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Return to Part IV of the course APMA0330
We recall the definition of a root multiplicity. A real or complex number α is called a root of multiplicity k of the
polynomial p(x) if there is a polynomial s(x) such that \( s(\alpha ) \ne 0 \)
and \( p(x) = \left( x- \alpha \right)^k s(x) . \) If k=1, then α is called a simple root.
If k ≥ 2, then α is called a multiple root.
Reduction of order for constant coefficient equations
We start with a constant coefficient linear homogeneous differential equation:
where \( \texttt{D}^0 \) is the identity operator, denoted by I. Then Eq.\eqref{EqReduction.1} can be rewritten in compact form
\begin{equation} \label{EqReduction.3}
L \left[ \texttt{D} \right] y = 0 .
\end{equation}
Suppose that the characteristic polynomial \( L \left( \lambda \right) = a_n \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_0 , \) corresponding to the operator L, has a repeated root r of multiplicity m.
That is, the characteristic polynomial can be factored \( L \left( \lambda \right) = \left( \lambda - r \right)^m M(\lambda ) , \) where M(λ) is a polynomial of degree n - m such that M(r) ≠ 0.
For example, the second order constant coefficient linear differential operator
\( L \left[ \texttt{D} \right] = a\,\texttt{D}^2 + b\,\texttt{D} +c \) has a repeated factor
if and only if the corresponding characteristic equation \( a\,\lambda^2 + b\,\lambda + c =0 \)
has a double root:
\[
\lambda_1 = \lambda_2 = -b/(2a) .
\]
In other words, the quadratic polynomial can be factored \( a\,\lambda^2 + b\,\lambda + c = a \left( \lambda - \lambda_1 \right)^2 \)
if and only if its discriminant \( b^2 - 4ac \) is zero. In this case we have only one solution of exponential form:
To find another linearly independent solution to the homogeneous equation \eqref{EqReduction.5} , we use the method of reduction of
order, credited to Jacob Bernoulli (1655--1705). Setting
The latter can be divided by nonzero term \( a\,y_1 (x) = a\,e^{-bx/(2a)} . \) This yields
\( v'' (x) =0, \) and after integration, we obtain
\( v (x) = C_1 + C_2 x, \) with arbitrary constants C_{1} and C_{2}.
Finally, substituting for v(x), we define the general solution
Therefore, these two functions form a fundamental set of solutions whenever γ is. ■
This method can be extended for arbitrary case when a linear constant coefficient differential operator has
a multiple root \( L \left[ \texttt{D} \right] = \left( \texttt{D} - \gamma \right)^m M \left[ \texttt{D} \right] \) of multiplicity m.
Then to such multiple correspond m linearly independent solutions:
The strict derivation of the above statement is made in annihilation section.
Theorem 1:
Let \( a_0 , a_1 , \ldots , a_n \) be n+1 real
(or complex) numbers with 𝑎_{n} ≠ 0, and y(x) be a n times
continuously differentiable function on some interval | 𝑎, b|. Then y(x) is a solution of the n-th order
linear differential equation with constant coefficients
\[
L \left[ \texttt{D} \right] y \equiv a_n \texttt{D}^n y + a_{n-1} \texttt{D}^{n-1} y + \cdots
+ a_1 \texttt{D} \, y+ a_0 \, y =0 , \qquad \texttt{D} = {\text d}/{\text d} x ,
\]
where \( \lambda_1 , \lambda_2 , \ldots , \lambda_m \) are distinct roots of the characteristic polynomial
\( \sum_{k=1}^n a_k \lambda^k \) with multiplicities \( m_1 , m_2 , \ldots , m_m , \)
respectively, and \( P_k (x) \) is a polynomial of degree \( m_k -1 . \)
⧫
For illustration,
we consider a particular case of second order differential equation---the motion for
undriven damped harmonic oscillator (coefficients μ and ω_{0} > 0 are assumed constants):
\[
\ddot{y} + 2\mu\,\dot{y} + \omega_0^2 y =0 ,
\]
subject to the initial conditions
\[
y(0) = y_0 , \qquad \dot{y}(0) = v_0 .
\]
We show that its solution can be obtained from the solution
are two distinct roots of the characteristic equation \( \lambda^2 + 2\mu\,\lambda + \omega_0^2 =0 . \)
Let assume, for simplicity, that \( \eta = \left( \mu^2 - \omega_0^2 \right)^{1/2} \ge 0 , \) that is
\( \mu \ge \omega_0 . \) Then rewriting the solution in the equivalent form:
Example:
Let us consider the differential equation
\[
y'' -4\,y' +4\,y =0 .
\]
The characteristic equation \( \lambda^2 - 4\,\lambda +4 = \left( \lambda -2 \right)^2 =0 \)
has a double root \( \lambda =2 . \) Hence we get one exponential solution
\( y_1 (x) = e^{2x} \) and another one is a multiple of the latter:
\( y_2 (x) = x\,e^{2x} . \) So the general solution becomes
The characteristic polynomial \( \lambda^2 + 6\,\lambda +9 = = \left( \lambda +3 \right)^2 \)
has one double root \( \lambda =-3 . \) Therefore, the general solution of the given homogeneous differential equation becomes
\[
y(x) = C_1 e^{-3x} + C_2 x\, e^{-3x} .
