Calculus ## G(y, z)=z\frac{\partial}{\partial z}F(y, z)-F(y, z) ##?

[Math100]

Homework Statement

Let $n\geq 1$ be an integer and let $F(y, z)=y^{n-1}\sqrt{y^2+z^2}$.
a) Calculate $G(y, z)=z\frac{\partial}{\partial z}F(y, z)-F(y, z)$, and solve the equation $G(y, z)=C$ for $z$, where $C$ is a constant and $y^{2n}\geq C^2$.
b) Let $C>0, a>0, A>0$ be constants, with $na<\pi/2$. Show that the solution $y(x)$ on the interval $-a\leq x\leq a$ of the first-order differential equation $C^{2}(\frac{dy}{dx})^2=y^2(y^{2n}-C^2), y(a)=A, y'(0)=0, y^{2n}\geq C^2$, is $y(x)=A(\frac{cos(na)}{cos(nx)})^{1/n}$. You may use the integral $\int\frac{dy}{y(y^{2n}-C^2)^{1/2}}=\frac{1}{nc}arccos(\frac{C}{y^{n}})+constant$, where $n>0, C>0$ are constants.

Relevant Equations

None.

a) Observe that $\frac{\partial}{\partial z}F(y, z)=y^{n-1}\cdot \frac{2z}{2\sqrt{y^2+z^2}}=\frac{zy^{n-1}}{\sqrt{y^2+z^2}}$.
This means $G(y, z)=\frac{z^2\cdot y^{n-1}}{\sqrt{y^2+z^2}}-y^{n-1}\cdot \sqrt{y^2+z^2}=\frac{z^2\cdot y^{n-1}}{\sqrt{y^2+z^2}}-\frac{y^{n-1}(y^2+z^2)}{\sqrt{y^2+z^2}}=\frac{-y^{n+1}}{\sqrt{y^2+z^2}}$.
Let $G(y, z)=C$.
Then $\frac{-y^{n+1}}{\sqrt{y^2+z^2}}=C\implies \sqrt{y^2+z^2}=\frac{-y^{n+1}}{C}$,
so $y^2+z^2=\frac{y^{2n+2}}{C^2}\implies z^2=\frac{y^{2n+2}}{C^2}-y^2$.
Thus $z=\pm\sqrt{\frac{y^{2n+2}-C^2y^2}{C^2}}=\pm\sqrt{\frac{y^2(y^{2n}-C^2)}{C^2}}$.
Therefore, $z=\pm\frac{y\cdot \sqrt{y^{2n}-C^2}}{C}$.

b) Consider the first-order differential equation $C^2(\frac{dy}{dx})^2=y^2(y^{2n}-C^2), y(a)=A, y'(0)=0, y^{2n}\geq C^2$.
Then $C^2(\frac{dy}{dx})^2=y^2(y^{2n}-C^2)$
$\frac{C^2(\frac{dy}{dx})^2}{y^2(y^{2n}-C^2)}=1$
$\frac{(\frac{dy}{dx})^2}{y^2(y^{2n}-C^2)}=\frac{1}{C^2}$
$\frac{dy}{dx}=\pm\sqrt{\frac{1}{C^2}\cdot [y^2(y^{2n}-C^2)]}$
$\frac{dy}{dx}=\pm\sqrt{\frac{1}{C^2}}\cdot \sqrt{y^2(y^{2n}-C^2)}$
$\int\frac{dy}{\sqrt{y^2(y^{2n}-C^2)}}=\int\sqrt{\frac{1}{C^2}}dx$ since $C>0$
$\int\frac{dy}{\sqrt{y^2}\cdot \sqrt{y^{2n}-C^2}}=\int\frac{\sqrt{C^2}}{C^2}dx$
$\int\frac{dy}{y\cdot \sqrt{y^{2n}-C^2}}=\int\frac{\sqrt{C^2}}{C^2}dx$.
Now we use the integral $\int\frac{dy}{y(y^{2n}-C^2)^{1/2}}=\frac{1}{nC}arccos(\frac{C}{y^{n}})+constant$ to get
$\frac{1}{nC}arccos(\frac{C}{y^{n}})+constant=\frac{1}{C}x$.
Let $K=constant$.
Then $arccos(\frac{C}{y^{n}})=nC(\frac{1}{c}x-K)\implies \frac{C}{y^{n}}=cos(nx-nCK)\implies y^{n}=\frac{C}{cos(nx-nCK)}$,
so $y=(\frac{C}{cos(nx-nCK)})^{1/n}=(C\cdot sec(nx-nCK))^{1/n}$.
Applying the initial condition of $y(a)=A$ produces $A=(\frac{C}{cos(na-nCK)})^{1/n}\implies A^{n}=\frac{C}{cos(na-nCK)}$,
so $C=A^{n}\cdot cos(na-nCK)$.
Note that $y'=\frac{1}{n}(C\cdot sec(nx-nCK))^{\frac{1-n}{n}}\cdot C\cdot sec(nx-nCK)\cdot tan(nx-nCK)$.
Now we have $y'(0)=0\implies 0=\frac{1}{n}(C\cdot sec(-nCk))^{\frac{1-n}{n}}\cdot c\cdot sec(-nCk)\cdot tan(-nCK)$.
Thus $0=\frac{1}{n}(C\cdot sec(nCK))^{\frac{1-n}{n}}\cdot C\cdot sec(nCK)\cdot tan(nCK)$ (since $sec(x)$ is an even function)
$0=\frac{1}{n}(C\cdot sec(nCK))^{\frac{1}{n}-1}(C\cdot sec(nCK))\cdot tan(nCK)$
$0=\frac{1}{n}(C\cdot sec(nCK))^{\frac{1}{n}}\cdot tan(nCK)$ for $n\neq 0, C>0$.
After simplifying the equation, we have $tan(nCK)=0\implies K=0$.
Hence, $C=A^{n}\cdot cos(na)$ and $y=(A^{n}\cdot cos(na)\cdot sec(nx))^{1/n}=A(cos(na)\cdot \frac{1}{cos(nx)})^{1/n}$.
Therefore, we have shown that the solution is $y(x)=A(\frac{cos(na)}{cos(nx)})^{1/n}$.

