# Theoretical question on the Variation of Parameters method

• Jun 24th 2011, 05:02 PM
Lancet
Theoretical question on the Variation of Parameters method
I have a question about the Variation of Parameters method... I have no idea whether this is an easy or difficult question to answer. :)

With the Variation of Parameters method, we start with this concept:

$y = uy_1 + vy_2$

$y' = u'y_1 + uy'_1 + v'y_2 + vy'_2$

$y' = u'y_1 + v'y_2 + uy'_1 + vy'_2$

Here, an assumption is made that:

$u'y_1 + v'y_2 = 0$

My question is why? I understand that it all works out, but I'd like to understand what the reason is that we can do this. Every source of information I've come across states this assumption, but not a single one I've found ever explains the reasoning for it.

Can anyone enlighten me? :)
• Jun 25th 2011, 06:21 AM
Jester
Re: Theoretical question on the Variation of Parameters method
Well let me try (remember try :-))

You have the ODE

$y'' + p(x)y' + q(x)y = f(x)$

to solve. You have the complimentary solution

$y = c_1 y_1 + c_2 y_2$.

To find a particular solution you try

$y_p = u(x) y_1 + v(x) y_2$

and see if you can find a $u$ and $v$ to satisfies your ODE with $f(x).$ As you said

$y' = u' y_1 + v' y_2 + uy_1' + v y_2'$.

Next (no assumption here)

$y'' = \left( u' y_1 + v' y_2\right)' + u y_1'' + v y_2'' + u' y_1' + v' y_2'$

Substituting into your ODE you have

$\left( u' y_1 + v' y_2\right)' + u y_1'' + v y_2'' + u' y_1' + v' y_2'$ $+ p \left( u' y_1 + v' y_2 + uy_1' + v y_2'\right) + q \left(u y_1 + v y_2 \right) = f$.

Now you know that some pieces together go to zero as both $y_1$ and $y_2$ satisfies the complementary ODE so

$\left( u' y_1 + v' y_2\right)' + u' y_1' + v' y_2'$ $+ p \left( u' y_1 + v' y_2 \right) = f$

or

$\left( u' y_1 + v' y_2\right)' + \left(y_1' + p y_1 \right) u' + \left(y_2' + p y_2 \right) v' = f\;\;\;(1)$.

At this point you have one equation for the two unknows $u$ and $v$ - equation (1) undetermined. You could guess a $v$ and try and solve (1) but that could prove to be difficult. As you only have a single equation, i.e eqn. (1), you could split the equation into two pieces. For example,

$\left( u' y_1 )' + \left(y_1' + p y_1 \right) u' = 0$.

$\left( v' y_2\right)' + \left(y_2' + p y_2 \right) v' = f$.

and each can be reduced to a linear first order ODE, however, by setting

$u' y_1 + v' y_2 = 0$

gives

$\left(y_1' + p y_1 \right) u' + \left(y_2' + p y_2 \right) v' = f\;\;\;(1)$.

What's nice about this split is you have two linear equations for the two unknows $u'$ and $v'$ - there's no $u''$ and $v''$! So solve for $u'$ and $v'$ and then integrate.
• Jun 25th 2011, 07:12 AM
Lancet
Re: Theoretical question on the Variation of Parameters method
Quote:

Originally Posted by Danny

$\left( u' y_1 + v' y_2\right)' + \left(y_1' + p y_1 \right) u' + \left(y_2' + p y_2 \right) v' = f\;\;\;(1)$.

I'm with you up to here.

Quote:

Originally Posted by Danny

At this point you have one equation for the two unknows $u$ and $v$ - equation (1) undetermined. You could guess a $v$ and try and solve (1) but that could prove to be difficult. As you only have a single equation, i.e eqn. (1), you could split the equation into two pieces. For example,

$\left( u' y_1 )' + \left(y_1' + p y_1 \right) u' = 0$.

$\left( v' y_2\right)' + \left(y_2' + p y_2 \right) v' = f$.

This is the first part I'm confused about, because I don't understand how you can split up the top equation in this manner.

To me, that's like having

$4x + 2y = 20$

and saying you can split it up into

$4x = 0$

$2y = 20$

...which to me seems erroneous.

Quote:

Originally Posted by Danny

and each can be reduced to a linear first order ODE, however, by setting

$u' y_1 + v' y_2 = 0$

gives

$\left(y_1' + p y_1 \right) u' + \left(y_2' + p y_2 \right) v' = f\;\;\;(1)$.

What's nice about this split is you have two linear equations for the two unknows $u'$ and $v'$ - there's no $u''$ and $v''$! So solve for $u'$ and $v'$ and then integrate.

But here we come back to the original question. *Why* is it that you can set

$u' y_1 + v' y_2 = 0$

To me, this comes back to the original issue. You're stating we can do it, and you're stating that it makes things neat - and that I get. But I still don't know *why* we can do it. What makes it valid to set that equal to zero?

If you have something like this:

$e^x(y) = 0$

You can state the reasons why $y$ must be equal to zero, which is that $e^x$ can never ever be zero, therefore $y$ must be equal to zero.

I'm looking for a similar set of reasons as to whey we can say

$u' y_1 + v' y_2 = 0$

p.s. Thanks for trying to help me understand this, I know it's not an easy question to answer! :)
• Jun 25th 2011, 08:19 AM
Jester
Re: Theoretical question on the Variation of Parameters method
Well, this is the way I look at it. We only need one solution - $u$ and $v$. There are actually and infinite number of solutions. For example, you have the equation

$4x + 2y = 20$.

There are an infinite number of solutions $(x,y)$ to this equation. By splitting as you have

$4x = 0, 2y = 20$ gives $(0,10)$ one solution.

That's all you need is one solution.
• Jun 25th 2011, 10:08 AM
Lancet
Re: Theoretical question on the Variation of Parameters method
Quote:

Originally Posted by Danny

There are an infinite number of solutions $(x,y)$ to this equation. By splitting as you have

$4x = 0, 2y = 20$ gives $(0,10)$ one solution.

That's all you need is one solution.

Interesting. Okay, so I think I'm starting to understand this...

When we solve this kind of ODE, we end up with

$y = C_1y_1 + C_2y_2 + y_p$

If there are an unlimited number of solutions for u and v, would it be accurate for me to guess that the reason we choose

$u'y_1 + v'y_2 = 0$

...is because that is the only assumption that will give us the end result where $y_1$ and $y_2$ match up with the homogeneous solution?

Or am I looking at this the wrong way?
• Jun 26th 2011, 07:15 AM
Jester
Re: Theoretical question on the Variation of Parameters method
The reason I chose

$u' y_1 + v' y_2 = 0,\;\;(1)$

is because

$\left( u' y_1 + v' y_2\right)' + \left(y_1' + p y_1 \right) u' + \left(y_2' + p y_2 \right) v' = f$.

reduces to

$\left(y_1' + p y_1 \right) u' + \left(y_2' + p y_2 \right) v' = f\;\;(2)$.

Equation (1) and (2) are two equations that only involve u' and v' only. There's no u'' and v''. Second, they are two linear equations for u' and v'. Solving for u' and v' then gives you something that only need to be integrated.