Why is the total derivative of a one dimensional solution to the wave equation zero?

323 Views Asked by At

Page 23 below is from Garrity's book "Electricity and Magnetism for Mathematicians". He is discussing how changing to a coordinate frame moving with some speed relative to the original coordinate frame results in our solution to the one dimensional wave equation having a corresponding change in speed.

However, in the discussion, he uses a fact I don't understand. If y(x,t) is our solution to the one dimensional wave equation, then why should 0 = dy/dt? Please see picture below. Thanks for the help!

It seems like dy/dt = 0 because we are restricting attention to just values of x and t that give us y0 = y(x,t). If this is the case, then our result is only valid for those special x and t values?

enter image description here

2

There are 2 best solutions below

0
On BEST ANSWER

It seems he's trying to derive the wave equation. He's doing that by "following" a point on a wavefront, and showing how the derivatives of are then related.

Here's what's happening: he has some function $y(x,t)$ which tells us the amplitute at a specific place $x$ and time $t$. Then he's asking us to follow, say, the node of a wave over time. This corresponds to setting $y(x,t)=y_0$. Then we no longer have $x$ and $t$ as independent variables; rather $x$ secretly depends on $t$, through the function $y$. Then we can take derivatives of both sides with respect to $t$. Clearly the derivative of the right side is $0$, which is what that first line says. Then the derivative of the second side is found using the chain rule. From this, and the fact that given these constraints $\partial x/\partial t$ is the velocity of the wave, you can derive the equation at the bottom.

All of this math is meant to constrain the form that $y$ can take, if we want it to represent a wave.

0
On

Don't ask me why, but he writes that $y(x,t)=y_0$ the the line $x=x_0+vt$. Since $y_0$ is a constant with respect to $t$, so must be $y(x_0+vt,t)$.