Characteristic equation (calculus)

Template:Short description In mathematics, the characteristic equation (or auxiliary equation) is an algebraic equation of degree Template:Mvar upon which depends the solution of a given Template:Nowrap beginTemplate:Mvarth-Template:Nowrap endorder differential equation or difference equation. The characteristic equation can only be formed when the differential equation is linear and homogeneous, and has constant coefficients. Such a differential equation, with Template:Mvar as the dependent variable, superscript Template:Math denoting n^th-derivative, and Template:Math as constants,

$a_n{y}^{(n)}+a_{n-1}{y}^{(n-1)}+\cdots +a_1{y}^{\prime }+a_{0}y=0,$

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 Template:Math are the roots from which the general solution can be formed. 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 (roots of the characteristic equation) also provide qualitative information about the behavior of the variable whose evolution is described by the dynamic equation. For a differential equation parameterized on time, the variable's evolution is stable if and only if the 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. The qualities of the Euler's characteristic equation were later considered in greater detail by French mathematicians Augustin-Louis Cauchy and Gaspard Monge.

Derivation

Starting with a linear homogeneous differential equation with constant coefficients Template:Math,

$a_n{y}^{(n)}+a_{n-1}{y}^{(n-1)}+\cdots +a_1{y}^{\prime }+a_{0}y=0,$

it can be seen that if Template:Math, each term would be a constant multiple of Template:Math. This results from the fact that the derivative of the exponential function Template:Math is a multiple of itself. Therefore, Template:Math, Template:Math, and Template:Math are all multiples. This suggests that certain values of Template:Mvar will allow multiples of Template:Math to sum to zero, thus solving the homogeneous differential equation. In order to solve for Template:Mvar, one can substitute Template:Math and its derivatives into the differential equation to get

$a_n{r}^n{e}^{rx}+a_{n-1}{r}^{n-1}{e}^{rx}+\cdots +a_{1}r{e}^{rx}+a_0{e}^{rx}=0$

Since Template:Math 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, Template:Mvar, in this characteristic equation, one can find the general solution to the differential equation. For example, if Template:Mvar 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, Template:Math, 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, Template:Mvar repeated roots, or Template:Mvar complex roots corresponding to general solutions of Template:Math, Template:Math, and Template:Math, 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

$y^{(5)}+y^{(4)}-4y^{(3)}-16y^{″}-20y^{\prime }-12y=0$

has the characteristic equation

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

By factoring the characteristic equation intoTemplate:Explain

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

one can see that the solutions for Template:Mvar are the distinct single root Template:Math and the double complex roots Template:Math. This corresponds to the real-valued general solution

$y(x)=c_1{e}^{3x}+e^x(c_2\cos x+c_3\sin x)+x{e}^x(c_4\cos x+c_5\sin x)$

with constants Template:Math.

Distinct real roots

The superposition principle for linear homogeneous says that if Template:Math are Template:Mvar linearly independent solutions to a particular differential equation, then Template:Math is also a solution for all values Template:Math. Therefore, if the characteristic equation has distinct real roots Template:Math, 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 Template:Math that is repeated Template:Mvar times, then it is clear that Template:Math is at least one solution. However, this solution lacks linearly independent solutions from the other Template:Math roots. Since Template:Math has multiplicity Template:Mvar, the differential equation can be factored into

${(\frac{d}{dx}-r_1)}^{k}y=0.$

The fact that Template:Math is one solution allows one to presume that the general solution may be of the form Template:Math, where Template:Math is a function to be determined. Substituting Template:Math gives

$(\frac{d}{dx}-r_1)u{e}^{r_{1}x}=\frac{d}{dx}\left(u{e}^{r_{1}x}\right)-r_{1}u{e}^{r_{1}x}=\frac{d}{dx}(u)e^{r_{1}x}+r_{1}u{e}^{r_{1}x}-r_{1}u{e}^{r_{1}x}=\frac{d}{dx}(u)e^{r_{1}x}$

when Template:Math. By applying this fact Template:Mvar times, it follows that

${(\frac{d}{dx}-r_1)}^{k}u{e}^{r_{1}x}=\frac{d^k}{d{x}^k}(u)e^{r_{1}x}=0.$

By dividing out Template:Math, it can be seen that

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

Therefore, the general case for Template:Math is a polynomial of degree Template:Math, so that Template:Math. Since Template:Math, the part of the general solution corresponding to Template:Math is

$y_{\mathrm{R}}(x)=e^{r_{1}x}(c_1+c_{2}x+\cdots +c_k{x}^{k-1}).$

Complex roots

If a second-order differential equation has a characteristic equation with complex conjugate roots of the form Template:Math and Template:Math, then the general solution is accordingly Template:Math. By Euler's formula, which states that Template:Math, this solution can be rewritten as follows:

$\begin{array}{cc}y(x)& =c_1{e}^{(a+bi)x}+c_2{e}^{(a-bi)x}\\ & =c_1{e}^{ax}(\cos bx+i\sin bx)+c_2{e}^{ax}(\cos bx-i\sin bx)\\ & =(c_1+c_2)e^{ax}\cos bx+i(c_1-c_2)e^{ax}\sin bx\end{array}$

where Template:Math and Template:Math are constants that can be non-real and which depend on the initial conditions. (Indeed, since Template:Math is real, Template:Math must be imaginary or zero and Template:Math must be real, in order for both terms after the last equals sign to be real.)

For example, if Template:Math, then the particular solution Template:Math is formed. Similarly, if Template:Math and Template:Math, then the independent solution formed is Template:Math. Thus by the superposition principle for linear homogeneous differential equations, a second-order differential equation having complex roots Template:Math 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