Proofs about the second-order linear differential equation?

[Math100]

Homework Statement

Let ## u_{1} ## and ## u_{2} ## be two linearly independent solutions of the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
(i) Let ## w=u_{2}/u_{1} ##. Show that ## w'=c/u_{1}^2 ## for some nonzero constant ## c ##.
(ii) Let ## y=-2u_{1}'/u_{1} ##. Show that ## y ## satisfies the Ricatti equation ## y'-\frac{y^2}{2}=f ##.
(iii) Show that ## w''/w'=-2u_{1}'/u_{1} ##, and prove that ## w ## satisfies the equation ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=f ##.
(iv) Hence find a general solution of the equation ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=2 ##. [You are not required to prove that your answer is the general solution.]

Relevant Equations

None.

Proof:

(i) Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u''+\frac{f}{2}u=0\implies r^2+\frac{f}{2}=0 ##, so ## r=\pm \sqrt{\frac{f}{2}}i ##.
This implies ## u_{1}=c_{1}cos(\sqrt{\frac{f}{2}}x) ## and ## u_{2}=c_{2}sin(\sqrt{\frac{f}{2}}x) ##.
Note that ## u_{1}'=-\sqrt{\frac{f}{2}}c_{1}sin(\sqrt{\frac{f}{2}}x), u_{2}'=\sqrt{\frac{f}{2}}c_{2}cos(\sqrt{\frac{f}{2}}x) ##.
Let ## w=\frac{u_{2}}{u_{1}} ##.
Then ## w'=\frac{u_{1}u_{2}'-u_{2}u_{1}'}{u_{1}^2}=\frac{c}{u_{1}^2} ##.
Observe that ## w'=\frac{c_{1}cos(\sqrt{\frac{f}{2}}x)\cdot \sqrt{\frac{f}{2}}c_{2}cos(\sqrt{\frac{f}{2}}x)-c_{2}sin(\sqrt{\frac{f}{2}}x)[-\sqrt{\frac{f}{2}}c_{1}sin(\sqrt{\frac{f}{2}}x)]}{u_{1}^2}=\frac{c}{u_{1}^2} ##.
Thus ## w'=\frac{\sqrt{\frac{f}{2}}c_{1}c_{2}cos^2(\sqrt{\frac{f}{2}}x)+\sqrt{\frac{f}{2}}c_{1}c_{2}sin^2(\sqrt{\frac{f}{2}}x)}{u_{1}^2}=\frac{c}{u_{1}^2} ##,
so ## w'=\frac{\sqrt{\frac{f}{2}}c_{1}c_{2}}{u_{1}^2}=\frac{c}{u_{1}^2} ##, where ## c=\sqrt{\frac{f}{2}}c_{1}c_{2} ##.
Therefore, ## w'=c/u_{1}^2 ## for some nonzero constant ## c ##.

(ii) Let ## y=-\frac{2u_{1}'}{u_{1}} ##.
Then ## y'=\frac{u_{1}(-2u_{1}'')-(-2u_{1}')(u_{1}')}{u_{1}^2}=\frac{-2u_{1}u_{1}''+2(u_{1}')^2}{u_{1}^2} ##.
Consider the Ricatti equation ## y'-\frac{y^2}{2}=f ##.
Observe that ## \frac{-2u_{1}u_{1}''+2(u_{1}')^2}{u_{1}^2}-\frac{1}{2}(-\frac{2u_{1}'}{u_{1}})^2=f ##
## \frac{-2u_{1}u_{1}''+2(u_{1}')^2}{u_{1}^2}-\frac{1}{2}(\frac{4(u_{1}')^2}{u_{1}^2}=f ##
## \frac{-2u_{1}u_{1}''+2(u_{1}')^2-2(u_{1}')^2}{u_{1}^2}=f ##
## \frac{-2u_{1}u_{1}''}{u_{1}^2}=f ##.
Since ## y'-\frac{y^2}{2}=\frac{-2u_{1}u_{1}''}{u_{1}^2}=f ##, it follows that ## -2u_{1}u_{1}''=fu_{1}^2\implies -2u_{1}''=fu_{1}\implies u_{1}''=-\frac{fu_{1}}{2}\implies u_{1}''+\frac{fu_{1}}{2}=0 ##.
Therefore, ## y ## satisfies the Ricatti equation ## y'-\frac{y^2}{2}=f ##.

