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I never fully understood the derivation of the method of variation of parameters. Consider the simple case $$y'' + p(x)y' + q(x)y = f(x)\,.$$

The homogeneous solution is $y_h=c_1y_1+c_2y_2$ and the particular solution that we guess is $y_p =u_1y_1+u_2y_2$ for some function $u_1(t)$ and $u_2(t)$. Next we take derivatives of the particular solution and substitute those back into the original ODE.

The part done next is the part I don't quite see the justification for. We assume that our functions $u_1$ and $u_2$ will satisfy the constraint $$u_1'y_1 + u_2'y_2=0\,,$$

and this particular constraint will yield a cleaner result when looking at $y'_p$: $$u_1'y_1'+u_2'y_2'=f(x)\,.$$

At least in the textbook I'm looking at, the justification for assuming that $u_1$ and $u_2$ satisfy $u_1'y_1 + u_2'y_2=0$ is simply omitted. How do we know that the solutions we're after satisfy that constraint? All other sources I look at just say something like, "okay we are going to impose this constraint. Now moving on ..."

Could someone give me a proper explanation as to why we can make this assumption?

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  • $\begingroup$ My guess would be that, like in algebra, you need a second equation to obtain a unique solution for 2 unknowns. $\endgroup$
    – Mike
    Commented Dec 22, 2012 at 21:35

3 Answers 3

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I quote here an answer I gave to a similar question. The notation is that used here.

This is closely tied to the method of osculating parameters. Suppose we wish to represent, with constant coefficients, some arbitrary function $u(x)$ with two linearly independent functions $u_1(x)$ and $u_2(x)$, $$u(x) = A u_1(x) + B u_2(x).$$ In general this can not be done. The best we can do is match the value of the function and its derivative at some point $x_0$, $$\begin{eqnarray*} u(x_0) &=& A u_1(x_0) + B u_2(x_0) \\ u'(x_0) &=& A u_1'(x_0) + B u_2'(x_0). \end{eqnarray*}$$ The conditions above determine the osculating parameters, the constants $A$ and $B$. $A$ and $B$ will be different depending on the point $x_0$. In general this fit will be poor at points far from $x_0$.

The method of variation of parameters involves finding the osculating parameters $A$ and $B$ at every point. That is, we let $A$ and $B$ be functions of $x$. The condition that they are the osculating parameters is that they satisfy $$\begin{eqnarray*} u_G(x) &=& A(x) u_1(x) + B(x) u_2(x) \\ u_G'(x) &=& A(x) u_1'(x) + B(x) u_2'(x), \end{eqnarray*}$$ just as above. For the second equation to hold it must be the case that $$A'(x)u_1(x) + B'(x)u_2(x) = 0.$$

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  • $\begingroup$ How does $$\int A'(x)u_1(x) + B'(x)u_2(x) dx = 0 \implies A'(x)u_1(x) + B'(x)u_2(x) = 0 $$? I got the integral from integration by parts of $u_G'(x) = A(x) u_1'(x) + B(x) u_2'(x)$. $\endgroup$
    – Lemon
    Commented Mar 29, 2014 at 8:23
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    $\begingroup$ @sidht: Simply take the derivative of $u_G$ and impose the quoted condition to find the last relation. $\endgroup$
    – user26872
    Commented Mar 29, 2014 at 23:29
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You can impose any constraint that you want. Now, you may or may not be able to find a solution to the differential equation that also satisfies your constraint...

In this case, you choose that constraint because it makes half the terms vanish before computing the second derivative. If it "works" in the sense that you indeed are able to find a solution to the original DE after imposing this constraint, then there's no reason to look back (or fret about it).

And, as you know, it does indeed work in that sense, and thus is valid.

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    $\begingroup$ You can impose any constraint that you want. Can you give me an example in algebra? I don't think I'll ever understand why if i view it Diff Eqtn mode $\endgroup$
    – Lemon
    Commented Dec 22, 2012 at 22:09
  • $\begingroup$ Suppose you want to find a solution of $x+y=1$, $x,y\in\mathbb{R}$. There are many! But you might choose to impose the constraint that $x$ and $y$ are integers. Under that constraint, you can still find a solution (many in fact), so that constraint 'works" or is "viable" in the sense that if you impose it, the solution set of the original equation subject to the constraint is nonempty. On the other hand, you might impose the constraint that $x$ and $y$ are even integers. In that case, there the solution set is empty. $\endgroup$
    – JohnD
    Commented Dec 22, 2012 at 23:49
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    $\begingroup$ It is important to note that at the outset of variation of parameters, there are two unknowns $u_1$ and $u_2$, but only one equation (the given ODE). So the problem of finding $u_1$, $u_2$ inherently has the freedom to impose one more equation on those unknowns, which you do. $\endgroup$
    – JohnD
    Commented Dec 22, 2012 at 23:53
  • $\begingroup$ But in the ODE, the particular solution (which is uniquely determined) and the homogeneous solution are unique no? I can't relate the two examples $\endgroup$
    – Lemon
    Commented Dec 23, 2012 at 0:09
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    $\begingroup$ The homogeneous solution is not uniquely determined: there is a two parameter family of solutions $c_1y_1(x)+c_2y_2(x)$ where $\{y_1(x),y_2(x)\}$ is a fundamental set of solutions. In some sense, the method of V.O.P. looks to exploit these two "degrees of freedom" by seeking a particular solution of the nonhomogeneous problem of the form $y_p=u_1y_1+u_2y_2$. So there's one equation (the ODE), but two "unknowns" ($u_1$ and $u_2$), and thus it is reasonable to impose a second equation involving $u_1$, $u_2$ in order to (uniquely) determine the unknowns. Hope that helps. $\endgroup$
    – JohnD
    Commented Dec 23, 2012 at 5:54
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Yes, I agree that the usual derivation with the rabbit out of the hat is not very compelling. Here is another one. We know that if one has a system of linear equations $$ \dot x=Ax+f(t), \quad A=(a_{ij})_{n\times n},\quad f\colon \mathbf R\to\mathbf R^n, $$ then if one has a fundamental matrix solution $\Phi(t)$ for the homogeneous system, then the method of variation of parameters is very straightforward and does not require any additional assumptions or constraints. To apply it you would need to solve $$ \Phi(t) \dot c(t)=f(t),\quad c(t)=(u_1,u_2) $$ Now rewrite your equation as a system of two equations and assume that you have a fundamental matrix solution, conclude that this is exactly what you have written already.

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