I'm practically blown away!! I'm at the phase of research where I'm trying to understand how PDEs are embedded in Machine learning loss function. Viola! Here I am consuming mathematical chocolates!
Agreed. I always thought Gilbert Strang at MIT is a great math teacher. Steve has proven to be just as good, if not better, than Prof Strang. Kudos to eigensteve 🙏🙏🙏
I am blown away at how easily understandable this derivation is. Every step is broken down into high school calculus concepts. We do not have time in our DE course to derive this but just to understand that the vertical acceleration is proportional to the curvature of the string is just so perfectly intuitive that it makes me wonder why it wasn’t explained to us this way to us before. Thank you so much for this thoughtful video.
This video series explains why it's harder and harder to resist binging UA-cam these days, any other series like this? The new videos are literally in sync with my PDE class oh my goat
Very intuitive and easy to understand. I also appreciated the emphasis on how it was not that easy to finally get comfortable with the manipulation of such an equation. Thank you very much for this video and for the whole channel 🙏
From a geometrical standpoint, the laplace equation means: "scalar field without local max and min"; heat equation means: "the change in one variable is proportional to the curvature in another"; and the wave equation means: "the curvature in one variable is proportional to the curvature in another". If you can imagine how the information change, you can easiliy derivide this partial differential equation.
I had the exact same block when starting to learn about PDEs. This derivation is so crystal clear about what assumptions are being made and why they are made.
Thanks Steve. For many years I have dealing in higher maths subjects and, honestly, this is one of the best lectures I have watched (or physically attended). Please, keep producing tuition material at this level of excellency.
Awesome intuitive approach to setting up the wave equation from F = ma. Reminds me of my General Physics course when I was reading the Young and Freedman text.
What I didn't get the first time I saw this derivation is why the length of the rope is dx (in the context of the mass). it's actually: sqrt(dx² + dy²) = dx•sqrt(1 + (dy/dx)²) = dx•sqrt(1+(y')²) But since we assume small oscillations all nonlinear terms are negligible so ds = dx•sqrt. It's similar to us saying cos(θ) = 1 and not 1 + θ²/2! + .... Hope this helps someone!
Very neat introduction to the wave equation, well done prof ! One could add - just for more fun - that those smart mathematicians from the 18th century wrote all their ‘papers’ in Latin and so the obvious symbol choice to represent speed was the letter c .. speed being “celeritas” in Latin. Funny how Latin even got into the most famous among all equations, even if Albert Einstein didn’t use c initially 😊
How we derive that c^2 = T / ro? In this video Steve is explaining how to derive the wave equation Utt=c^2 * Uxx - correct? From F=ma Steve derives Utt = T / ro * Uxx, and then at 25:15 he just says “where c squared is equal to T (tension) divided by ro(linear density)”. Where that comes from? How speed comes into the equation?
I’m just guessing here but Tension is measured in Newtons (N) which can be expressed by (kg)(m)/s^2. Linear density is measured in kg/m. If you divide the units for tension over the units of linear density you get ((kg)(m)/s^2)/(kg/m). If you simplify this to (kg)(m)/s^2 x m/kg you can cancel out the kg. Then you are left with m^2/s^2 which is the square of velocity (m/s). So if c represents velocity then c^2 is the velocity squared solved by the units.
A very nice lecture, thank you! I have a minor comment though regarding the derivation. When considering the force equilibrium of the infinitesimal element, I am afraid the equilibrium in not maintained if the two tangential forces T at the ends are identical. What must be identical to prevent horizontal movement are the horizontal projections of these forces, say “N”. When these horizontal forces are identical, the vertical projections of the tensile forces T, which we can call F, are equal N*tan(theta). And it is the difference between these vertical forces: N*[tan(theta+dtheta)-tan(theta)], which equals the Newton’s inertia forces “m*a”. This is just a minor fix which removes the weak arguments (time 20:20) about sine being roughly equal to tan, which is equal to the angle itself, and cosine being roughly equal to one for small angles theta.
