Difference between revisions of "Reduction of the Order"

From Department of Mathematics at UTSA
Jump to navigation Jump to search
(Created page with "* [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University")
 
Line 1: Line 1:
 +
'''Reduction of order''' is a technique in [[mathematics]] for solving second-order linear [[Ordinary differential equation|ordinary]] [[differential equation]]s.  It is employed when one solution <math>y_1(x)</math> is known and a second [[Linear independence|linearly independent]] solution <math>y_2(x)</math> is desired. The method also applies to ''n''-th order equations. In this case the [[ansatz]] will yield an (''n''−1)-th order equation for <math>v</math>.
 +
 +
== Second-order linear ordinary differential equations==
 +
 +
===An example===
 +
 +
Consider the general, homogeneous, second-order linear constant coefficient ordinary differential equation. (ODE)
 +
 +
:<math> a y''(x) + b y'(x) + c y(x) = 0,</math>
 +
 +
where <math>a, b, c</math> are real non-zero coefficients. Two linearly independent solutions for this ODE can be straightforwardly found using [[characteristic equation (calculus)|characteristic equations]] except for the case when the [[discriminant]], <math>b^2 - 4 a c</math>, vanishes. In this case,
 +
 +
:<math> a y''(x) + b y'(x) + \frac{b^2}{4a} y(x) = 0,</math>
 +
 +
from which only one solution,
 +
 +
:<math>y_1(x) = e^{-\frac{b}{2a} x},</math>
 +
 +
can be found using its characteristic equation.
 +
 +
The method of reduction of order is used to obtain a second linearly independent solution to this differential equation using our one known solution. To find a second solution we take as a guess
 +
 +
:<math>y_2(x) = v(x) y_1(x)</math>
 +
 +
where <math>v(x)</math> is an unknown function to be determined. Since <math>y_2(x)</math> must satisfy the original ODE, we substitute it back in to get
 +
 +
:<math> a \left( v'' y_1 + 2 v' y_1' + v y_1'' \right) + b \left( v' y_1 + v y_1' \right) + \frac{b^2}{4a} v y_1 = 0.</math>
 +
 +
Rearranging this equation in terms of the derivatives of <math>v(x)</math> we get
 +
 +
:<math> \left(a y_1 \right) v'' + \left( 2 a y_1' + b y_1 \right) v' + \left( a y_1'' + b y_1' + \frac{b^2}{4a} y_1 \right) v = 0.</math>
 +
 +
Since we know that <math>y_1(x)</math> is a solution to the original problem, the coefficient of the last term is equal to zero. Furthermore, substituting <math>y_1(x)</math> into the second term's coefficient yields (for that coefficient)
 +
 +
:<math>2 a \left( - \frac{b}{2a} e^{-\frac{b}{2a} x} \right) + b e^{-\frac{b}{2a} x} = \left( -b + b \right) e^{-\frac{b}{2a} x} = 0.</math>
 +
 +
Therefore, we are left with
 +
 +
:<math> a y_1 v'' = 0.</math>
 +
 +
Since <math>a</math> is assumed non-zero and <math>y_1(x)</math> is an [[exponential function]] (and thus always non-zero), we have
 +
 +
:<math> v'' = 0.</math>
 +
 +
This can be integrated twice to yield
 +
 +
:<math> v(x) = c_1 x + c_2</math>