(iii) Let ## w'=\frac{c}{u_{1}^2} ## for some nonzero constant ## c ##.
Then ## w''=(cu_{1}^{-2})'=-2cu_{1}^{-3}u_{1}'=-\frac{2cu_{1}'}{u_{1}^3} ## by the chain rule.
Thus ## \frac{w''}{w'}=\frac{-2cu_{1}'}{u_{1}^3}\cdot \frac{u_{1}^2}{c}=\frac{-2u_{1}'}{u_{1}} ##.
Therefore, ## y=\frac{-2u_{1}'}{u_{1}}=\frac{w''}{w'} ## (by part ii) and ## w ## satisfies the equation ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=f ##.

(iv) By parts (ii) and (iii), we have that ## y=\frac{w''}{w'}=\frac{-2u_{1}'}{u_{1}} ##.
Then ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=2\implies y'-\frac{1}{2}y^2=2 ##.
Now we have ## \frac{dy}{dx}=2+\frac{y^2}{2} ##.
Observe that ## \frac{dy}{dx}=\frac{4+y^2}{2} ##
## \frac{dy}{4+y^2}=\frac{dx}{2} ##
## \int \frac{dy}{4+y^2}=\int \frac{dx}{2} ##
## \frac{1}{2}\cdot tan^{-1}(\frac{y}{2})=\frac{x}{2}+c ##
## tan^{-1}(\frac{y}{2})=x+c ##
## \frac{y}{2}=tan(x+c) ##
## y=2tan(x+c) ##.
Therefore, a general solution of the equation ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=2 ## is ## y=2tan(x+c) ##.

 

[mathhabibi]

I am assuming you are trying to find out whether your work is correct or not. If ##f(x)## is non-constant then your solution doesn't work. See why by substituting any non-constant function, x for example.

 

[Math100]

I am assuming you are trying to find out whether your work is correct or not. If ##f(x)## is non-constant then your solution doesn't work. See why by substituting any non-constant function, x for example.

Are you talking about the general solution? To which part of the problem are you referring to?

 

[mathhabibi]

Are you talking about the general solution? To which part of the problem are you referring to?

At the very beginning, you say that ##u_1=c_1\cos(\sqrt{\frac{f}2}x)##. I was saying that this is true only when ##f## is constant.

 

[Math100]

At the very beginning, you say that ##u_1=c_1\cos(\sqrt{\frac{f}2}x)##. I was saying that this is true only when ##f## is constant.

Okay, I see that now. Then what's the error here? Is the characteristic equation wrong?

 

[mathhabibi]

Okay, I see that now. Then what's the error here? Is the characteristic equation wrong?

Your solutions ##u_1## and ##u_2## are generally incorrect. If ##f## is constant then you're fine.

 

[fresh_42]

I would write ##u_i =-\dfrac{2}{f}u_i''## so ##\omega =\dfrac{u_2}{u_1}=\dfrac{u_2''}{u_1''}## and start with both differentiations of ##\omega ## since you have to do this anyway.

 

[Mark44]

My comments apply to part (i) only.

Homework Statement: Let ## u_{1} ## and ## u_{2} ## be two linearly independent solutions of the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
(i) Let ## w=u_{2}/u_{1} ##. Show that ## w'=c/u_{1}^2 ## for some nonzero constant ## c ##.
<snip>

(i) Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u''+\frac{f}{2}u=0\implies r^2+\frac{f}{2}=0 ##, so ## r=\pm \sqrt{\frac{f}{2}}i ##.

1) Since f is a function, you should write this explicitly as f(x). Not doing so possibly led to the error that @mathhabibi points out.
2) You apparently are assuming a solution of the form ##y = e^{rx}##. This is something that you should state.