Very nice presentation. I struggle with your interpretation of speed c. You see c in the guitar-cord example as the speed of the string in the y-direction. In the classical wave equation, however, the speed c is the speed in the x-direction, as in a waterwave or in a em-wave (i.e. how fast the wave runs)
Thanks for making this content openly available! It has certainly been extremely helpful while brushing up my memory on these concepts. I did have a question: Couldn't we skip the sin(theta) ~ tan(theta) step altogether by utilizing the requirement that the x-component of the tension at points x and x+dx must be equal (in opposite directions)? At either point, we have tan(theta) = T_y/T_x. Solving for T_y, we have T_y = T_x*tan(theta). Again, T_x is the same at both x and x+dx (save for the minus sign), so it can be factored out when calculating the net vertical force, F = T_y(x + dx) + T_y(x) = T_x*[tan(theta + dtheta) - tan(theta)]. Thanks again!
That was a very nice video. I will refer my students to this video in my PDE class. There is something that I don't quite get though. It might be just a detail and probably an approximation matter (I'm a mathematician by formation, so I do have a hard time sometimes with getting the right approximations). When we pitch a string of a guitar to move it to its original shape, say h(x), and then release it, do we assume that the length of the string does not change? I am asking this question because I have a hard time seeing how the length L would remain constant. For instance, in the picture you draw of a guitar string attached at x = 0 and x = L, the string must be longer than L to get that curved shape. Do we allow that elasticity so that there is a change in the value of the tension in the string?
You are not only a great professor, but also a great person!!! Great great Thanks and God Bless!!! One quick Q: why that "c" is the speed of wave? any derivation reference? Thank you!
@@Tyokok Dimensional analysis. Tension is just force (Kg .m .s^-2). Linear density (Kg .m^-1). Now when you divide both, the units you are left with is that of speed squared. Clearly the constant is indicative of some speed. I can only see two kinds of it, one along the medium, another is that of string's up-down vibration which decays in the process. But speed of wave stays constant till the end, just like on whipping the long rope, the bump moves with same speed till end. Anyone can correct me.
@@ananthakrishnank3208 Sorry for late reply and thank you so much for your comment! I think you are right. From dimensional perspective, it's of the unit of distance/time, and from his solving derivation this is the speed of wave traveling.
why can you ignore the very small vertical distance when you say rho*delta x = mass, but you can't ignore the very small vertical distance when you assign non-zero angles?
I am looking for a video on how to generate the Energy Functional of a partial diff equation. How to generate the energy function of a wave equation. Is this video about that?
I think you don't actually have to assume that gravity is negligible, for this to work. Because all gravity would do is increase the tension in the string, and change the resting position of the string a small amount. And so the tension value already includes the effect of gravity. And the small change in the resting position is included in the selection of u(x) = 0 when it is at rest.
As for now, I am not convinced on neglecting cosine components of T, saying theta is close to zero. Just as I type, maybe I get it. For angles x1 = 0.01 to x2 = 0.02 (tending to x1), we are kind of looking for sin(x2) - sin(x1). From differentiation we know that sin(x2) - sin(x1) = dx * cos (x1). So analogously for cosine components, cos(x2) - cos(x1) = dx * sin(x1) (approximately zero for very small x). So, this way, it makes sense to only include the sine components. But now why this is not convincing is, for x = 10.00 to x = 10.01, I cannot justify the neglection of cosine component. Ooh. I get it. First of all, when we visualize the movement of a guitar string, we see that it's as if the string just moved a tiny bit up, even when its up it looks still flat. In that case, theta is obviously close to zero. So I should have not even considered x >> 0 case. Thanks for the lecture! :)
That’s an interesting lecture that makes me to revisit my knowledge of physics again. The tangent of angle is equal to the Uxx (x,t) is kind of tricky which I need to do some revision. Do you have any idea about the speed of the wave if it is a probability wave in quantum physics?
If you want to derive the net force on a small segment of the string, then you have to add all of the forces acting on each infinitesimal part, correct? But instead you subtract two forces from each other. This must be the result of the integration. Therefore, I believe that something is missing in this derivation.
That is exactly what he’s doing. He defines a differential unit, x to x+dx, and he assumes that the only forces acting on this section are the tensions of the string of the unit cell either side of this section…then he sums those forces, which since they are acting in opposite directions becomes a difference.
I don’t think you ended up explaining why this is considered a hyperbolic differential equation. I would love to understand what types of differential equations are elliptic, parabolic and hyperbolic.
