Justifying the Method of Undetermined Coefficients

[MarkFL]

The annihilator method can be used to derived the entries in the following table for the method of undetermined coefficients, familiar to all students of ordinary differential equations:

Type	$g(x)$	$y_p(x)$
(I)	$p_n(x)=a_nx^n+\cdots+a_1x+a_0$	$x^sP_n(x)=x^s\left(A_nx^n+\cdots+A_1x+A_0 \right)$
(II)	$ae^{\alpha x}$	$x^sAe^{\alpha x}$
(III)	$a\cos(\beta x)+b\sin(\beta x)$	$x^s\left(A\cos(\beta x)+B\sin(\beta x) \right)$
(IV)	$p_n(x)e^{\alpha x}$	$x^sP_n(x)e^{\alpha x}$
(V)	$p_n(x)\cos(\beta x)+q_m\sin(\beta x)$ where $q_m(x)=b_mx^m+\cdots+b_1x+b_0$	$x^s\left(P_N(x)\cos(\beta x)+Q_N(x)\sin(\beta x) \right)$ where $Q_N(x)=B_Nx^N+\cdots+B_1x+B_0$ and $N=\max(n,m)$
(VI)	$ae^{\alpha x}\cos(\beta x)+be^{\alpha x}\sin(\beta x)$	$x^s\left(Ae^{\alpha x}\cos(\beta x)+Be^{\alpha x}\sin(\beta x) \right)$
(VII)	$p_ne^{\alpha x}\cos(\beta x)+q_me^{\alpha x}\sin(\beta x)$	$x^se^{\alpha x}\left(P_N(x)\cos(\beta x)+Q_N(x)\sin(\beta x) \right)$ where $N=\max(n,m)$

Notes:

The non-negative integer $s$ is chosen to be the smallest integer so that no term in the particular solution $y_p(x)$ is a solution to the corresponding homogeneous solution $L[y](x)=0$.

$P_n(x)$ must include all its terms even if $p_n(x)$ has some terms that are zero.

To show this, it suffices to work with type VII functions--that is, functions of the form:

(1) \(\displaystyle g(x)=p_n(x)e^{\alpha x}\cos(\beta x)+q_m(x)e^{\alpha x}\sin(\beta x)\)

where $p_n$ and $q_m$ are polynomials of degrees $n$ and $m$ respectively--since the other types listed in the table are just special cases of (1).

Consider the inhomogeneous equation:

(2) \(\displaystyle L[y](x)=g(x)\)

where $L$ is the linear operator:

(3) \(\displaystyle L\equiv a_nD^{n}+a_{n-1}D^{n-1}+\cdots+a_1D+a_0\)

with $a_n,\,a_{n-1},\,\cdots\,a_0$ constants, and $g(x)$ as given in equation (1). Let $N=\max(n,m)$.

Now, we need to find an annihilator for $g$. If we consider the function:

\(\displaystyle f(x)=e^{\alpha x}\sin(\beta x)\)

we find that:

\(\displaystyle f'(x)=e^{\alpha x}\left(\alpha\sin(\beta x)+\beta\cos(\beta x) \right)\)

\(\displaystyle f''(x)=e^{\alpha x}\left(\left(\alpha^2-\beta^2 \right)\sin(\beta x)+2\alpha\beta\cos(\beta x) \right)\)

If we observe that:

\(\displaystyle f''(x)-2\alpha f'(x)+\left(\alpha^2+\beta^2 \right)f(x)=0\)

then we may state that:

\(\displaystyle D^2-2\alpha D+\alpha^2+\beta^2=(D-\alpha)^2+\beta^2\)

annihilates $f(x)$, and so we conclude that:

\(\displaystyle A\equiv\left((D-\alpha)^2+\beta^2 \right)^{N+1}\)

annihilates $g$.

Now, we need to find the auxiliary equation associated with:

\(\displaystyle AL[y]=0\)

\(\displaystyle \left((D-\alpha)^2+\beta^2 \right)^{N+1}\left(a_nD^n+a_{n-1}D^{n-1}+\cdots+a_0 \right)=0\)

Suppose $0\le s$ is the multiplicity of the roots $\alpha\pm\beta i$ of the auxiliary equation associated with $L[y]=0$, and $r=2_{2s+1}\,\cdots\,r_n$ are the remaining roots, then we have:

(4) \(\displaystyle \left((r-\alpha)^2+\beta^2 \right)^{s+N+1}\left(r-r_{2s+1} \right)\,\cdots\,\left(r-r_n \right)=0\)

Now, as the solution to \(\displaystyle AL[y]=0\) can be written in the form:

\(\displaystyle y(x)=y_h(x)+y_p(x)\)

and we must have:

\(\displaystyle y_h(x)=p_{s+1}(x)e^{\alpha x}\left(\cos(\beta x)+\sin(\beta x) \right)+\sum_{k=2s+1}^n e^{k}\)

then we may conclude that:

\(\displaystyle y_p(x)=x^se^{\alpha x}\left(P_N(x)\cos(\beta x)+Q_N\sin(\beta x) \right)\)

Questions and comments should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-justifying-method-undetermined-coefficients-4840.html

 

[MarkFL]

This topic is for questions and comments pertaining to this tutorial:

http://mathhelpboards.com/math-notes-49/justifying-method-undetermined-coefficients-4839.html

 

[Ackbach]

Good stuff! Just a couple of minor edits I'd recommend:

1. In the table, in the $y_{p}(x)$ column for Type I's, an $x^{s}$ seems to have become an $x^{2}$.

2. I would rewrite Equation (3) as follows (you haven't really used operator notation, but have included the test function in your definition of $L$, which is not usual):
$$(3) \quad L[y] \equiv a_nD^{n}+ a_{n-1}D^{n-1}+ \cdots+ a_{1}D+ a_0.$$
You do this later on, so this is more of a consistency thing, I think.

 

[MarkFL]

Thank you! For some reason, I want to enter a 2 instead of an s. I appreciate you catching this!

Your suggestion of being consistent with operator notation is an excellent one.

I have made both edits. (Sun)

 

[Ackbach]

Hehe. Actually, I'm not sure I was consistent just then! You could either write
$$L \equiv a_{n}D^{n}+a_{n-1}D^{n-1}+ \dots + a_{1}D + a_{0},$$
or
$$L[y] = \left( a_{n}D^{n}+a_{n-1}D^{n-1}+ \dots + a_{1}D + a_{0} \right)y.$$

 

[MarkFL]

I prefer the first notation. This was a group project taken from my old ODE textbook, and the original notation came from there (notice how I am "passing the buck?"). (Rofl)

I truly appreciate your suggestions, and feel the post has been improved because of them. (Rock)