Characteristic equation (calculus)

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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 a 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:

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

where $c_{{1}}\,$ and $c_{{2}}\,$ are constants that can be 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 $c_{{1}}={\tfrac {1}{2}}i$ and $c_{{2}}=-{\tfrac {1}{2}}i$ , then the independent solution formed is $y_{{2}}(x)=e^{{ax}}\sin bx\,$ . Thus by the superposition principle for linear homogeneous differential equations 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.

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