I think going the step to set sin(...) = tan(...) is unnecessary. You have the force T going to the left at and the same force going to the right at the other end of the segment of the string. You can assume that the "horizontal" component of the force is equal at both ends. Otherwise, the hole segment of string would start to move sideways. Set that force equal to "T" (because of small angles and cos being equal to 1) and the need to argue that sin=tan will disappear. To me that seems a bit tidier and easier to follow.
Hello, first thank you so much for these amazing videos. I am very rusty at math, but could somebody tell me why second order linear ODEs have exponential or sines solutions? I would really appreciate it.
Because derivatives of sines and exponents to the base e are also sines and exponents to the base e of the same order. This makes the solutions just combinations of the same sines and exponentials.
θ, when you consider θ at equilibrium, large θ is going to imply large amplitude, so does Δθ, bc θ is zero at top/bottom. Some textbook writes small amplitude.
X could be expressed as function of t for each particular U. Or more generally X is a function of two variables - t and U. Because that equation has 3 variable - You could express any variable as function of two others.
You are right that model building is practically inexistent in both mathematics and Physics curricula. Do yourself a favor and read G. Strang's "Applied Mathematics" and Enzio Tonti's papers. Read all the stuff about across and through variables and how this equation is just the continuous version of a discrete electric circuit. Then, students will also get a better understanding of the meaning of curl and div operators.
really great video but currently i struggle at one point: where you let dx ->0 to me, that would just let the term 1/dx grow infinitely large but instead you „define“ this as Uxx and consider this clear. unfortunately i cant follow that step, so could you please explain that step in some more detail?
It's a logic point. As dx->0, so does du->0, in the limit du/dx is finite. Go back to fundamental theory of calculus (FTC) for the complete story (which is tedious).
It just felt very dry and unmotivated. I don’t think I intuitively understood the assumptions and I struggled with the partial derivatives and what they meant physically.
@11:30 me too, i have much better math skills (and was able to do the cosine thing now -insulated boundaries, but idk always found PDE hard as self study😢) fingers crossed!
The fact that I can access this high quality of a lecture for free is astonishing..
I'm practically blown away!!
I'm at the phase of research where I'm trying to understand how PDEs are embedded in Machine learning loss function. Viola! Here I am consuming mathematical chocolates!
same , i love this kind of detail approach to concepts , with implementation of human logic and human intuition at grond level.
Agreed. I always thought Gilbert Strang at MIT is a great math teacher. Steve has proven to be just as good, if not better, than Prof Strang. Kudos to eigensteve 🙏🙏🙏
I am blown away at how easily understandable this derivation is. Every step is broken down into high school calculus concepts. We do not have time in our DE course to derive this but just to understand that the vertical acceleration is proportional to the curvature of the string is just so perfectly intuitive that it makes me wonder why it wasn’t explained to us this way to us before. Thank you so much for this thoughtful video.
I really love when teachers share their own history of struggles with a topic.
This video series explains why it's harder and harder to resist binging UA-cam these days, any other series like this? The new videos are literally in sync with my PDE class oh my goat
Very intuitive and easy to understand. I also appreciated the emphasis on how it was not that easy to finally get comfortable with the manipulation of such an equation.
Thank you very much for this video and for the whole channel 🙏
From a geometrical standpoint, the laplace equation means: "scalar field without local max and min"; heat equation means: "the change in one variable is proportional to the curvature in another"; and the wave equation means: "the curvature in one variable is proportional to the curvature in another". If you can imagine how the information change, you can easiliy derivide this partial differential equation.
I had the exact same block when starting to learn about PDEs. This derivation is so crystal clear about what assumptions are being made and why they are made.
Thanks Steve. For many years I have dealing in higher maths subjects and, honestly, this is one of the best lectures I have watched (or physically attended). Please, keep producing tuition material at this level of excellency.
I can't stop loving this person.
I'm taking PDE this semester and your PDE playlist has been awesome. Thanks prof.
I do envy the new generation who can study PDE with high quality reference like this. It took me years to think through some of the concepts.
Awesome intuitive approach to setting up the wave equation from F = ma. Reminds me of my General Physics course when I was reading the Young and Freedman text.