 

[fresh_42]

You should add the possibility $C=0 \Rightarrow y=0$ in a). I would finish part a) slightly differently. We have in case $C\neq 0$
\begin{align*}
G(y,z)=C &\Rightarrow - \dfrac{y^{n+1}}{\sqrt{y^2+z^2}}=C \\
&\Rightarrow z=\pm \dfrac{y}{C}\sqrt{y^{2n}-C}=\pm \dfrac{y}{C}\sqrt{(y^n-C)(y^n+C)}\in \mathbb{R}
\end{align*}
Note that we lost the sign when we squared the equation, so we will have to keep this in mind if not even discuss it. Since $\sqrt{y^2+z^2}\geq 0$ we have that
$$
-\dfrac{y^{n+1}}{C}=\sqrt{y^2+z^2 }\geq 0.
$$
Now, b). I think you should make your life easier. As long as you do not determine the constant via the initial values you are allowed to use the constant as a bin where you keep one variable name even if it changes value. Just make sure you do not put functions or variables into it. We also should keep the known constants since they are part of the problem description. But all other constants are simply $K.$ Hence we have
\begin{align*}
y'=\dfrac{dy}{dx}&= \pm \dfrac{y}{C}\sqrt{y^{2n}-C^2} \\
&\Rightarrow \dfrac{dy}{y\sqrt{y^{2n}-C^2}}=\pm \dfrac{dx}{C}\\
&\Rightarrow \int \dfrac{dy}{y\sqrt{y^{2n}-C^2}}=\pm \int \dfrac{dx}{C}\\
&\Rightarrow \dfrac{1}{nC}\operatorname{arccos}\dfrac{C}{y^n}=\pm \dfrac{x}{C} + K\\
&\Rightarrow \operatorname{arccos}\dfrac{C}{y^n}=\pm nx +K\\
&\Rightarrow \dfrac{C}{y^n}= \cos(\pm nx+K)=\cos(nx+K)\\
&\Rightarrow y^n=\dfrac{C}{\cos(nx+K)}
\end{align*}
Differentiation yields $ny^{n-1}y'=-nC\dfrac{\sin(nx+K)}{\cos^2(nx+K)}$ and from $y'(0)=0$ we get $\sin K=0$ and from $0=x\in [-a,a]\subseteq [-\pi/2n\, , \,\pi/2n]$ that $K=0.$ We thus have
$$
y^n(x)=\dfrac{C}{\cos(nx)}
$$
There is no need to use the secant function or even to draw the root. We also know that $y(a)=A.$ So
\begin{align*}
A^n=\dfrac{C}{\cos na} &\Longrightarrow C=A^n\cos (na) \Longrightarrow y^n(x)=A^n\dfrac{\cos(na)}{\cos(nx)}
\end{align*}
I think you generally make your mathematical life more complicated than it is necessary. That has the disadvantage that it costs time and is a possible source of calculation errors. Plus you need more trig functions and thus more formulas.

The basic principles of my calculation were:

1. Do not resolve equations if it is not necessary.
   This means, I worked with $y^n=f(x)$ instead of $y=f^{1/n}(x)=\sqrt[n]{f(x)}.$ It simplifies the algebra and even differentiation was a lot easier.
2. Use what you have in the problem statement when you need it.
   This means I didn't care about any integration constants. We have had two integrations, and a factor $n$. They all are simply $K$ and only at the end, I used the given initial values to determine $K$ and not prior to the point when I needed it.
3. And last but not least: I scribbled a lot into my notebook and I used your results (after I checked them). What looks so simple here looks like a mess in my notebook. This means: only write down the necessary steps in your final solution.