This implies ## u_{1}=c_{1}cos(\sqrt{\frac{f}{2}}x) ## and ## u_{2}=c_{2}sin(\sqrt{\frac{f}{2}}x) ##.
Note that ## u_{1}'=-\sqrt{\frac{f}{2}}c_{1}sin(\sqrt{\frac{f}{2}}x), u_{2}'=\sqrt{\frac{f}

The error pointed out by @mathhabibi is that both ##u_1(x)## and ##u_2(x)## are functions that involve f(x), when you differentiate them, you need to use the chain rule.

 

[Math100]

My comments apply to part (i) only.
1) Since f is a function, you should write this explicitly as f(x). Not doing so possibly led to the error that @mathhabibi points out.
2) You apparently are assuming a solution of the form ##y = e^{rx}##. This is something that you should state.
The error pointed out by @mathhabibi is that both ##u_1(x)## and ##u_2(x)## are functions that involve f(x), when you differentiate them, you need to use the chain rule.

I noticed the issue. Since both of the functions ## u_{1}(x), u_{2}(x) ## were incorrect, then how should I solve this differential equation and find these correct functions?

 

[Math100]

Your solutions ##u_1## and ##u_2## are generally incorrect. If ##f## is constant then you're fine.

I forgot the fact that ## f ## is a function, not a constant.

 

[mathhabibi]

I forgot the fact that ## f ## is a function, not a constant.

A week or two ago, I was trying to solve the first differential equation in your post, except that ##p=\frac f2##. According to Wolfram alpha, it is a Riccati differential equation, so it is probably unsolvable. If you find a solution to it, then get ready for fame.

 

[Math100]

I would write ##u_i =-\dfrac{2}{f}u_i''## so ##\omega =\dfrac{u_2}{u_1}=\dfrac{u_2''}{u_1''}## and start with both differentiations of ##\omega ## since you have to do this anyway.

I see where you got the ## -\frac{2}{f}u_{i}''=u_{i} ## from. But I don't understand why/how does ## w=\frac{u_{2}}{u_{1}}=\frac{u_{2}''}{u_{1}''} ##. And by starting both differentiations of ## \omega ##, do you mean to take up to second derivatives of ## \omega ##?

 

[fresh_42]

I see where you got the ## -\frac{2}{f}u_{i}''=u_{i} ## from. But I don't understand why/how does ## w=\frac{u_{2}}{u_{1}}=\frac{u_{2}''}{u_{1}''} ##. And by starting both differentiations of ## \omega ##, do you mean to take up to second derivatives of ## \omega ##?

##\omega =u_2/u_1## is a definition, and ## -\frac{2}{f}u_{i}''=u_{i} ## is the fact that they solve the differential equation. I only substituted the two into the definition of ##\omega ##. Of course, it could be that ##f(x)=0## somewhere, but I think we should first tackle the problem generically and bother about exceptions later on. Differentiating ##\omega ## yields new equations in which we could substitute former ones.

I don't think that you should try to solve the equation, especially as we don't know anything about ##f(x).## And it is not explicitly required. You are only asked about certain behaviors of flows (solutions). You can solve differential equations if you encounter easy ones e.g. ## v''+v=0## or ##v=v'.##

The general route to the proof should be:
- use as many pieces of information as you have
- calculate everything you have to calculate anyway at the beginning (for later use)
- ignore exceptions like ##f(x)=0## and deal with them at the end
- ##\omega ## is the quotient of two solutions; use this to get rid of common factors like ##-2/f.##
- draw solutions if necessary

 

[WWGD]

I'm a little stale in this area, but, have you considered using characteristic curves?

 

[Math100]

I'm a little stale in this area, but, have you considered using characteristic curves?

I'm working on this proof, almost done.

 

[fresh_42]

I'm a little stale in this area, but, have you considered using characteristic curves?

Yes, writing an article about the meaning of differential equations in general was a lot easier.

 

[Math100]

##\omega =u_2/u_1## is a definition, and ## -\frac{2}{f}u_{i}''=u_{i} ## is the fact that they solve the differential equation. I only substituted the two into the definition of ##\omega ##. Of course, it could be that ##f(x)=0## somewhere, but I think we should first tackle the problem generically and bother about exceptions later on. Differentiating ##\omega ## yields new equations in which we could substitute former ones.

