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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

  1. ^ a b c d e f g 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. ^ a b c Smith, David Eugene. "History of Modern Mathematics: Differential Equations". University of South Florida. 
  3. ^ a b Chu, Herman; Shah, Gaurav; Macall, Tom. "Linear Homogeneous Ordinary Differential Equations with Constant Coefficients". eFunda. Retrieved 1 March 2011. 
  4. ^ a b c d e Cohen, Abraham (1906). An Elementary Treatise on Differential Equations. D. C. Heath and Company. 
  5. ^ Dawkins, Paul. "Differential Equation Terminology". Paul's Online Math Notes. Retrieved 2 March 2011.