Thank you so much Steve, its like reading a very huge book in a short moment.
This is next level lecture. Love your videos. 👏
Great explanation, professor! I'm looking forward to see the upcoming videos!
I don't know how to say thank you to making my nightmare to day dream,
Wish i was your student and learn this things directly in your class
These are awesome Steve, great work.
What I didn't get the first time I saw this derivation is why the length of the rope is dx (in the context of the mass).
it's actually:
sqrt(dx² + dy²) = dx•sqrt(1 + (dy/dx)²) = dx•sqrt(1+(y')²)
But since we assume small oscillations all nonlinear terms are negligible so ds = dx•sqrt.
It's similar to us saying cos(θ) = 1 and not 1 + θ²/2! + ....
Hope this helps someone!
The physics teacher we all need
Fantastic intro
Played the intro a couple of times - Nice segue!
these lectures will keep on giving into the future. you are doing a great service.
some professors should also take your classes. 😂
Beautiful Lecture and Wonderful Lecture series!
Excellent explaination! Correct pace.
Very neat introduction to the wave equation, well done prof !
One could add - just for more fun - that those smart mathematicians from the 18th century wrote all their ‘papers’ in Latin and so the obvious symbol choice to represent speed was the letter c .. speed being “celeritas” in Latin.
Funny how Latin even got into the most famous among all equations, even if Albert Einstein didn’t use c initially 😊
Thank you for this playlist. Your videos are helping me a lot in my PDE class.
First, second, third times watching this: _crickets_
Fourth time watching this: "YOOO THAT DERIVATION IS SO COOL!"
I did find it interesting and fun. Thank you for simplifying the concept.
Love you, Steve!
Blessed!
Excellent explanation. Crystal clear. Thank you
what a beautiful lecture!
How we derive that c^2 = T / ro?
In this video Steve is explaining how to derive the wave equation Utt=c^2 * Uxx - correct?
From F=ma Steve derives Utt = T / ro * Uxx, and then at 25:15 he just says “where c squared is equal to T (tension) divided by ro(linear density)”. Where that comes from? How speed comes into the equation?
I’m just guessing here but Tension is measured in Newtons (N) which can be expressed by (kg)(m)/s^2. Linear density is measured in kg/m. If you divide the units for tension over the units of linear density you get ((kg)(m)/s^2)/(kg/m). If you simplify this to (kg)(m)/s^2 x m/kg you can cancel out the kg. Then you are left with m^2/s^2 which is the square of velocity (m/s). So if c represents velocity then c^2 is the velocity squared solved by the units.
Thank you very much for your fantastic lecture and your hard work. I love it.
Excellent and practical video on the topic.
A very nice lecture, thank you! I have a minor comment though regarding the derivation. When considering the force equilibrium of the infinitesimal element, I am afraid the equilibrium in not maintained if the two tangential forces T at the ends are identical. What must be identical to prevent horizontal movement are the horizontal projections of these forces, say “N”. When these horizontal forces are identical, the vertical projections of the tensile forces T, which we can call F, are equal N*tan(theta). And it is the difference between these vertical forces: N*[tan(theta+dtheta)-tan(theta)], which equals the Newton’s inertia forces “m*a”. This is just a minor fix which removes the weak arguments (time 20:20) about sine being roughly equal to tan, which is equal to the angle itself, and cosine being roughly equal to one for small angles theta.
Very nice presentation. I struggle with your interpretation of speed c. You see c in the guitar-cord example as the speed of the string in the y-direction. In the classical wave equation, however, the speed c is the speed in the x-direction, as in a waterwave or in a em-wave (i.e. how fast the wave runs)
Amazing video! Thank you 🙏🏻
wow this is walter lewin's level of lecture, thankyou sir
Thanks for making this content openly available! It has certainly been extremely helpful while brushing up my memory on these concepts.
I did have a question:
Couldn't we skip the sin(theta) ~ tan(theta) step altogether by utilizing the requirement that the x-component of the tension at points x and x+dx must be equal (in opposite directions)? At either point, we have tan(theta) = T_y/T_x. Solving for T_y, we have T_y = T_x*tan(theta). Again, T_x is the same at both x and x+dx (save for the minus sign), so it can be factored out when calculating the net vertical force, F = T_y(x + dx) + T_y(x) = T_x*[tan(theta + dtheta) - tan(theta)].
