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- TL;DR
- How to show that if the usual constraint in the method of variation of parameters is replaced by a more general one, the particular solution doesn't change?
Hi folks,
I decided to brush up my knowledge of Green's functions and Differential Equations, and came across this chapter,
https://people.uncw.edu/hermanr/mat463/odebook/book/greens.pdf
of some unknown book (if anybody recognizes it, I'd like to know the title). In 8.3 (page 265) the well-known method of "Variation of Parameters" is treated for 2nd order differential equations in self-adjoint form (which can easily be made more general):
$$\frac{d}{dx}\Bigl(p(x) \frac{dy}{dx}\Bigr) + q(x)y(x)=f(x)$$
So we use for our particular solution the Ansatz
$$y_p(x) = c_1(x)y_1(x)+c_2(x)y_2(x)$$
and plug it in. The first derivative becomes
$$y'_p(x) = c_1(x)y'_1(x)+c_2(x)y'_2(x) + c'_1(x)y_1(x)+c'_2(x)y_2(x)$$
Then follows the notorious constraint which confuses so many students:
$$c'_1(x)y_1(x)+c'_2(x)y_2(x) \equiv 0 \ \ \ \ \ \ \ \ (1)$$
I understand this part: we introduced 2 new functions c_i(x) (i = 1,2), and with only 1 constraint (the original ODE) we will not get a unique solution. If we would have imposed e.g.
$$c_1(x)y'_1(x)+c'_2(x)y'_2(x) \equiv 0$$
we would obtain an equation with 2nd order derivatives for c_i(x) (i = 1,2), which would give us the same kind of problem for these coefficients. So that wouldn't work (but I'm not sure if this would be consistent...) So that motivates us to impose a condition on the derivatives of c_i(x). So we impose the constraint (1) and this leads to two equations for only the two c'_i(x), which is easily solved. An integration then gives us the c_i(x), and the particular solution coincides with the one we would obtain by the Green's function. So far, so good.
My question is about an exercise on page 302 of the notes: exercise 2. It says that if we now impose (this is a restriction, not a mere relabeling!)
$$c'_1(x)y_1(x)+c'_2(x)y_2(x) \equiv h(x) \ \ \ \ \ \ \ \ (2)$$
for some function h(x), "show that one gets the same particular solution". And this is where I get stuck.
So I just redid the calculation but with constraint (1) replaced by (2). I will show my work how the usual results change. With constraint (2) the equation (8.10) is changed as follows:
$$[p(x)h(x)]' + c'_1(x)y'_1(x)+c'_2(x)y'_2(x) = \frac{f(x)}{p(x)}$$
i.e. one gets an extra total derivative [p(x)h(x)]' on the left hand side. Along with the constraint (2) we can solve for c'_i(x) (i = 1,2), in which I denoted the Wronskian with W and drop the arguments for notation sake (every function depends on x):
$$
\begin{split}
c'_1 = W^{-1}\Bigl[hy_2' - (\frac{f}{p} - [ph]')y_2\Bigr] \\
c'_2 = W^{-1}\Bigl[-hy_1' + (\frac{f}{p} - [ph]')y_1 \Bigr]
\end{split}
$$
So our constraint (2) changes these equations (eqn. 8.12 in the notes) as follows:
$$
\begin{split}
c'_1 \rightarrow c'_1 + \frac{hy_2'}{W} +\frac{(ph)'y_2'}{W} \\
c'_2 \rightarrow c'_2 - \frac{hy_1'}{W} - \frac{(ph)'y_1'}{W}
\end{split}
$$
and indeed, the extra terms disappear if we apply the constraint (1) (i.e. h(x) \equiv 0). So if we now integrate both equations, the coefficients themselves are now changed with respect to constraint (1) as follows:
$$\begin{split}
c_1(x) \rightarrow c_1(x) + \int_{x_0}^x \frac{h(\xi)y_2'(\xi)}{W(\xi)} d\xi +\int_{x_0}^x \frac{[p(\xi)h(\xi)]'y_2(\xi)}{W(\xi)} d\xi \\
c_2(x) \rightarrow c_2(x) - \int_{x_0}^x \frac{h(\xi)y_1'(\xi)}{W(\xi)} d\xi - \int_{x_0}^x \frac{[p(\xi)h(\xi)]'y_1(\xi)}{W(\xi)} d\xi \end{split}
$$
If our particular solution is kept unchanged, this means that our coefficients merely change by a constant. So I guess in both expression the two extra integrals which show up due to constraint (2) have to cancel. Or maybe they combine to merely a constant, such that in the total solution the constant coefficients in front of the homogenous part merely changes, but that's not what the exercise says). But I don't see how; some partial integration with the triple product under the integral sign doesn't do it, and the condition p(x)W(x) = cnst. which is mentioned in the notes in equation (8.13) also doesn't help me. If I plug these coefficients into the expansion of the particular solution, I also don't see how terms from both coefficients could cancel against each other.
So my question: how do I show this? Am I right to consider constraint (1) as a particular "gauge choice" of constraint (2), and that both the coefficients merely change by some constants? Or am I misunderstanding the core of the method?