 +
 +
where <math>c_1, c_2</math> are constants of integration. We now can write our second solution as
 +
 +
:<math> y_2(x) = ( c_1 x + c_2 ) y_1(x) = c_1 x y_1(x) + c_2 y_1(x).</math>
 +
 +
Since the second term in <math>y_2(x)</math> is a scalar multiple of the first solution (and thus linearly dependent) we can drop that term, yielding a final solution of
 +
 +
:<math> y_2(x) = x y_1(x) = x e^{-\frac{b}{2 a} x}.</math>
 +
 +
Finally, we can prove that the second solution <math>y_2(x)</math> found via this method is linearly independent of the first solution by calculating the [[Wronskian]]
 +
 +
:<math>W(y_1,y_2)(x) = \begin{vmatrix} y_1 & x y_1 \\ y_1' & y_1 + x y_1' \end{vmatrix} = y_1 ( y_1 + x y_1' ) - x y_1 y_1' = y_1^{2} + x y_1 y_1' - x y_1 y_1' = y_1^{2} = e^{-\frac{b}{a}x} \neq 0.</math>
 +
 +
Thus <math>y_2(x)</math> is the second linearly independent solution we were looking for.
 +
 +
===General method===
 +
 +
Given the general non-homogeneous linear differential equation
 +
 +
:<math>y'' + p(t)y' + q(t)y = r(t)</math>
 +
 +
and a single solution <math>y_1(t)</math> of the homogeneous equation [<math>r(t)=0</math>], let us try a solution of the full non-homogeneous equation in the form:
 +
 +
:<math>y_2 = v(t)y_1(t)</math>
 +
 +
where <math>v(t)</math> is an arbitrary function.  Thus
 +
 +
:<math>y_2' = v'(t)y_1(t) + v(t)y_1'(t)</math>
 +
 +
and
 +
 +
:<math>y_2'' = v''(t)y_1(t) + 2v'(t)y_1'(t) + v(t)y_1''(t).</math>
 +
 +
If these are substituted for <math>y</math>, <math>y'</math>, and <math>y''</math> in the differential equation, then
 +
 +
:<math>y_1(t)\,v'' + (2y_1'(t)+p(t)y_1(t))\,v' + (y_1''(t)+p(t)y_1'(t)+q(t)y_1(t))\,v = r(t).</math>
 +
 +
Since <math>y_1(t)</math> is a solution of the original homogeneous differential equation, <math>y_1''(t)+p(t)y_1'(t)+q(t)y_1(t)=0</math>, so we can reduce to
 +
 +
:<math>y_1(t)\,v'' + (2y_1'(t)+p(t)y_1(t))\,v' = r(t)</math>
 +
 +
which is a first-order differential equation for <math>v'(t)</math> (reduction of order).  Divide by <math>y_1(t)</math>, obtaining
 +
 +
:<math>v''+\left(\frac{2y_1'(t)}{y_1(t)}+p(t)\right)\,v'=\frac{r(t)}{y_1(t)}.</math>
 +
 +
The [[integrating factor]] is <math>\mu(t)=e^{\int(\frac{2y_1'(t)}{y_1(t)}+p(t))dt}=y_1^2(t)e^{\int p(t) dt}</math>.
 +
 +
Multiplying the differential equation by the integrating factor <math>\mu(t)</math>, the equation for <math>v(t)</math> can be reduced to
 +
:<math>\frac{d}{dt}\left(v'(t) y_1^2(t) e^{\int p(t) dt}\right) = y_1(t)r(t)e^{\int p(t) dt}.</math>
 +
 +
After integrating the last equation, <math>v'(t)</math> is found, containing one constant of integration. Then, integrate <math>v'(t)</math> to find the full solution of the original non-homogeneous second-order equation, exhibiting two constants of integration as it should:
 +
 +
:<math>y_2(t) = v(t)y_1(t).</math>
 +
 +
 +
==Resources==
 +
* [https://en.wikipedia.org/wiki/Reduction_of_order Reduction of order], Wikipedia
 
