In mathematics, the exponential response formula (ERF), also known as exponential response and complex replacement, is a method used to find a particular solution of a non-homogeneous linear ordinary differential equation of any order. The exponential response formula is applicable to non-homogeneous linear ordinary differential equations with constant coefficients if the function is polynomial, sinusoidal, exponential or the combination of the three. The general solution of a non-homogeneous linear ordinary differential equation is a superposition of the general solution of the associated homogeneous ODE and a particular solution to the non-homogeneous ODE. Alternative methods for solving ordinary differential equations of higher order are method of undetermined coefficients and method of variation of parameters.
The ERF method of finding a particular solution of a non-homogeneous differential equation is applicable if the non-homogeneous equation is or could be transformed to form
\gamma1t | |
f(t)=B | |
1e |
\gamma2t | |
+B | |
2e |
+ … +Bn
\gammant | |
e |
B,\gamma
f(t)
Complex replacement is a method of converting a non-homogeneous term of equation into a complex exponential function, which makes a given differential equation a complex exponential.
Consider differential equation
y''+y=\cos(t)
To make complex replacement, Euler's formula can be used;
\begin{align} \cos(t)&=\operatorname{Re}(ei)=\operatorname{Re}(\cos(t)+i\sin(t))\\ \sin(t)&=\operatorname{Im}(ei)=\operatorname{Im}(\cos(t)+i\sin(t)) \end{align}
Therefore, given differential equation changes to
z''+z=eit
z(t)
Complex replacement is used for solving differential equations when the non-homogeneous term is expressed in terms of a sinusoidal function or an exponential function, which can be converted into a complex exponential function differentiation and integration. Such complex exponential function is easier to manipulate than the original function.
When the non-homogeneous term is expressed as an exponential function, the ERF method or the undetermined coefficients method can be used to find a particular solution. If non-homogeneous terms can not be transformed to complex exponential function, then the Lagrange method of variation of parameters can be used to find solutions.
The differential equations are important in simulating natural phenomena. In particular, there are numerous phenomena described as high order linear differential equations, for example the spring vibration, LRC circuit, beam deflection, signal processing, control theory and LTI systems with feedback loops.
Mathematically, the system is time-invariant if whenever the input
f(t)
x(t)
f(t-a)
x(t-a)
When the time-invariant system is also linear, it is called a linear time-invariant system (LTI system). Most of these LTI systems are derived from linear differential equations, where the non-homogeneous term is called the input signal and solution of the non-homogeneous equations is called the response signal. If the input signal is given exponentially, the corresponding response signal also changes exponentially.
Considering the following
n
an
dny | |
dtn |
+an-1
dn-1y | |
dtn-1 |
+ … +a1
dy | |
dt |
+a0y=f(t) (1)
and denoting
L=an
n+a | |
D | |
n-1 |
Dn-1+ … +a1D1+a0I,
Dk=
dk | |
dtk |
(k=1,2,\ldots,n),
where
a0,\ldots,an
L
L
P(s)=an
n+a | |
s | |
n-1 |
sn-1+ … +a0
D
L=P(D)
P(D)=an
n+a | |
D | |
n-1 |
Dn-1+ … +a0I
Therefore, the equation (1) can be written as
P(D)y=f(t) (2)
Considering LTI differential equation above, with exponential input
f(t)=Be\gamma
B
\gamma
provide only that
P(\gamma) ≠ 0
Proof: Due to linearity of operator
P(D)
P(D)(yp)=P(D)\left(
Be\gamma | \right)= | |
P(\gamma) |
B | |
P(\gamma) |
P(D)(e\gamma) (3)
On the other hand, since
P(D)\left(e\gamma\right)=P(\gamma)e\gamma,
substituting this into equation (3), produces
P(D)(yp)=P(D)\left(
Be\gamma | \right)= | |
P(\gamma) |
B | |
P(\gamma) |
P(D)\left(e\gamma\right)=
B | |
P(\gamma) |
P(\gamma)e\gamma=Be\gamma.
Therefore,
yp
Thus, the above equation for a particular response
yp
In particular, in case of
P(\gamma)=0
yp=
Bte\gamma | |
P'(\gamma) |
, P'(\gamma) ≠ 0
and is called the resonant response formula.
Let's find the particular solution to 2nd order linear non-homogeneous ODE;
2x''+x'+x=1+2et+e-t\cos(t).
The characteristic polynomial is
P(s)=2s2+s+1
f(t)=1+2et+e-t\cos(t)
f(t)=f1(t)+f2(t)+f3(t),f1(t)=1,f2(t)=2et,
-t | |
f | |
3(t)=e |
\cos(t).
Then, the particular solutions corresponding to
f1(t),f2(t)
f3(t)
First, considering non-homogeneous term,
f1(t)=1
0 ⋅ t | |
f | |
1(t)=1=e |
,\gamma=0
P(\gamma)=P(0)=1 ≠ 0
from the ERF, a particular solution corresponding to
f1(t)
x1p=
f1(t) | |
P(0) |
=
1 | |
1 |
=1
Similarly, a particular solution can be found corresponding to
f2(t)
x2p=
f2(t) | |
P(1) |
=
2et | |
4 |
=
et | |
2 |
.
Let's find a particular solution to DE corresponding to 3rd term;
2x''+x'+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:
2z''+z'+z=e(-1+i)t.
