Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So I never fully understood the derivation of the method of variation of parameters.

Consider the simplest case $$y'' + p(x)y' + q(x)y = f(x)$$

And the homogenous solutions is $y_h=c_1y_1+c_2y_2$ and the guess particular solution is $y_p =u_1y_1+u_2y_2$

What is often done is then we take derivatives of the particular solution and substitute back into the original ODE. What is also often done is that we set a constraint $$u_1'y_1 + u_2'y_2=0$$

and this particular constraint will yield $$u_1'y_1'+u_2'y_2'=f(x)$$

And what is often omitted is the explanation for $u_1'y_1 + u_2'y_2=0$. I am going through a book by Nagle and the writer would just throw this out of nowhere and forces me to accept it without fully understanding why we can do this and how do we know the solutions satisfies that particular constraint.

Going through other sources (probably not reliable), I've found that it has something to with the core concept of simple algebra. Like what is known and what is unknown, though it still confounded me...

It was said that it took Lagrange (the creator) a long time to figure out this method. So could someone give me a proper explanation as to why this is true? All the other sources just says "okay we are going to impose this constraint, next moving on..."

share|improve this question
    
My guess would be that, like in algebra, you need a second equation to obtain a unique solution for 2 unknowns. –  Mike Dec 22 '12 at 21:35
    
This has a natural interpretation and is related to the method of osculating parameters. See this answer to a very similar question. –  user26872 Dec 22 '12 at 22:47

3 Answers 3

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.

share|improve this answer
1  
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 –  jip Dec 22 '12 at 22:09
    
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. –  JohnD Dec 22 '12 at 23:49
    
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. –  JohnD Dec 22 '12 at 23:53
    
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 –  jip Dec 23 '12 at 0:09
1  
@JohnD Don't know about OP, but this answer was immensely helpful for me. Thanks! –  Asad Dec 7 '13 at 23:57

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 $$A'(x)u_1(x) + B'(x)u_2(x) = 0.$$

share|improve this answer
    
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 $u_G'(x) = A(x) u_1'(x) + B(x) u_2'(x)$. –  jip Mar 29 at 8:23
    
@sidht: Simply take the derivative of $u_G$ and impose the quoted condition to find the last relation. –  user26872 Mar 29 at 23:29

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.