Characteristic equation (calculus)

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

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

will have a characteristic equation of the form

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

whose solutions ${\displaystyle r_{1},r_{2},\ldots ,r_{n}}$ are the roots from which the general solution can be formed. Analogously, a linear difference equation of the form

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

has characteristic equation

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

discussed in more detail at Linear difference equation#Solution of homogeneous case.

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 (absolute value) 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.

Video Characteristic equation (calculus)

Derivation

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

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

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

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

Since ${\displaystyle e^{rx}\,}$ can never equate to zero, it can be divided out, giving the characteristic equation

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

By solving for the roots, ${\displaystyle r\,}$, in this characteristic equation, one can find the general solution to the differential equation. For example, if ${\displaystyle r\,}$ is found to equal to 3, then the general solution will be ${\displaystyle y(x)=ce^{3x}\,}$, where ${\displaystyle c\,}$ is an arbitrary constant.

Maps Characteristic equation (calculus)

Formation of the general solution

Solving the characteristic equation for its roots, ${\displaystyle r_{1},\ldots ,r_{n}}$, 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, ${\displaystyle h\,}$ repeated roots, and/or ${\displaystyle k\,}$ complex roots corresponding to general solutions of ${\displaystyle y_{D}(x)\,}$, ${\displaystyle y_{R_{1}}(x),\ldots ,y_{R_{h}}(x)}$, and ${\displaystyle y_{C_{1}}(x),\ldots ,y_{C_{k}}(x)}$, respectively, then the general solution to the differential equation is

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

Example

The linear homogeneous differential equation with constant coefficients

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

has the characteristic equation

${\displaystyle r^{5}+r^{4}-4r^{3}-16r^{2}-20r-12=0\,}$.

By factoring the characteristic equation into

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

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

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

with constants ${\displaystyle c_{1},\ldots ,c_{5}}$.

Distinct real roots

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

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

Repeated real roots

If the characteristic equation has a root ${\displaystyle r_{1}\,}$ that is repeated ${\displaystyle k\,}$ times, then it is clear that ${\displaystyle y_{p}(x)=c_{1}e^{r_{1}x}}$ is at least one solution. However, this solution lacks linearly independent solutions from the other ${\displaystyle k-1\,}$ roots. Since ${\displaystyle r_{1}\,}$ has multiplicity ${\displaystyle k\,}$, the differential equation can be factored into

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

The fact that ${\displaystyle y_{p}(x)=c_{1}e^{r_{1}x}}$ is one solution allows one to presume that the general solution may be of the form ${\displaystyle y(x)=u(x)e^{r_{1}x}\,}$, where ${\displaystyle u(x)\,}$ is a function to be determined. Substituting ${\displaystyle ue^{r_{1}x}\,}$ gives

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

when ${\displaystyle k=1\,}$. By applying this fact ${\displaystyle k\,}$ times, it follows that

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

By dividing out ${\displaystyle e^{r_{1}x}\,}$, it can be seen that

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

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

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

Complex roots

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

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

where ${\displaystyle c_{1}\,}$ and ${\displaystyle c_{2}\,}$ are constants that can be non-real and which depend on the initial conditions. (Indeed, since ${\displaystyle y(x)}$ is real, ${\displaystyle c_{1}-c_{2}\,}$ must be imaginary or 0 and ${\displaystyle c_{1}+c_{2}\,}$ must be real, in order for both terms after the last equality sign to be real.)

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

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.

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References

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