I have used the variation of parameters method (and have been taught it, although not hugely in depth) and I was wondering if I've understood the intuition behind it. In particular I've been thinking about the method for second order ODEs: $$a_{2}(x)y''(x)+a_{1}(x)y'(x)+a_{0}(x)y(x)=f(x)$$

Does the motivation for considering a particular solution of the form $$y_{p}(x)=u_{1}(x)y_{1}(x)+u_{2}(x)y_{2}(x)$$ (where $y_{1},\;y_{2}$ are solutions to the corresponding homogeneous equation) because $(y_{p}/y_{1})\neq\text{ constant}$ and likewise $(y_{p}/y_{2})\neq\text{ constant}$, as otherwise we would just obtain the complementary solution again. This suggests that both are instead functions of $x$, i.e. $(y_{p}/y_{1})=u_{1}(x)$ and $(y_{p}/y_{2})=u_{2}(x)$, leading to the form of the ansatz I gave above?

Secondly, is the reason why we place a further constraint on the form of $y_{p}$ (other that it be a solution to the original ODE) because, in principal, there will be an infinite number of particular solutions of the form $$y_{p}(x)=u_{1}(x)y_{1}(x)+u_{2}(x)y_{2}(x)$$ but we only require one particular solution, and so by imposing an additional constraint we have two equations for the two unknowns $u_{1}$ and $u_{2}$, thus enabling us to uniquely determine a solution for each of them and subsequently enabling us to find a general solution to the original ODE?!

  • nice question. i don't know an answer to your question though i have been using useful method a lot. i will look into the history of this. – abel Mar 10 '15 at 15:59
  • Thanks :) Likewise, and it's been bugging me that I don't have a deeper understanding of it. Appreciate you looking into it! – Perpetual learner Mar 10 '15 at 16:01
  • as with everything in math, it seems to have originated with euler and used by lagrange. i am looking at en.wikipedia.org/wiki/Variation_of_parameters – abel Mar 10 '15 at 16:09
  • 1
    These guys are Math Gods, nothing gets past them! I had a look at the Wiki page, but I can't see any real motivation on there for the parts I'm trying to justify, unfortunately :( – Perpetual learner Mar 10 '15 at 16:16
  • For myself I usually explain that the second part is the trick that massively simplifies finding any particular solution of equation and replaces it with something as simple as solving system of linear equations. – Evgeny Mar 11 '15 at 18:33

Here is one way of getting the variation of parameters formulas. Firstly, some motivation mixed with a bit of jargon:

We think of $a_{2}(x)y''(x)+a_{1}(x)y'(x)+a_{0}(x)y(x)$ as a result of applying to $y(x)$ an operation $D=a_{2}(x)\frac{d^2}{dx^2 }+a_{1}(x)\frac{d}{dx}+a_{0}(x) Id$. This is a linear operation (aka a linear differential operator), meaning that $D(c_1y_1(x)+c_2y_2(x))=c_1 D(y_1(x))+c_2 D((y_2(x))$. Our equation is then $D (y(x))=f(x)$.

Linearity ensures that if $y_a(x)$ and $y_b(x)$ solve $Dy_a=f_a$ and $Dy_b=f_b$ then $y_a+y_b$ solves $Dy=f_a+f_b$. So if we can decompose our inhomogeneous term $f(x)$ as a sum $f=f_a+f_b$ and solve the two "pieces" $Dy=f_a$ and $Dy=f_b$ we would be done - just add the two solutions and get a solution for $f$.

The key idea now (this is known as Green's function method, or perhaps sometimes Duhamel's principle) is to think of $f$ as being a "sum" of a continuum of delta functions $f(x)=\int f(t)\delta(x-t) dt$. Thus we want to solve $Dy(x)= \delta(x-t)$ for each $t$ (these solutions are the Green's functions for our equation), to get solutions $y_t(x)=y(t,x)$, and then write our solution to $Dy=f$ as the "sum" over t $y(x)=\int f(t) y_t (x) dt = \int f(t) y (t , x) dt $.

Continuing in this way (without paying much attention to technical details or introducing the theory of distributions needed to provide such details and make the above rigorous), we are now tasked with finding $y_t(x)$ i.e. solving $Dy_t(x)=\delta(x-t)$. The solution is of course not unique, but is only unique up to adding $y_h$ - an arbitrary solution of the homogeneous equation $Dy=0$. Note that in fact for $x>t$ and for $x<t$ we have $\delta(x-t)=0$, and our $y_t$ will coincide with some $y_h$. But it can not be the same $y_h$, or we won't get $\delta(x-t)$, just $0$ everywhere. So we need to jump at $x=t$. Namely, we can start with $y=0$ for $x<t$ and continue as some $y_h$ after. Now, any $y_h$ is a combination of fixed homogeneous solutions $y_1$ and $y_2$, i.e. $y_h(x)=c_1y_1(x)+c_2y_2(x)$. So our Green's function $y_t=0$ for $x<t$ and $y_t(x)=c_1y_1(x)+c_2y_2(x)$ for $x>t$. What should $c_1, c_2$ be? We want $Dy_t(x)= a_{2}(x)y_t''(x)+a_{1}(x)y_t'(x)+a_{0}(x)y_t(x)= \delta(x-t)$. So $y_t$ should be continuous, so that $y_t'$ has only a step discontinuity at $x=t$, and $y_t''$ only a delta (and not $\delta'(x-t)$). Continuity imposes $c_1y_1(t)+c_2y_2(t)=0$. We also want to have a unit delta jump and not bigger or smaller. The size of the jump is controlled by $a_2 \frac{d^2}{dx^2}$ at $x=t$ and so is $a_2(t)[ c_1y'_1(t)+c_2y'_2(t)]$. Equating it to 1 imposes $ c_1y'_1(t)+c_2y'_2(t)=\frac{1}{a_2(t)}$

To summarize:

The Green's function at $t$ is [$0$ for $x<t$, then $c_1(t)y_1(x)+c_2(t)y_2(x)$ for $x>t$]

with $c_1(t), c_2(t)$ subject to

$c_1(t)y_1(t)+c_2(t)y_2(t)=0$

$ c_1y'_1(t)+c_2y'_2(t)=\frac{1}{a_2(t)}$

Compare with the variation of parameters: $c_1=\frac{u_1'}{f}, c_2=\frac{u_2'}{f}$

And the solution to $Dy=f$ is

$y(x)=\int f(t) y_t (x) dt = \int f(t) y (t , x) dt $

Plugging in, we get

$y(x)=\int_{x>t} f(t) (c_1(t)y_1(x))+c_2(t)y_2(x)) dt=[\int_{-\infty}^{x} f(t)c_1(t)) dt] y_1(x)+[\int_{-\infty}^{x} f(t)c_2(t))dt] y_2(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$.

So we recover $y$ as a combination of $y_1$ and $y_2$ with function coefficients, and we know exactly which relations the derivatives of these coefficients should satisfy (they have to combine to give Green's functions times $f(x)$). Writing out these relations we get the same formulas as for the variation of parameters formulas, which we can then justify independently in an ad hoc manner.

Of course this generalizes in a straightforward manner to higher order linear equations - the $c_i$s will have more linear `matching of derivatives' constraints to satisfy to give a Green's function, and the rest is the same.


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