Characteristic equation (calculus)

In mathematics, the characteristic equation (or auxiliary equation^[1]) is an algebraic equation of degree $n \,$ on which depends the solutions of a given $n\,$ ^th-order differential equation.^[2] The characteristic equation can only be formed when the differential equation is linear, homogeneous, and has constant coefficients.^[1] Such a differential equation, with $y \,$ as the dependent variable and $a_{n}, a_{n-1}, \ldots , a_{1}, a_{0}$ 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

where $r^{n}, r^{n-1}, \ldots ,r$ are the roots from which the general solution can be formed.^[1]^[3]^[4] This 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]^[4]

Derivation

Starting with a linear homogeneous differential equation with constant coefficients $a_{n}, a_{n-1}, \ldots , a_{1}, a_{0}$ ,

a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_{1}y^{'} + 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.^[3] 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 equate to 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]^[4] For example, if $r \,$ is found to equal to 3, then the general solution will be $y(x) = ce^{3x} \,$ , where $c \,$ is an arbitrary constant.

Formation of the general solution

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

(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 root $r_{2,3,4,5} = -1 \pm i$ . This corresponds to the real-valued general solution with constants $c_{1} , \ldots , c_{5}$ of

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) \,

Solving the characteristic equation for its roots, $r_{1}, \ldots , r_{n}$ , allows one to find the general solution of the differential equation. The roots may be real and/or complex, as well as distinct and/or repeated. If a characteristic equation has parts with distinct real roots, $h \,$ repeated roots, and/or $k \,$ complex roots corresponding to general solutions of $y_{D}(x) \,$ , $y_{R_{1}}(x), \ldots , y_{R_{h}}(x)$ , and $y_{C_{1}}(x), \ldots , y_{C_{k}}(x)$ , respectively, then the general solution to the differential equation is

y(x) = y_{D}(x) + y_{R_{1}}(x) + \cdots + y_{R_{h}}(x) + y_{C_{1}}(x) + \cdots + y_{C_{k}}(x)

Distinct real roots

The superposition principle for linear homogeneous differential equations with constant coefficients says that if $u_{1}, \ldots , u_{n}$ are $n \,$ linearly independent solutions to a particular differential equation, then $c_{1}u_{1} + \cdots + c_{n}u_{n}$ is also a solution for all values $c_{1}, \ldots , c_{n}$ .^[1]^[5] Therefore, if the characteristic equation has distinct real roots $r_{1}, \ldots , r_{n}$ , then a general solution will be of the form

y_{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}(ue^{r_{1}x}) - 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

However, this is the case if and only if $u(x) \,$ is a polynomial of degree $k-1 \,$ , so that $u(x) = c_{1} + c_{2}x + c_{3}x^2 + \cdots + c_{k}x^{k-1}$ .^[4] Since $y(x) = ue^{r_{1}x} \,$ , the part of the general solution corresponding to $r_{1}$ is

y_{R}(x) = e^{r_{1}x}(c_{1} + c_{2}x + \cdots + c_{k}x^{k-1})

Complex roots

If the characteristic equation has complex roots of the form $r_{1} = a + bi$ and $r_{2} = a - bi$ , then the general solution is accordingly $y(x) = c_{1}e^{(a + bi)x} + c_{2}e^{(a - bi)x} \,$ . However, by Euler's formula, which states that $e^{i \theta } = \cos \theta + i \sin \theta \,$ , this solution can be rewritten as follows:

\begin{array}{rcl} 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 ) \\ &=& (c_{1} + c_{2})e^{ax} \cos bx + i(c_{1} - c_{2})e^{ax} \sin bx \end{array}

where $c_{1} \,$ and $c_{2} \,$ are constants that can be complex.^[4]

Note that if $c_{1} = c_{2} = \tfrac{1}{2}$ , then the particular solution $y_{1}(x) = e^{ax} \cos bx \,$ is formed.

Similarly, if $c_{1} = \tfrac{1}{2i}$ and $c_{2} = - \tfrac{1}{2i}$ , then the independent solution formed is $y_{2}(x) = e^{ax} \sin bx \,$ . Thus by the superposition principle for linear homogeneous differential equations with constant coefficients, the part of a differential equation having complex roots $r = a \pm bi \,$ will result in the following general solution: $y_{C}(x) = e^{ax}(c_{1} \cos bx +c_{2} \sin bx ) \,$

References

↑ 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Edwards, C. Henry; Penney, David E. "3". Differential Equations: Computing and Modeling. David Calvis. Upper Saddle River, New Jersey: Pearson Education. pp. 156–170. ISBN 978-0-13-600438-7.
↑ 2.0 2.1 2.2 Smith, David Eugene. "History of Modern Mathematics: Differential Equations". University of South Florida.
↑ 3.0 3.1 Chu, Herman; Shah, Gaurav; Macall, Tom. "Linear Homogeneous Ordinary Differential Equations with Constant Coefficients". eFunda. Retrieved 1 March 2011.
↑ 4.0 4.1 4.2 4.3 4.4 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.