Thanks again!
also i hope you plan to do this all the way to the schrödinger equation :)
That would be epic! Please do sir!🤩🥳🤓
I’m working up to it. Might take a little while. Navier-Stokes equations will be sooner.
@@Eigensteve 😍😍
That was a very nice video. I will refer my students to this video in my PDE class.
There is something that I don't quite get though. It might be just a detail and probably an approximation matter (I'm a mathematician by formation, so I do have a hard time sometimes with getting the right approximations). When we pitch a string of a guitar to move it to its original shape, say h(x), and then release it, do we assume that the length of the string does not change? I am asking this question because I have a hard time seeing how the length L would remain constant. For instance, in the picture you draw of a guitar string attached at x = 0 and x = L, the string must be longer than L to get that curved shape. Do we allow that elasticity so that there is a change in the value of the tension in the string?
OMG the guitar is so cool!!
Thanks!! So much fun to play
You are not only a great professor, but also a great person!!! Great great Thanks and God Bless!!!
One quick Q: why that "c" is the speed of wave? any derivation reference? Thank you!
@@sorry4all thanks for reply. not sure if c is constant. no what it is or not, what's the physics meaning of it?
@@Tyokok Dimensional analysis. Tension is just force (Kg .m .s^-2). Linear density (Kg .m^-1). Now when you divide both, the units you are left with is that of speed squared. Clearly the constant is indicative of some speed. I can only see two kinds of it, one along the medium, another is that of string's up-down vibration which decays in the process. But speed of wave stays constant till the end, just like on whipping the long rope, the bump moves with same speed till end. Anyone can correct me.
@@ananthakrishnank3208 Sorry for late reply and thank you so much for your comment! I think you are right. From dimensional perspective, it's of the unit of distance/time, and from his solving derivation this is the speed of wave traveling.
why can you ignore the very small vertical distance when you say rho*delta x = mass, but you can't ignore the very small vertical distance when you assign non-zero angles?
I expect a full concert in the next video
Haha yeah… I’m not going to quit my day job…
I am looking for a video on how to generate the Energy Functional of a partial diff equation. How to generate the energy function of a wave equation. Is this video about that?
I think you don't actually have to assume that gravity is negligible, for this to work. Because all gravity would do is increase the tension in the string, and change the resting position of the string a small amount. And so the tension value already includes the effect of gravity. And the small change in the resting position is included in the selection of u(x) = 0 when it is at rest.
Very good teaching. From thai student
Nice video! What song were u playing in the beginning?
You mentioned the wave being infinitesimal and the effect on entropy, any clues where I can follow up on that idea?
You are the best ever !
Amazing, Thank you
Nice video and presentation.
Are we talking about mechanical wave or not?
Happy 2024, thank you so much for this excellent lecture. 🎉
As for now, I am not convinced on neglecting cosine components of T, saying theta is close to zero.
Just as I type, maybe I get it. For angles x1 = 0.01 to x2 = 0.02 (tending to x1), we are kind of looking for sin(x2) - sin(x1). From differentiation we know that sin(x2) - sin(x1) = dx * cos (x1). So analogously for cosine components, cos(x2) - cos(x1) = dx * sin(x1) (approximately zero for very small x).
So, this way, it makes sense to only include the sine components. But now why this is not convincing is, for x = 10.00 to x = 10.01, I cannot justify the neglection of cosine component.
Ooh. I get it. First of all, when we visualize the movement of a guitar string, we see that it's as if the string just moved a tiny bit up, even when its up it looks still flat. In that case, theta is obviously close to zero. So I should have not even considered x >> 0 case.
Thanks for the lecture! :)
That’s an interesting lecture that makes me to revisit my knowledge of physics again. The tangent of angle is equal to the Uxx (x,t) is kind of tricky which I need to do some revision.
Do you have any idea about the speed of the wave if it is a probability wave in quantum physics?
Bravo👏🏼🎓
Thank you!!
Beautiful!
Thank you, this was helpful!
❤ thanks.
God bless you ❤️.
Is it true for big deflections?
The derivation starts at 12:30 if you want to jump straight to it
You are a legend
love it!
