Question

Handle with care:\[\frac{\partial u}{\partial t}+c\frac{\partial u}{\partial x}+au=0.\]Let us solve for u(t,x)!

Answer 1

I actually want to re-do this to refresh my memory.

Answer 2

If I recall correctly, we need to perform a substitution. That is,\[\xi=x-ct.\]

Answer 3

This is so that the equation can be simplified by having the plane be moving along with the "wave" being described. Something akin to Einstein's relativistic concepts.

Answer 4

\[u(t,x)=v(t,x-ct)=v(t,\xi).\]We now need to find how to express the partial differentials in terms of the new function v.

Answer 5

Would separation of variables work? \[(af(x)+cf'(x))/f(x) = - g'(t)/g(t), u(x, t) = g(t)*f(x)\]

Answer 6

Or only for the heat equation?

Answer 7

\[\frac{\partial u}{\partial t}=\frac{\partial v}{\partial t}-c\frac{\partial v}{\partial \xi}\]\[\frac{\partial u}{\partial x}=\frac{\partial v}{\partial \xi}\]

Answer 8

Hmm, let me think about it.

Answer 9

I'm a bit confused with what you're trying to say with that equation up there. Is that, on the very right, the convolution of those two functions?

Answer 10

No, multiplication.

Answer 11

Substituting you can get \[\frac{\partial v}{\partial t} = -au\]

Answer 12

Ah, I see what you're doing now!

Answer 13

We could do\[\frac{\partial v}{\partial t}+av=0\]\[\frac{\partial}{\partial t}\left [ e^{at}v \right ]=0\]Since it's now basically an ODE.

Answer 14

\[e^{at}v=f(\xi)\]\[v(t,\xi)=f(\xi)e^{-at}\]

Answer 15

I'm not sure about that because the RHS has the original function \[u\] and the LHS has modified version \[v\] with the coordinate transformation.

Answer 16

Can't we replace the u on the RHS with the new function v?

Answer 17

If we can, then by re-substituting we reach\[u(t,x)=f(x-ct)e^{-at}\]to be a solution, which I'm not too sure about. I'll recheck.

Answer 18

I substituted that back into the original equation and it checked out for me

Answer 19

Phew! I got carried away reviewing all my notes just now; it's quite an intriguing subject!

Answer 20

\[\frac{\partial u}{\partial t} = -(af(x - ct)e^{-at} + cf'(x - ct)e^{-at})\] and \[\frac{\partial u}{\partial x} = f'(x - ct)e^{-at}\]

Answer 21

It really is! That's why I want to go to school for Math/CS

Answer 22

Wait, are you not in college? o.o

Answer 23

Just a lowly high school student. But I enjoy math.

Answer 24

Oh my! I am a graduate student. Haha, you must be really, really smart!

Answer 25

Lol, Well thank you. I'm sure you are too, I mean graduate school is pretty tough.

Answer 26

They don't teach anything beyond Calculus I in American high schools (excluding outlier schools). Where are you from?

Answer 27

I'm from Indiana. I actually taught myself most of what I know about advanced mathematics. Mainly from lecture notes that professors posted from their classes. Right now I'm taking Calculus 1 in school.

Answer 28

I did take a few college dual credit trigonometry and college algebra classes earlier in high school.

Answer 29

I see! Long live the open course-ware, huh? :) I wish I would've taught myself more interesting things whilst still a high school student. You're doing a great thing!

Answer 30

Well, I've always enjoyed learning new things.

Answer 31

Mathematica for the folks who cannot integrate x with pencil and paper.