonant case of the ERF gives

${\displaystyle \begin{array}{cc}{y}_p& =\mathrm{Re}\left(\frac{F_{0}t{e}^{i\omega t}}{P^{\prime }(\gamma )}\right)\\ & =\mathrm{Re}\left(\frac{F_{0}t(\cos (\omega t)+i\sin (\omega t))}{2mi\omega }\right)\\ & =\mathrm{Re}\left(\frac{-F_{0}t(i\cos (\omega t)-\sin (\omega t))}{2m\omega }\right)\\ & =\frac{F_{0}t\sin (\omega t)}{2m\omega }\\ & =\frac{F_0}{2\sqrt{km}}t\sin (\omega t).\end{array}}$

Electrical circuits

Considering the electric current flowing through an electric circuit, consisting of a resistance (${\displaystyle R}$), a capacitor (${\displaystyle C}$), a coil wires (${\displaystyle L}$), and a battery (${\displaystyle E}$), connected in series.

This system is described by an integral-differential equation found by Kirchhoff called Kirchhoff’s voltage law, relating the resistor ${\displaystyle R}$, capacitor ${\displaystyle C}$, inductor ${\displaystyle L}$, battery ${\displaystyle E}$, and the current ${\displaystyle I}$ in a circuit as follows,

${\displaystyle L{I}^{\prime }(t)+RI(t)+\frac{1}{C}{\int }_{t_0}^{t}I(s)ds={\int }_{t_0}^{t}E(s)ds}$

Differentiating both sides of the above equation, produces the following ODE.

${\displaystyle L{I}^{″}(t)+R{I}^{\prime }(t)+\frac{1}{C}I(t)=E(t)}$

Now, assuming ${\displaystyle E(t)=E_0\sin ({\omega }_{0}t)}$, where ${\displaystyle {\omega }_0=\sqrt{\frac{1}{LC}}}$. (${\displaystyle {\omega }_0}$ is called resonance frequency in LRC circuit). Under above assumption, the output (particular solution) corresponding to input ${\displaystyle E(t)}$ can be found. In order to do it, given input can be converted in complex form:

${\displaystyle E(t)=E_0\sin ({\omega }_{0}t)=\mathrm{Im}(E_0{e}^{i{\omega }_{0}t})}$

The characteristic polynomial is ${\displaystyle P(s)=L{s}^2+Rs+\frac{1}{s}}$, where ${\displaystyle P(i{\omega }_0)=i{\omega }_{0}R\ne 0}$. Therefore, from the ERF, a particular solution can be obtained as follows;

${\displaystyle \begin{array}{cc}{I}_p& =\mathrm{Im}\left(\frac{E_0{e}^{i{\omega }_{0}t}}{P(i{\omega }_0)}\right)\\ & =\mathrm{Im}\left(\frac{E_0{e}^{i{\omega }_{0}t}}{i{\omega }_{0}R}\right)\\ & =\mathrm{Im}\left(\frac{-E_0(i\cos ({\omega }_{0}t)-\sin ({\omega }_{0}t))}{{\omega }_{0}R}\right)\\ & =\frac{-E_0\cos ({\omega }_{0}t)}{{\omega }_{0}R}\end{array}}$

Complex gain and phase lag

Considering the general LTI system

${\displaystyle P(D)x=Q(D)f(t)}$

where ${\displaystyle f(t)}$ is the input and ${\displaystyle P(D),Q(D)}$ are given polynomial operators, while assuming that ${\displaystyle P(s)\ne 0}$. In case that ${\displaystyle f(r)=F_0\cos (\omega t)}$, a particular solution to given equation is

${\displaystyle x_p(t)=\mathrm{Re}\left(F_0\frac{Q(i\omega )}{P(i\omega )}{e}^{i\omega t}\right).}$

Considering the following concepts used in physics and signal processing mainly.

• The amplitude of the input is ${\displaystyle F_0}$. This has the same units as the input quantity.
• The angular frequency of the input is ${\displaystyle \omega }$. It has units of radians/time. Often it will be referred to it as frequency, even though technically frequency should have units of cycles/time.
• The amplitude of the response is ${\displaystyle A=F_0|Q(i\omega )/P(i\omega )|}$. This has the same units as the response quantity.
• The gain is ${\displaystyle g(\omega )=|Q(i\omega )/P(i\omega )|}$. The gain is the factor that the input amplitude is multiplied by to get the amplitude of the response. It has the units needed to convert input units to output units.
• The phase lag is ${\displaystyle \varphi =-\mathrm{Arg}(Q(i\omega )/P(i\omega ))}$. The phase lag has units of radians, i.e. it’s dimensionless.
• The time lag is ${\displaystyle \varphi /\omega }$. This has units of time. It is the time that peak of the output lags behind that of the input.
• The complex gain is ${\displaystyle Q(i\omega )/P(i\omega )}$. This is the factor that the complex input is multiplied by to get the complex output.