If you want to derive the net force on a small segment of the string, then you have to add all of the forces acting on each infinitesimal part, correct? But instead you subtract two forces from each other. This must be the result of the integration. Therefore, I believe that something is missing in this derivation.
That is exactly what he’s doing. He defines a differential unit, x to x+dx, and he assumes that the only forces acting on this section are the tensions of the string of the unit cell either side of this section…then he sums those forces, which since they are acting in opposite directions becomes a difference.
I don’t think you ended up explaining why this is considered a hyperbolic differential equation. I would love to understand what types of differential equations are elliptic, parabolic and hyperbolic.
Good point. I will be making a video on this soon I hope.
I think going the step to set sin(...) = tan(...) is unnecessary.
You have the force T going to the left at and the same force going to the right at the other end of the segment of the string. You can assume that the "horizontal" component of the force is equal at both ends. Otherwise, the hole segment of string would start to move sideways. Set that force equal to "T" (because of small angles and cos being equal to 1) and the need to argue that sin=tan will disappear. To me that seems a bit tidier and easier to follow.
i don't see the difference. in either case, you heavily lean on small angles
@@fahrenheit2101 To me that seems a bit tidier and easier to follow. But it assumes that there is no sideways motion of the particles.
Thank you so much!
Hi. Can someone explain how the limit becomes the second derivative. I would reallt appreciate this, as then i would fully understand. Lee
Look for the definition of a derivative - it's exactly what we have here.
Thank you!
I'm loving this pde kind of mathematics
Im sorry, but im not thinking any of this is easy man. This stuff was made by great minds.
absolutely true
Hello, first thank you so much for these amazing videos.
I am very rusty at math, but could somebody tell me why second order linear ODEs have exponential or sines solutions? I would really appreciate it.
Because derivatives of sines and exponents to the base e are also sines and exponents to the base e of the same order. This makes the solutions just combinations of the same sines and exponentials.
@@dennisgawera8788 Thank you so much!
High quality
crazy good
What is deflection
θ, when you consider θ at equilibrium, large θ is going to imply large amplitude, so does Δθ, bc θ is zero at top/bottom. Some textbook writes small amplitude.
Hi, Steve. Thanks fr the great video. A quiz here, isn't X a function of time?
X could be expressed as function of t for each particular U. Or more generally X is a function of two variables - t and U.
Because that equation has 3 variable - You could express any variable as function of two others.
You are right that model building is practically inexistent in both mathematics and Physics curricula. Do yourself a favor and read G. Strang's "Applied Mathematics" and Enzio Tonti's papers. Read all the stuff about across and through variables and how this equation is just the continuous version of a discrete electric circuit. Then, students will also get a better understanding of the meaning of curl and div operators.
Thanks! I get it!
super clever :)
really great video but currently i struggle at one point: where you let dx ->0
to me, that would just let the term 1/dx grow infinitely large but instead you „define“ this as Uxx and consider this clear. unfortunately i cant follow that step, so could you please explain that step in some more detail?
It's a logic point. As dx->0, so does du->0, in the limit du/dx is finite. Go back to fundamental theory of calculus (FTC) for the complete story (which is tedious).
The answer below is the right idea. We are essentially using the definition of a derivative, which has some assumptions involved.
@@Eigensteve damn, 20 years ago i would have remembered 😭 its all too long ago, but thanks for responding.
What part of this explanation you did not understand when you were a teenager?
It just felt very dry and unmotivated. I don’t think I intuitively understood the assumptions and I struggled with the partial derivatives and what they meant physically.
😂I love you.
LC oscillators in electric circuits too. This is not hyperbole!
@11:30 me too, i have much better math skills (and was able to do the cosine thing now -insulated boundaries, but idk always found PDE hard as self study😢) fingers crossed!
🥲
❤️❤️
i love you
nothing new. it looks like for high school levels
You still need to learn more mathematical physics, some of your fundamentals are still not clear to you 😢
I FEEL SAFE WITH THIS DERIVATION. AS SAFE AS KELSEY'S NOOTZ.
Thank you!
Thank you!!
What is deflection
I think it means the maximum deviation of the string from its rest position. Basically the highest point of that string you see in the plot.