* [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University
 
* [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University

Revision as of 14:50, 20 October 2021

Reduction of order is a technique in mathematics for solving second-order linear ordinary differential equations. It is employed when one solution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1(x)} is known and a second linearly independent solution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2(x)} is desired. The method also applies to n-th order equations. In this case the ansatz will yield an (n−1)-th order equation for .

Second-order linear ordinary differential equations

An example

Consider the general, homogeneous, second-order linear constant coefficient ordinary differential equation. (ODE)

where are real non-zero coefficients. Two linearly independent solutions for this ODE can be straightforwardly found using characteristic equations except for the case when the discriminant, , vanishes. In this case,

from which only one solution,

can be found using its characteristic equation.

The method of reduction of order is used to obtain a second linearly independent solution to this differential equation using our one known solution. To find a second solution we take as a guess

where is an unknown function to be determined. Since must satisfy the original ODE, we substitute it back in to get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \left( v'' y_1 + 2 v' y_1' + v y_1'' \right) + b \left( v' y_1 + v y_1' \right) + \frac{b^2}{4a} v y_1 = 0.}

Rearranging this equation in terms of the derivatives of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v(x)} we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(a y_1 \right) v'' + \left( 2 a y_1' + b y_1 \right) v' + \left( a y_1'' + b y_1' + \frac{b^2}{4a} y_1 \right) v = 0.}

Since we know that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1(x)} is a solution to the original problem, the coefficient of the last term is equal to zero. Furthermore, substituting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1(x)} into the second term's coefficient yields (for that coefficient)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 a \left( - \frac{b}{2a} e^{-\frac{b}{2a} x} \right) + b e^{-\frac{b}{2a} x} = \left( -b + b \right) e^{-\frac{b}{2a} x} = 0.}

Therefore, we are left with

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a y_1 v'' = 0.}

Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} is assumed non-zero and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1(x)} is an exponential function (and thus always non-zero), we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v'' = 0.}

This can be integrated twice to yield

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v(x) = c_1 x + c_2}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_1, c_2} are constants of integration. We now can write our second solution as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2(x) = ( c_1 x + c_2 ) y_1(x) = c_1 x y_1(x) + c_2 y_1(x).}

Since the second term in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2(x)} is a scalar multiple of the first solution (and thus linearly dependent) we can drop that term, yielding a final solution of

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2(x) = x y_1(x) = x e^{-\frac{b}{2 a} x}.}

Finally, we can prove that the second solution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2(x)} found via this method is linearly independent of the first solution by calculating the Wronskian

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(y_1,y_2)(x) = \begin{vmatrix} y_1 & x y_1 \\ y_1' & y_1 + x y_1' \end{vmatrix} = y_1 ( y_1 + x y_1' ) - x y_1 y_1' = y_1^{2} + x y_1 y_1' - x y_1 y_1' = y_1^{2} = e^{-\frac{b}{a}x} \neq 0.}

Thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2(x)} is the second linearly independent solution we were looking for.

General method

Given the general non-homogeneous linear differential equation

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'' + p(t)y' + q(t)y = r(t)}

and a single solution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1(t)} of the homogeneous equation [Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r(t)=0} ], let us try a solution of the full non-homogeneous equation in the form:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2 = v(t)y_1(t)}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v(t)} is an arbitrary function. Thus

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2' = v'(t)y_1(t) + v(t)y_1'(t)}

and

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2'' = v''(t)y_1(t) + 2v'(t)y_1'(t) + v(t)y_1''(t).}

If these are substituted for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y''} in the differential equation, then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1(t)\,v'' + (2y_1'(t)+p(t)y_1(t))\,v' + (y_1''(t)+p(t)y_1'(t)+q(t)y_1(t))\,v = r(t).}

Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1(t)} is a solution of the original homogeneous differential equation, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1''(t)+p(t)y_1'(t)+q(t)y_1(t)=0} , so we can reduce to

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1(t)\,v'' + (2y_1'(t)+p(t)y_1(t))\,v' = r(t)}

which is a first-order differential equation for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v'(t)} (reduction of order). Divide by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1(t)} , obtaining

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v''+\left(\frac{2y_1'(t)}{y_1(t)}+p(t)\right)\,v'=\frac{r(t)}{y_1(t)}.}

The integrating factor is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(t)=e^{\int(\frac{2y_1'(t)}{y_1(t)}+p(t))dt}=y_1^2(t)e^{\int p(t) dt}} .

Multiplying the differential equation by the integrating factor Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(t)} , the equation for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v(t)} can be reduced to

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dt}\left(v'(t) y_1^2(t) e^{\int p(t) dt}\right) = y_1(t)r(t)e^{\int p(t) dt}.}

After integrating the last equation, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v'(t)} is found, containing one constant of integration. Then, integrate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v'(t)} to find the full solution of the original non-homogeneous second-order equation, exhibiting two constants of integration as it should:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2(t) = v(t)y_1(t).}


Resources

Retrieved from "https://mathresearch.utsa.edu/wiki/index.php?title=Reduction_of_the_Order&oldid=2742"