Any help is greatly appreciated! :)
I decided to brush up my knowledge of Green's functions and Differential Equations, and came across this chapter,
https://people.uncw.edu/hermanr/mat463/odebook/book/greens.pdf
of some unknown book (if anybody recognizes it, I'd like to know the title). In 8.3 (page 265) the well-known method of "Variation of Parameters" is treated for 2nd order differential equations in self-adjoint form (which can easily be made more general):
$$\frac{d}{dx}\Bigl(p(x) \frac{dy}{dx}\Bigr) + q(x)y(x)=f(x)$$
So we use for our particular solution the Ansatz
$$y_p(x) = c_1(x)y_1(x)+c_2(x)y_2(x)$$
and plug it in. The first derivative becomes
$$y'_p(x) = c_1(x)y'_1(x)+c_2(x)y'_2(x) + c'_1(x)y_1(x)+c'_2(x)y_2(x)$$
Then follows the notorious constraint which confuses so many students:
$$c'_1(x)y_1(x)+c'_2(x)y_2(x) \equiv 0 \ \ \ \ \ \ \ \ (1)$$
I understand this part: we introduced 2 new functions c_i(x) (i = 1,2), and with only 1 constraint (the original ODE) we will not get a unique solution. If we would have imposed e.g.
$$c_1(x)y'_1(x)+c'_2(x)y'_2(x) \equiv 0$$
we would obtain an equation with 2nd order derivatives for c_i(x) (i = 1,2), which would give us the same kind of problem for these coefficients. So that wouldn't work (but I'm not sure if this would be consistent...) So that motivates us to impose a condition on the derivatives of c_i(x). So we impose the constraint (1) and this leads to two equations for only the two c'_i(x), which is easily solved. An integration then gives us the c_i(x), and the particular solution coincides with the one we would obtain by the Green's function. So far, so good.
My question is about an exercise on page 302 of the notes: exercise 2. It says that if we now impose (this is a restriction, not a mere relabeling!)
$$c'_1(x)y_1(x)+c'_2(x)y_2(x) \equiv h(x) \ \ \ \ \ \ \ \ (2)$$
for some function h(x), "show that one gets the same particular solution". And this is where I get stuck.
So I just redid the calculation but with constraint (1) replaced by (2). I will show my work how the usual results change. With constraint (2) the equation (8.10) is changed as follows:
$$[p(x)h(x)]' + c'_1(x)y'_1(x)+c'_2(x)y'_2(x) = \frac{f(x)}{p(x)}$$
i.e. one gets an extra total derivative [p(x)h(x)]' on the left hand side. Along with the constraint (2) we can solve for c'_i(x) (i = 1,2), in which I denoted the Wronskian with W and drop the arguments for notation sake (every function depends on x):
$$
\begin{split}
c'_1 = W^{-1}\Bigl[hy_2' - (\frac{f}{p} - [ph]')y_2\Bigr] \\
c'_2 = W^{-1}\Bigl[-hy_1' + (\frac{f}{p} - [ph]')y_1 \Bigr]
\end{split}
$$
So our constraint (2) changes these equations (eqn. 8.12 in the notes) as follows:
$$
\begin{split}
c'_1 \rightarrow c'_1 + \frac{hy_2'}{W} +\frac{(ph)'y_2'}{W} \\
c'_2 \rightarrow c'_2 - \frac{hy_1'}{W} - \frac{(ph)'y_1'}{W}
\end{split}
$$
and indeed, the extra terms disappear if we apply the constraint (1) (i.e. h(x) \equiv 0). So if we now integrate both equations, the coefficients themselves are now changed with respect to constraint (1) as follows:
$$\begin{split}
c_1(x) \rightarrow c_1(x) + \int_{x_0}^x \frac{h(\xi)y_2'(\xi)}{W(\xi)} d\xi +\int_{x_0}^x \frac{[p(\xi)h(\xi)]'y_2(\xi)}{W(\xi)} d\xi \\
c_2(x) \rightarrow c_2(x) - \int_{x_0}^x \frac{h(\xi)y_1'(\xi)}{W(\xi)} d\xi - \int_{x_0}^x \frac{[p(\xi)h(\xi)]'y_1(\xi)}{W(\xi)} d\xi \end{split}
$$
If our particular solution is kept unchanged, this means that our coefficients merely change by a constant. So I guess in both expression the two extra integrals which show up due to constraint (2) have to cancel. Or maybe they combine to merely a constant, such that in the total solution the constant coefficients in front of the homogenous part merely changes, but that's not what the exercise says). But I don't see how; some partial integration with the triple product under the integral sign doesn't do it, and the condition p(x)W(x) = cnst. which is mentioned in the notes in equation (8.13) also doesn't help me. If I plug these coefficients into the expansion of the particular solution, I also don't see how terms from both coefficients could cancel against each other.
So my question: how do I show this? Am I right to consider constraint (1) as a particular "gauge choice" of constraint (2), and that both the coefficients merely change by some constants? Or am I misunderstanding the core of the method?
Any help is greatly appreciated! :)
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