Textbook 'The Physics of Waves': Derive Complex Solution Form for SHM

[brettng]

TL;DR Summary: Question on deriving the complex irreducible solution form for simple harmonic motions based on time translation invariant.

Reference textbook “The Physics of Waves” in MIT website:
https://ocw.mit.edu/courses/8-03sc-...es-fall-2016/resources/mit8_03scf16_textbook/

Chapter 1 - Section 1.5 [Pages 21 & 22] (see attached file)

Question: In the content, it states that we are differentiating both sides of (1.74) with respect to a, which results in (1.82). However, when obtaining (1.85) from (1.83), the integration is done with respect to t. Grateful if you could explain explicitly why the mathematics are valid here.

 

[jambaugh]

If two functions of many variables are equal then so are their partial derivatives, (which are again two equal functions of many variables). If two function of many variables are equal then their integral with respect to one of the variables are equal up to a constant w.r.t. that variable, i.e. a function of the remaining variables.
And, this "constant" of integration can be solved for given an initial/boundary condition.

 

[brettng]

Thank you for your reply. But how do we usually write the mathematics explicitly?

Differentiate $z(t+a)=h(a)z(t)$ with respect to $a$, and set $a=0$.
$$\frac {\partial z(t+0)} {\partial a}=\frac {dh(0)} {da}z(t)$$
$$\frac {\partial z(t)} {\partial a}=Hz(t)$$
At this stage, the derivative of the function $z(t)$ is still with respect to $a$. And I have no idea how to write the next step in order to obtain equation (1.85). Grateful if you could help please.

 

[renormalize]

Differentiate $z(t+a)=h(a)z(t)$ with respect to $a$, and set $a=0$.
$$\frac {\partial z(t+0)} {\partial a}=\frac {dh(0)} {da}z(t)$$

This means setting $a=0$ after differentiating, not before. What you've written is $\frac{\partial z(t)}{\partial a}$, which of course vanishes because $z(t)$ is not a function of $a$.
Instead, differentiating and then setting $a=0$ gives:$$\frac{\partial z(t+a)}{\partial a}\Biggl|_{a=0}=\frac{\partial z(t+a)}{\partial t}\Biggl|_{a=0}=\frac{\partial z(t)}{\partial t}$$ which leads to the book's result.

 

[brettng]

This means setting $a=0$ after differentiating, not before. What you've written is $\frac{\partial z(t)}{\partial a}$, which of course vanishes because $z(t)$ is not a function of $a$.
Instead, differentiating and then setting $a=0$ gives:$$\frac{\partial z(t+a)}{\partial a}\Biggl|_{a=0}=\frac{\partial z(t+a)}{\partial t}\Biggl|_{a=0}=\frac{\partial z(t)}{\partial t}$$ which leads to the book's result.

Could you elaborate more about the 1st equal sign?

Because to me, I am thinking a $t$-$a$-$z$ 3D geometry (similar to $x$-$y$-$z$ plane), that $\frac{\partial z(t+a)}{\partial a}\Biggl|_{a=0}$ means the “slope in $a$ direction” of graph $z(t+a)$ evaluated at $a=0$.

However, your middle formula is the “slope in $t$ direction” of graph $z(t+a)$ evaluated at $a=0$.

It seems that these 2 slopes are orthogonal. So, may I know why they are equal in this sense? In other words, why we can change $a$ to $t$ directly in the 1st equal sign?

 

[renormalize]

Could you elaborate more about the 1st equal sign?

The equality follows simply from the chain rule and it holds for all values of $t,a$. Let $u=t+a$. Then the two derivatives are:$$\frac{\partial z(t+a)}{\partial a}=\frac{\partial z(u)}{\partial a}=\frac{dz(u)}{du}\frac{\partial u}{\partial a}=\frac{dz(u)}{du}\frac{\partial(t+a)}{\partial a}=\frac{dz(u)}{du}\tag{1a}$$and:$$\frac{\partial z(t+a)}{\partial t}=\frac{\partial z(u)}{\partial t}=\frac{dz(u)}{du}\frac{\partial u}{\partial t}=\frac{dz(u)}{du}\frac{\partial(t+a)}{\partial t}=\frac{dz(u)}{du}\tag{1b}$$Therefore:$$\frac{\partial z(t+a)}{\partial a}=\frac{dz(u)}{du}=\frac{\partial z(t+a)}{\partial t}\tag{2}$$as asserted.

 

[brettng]

The equality follows simply from the chain rule and it holds for all values of $t,a$. Let $u=t+a$. Then the two derivatives are:$$\frac{\partial z(t+a)}{\partial a}=\frac{\partial z(u)}{\partial a}=\frac{dz(u)}{du}\frac{\partial u}{\partial a}=\frac{dz(u)}{du}\frac{\partial(t+a)}{\partial a}=\frac{dz(u)}{du}\tag{1a}$$and:$$\frac{\partial z(t+a)}{\partial t}=\frac{\partial z(u)}{\partial t}=\frac{dz(u)}{du}\frac{\partial u}{\partial t}=\frac{dz(u)}{du}\frac{\partial(t+a)}{\partial t}=\frac{dz(u)}{du}\tag{1b}$$Therefore:$$\frac{\partial z(t+a)}{\partial a}=\frac{dz(u)}{du}=\frac{\partial z(t+a)}{\partial t}\tag{2}$$as asserted.

Thank you so much for your help!!!!!!