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?