\]
To satisfy the initial conditions, we have to choose arbitrary constants that are solutions of the system of equations:
We generalized considered above two examples by introducing the homogeneous
differential equation
\[
y'' -2\alpha\,y' + \alpha^2 y =0 ,
\]
which has a solution \( y(x) = e^{\alpha\, x} . \)
Letting, \( y(x) = e^{\alpha\, x} u(x) , \) we
obtain its derivatives
\( y' =\alpha\, y + e^{\alpha\, x} u' (x) , \) and
\( y'' =\alpha^2\, y + 2\,e^{\alpha\, x} u (x) + e^{\alpha\, x} u'' (x) . \)
Substituting into the equation and cancelling the common exponential multiple
yields
ntegrating twice gives\( u = A + B\,x . \) Hence,
the second linearly independent solution becomes
\[
y(x) = x\, e^{\alpha\, x} .
\]
Example:
Let us consider the fourth order linear differential operator:
\[
L \left[ \texttt{D} \right] y \equiv \texttt{D}^4 y - 8\, \texttt{D}^{3} y + 18\,
\texttt{D}^2 \, y -27 \, y =0 , \qquad \texttt{D} = {\text d}/{\text d} x .
\]
has one simple root %lambda; = −1 ≠ 0 and a triple root λ = 3.
Therefore, the general solution to the fourth order differential equation \( L \left[ \texttt{D} \right] y =0 \)
is
where C_{k}, \( k=1,2,3,4 ,\) are arbitrary constants.
■
Variable coefficient equations
The reduction of order technique, which applies to
arbitrary linear differential equations, allows us to go
beyond equations with constant coefficients,
provided that we
already know one solution. For sake of clarity, we start with a second order linear differential equation with
variable coefficients:
Here \( W[x] = W\left( x_0 \right) \exp \left\{ -\int a_1 (x)/a_2 (x)\,{\text d}x \right\} \) is the Wronskian (arbitrary multiplier is ommited) of the operator \eqref{EqReduction.8}.
⧫
Example 4:
Consider the Chebyshev equation
\[
\left( 1 - x^2 \right) y'' -x\,y' +y =0 .
\]
It is easy to veryfy that this equation has a linear function solution φ(x) = x. However, it is very difficult to guess a solution for the adjoint differential equation
To get rid of a trouble maker, the leading coefficient 𝑎_{2} in Eq.\eqref{EqReduction.7}, we divide every term in this equation to obtain the second order differential equation in normal form:
\[
y'' +p(x)\,y' +q(x)\,y =0 ,
\]
where p(x) and q(x) are some continuous functions on some interval |a,b|. Suppose that we know one
its solution \( y = y_1 (x) \ne 0. \) This means that
Then we seek another linearly independent solution in the form \( y(x) = v(x)\, y_1 (x) , \)
where unknown function v(x) s determined upon substitution of this form into the given differential equation.
First, we use the product rule to obtain
This is a second order differential equation where dependent function is missing. If we set \( u= v' , \)
we reduce it to a first order differential equation:
\[
\ln u = - 2\int \frac{y'_1 (x)}{y_1 (x)} \,{\text d}x - \int p(x) \, {\text d} x = \ln y_1^{-2} - \int p(x) \, {\text d} x .
\]
This allows us to determine u(x) explicitly:
\[
v' = u = y_1^{-2} \, \exp \left\{ - \int p(x) \, {\text d} x \right\} ,
\]
where we dropped an arbitrary constant because we are after another linearly independent solution. Next integration yields
\[
v = \int y_1^{-2} \, \exp \left\{ - \int p(x) \, {\text d} x \right\} {\text d} x \qquad\Longrightarrow \qquad
y_2 (x) = y_1 (x) \, \int y_1^{-2} \, \exp \left\{ - \int p(x) \, {\text d} x \right\} {\text d} x .
\]
Theorem 3:
Let φ_{1}(x be a solution of the second order homogeneous equation \( a_2 (x)\,y'' + a_1 (x)\,y' + a_0 (x)\,y =0 . \) Then the second linearly independent solution to Eq.\eqref{EqReduction.7} is
where ψ_{1}(x) is a solution of the adjoint homogeneous equation \( L^{\ast} \left[ x, \texttt{D} \right] \psi = 0 , \) and \( W[x] = \exp \left\{ -\int a_1 (x)/a_2 (x)\,{\text d}x \right\} \) is the Wronskian (arbitrary multiplier is ommited) of the operator \eqref{EqReduction.8}.
⧫
Example 5:
Find a second linearly independent solution to the
differential equation
Of course, the constant multiple (1/2) could be dropped.
■
One solution of \( x^2 y'' - \left( x^2 + 2x \right) y' + \left( x+2 \right) y = 0 \) is y(x) = x, Use the method of reduction of order to find a 2nd solution and hence write down the general solution.
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