Applying the exponential response formula (ERF), produces
\begin{align} zp&=
e(-1+i)t | \\ &= | |
P(-1+i) |
ie(-1+i)t | |
3 |
&&P(-1+i)=2(-1+i)2+(-1+i)+1=-3i \end{align}
and the real part is
x3p=-
1 | |
3 |
e-t\sin(t).
Therefore, the particular solution of given equation,
xp
xp=x1p+x2p+x3p=1+
et | - | |
2 |
1 | |
3 |
e-t\sin(t).
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
y''+y=et
λ2+1=0,λ=\pmi
f(t)=Be\gamma,B=1,\gamma=1
\gamma=1
yp(t)=Ae\gamma
A
λ2Aeλ+Aeλ=eλ
therefore
A=
1 | |
λ2+1 |
.
The particular solution can be found in form:[1]
yp(t)=Aeλ=
eλ | |
λ2+1 |
.
P(s)=s2+1
f(t)=Be\gamma,B=1,\gamma=1
yp(t)=
eλ | |
P(λ) |
=
eλ | |
λ2+1 |
.
The exponential response formula method was discussed in case of
P(\gamma) ≠ 0
P(\gamma)=0,P'(\gamma) ≠ 0
In the case of
P(\gamma)=P'(\gamma)= … =P(k-1)(\gamma)=0,Pk(\gamma) ≠ 0
Let
P(D)
P(m)
m
P(D)y=Be\gamma
\gamma
P(\gamma) ≠ 0
\gammat | |
y | |
p(t)=\tfrac{Be |
P(\gamma)=0
P'(\gamma) ≠ 0
yp(t)=\tfrac{Bte\gamma
P(\gamma)=P'(\gamma)= … =P(k-1)(\gamma)=0
Pk(\gamma) ≠ 0
Above equation is called generalized exponential response formula.
To find a particular solution of the following ODE;
y'''-3y'+2y=6et.
P(s)=s3-3s+2
By the calculating, we get the following:
P(1)=0,P'(1)=0,P''(1)=6 ≠ 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
y | = | ||||
|
6t2et | |
6 |
=t2et.
Object hanging from a spring with displacement
d
m
d2x | +r | |
dt2 |
dx | |
dt |
+kx=F(t),
where
F(t)
Now, assuming drag is neglected and
F(t)=F0\cos(\omegat)
\omega=\sqrt{\tfrac{k}{m}}
m
d2x | |
dt2 |
+kx=F(t).
Then, a particular solution is
xp=
F0 | |
2\sqrt{km |
Applying complex replacement and the ERF: if
zp
m
d2z | |
dt2 |
+kz=F0ei\omega,
then
xp=\operatorname{Re}(zp)
The characteristic polynomial is
P(s)=ms2+k
\gamma=i\omega
P(\gamma)=0
P'(s)=2ms
P'(\gamma)=P'(i\omega)=2m\omegai ≠ 0
\begin{align} yp&=\operatorname{Re}\left(
F0tei\omega | |
P'(\gamma) |
\right)\\[4pt] &=\operatorname{Re}\left(
F0t(\cos(\omegat)+i\sin(\omegat)) | |
2mi\omega |
\right)\\[4pt] &=\operatorname{Re}\left(
-F0t(i\cos(\omegat)-\sin(\omegat)) | |
2m\omega |
\right)\\[4pt] &=
F0t\sin(\omegat) | \\[4pt] &= | |
2m\omega |
F0 | |
2\sqrt{km |
Considering the electric current flowing through an electric circuit, consisting of a resistance (
R
C
L
E
This system is described by an integral-differential equation found by Kirchhoff called Kirchhoff’s voltage law, relating the resistor
R
C
L
E
I
LI'(t)+RI(t)+ | 1 |
C |
t | |
\int | |
t0 |
I(s)ds=
t | |
\int | |
t0 |
E(s)ds
Differentiating both sides of the above equation, produces the following ODE.
LI''(t)+RI'(t)+ | 1 |
C |
I(t)=E(t)
Now, assuming
E(t)=E0\sin(\omega0t)
\omega0=\sqrt{\tfrac{1}{LC}}
\omega0
E(t)
E(t)=E0\sin(\omega0t)=\operatorname{Im}(E0
i\omega0t | |
e |
)
The characteristic polynomial is
P(s)=
| ||||
Ls |
P(i\omega0)=i\omega0R ≠ 0
\begin{align} Ip&=\operatorname{Im}\left(
| |||||||||||
P(i\omega0) |
\right)\\ &=\operatorname{Im}\left(
| |||||||||||
i\omega0R |
\right)\\ &=\operatorname{Im}\left(
-E0(i\cos(\omega0t)-\sin(\omega0t)) | |
\omega0R |
\right)\\[4pt] &=
-E0\cos(\omega0t) | |
\omega0R |
\end{align}
Considering the general LTI system
P(D)x=Q(D)f(t)
where
f(t)
P(D),Q(D)
P(s) ≠ 0
f(r)=F0\cos(\omegat)
xp(t)=\operatorname{Re}\left(F0
Q(i\omega) | |
P(i\omega) |
ei\omega\right).
Considering the following concepts used in physics and signal processing mainly.
F0
\omega
A=F0|Q(i\omega)/P(i\omega)|
g(\omega)=|Q(i\omega)/P(i\omega)|
\phi=-\operatorname{Arg}(Q(i\omega)/P(i\omega))
\phi/\omega
Q(i\omega)/P(i\omega)