Characteristic equation (calculus)

In

dependent variable, superscript

(n)

denoting n^th-derivative

, and

a n, a n - 1, ..., a 1, a 0

as constants,

a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+\cdots +a_{1}y'+a_{0}y=0,

will have a characteristic equation of the form

a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots +a_{1}r+a_{0}=0

whose solutions $r 1, r 2, ..., r n$ are the roots from which the

general solution can be formed.^[1]^[5]^[6]

Analogously, a linear difference equation of the form

y_{t+n}=b_{1}y_{t+n-1}+\cdots +b_{n}y_{t}

has characteristic equation

r^{n}-b_{1}r^{n-1}-\cdots -b_{n}=0,

discussed in more detail at Linear recurrence with constant coefficients.

The characteristic roots (

real part of each root is negative. For difference equations, there is stability if and only if the modulus of each root is less than 1. For both types of equation, persistent fluctuations occur if there is at least one pair of complex

roots.

The method of integrating linear ordinary differential equations with constant coefficients was discovered by Leonhard Euler, who found that the solutions depended on an algebraic 'characteristic' equation.^[2] The qualities of the Euler's characteristic equation were later considered in greater detail by French mathematicians Augustin-Louis Cauchy and Gaspard Monge.^[2]^[6]

Derivation

Starting with a linear homogeneous differential equation with constant coefficients $a n, a n - 1, ..., a 1, a 0$ ,

a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+\cdots +a_{1}y^{\prime }+a_{0}y=0,

it can be seen that if $y (x) = e rx$ , each term would be a constant multiple of $e rx$ . This results from the fact that the derivative of the exponential function $e rx$ is a multiple of itself. Therefore, $y' = re rx$ , $y ″ = r 2 e rx$ , and $y (n) = r n e rx$ are all multiples. This suggests that certain values of $r$ will allow multiples of $e rx$ to sum to zero, thus solving the homogeneous differential equation.^[5] In order to solve for $r$ , one can substitute $y = e rx$ and its derivatives into the differential equation to get

a_{n}r^{n}e^{rx}+a_{n-1}r^{n-1}e^{rx}+\cdots +a_{1}re^{rx}+a_{0}e^{rx}=0

Since $e rx$ can never equal zero, it can be divided out, giving the characteristic equation

a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots +a_{1}r+a_{0}=0.

By solving for the roots, $r$ , in this characteristic equation, one can find the general solution to the differential equation.^[1]^[6] For example, if $r$ has roots equal to 3, 11, and 40, then the general solution will be $y(x)=c_{1}e^{3x}+c_{2}e^{11x}+c_{3}e^{40x}$ , where $c_{1}$ , $c_{2}$ , and $c_{3}$ are arbitrary constants which need to be determined by the boundary and/or initial conditions.

Formation of the general solution

Solving the characteristic equation for its roots, $r 1, ..., r n$ , allows one to find the general solution of the differential equation. The roots may be real or complex, as well as distinct or repeated. If a characteristic equation has parts with distinct real roots, $h$ repeated roots, or $k$ complex roots corresponding to general solutions of $y D (x)$ , $y R 1 (x), ..., y R h (x)$ , and $y C 1 (x), ..., y C k (x)$ , respectively, then the general solution to the differential equation is

y(x)=y_{\mathrm {D} }(x)+y_{\mathrm {R} _{1}}(x)+\cdots +y_{\mathrm {R} _{h}}(x)+y_{\mathrm {C} _{1}}(x)+\cdots +y_{\mathrm {C} _{k}}(x)

Example

The linear homogeneous differential equation with constant coefficients

y^{(5)}+y^{(4)}-4y^{(3)}-16y''-20y'-12y=0

has the characteristic equation

r^{5}+r^{4}-4r^{3}-16r^{2}-20r-12=0

By factoring the characteristic equation into^{[further explanation needed]}

(r-3)(r^{2}+2r+2)^{2}=0

one can see that the solutions for $r$ are the distinct single root $r 1 = 3$ and the double complex roots $r 2,3,4,5 = 1 \pm i$ . This corresponds to the real-valued general solution

y(x)=c_{1}e^{3x}+e^{x}(c_{2}\cos x+c_{3}\sin x)+xe^{x}(c_{4}\cos x+c_{5}\sin x)

with constants $c 1, ..., c 5$ .

Distinct real roots

The

linearly independent solutions to a particular differential equation, then $c 1 u 1 + \dots + c n u n$ is also a solution for all values $c 1, ..., c n$ .^[1]^[7]

Therefore, if the characteristic equation has distinct real roots $r 1, ..., r n$ , then a general solution will be of the form

$y_{\mathrm {D} }(x)=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+\cdots +c_{n}e^{r_{n}x}$

Repeated real roots

If the characteristic equation has a root $r 1$ that is repeated $k$ times, then it is clear that $y p (x) = c 1 e r 1 x$ is at least one solution.[1] However, this solution lacks linearly independent solutions from the other $k - 1$ roots. Since $r 1$ has multiplicity $k$ , the differential equation can be factored into^[1]

