We consider the equations:

$(p_1y')'+q_1y=0 ...(E_1)$ ,$(p_2y')'+q_2y=0 ...(E_2)$ $p_1,p_2 \in C^1([0,1],(0,\infty)) ; q_1,q_2 \in C([0,1],\mathbb{R})$ $y_1$and $y_2$ are respectivly solutions of $E_1$ and $E_2$.

How do you prove that :

$\int_{t_1}^{t_2}(q_1-q_2)y_1^2 + (p_1-p_2)(y_1')^2 dt +\int_{t_1}^{t_2}p_2(\frac{y_1'y_2-y_1y_2'}{y_2})^2dt$ =$[\frac{y_1} y_2(p_1y_1'y_2-p_2y_1y_2')]_{t_1}^{t_2}$ please help me please.


We are given:

$$\tag 1 (p_1y')'+ q_1y = 0,$$

$$\tag 2 (p_2y')'+ q_2y = 0.$$

where, $p_1,p_2 \in C^1([0,1],(0,\infty)) ; q_1,q_2 \in C([0,1],\mathbb{R})$ and, $y_1$and $y_2$ are respectively solutions of $(1)$ and $(2)$.

Note, subtracting $(2)$ from $(1)$ and integrating both sides, yields (we need this for a substitution later, and could have done it then, but this seemed cleaner for the write-up):

$$\tag 3 \left[ p_1y'_1y_2 - p_2y_1y_2' \right]_{t_1}^{t_2} = \int_{t_1}^{t_2}(q_1-q_2)y_1y_2~dt.$$

We are asked to prove:

$$\tag 4 \displaystyle \int_{t_1}^{t_2}(q_1-q_2)y_1^2 + (p_1-p_2)(y_1')^2 dt +\int_{t_1}^{t_2}p_2(\frac{y_1'y_2-y_1y_2'}{y_2})^2dt = \left[\frac{y_1}{y_2}(p_1y_1'y_2-p_2y_1y_2')\right]_{t_1}^{t_2}$$

Lets see if we can figure out a potential approach.

The left-hand-side (LHS) of $(4)$ is rather ominous looking, so lets see if we can do anything with the RHS.

Lets take the derivative of the RHS and see if that bears fruit (because I know how to do derivatives and don't have a clue on how to solve those integrals without further information).

So, we want to find:

$\tag 5 \displaystyle \frac{d}{dt} \left[\frac{y_1}{y_2}(p_1y_1'y_2-p_2y_1y_2')\right]$

$\tag 6 \displaystyle = \frac{y_1}{y_2} \left[y_2 \frac{d}{dt} (p_1y'_1) - y_1 \frac{d}{dt} (p_2 y'_2) + (p_1 - p_2)y'_1y'_2 \right] + \frac{y'_1y_2 - y_1y'_2}{y_2^2} (p_1y'_1y_2-p_2y_1y'_2)$

Substituting from $(3)$ into $(6)$ (note - this idea comes from the Sturm approach), yields:

$= \displaystyle \frac{u}{v} \left[(q_1 - q_2)y_1y_2 + (p_1 - p_2)y'_1y'_2 \right] + p_1(y'_1)^{2} - (p_1 + p_2)y_1y'_1 \frac{y'_2}{y_2} + p_2y_1^2 \frac{y_2'^2}{y_2^2}$

$\tag 7 = \displaystyle (q_1-q_2)y_1^2 + (p_1-p_2){y_1'}^2 + p_2 \left[y_1'-y_1 \frac{y_2'}{y_2}\right]^2.$

So, we have:

$\tag 8 \displaystyle \frac{d}{dt} \left[\frac{y_1}{y_2}(p_1y_1'y_2-p_2y_1y_2')\right] = (q_1-q_2)y_1^2 + (p_1-p_2){y_1'}^2 + p_2 \left[y_1'-y_1 \frac{y_2'}{y_2}\right]^2$

Now, lets integrate both sides of $(8)$ and swap sides of our expressions, yielding:

$$\tag 9 \displaystyle \int_{t_1}^{t_2} \left[(q_1-q_2)y_1^2 + (p_1-p_2)(y_1')^2\right] dt +\int_{t_1}^{t_2}p_2(\frac{y_1'y_2-y_1y_2'}{y_2})^2dt = \left[\frac{y_1}{y_2}(p_1y_1'y_2-p_2y_1y_2')\right]_{t_1}^{t_2}$$


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  • $\begingroup$ Great work here/of great value! $\endgroup$ – amWhy Apr 28 '13 at 0:23
  • $\begingroup$ And you obviously "had it in you," after all! $\endgroup$ – amWhy Apr 28 '13 at 0:27

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