I don't think that you should try to solve the equation, especially as we don't know anything about ##f(x).## And it is not explicitly required. You are only asked about certain behaviors of flows (solutions). You can solve differential equations if you encounter easy ones e.g. ## v''+v=0## or ##v=v'.##

The general route to the proof should be:
- use as many pieces of information as you have
- calculate everything you have to calculate anyway at the beginning (for later use)
- ignore exceptions like ##f(x)=0## and deal with them at the end
- ##\omega ## is the quotient of two solutions; use this to get rid of common factors like ##-2/f.##
- draw solutions if necessary

Okay, here's my revised proof on part (i).

Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u_{i}''+\frac{fu}{2}=0\implies u_{i}''=-\frac{f}{2}u ##, so ## -\frac{2}{f}u_{i}''=u_{i} ##.
Given that ## w=\frac{u_{2}}{u_{1}} ##, we have ## w'=\frac{u_{1}u_{2}'-u_{2}u_{1}'}{u_{1}^2} ##.
Since ## u_{1} ## and ## u_{2} ## are linearly independent, it follows that their Wronskian is nonzero.
Note that ## W=u_{1}u_{2}'-u_{2}u_{1}'\neq 0 ##.
This implies ## W'=(u_{1}u_{2}'-u_{2}u_{1}')'=u_{1}u_{2}''+u_{1}'u_{2}'-u_{2}u_{1}''-u_{2}'u_{1}'=u_{1}u_{2}''-u_{2}u_{1}'' ##.
Observe that ## W'=u_{1}u_{2}''-u_{2}u_{1}''=u_{1}(-\frac{1}{2}f\cdot u_{2})-u_{2}(-\frac{1}{2}f\cdot u_{1})=0 ##.
Thus, ## W=\frac{u_{2}}{u_{1}}=u_{1}u_{2}'-u_{2}u_{1}' ## is a nonzero constant ## c ##.
Therefore, ## w'=\frac{c}{u_{1}^2} ## for some nonzero constant ## c ##.

 

[Math100]

I'm a little stale in this area, but, have you considered using characteristic curves?

What are characteristic curves?

 

[fresh_42]

Okay, here's my revised proof on part (i).

Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u_{i}''+\frac{fu}{2}=0\implies u_{i}''=-\frac{f}{2}u ##, so ## -\frac{2}{f}u_{i}''=u_{i} ##.
Given that ## w=\frac{u_{2}}{u_{1}} ##, we have ## w'=\frac{u_{1}u_{2}'-u_{2}u_{1}'}{u_{1}^2} ##.
Since ## u_{1} ## and ## u_{2} ## are linearly independent, it follows that their Wronskian is nonzero.
Note that ## W=u_{1}u_{2}'-u_{2}u_{1}'\neq 0 ##.
This implies ## W'=(u_{1}u_{2}'-u_{2}u_{1}')'=u_{1}u_{2}''+u_{1}'u_{2}'-u_{2}u_{1}''-u_{2}'u_{1}'=u_{1}u_{2}''-u_{2}u_{1}'' ##.
Observe that ## W'=u_{1}u_{2}''-u_{2}u_{1}''=u_{1}(-\frac{1}{2}f\cdot u_{2})-u_{2}(-\frac{1}{2}f\cdot u_{1})=0 ##.
Thus, ## W=\frac{u_{2}}{u_{1}}=u_{1}u_{2}'-u_{2}u_{1}' ## is a nonzero constant ## c ##.
Therefore, ## w'=\frac{c}{u_{1}^2} ## for some nonzero constant ## c ##.

The line before the last should be

Thus, ## W=u_{1}u_{2}'-u_{2}u_{1}' ## is a nonzero constant ## c ##.

since it is only the numerator of ##\omega .##

 

[Math100]

The line before the last should be

since it is only the numerator of ##\omega .##

And also ## w ## instead of the Wronskian ## W ##.

 

[fresh_42]

And also ## w ## instead of the Wronskian ## W ##.

And you didn't use a division by ##f(x)## so it should work in all cases where ##f(x)## has zeros, too!

 

[WWGD]

Yes, writing an article about the meaning of differential equations in general was a lot easier.

Any Weak/Distributional solutions?