$\left({\frac {d}{dx}}-r_{1}\right)^{k}y=0.$

The fact that $y p (x) = c 1 e r 1 x$ is one solution allows one to presume that the general solution may be of the form $y (x) = u (x) e r 1 x$ , where $u (x)$ is a function to be determined. Substituting $ue r 1 x$ gives

$\left({\frac {d}{dx}}-r_{1}\right)\!ue^{r_{1}x}={\frac {d}{dx}}\left(ue^{r_{1}x}\right)-r_{1}ue^{r_{1}x}={\frac {d}{dx}}(u)e^{r_{1}x}+r_{1}ue^{r_{1}x}-r_{1}ue^{r_{1}x}={\frac {d}{dx}}(u)e^{r_{1}x}$

when $k = 1$ . By applying this fact $k$ times, it follows that

$\left({\frac {d}{dx}}-r_{1}\right)^{k}ue^{r_{1}x}={\frac {d^{k}}{dx^{k}}}(u)e^{r_{1}x}=0.$

By dividing out $e r 1 x$ , it can be seen that

${\frac {d^{k}}{dx^{k}}}(u)=u^{(k)}=0.$

Therefore, the general case for $u (x)$ is a polynomial of degree $k - 1$ , so that $u (x) = c 1 + c 2 x + c 3 x 2 + \dots + c k x k -1$ .^[6] Since $y (x) = ue r 1 x$ , the part of the general solution corresponding to $r 1$ is

$y_{\mathrm {R} }(x)=e^{r_{1}x}\!\left(c_{1}+c_{2}x+\cdots +c_{k}x^{k-1}\right).$

Complex roots

If a second-order differential equation has a characteristic equation with complex conjugate roots of the form $r 1 = a + bi$ and $r 2 = a - bi$ , then the general solution is accordingly $y (x) = c 1 e (a + bi) x + c 2 e (a - bi) x$ . By Euler's formula, which states that $e iθ = cos θ + i sin θ$ , this solution can be rewritten as follows:

${\begin{aligned}y(x)&=c_{1}e^{(a+bi)x}+c_{2}e^{(a-bi)x}\\&=c_{1}e^{ax}(\cos bx+i\sin bx)+c_{2}e^{ax}(\cos bx-i\sin bx)\\&=\left(c_{1}+c_{2}\right)e^{ax}\cos bx+i(c_{1}-c_{2})e^{ax}\sin bx\end{aligned}}$

where $c 1$ and $c 2$ are constants that can be non-real and which depend on the initial conditions.^[6] (Indeed, since $y (x)$ is real, $c 1 - c 2$ must be imaginary or zero and $c 1 + c 2$ must be real, in order for both terms after the last equals sign to be real.)
For example, if $c1 = c2 = .mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num{display:block;line-height:1em;margin:0.0em 0.1em;border-bottom:1px solid}.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0.1em 0.1em}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}⁠1/2⁠$ , then the particular solution $y 1 (x) = e ax cos bx$ is formed. Similarly, if $c 1 = ⁠ 1 / 2 i ⁠$ and $c 2 = - ⁠ 1 / 2 i ⁠$ , then the independent solution formed is $y 2 (x) = e ax sin bx$ . Thus by the superposition principle for linear homogeneous differential equations, a second-order differential equation having complex roots $r = a \pm bi$ will result in the following general solution:

$y_{\mathrm {C} }(x)=e^{ax}(C_{1}\cos bx+C_{2}\sin bx)$

This analysis also applies to the parts of the solutions of a higher-order differential equation whose characteristic equation involves non-real complex conjugate roots.

See also

Characteristic polynomial

References

^
ISBN 978-0-13-600438-7
.

^ ^a ^b ^c Smith, David Eugene. "History of Modern Mathematics: Differential Equations". University of South Florida.

^ Baumol, William J. (1970). Economic Dynamics (3rd ed.). p. 172.

ISBN 9780070107809
.

^ ^a ^b Chu, Herman; Shah, Gaurav; Macall, Tom. "Linear Homogeneous Ordinary Differential Equations with Constant Coefficients". eFunda. Retrieved 1 March 2011.

^ ^a ^b ^c ^d ^e Cohen, Abraham (1906). An Elementary Treatise on Differential Equations. D. C. Heath and Company.

^ Dawkins, Paul. "Differential Equation Terminology". Paul's Online Math Notes. Retrieved 2 March 2011.

Retrieved from "https://en.wikipedia.org/w/index.php?title=Characteristic_equation_(calculus)&oldid=1290949036"

[edwards-1] 
ISBN 978-0-13-600438-7
.

[smith-2] Smith, David Eugene. "History of Modern Mathematics: Differential Equations". University of South Florida.

[3] Baumol, William J. (1970). Economic Dynamics (3rd ed.). p. 172.

[4] ISBN 9780070107809
.

[eFunda-5] Chu, Herman; Shah, Gaurav; Macall, Tom. "Linear Homogeneous Ordinary Differential Equations with Constant Coefficients". eFunda. Retrieved 1 March 2011.

[cohen-6] Cohen, Abraham (1906). An Elementary Treatise on Differential Equations. D. C. Heath and Company.

[dawkins-7] Dawkins, Paul. "Differential Equation Terminology". Paul's Online Math Notes. Retrieved 2 March 2011.

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