 

[Math100]

You should add the possibility $C=0 \Rightarrow y=0$ in a). I would finish part a) slightly differently. We have in case $C\neq 0$
\begin{align*}
G(y,z)=C &\Rightarrow - \dfrac{y^{n+1}}{\sqrt{y^2+z^2}}=C \\
&\Rightarrow z=\pm \dfrac{y}{C}\sqrt{y^{2n}-C}=\pm \dfrac{y}{C}\sqrt{(y^n-C)(y^n+C)}\in \mathbb{R}
\end{align*}
Note that we lost the sign when we squared the equation, so we will have to keep this in mind if not even discuss it. Since $\sqrt{y^2+z^2}\geq 0$ we have that
$$
-\dfrac{y^{n+1}}{C}=\sqrt{y^2+z^2 }\geq 0.
$$
Now, b). I think you should make your life easier. As long as you do not determine the constant via the initial values you are allowed to use the constant as a bin where you keep one variable name even if it changes value. Just make sure you do not put functions or variables into it. We also should keep the known constants since they are part of the problem description. But all other constants are simply $K.$ Hence we have
\begin{align*}
y'=\dfrac{dy}{dx}&= \pm \dfrac{y}{C}\sqrt{y^{2n}-C^2} \\
&\Rightarrow \dfrac{dy}{y\sqrt{y^{2n}-C^2}}=\pm \dfrac{dx}{C}\\
&\Rightarrow \int \dfrac{dy}{y\sqrt{y^{2n}-C^2}}=\pm \int \dfrac{dx}{C}\\
&\Rightarrow \dfrac{1}{nC}\operatorname{arccos}\dfrac{C}{y^n}=\pm \dfrac{x}{C} + K\\
&\Rightarrow \operatorname{arccos}\dfrac{C}{y^n}=\pm nx +K\\
&\Rightarrow \dfrac{C}{y^n}= \cos(\pm nx+K)=\cos(nx+K)\\
&\Rightarrow y^n=\dfrac{C}{\cos(nx+K)}
\end{align*}
Differentiation yields $ny^{n-1}y'=-nC\dfrac{\sin(nx+K)}{\cos^2(nx+K)}$ and from $y'(0)=0$ we get $\sin K=0$ and from $0=x\in [-a,a]\subseteq [-\pi/2n\, , \,\pi/2n]$ that $K=0.$ We thus have
$$
y^n(x)=\dfrac{C}{\cos(nx)}
$$
There is no need to use the secant function or even to draw the root. We also know that $y(a)=A.$ So
\begin{align*}
A^n=\dfrac{C}{\cos na} &\Longrightarrow C=A^n\cos (na) \Longrightarrow y^n(x)=A^n\dfrac{\cos(na)}{\cos(nx)}
\end{align*}
I think you generally make your mathematical life more complicated than it is necessary. That has the disadvantage that it costs time and is a possible source of calculation errors. Plus you need more trig functions and thus more formulas.

The basic principles of my calculation were:

1. Do not resolve equations if it is not necessary.
   This means, I worked with $y^n=f(x)$ instead of $y=f^{1/n}(x)=\sqrt[n]{f(x)}.$ It simplifies the algebra and even differentiation was a lot easier.
2. Use what you have in the problem statement when you need it.
   This means I didn't care about any integration constants. We have had two integrations, and a factor $n$. They all are simply $K$ and only at the end, I used the given initial values to determine $K$ and not prior to the point when I needed it.
3. And last but not least: I scribbled a lot into my notebook and I used your results (after I checked them). What looks so simple here looks like a mess in my notebook. This means: only write down the necessary steps in your final solution.

This is much easier!

 

[Math100]

Also, isn't it $ny^{n-1}y'=nC\frac{sin(nx+K)}{cos^2(nx+K)}$?

 

[fresh_42]

Also, isn't it $ny^{n-1}y'=nC\frac{sin(nx+K)}{cos^2(nx+K)}$?

Let's see.
$$
(\cos^{-1}(nx))' = (-1)\cdot \cos^{-2}(nx)\cdot (\cos(nx))'=(-1)\cdot \cos^{-2}(nx)\cdot (-\sin (nx))\cdot n
$$
You are right. I made a sign error. But it makes no difference since it equals zero anyway.

 

[Math100]

Let's see.
$$
(\cos^{-1}(nx))' = (-1)\cdot \cos^{-2}(nx)\cdot (\cos(nx))'=(-1)\cdot \cos^{-2}(nx)\cdot (-\sin (nx))\cdot n
$$
You are right. I made a sign error. But it makes no difference since it equals zero anyway.

Correct, it's pointless because it equals $0$ anyways. I just feel relieved a lot more since I don't have to include the $sec(x)$ function on part b), and it makes the proof a lot shorter than before.

 

[fresh_42]

Correct, it's pointless because it equals $0$ anyways. I just feel relieved a lot more since I don't have to include the $sec(x)$ function on part b), and it makes the proof a lot shorter than before.

[Smart Experimentalist]: 'Yeah, it is reminiscent of what distinguishes the good theorists from the bad ones. The good ones always make an even number of sign errors, and the bad ones always make an odd number.'"-Anthony Zee, Quantum Field Theory in a Nutshell

from https://www.physicsforums.com/threads/great-and-favorite-quotes.12079/