PDE 4 | Transport equation: general solution

Hello and welcome to the second video
on the ‘Transport equation’. In this video we’re going to solve
the Transport equation. Remember that in last video we essentially derived the Transport
equation as being the simplest partial differential equation ( PDE )
which has, as a solution, a traveling wave. So, wave that looks like this,
like a water wave or something like that, which is moving with a
speed ‘c’. Now, let’s imagine that the two of us are out looking at this wave, and ‘you’ are standing on the ground,
watching the wave move forward with the speed of ‘c’.
(there you are.) And ‘I’ am standing on a cart, and that cart is moving forward with a speed of ‘c’. (Same speed is the wave.) It should be pretty obvious that I
will see something that looks a lot different than what you see. I will see a wave
which is, essentially, not moving at all. And you will see a wave, which is the same
wave that we’ve been talking about along, ( one moving with speed ‘c’.) Let’s kind of write that down.
We’re going to make this ‘mathematically precise’. So: ” I see.. ” ” a stationary wave. ” I see a stationary wave, and: ” You see.. ” a ” wave with speed ‘c’. ” Alright, you also see the Transport equation. You look at ‘u’ and you say:
“Okay, this wave satisfies the Transport equation.” Now the question here (and this is
essentially the ‘main’ question, the key question) : what PDE do I see? ” What PDE do I see ? ” And the answer to this will be the
simplification that we want to make. Okay let’s make this a little bit more precise, mathematically. Let’s do that by giving ‘me’ a coordinate
system, which is the coordinate system on the cart. Okay so here is the the coordinate system
that we’ve been using all along. (the coordinate system where I draw this wave) Now let’s imagine that we have another
coordinate system, and let’s attach it to the cart. Okay, so I will try to draw this.. so that it looks like it’s
attached to the cart. (Here it is.. ..and it’s attached to the cart.) Now suppose that at an ‘initial’ time (so, say.. ” time 0 “..) we both agree on ‘a point’. Okay and we will say that
this is the ‘same’ point. (it’s just the one dimensional coordinate system) This is the ‘same’ point:
let’s call it ‘x’. Now we wait a little while by a ” time t ” (so, some time has passed) and we look at that point again and we ask: if it still looks
like the same point to both of us? And here it is for you: ‘x’ But an amount of time has passed,
and my cart is moving forward. So here’s my cart, right here, it’s traveled some distance. And my coordinate system is
still attached to the cart. So, I will say it doesn’t look like it’s the same point anymore, in ‘my’ coordinate system, ‘Here’ is where it would be if it was still in the same place in my coordinate system.
But it’s not there anymore it’s right back ‘here’, because we’re still talking about the same point. And so it’s at: ” x – c*t “, because my cart is moving forward with a speed of ‘c’,
so I have to ‘back it up’, back-up the coordinate measurement by ‘c*t’ in order to talk about the ‘same’ point. And so the moving coordinate, which is the coordinate in on the cart, called just a: ” “moving” coordinate “, I will call it the Greek letter ‘Xi’ ( ”xi” )
( pronounced: “sea” or “zee” ) I know that’s sort of a hard one to say and write,
but it’s (kind of) ‘convention’. So in other words, what I want to do at this point,
is I want to rewrite the Transport equation, (which is the differential equation that ‘you’ see) in terms of the coordinates ”xi” and ‘t’ (time), and this will be the PDE that I ‘see’. So we both look at the same wave, and what you ‘see’ is a wave: ” u(x,t) ” And, in the ‘moving’ coordinates, I ‘see’ the same wave,
(that’s why I have an equality ( ‘=’ ) here) but I’m measuring it in ‘my’ coordinates, and so we’ll call it: ” v(‘xi’,t) “. (these are ‘my’ coordinates) Now let’s see what differential equation v(‘xi’,t) satisfies,
just based on the fact that we know u(x,t) satisfies the Transport equation? So here we’re going to have to use the ‘chain-rule’ (so brace yourself for some chain-rule computation.) Let’s compute each of these partial derivatives,
which are in the Transport equation, and plug-in their versions of partial
derivatives of v(‘xi’,t). So, what is: ‘u_t’? Well ‘u_t’ is ( because ‘u’=’v’ )
is going to be the partial derivative of ‘v’ with respect to ”xi”.. first.. times.. partial of.. ”xi” with respect to ‘t’ plus.. partial of ‘v’ with respect to,
the second coordinate, ‘t’, times.. the partial.. with respect to ‘t’. So this is the chain-rule:.
to take the derivative of ‘v(‘xi’,t)’ w.r.t. ‘t’.. I have to take the derivative w.r.t. the 1st component.. and then the derivative of the 1st component w.r.t. ‘t’.. plus, the derivative w.r.t. the 2nd component.. and then the derivative of the 2nd component w.r.t. ‘t’. Similarly, I would have.. ‘u_x’… equals partial ‘v’ partial ”xi”.. times.. partial ”xi”, partial ‘x’, plus.. partial ‘v’.. partial ‘t’, times.. partial ‘t’.. partial ‘x’. Now let’s simplify this little bit. Some of these things are kind of ‘silly ‘to write down. For instance: ” partial ‘t’.. partial ‘x’..” Why would I write that down?
(It doesn’t depend on ‘x’, right? so that’s ‘0’) So let’s get rid of this (this is ‘0’ over here)… ..so let’s erase it. (Rewind the video if you want to see that again) and: ” partial ‘t’.. partial ‘t’ ..” (that’s just ‘1’), and, what about these guys here?
Well, ” partial ”xi”..” with respect to ‘t’.. well, ”xi” is this ‘moving’ coordinate.. and so the partial derivative w.r.t. ‘t’ is: “- c” (So, let’s write that in..) And ” partial ”xi” ” with respect to ‘x’? ..that’s just ‘1’. Okay, so we’ve made our simplifications. Now, let’s take each of these guys and let’s plug them into the
Transport equation. So let’s take this guy and plug it in for ‘u_t’ in the
Transport equation, and.. let’s take this guy and plug it in and for ‘u_x’ in the
Transport equation. So the Transport equation:
“u_t + c*u_x”.. becomes.. in terms of ‘these’ partial derivatives: ” -c*(v_’xi’) + v_t ” (that’s just the ‘u_t’ term, here) plus: ” c*(v_’xi’) ” (this right here is the ‘u_x’ term). And what you notice here is that these two terms cancel one another out. And so this guy becomes ‘v_t’. Right that’s very nice and simple,
so what PDE do I see? I see.. ” v_t=0 ” All right, so that’s the answer to this
question: “what PDE do I see?” Now, let’s solve that: So, just to summarize.. we have in our usual coordinates: ” Usual coords : “.. ” (x,t) ” We have.. the ‘wave’.. ” u(x,t) ” and the Transport equation:
” u_t + c*u_x=0 “. Now, on the other hand,
we have (in) the ‘moving’ coordinates.. ” (‘xi’,t) “.. we have the same ‘wave’ that we write as a function now of ”xi” and ‘t’.. (I write “v(‘xi’,t) ” because it may be not the same function) and that, we find satisfies: ” v_t=0 “. Now the procedure will be to solve ‘this’. (So) the procedure is.. ” solve “.. ” v_t=0 ” Then ” use it “.. ” to find “.. ” u(x,t) “. So let me just show you how this works. The first thing is, is that solving ” v_t=0 ” is very simple. Alright, if: ” v_t=0 “, what does that mean? That means that ‘v’ is a function which
does not depend on ‘t’. (that’s what it means for the partial derivative to be ‘0’) So: ” ‘v’ is independent of ‘t’ ” What that means is that
‘v’ (=F(‘xi’;t=const)=F(‘xi’) ) can be: ‘any function at all’, depending only on ”xi”. Now you may object, and you may say:
” Well, ”xi” is a function of ‘t’ “. (because, remember ”xi” was equal to: ” x – c*t “) But I’m not thinking of ”xi” as being something
depending on ‘t’ when I write down ‘v’. I’m thinking of ”xi” as being a new coordinate
which ‘I’ use (it’s the coordinate on my cart). I don’t think about the coordinates on my
cart as depending on ‘time’, I think about them just as being a measurement of ‘distance’ from my cart. That means that ‘v’ is independent of ‘t’,
so ‘v is a function, any function at all of ”xi”. (any ‘nice’ function). And we know that’s the
‘general solution’ to this ‘stationary’ wave equation. Well, then we can use this solution to go back and find ‘u’.
So we know that ” v(‘xi’,t) ” is equal to.. ” u(x,t) ” and so what that means then is: ” u(x,t) ” is equal to.. ” F(‘xi’) “,
(that’s our solution) and since ”xi” is equal to: ” x – c*t “, that means that in my old coordinates I can substitute back in the expression for ”xi”, (and so this is the final result.) So, what’s the conclusion? ” Conclusion “.. the ” General solution “.. ” of u_t + c*u_x”.. (the Transport equation) ” is F(x – c*t) “.. ” where F is ” a ‘nice’ function, as before and we’re going to take ‘nice’ to mean just once differentiable, because I have to take its derivative in
order to plug it into the Transport equation to verify that it’s a solution. (so it should be differentable one-time) So in other words:
its first derivative is continuous, and I can compute it. (it has a first derivative)

Bernard Jenkins


  1. Yes. If c is negative then the wave would be moving in the opposite direction. For all
    practical purposes c is the velocity of the wave.

  2. wow v_t = 0 makes a lot of sense. … i mean you're traveling exactly the same speed with the wave… so at every second, the wave looks the same as the previous second.

  3. If c is not a constant and it was function of tiem c(t), how does this affects the solution? Would the trasnport equation be the same? Thank you,

  4. Why do you say 'it is once differentiable – in other words ist first derivative is contiuous and I can compute it'? One seems to always use 'C1' equivalently to simply 'differentiable'.. Why can we generalize that?

  5. I like how you kept this derivation very abstract. Now I can say I fully understand it. Thanks a lot for making the video. Have my like! 🙂

  6. EXCELLENT!!! Why don't exist so much teachers like you? thank you very much for sharing it

  7. Around 12.00, you say but xi is not dependent on t, xi is a coordinate measuring the distance of the cart from the original position. However the distance of the cart depends on t, quite explicitly. The end part, once you are solving v(t) is not coherent imo

  8. This video is perfect for a student studying PDEs, trying to grasp this concept

  9. If the wave is moving at the speed of light then both people must measure the wave travelling at the speed of light

  10. Thank you so much!!!! You are the best teacher and life saver! I wish you can be my professor in real life

  11. Does anyone have a recommendation for material on the non-uniform transport equation where c=c(x)

  12. Apreciate the work you have done on these videos they are really helpful!

  13. Such a good series so far, very pedagogical and gives you that deep understanding and intuition that makes you want to watch the next video to find out more. Definitely sets you up nicely for the method of characteristics. Good thing I found this channel so I dont have to waste more time in class. Tbh, the classical lecture format is ass, because even if you get that 1 in 20 great lecturer, you still cant pause and rewind when you need a little extra time to think. Thank you for these videos!

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