Thanks Dr. Brunton. I have finished 3 of your playlist. Vector Calc-PDE is the forth one I am watching. You are very passionate about these math theories. Your explanation are very easy to follow. Just wanted to thank you for your effort. These playlists are going benefit many future generations to come.
Fun fact about history of music and science. Equal temperament, the way we divide octaves in notes in multiple log_2(1/12) was rediscovered in mid1500 by Vincenzo Galilei. He's Galileo father
Thank you Dr.Brunton , I am watching your videos from Iran and wish all the best for you . Your lessons really helped me to pass the mathematics successfully . And i would like to watch your Data analyze videos ❤
Ce valeureux Professeur est génial, il a le don d'enseigner et de simplifier les concepts qu'on prenait parfois pour des citadelles impénétrables . Un grand Merci pour vous cher Monsieur . may God Bless you , I know it's hard, but. you have to publish more for the best of your thirsty and faithful audience, . Thanks,
for the negative sign, similar to the heat equation video, diffision was negative bc it was the state returning to equillibrium (exuding heat to the env), similarly the string will be returning to equillibrium in a non-preturbed state (at rest) at least kinda how i think of it, might help others with sign of lambdas
Frequency: number of waves passing by a specific point per second. Period: time it takes for one wave cycle to complete. The relation between frequency (f) and time period (T) is given by f=1/T. Notice that (f) increases when L is shortened.
the step to eliminate the sin solution part is not clear. and the constant c is employed twice in 2 different uses- But that's nitpicking. great lecture
He removed the sin part because sin(c£t) when t= 0 is equal to zero. Sin (0) = 0 . So we removed it . Because according to initial condition when t= 0 , U(x,0) = f(x).
Maybe the sin term in the general solution for G(t) should not have been dropped off? the coefficient associated with that term will be determined by a 2nd initial condition, i.e., u"(x,0).
At 28:08 he assumed implicitly that dU/dt (x,0) =0 which means the initial velocity is zero. So that's an extra initial condition that was not mentioned at the beginning.
Hi Steve, or anyone, one more question at 28:05 please! You drop sin(c*lambda*t) in G(t) by arguing at Initial Condition when t=0 we have f(x) not zero. But that doesn't give you reason to drop sin(...). Unlike in F(x) you drop cos(lambda*x) because Boundary Condition requires zero so you drop all that's non-zero. And why your original assumption of separation of F(x)G(t) is valid? Any intuition? Many thanks if anyone can advice!
I was having the same question, and I read @khaledqaraman 's comment, saying "At 28:08 he assumed implicitly that dU/dt (x,0) =0 which means the initial velocity is zero. So that's an extra initial condition that was not mentioned at the beginning." Now taking derivative of the cos you get sin so that is why you drop the sin in G(t). Hope this helps.
18:00ish the solutions with sin cos or e, are all equivalents. No need to say only cos? A sin and cos are the same with a dephase angle. Its just a matter of taste really.
At 28:00, I don't think I follow why Steve ignored the sin() part of G just because the Initial condition is equal to zero. I think we need to solve for the coefficient of the sine part of G just like we did for F. Because both G and F have the form 'A*cos() + B*sin()' we, really need 4 givens (2 initial and 2 boundary conditions) instead of 3. I added my own, setting Ut (the time derivative of U at time 0) equal to 0 and then it followed that the coefficient of the sine part of G had to be zero to satisfy that. I think that is the right way to do it... What do you think?
I was having the same question, and I read @khaledqaraman 's comment, saying "At 28:08 he assumed implicitly that dU/dt (x,0) =0 which means the initial velocity is zero. So that's an extra initial condition that was not mentioned at the beginning." Now taking derivative of the cos you get sin so that is why you drop the sin in G(t). Hope this helps.
Hello, thank you for this incredible video! I am a high school student from Korea, and I’m trying to better understand wave equations. I’ve learned that for non-standing waves, the wave function can be expressed as y=Asin(kx−ωt+ϕ). Could you explain how this form of the wave function is derived? I might have misunderstood something, so I would really appreciate your explanation. Also, I’m not very confident in my English, so I apologize if my question is unclear. Thank you so much! :D
I would recommend you to refer Linear algebra to understand that point. Once you understand eigenvalues it will be easy to understand eigenfunction. It is a bit tough but very beautiful.
Professor please show me that when a unit mass as a wave propagate and transfer energy to the mass energy is kept constant. I can find particle velocity and shear strain for a shear wave and the displacement at a particular point for any time t but I don’t get the total energy of at the point does not main the same value. As shear strain is directly related to the particle velocity, is it that I have to consider either particle velocity or shear strain plus displacement related velocity in the perpendicular direction of displacement. Please help me.
Newton wanted to apply music theory to his prism spectrum. He could "see" 6 colours. Red orange yellow green blue and the darker blue that he called Violet. But diatonic scale A-G is 7 notes. So he invented "indigo" to appear between blue and violet. Musical string analogy achieved 👍
This is Oh My God, breath taking, amazing lecture!!! NEVER saw a professor explain this clear!!! Hi Steve, why you call lambda square Eigenvalue? How does this relate to matrix Eigenvalue? Thank you so much again for such vivid elegant explanation of wave equation video!
If you think of a differential operator D, applying to a function and setting a eigenvalue problem is: D(y) = a*y where "a" is a scalar and "y" is a real-value function. Solving for "y" gives y=e^(ax), so you can see that e^(ax) is an eigenvector or "eigenfunction", meanwhile "a" is it's eigen value. In this case, the eigenvalues are infinitly many because it's a partial differential equation, meaning that it's has infinite solution. In a normal ODE, has finite many of them, so there is finite quantity of solutions.
it turned out to be that we got a vector space with orthonormal basis of infinite dimension that has infinite amount eigenfunctions and their corresponding eigenvalues… just like in quantum physics
Would lambda be the eigen vectors and Bn be the eigen values? When I imagine an infinite sum of frequencies forming a solution, I think of each frequency as the eigen vector and Bn is the correct weight. I may be confusing eigen vectors for Fourier basis functions...
A linear combination of eigenvector don't need to be weigthed by its eigenvalues. In this case, the sines are eigenvector or "eigenfunction" of the differential operator, lambdas are the eigenvalues, and the Bs are the unique weights that can form the initial distribution with the fourier series.
The only function which can acept non related argument is the constant function, because the other case is for any f, g: R to R such that f(x) = g(y), means that x = f^-1(g(y)) or viceversa, which can't be because x and y are not related by any function.
The solution of wave eq. is too ugly here and it presented in a weak way. There are far better and cleaner ways of defining the solution analytically! Such a pity!
@@demr04 The way of presenting the solution in comparison with others who did the same. Up to this point, almost everything was smooth and pretty. I think he needs to improve it.
this is a brilliant presentation by a master teacher. He has put so much work into it and then gives it to the community for free. He deserves our respect
Can you solve this question? I couldn't solve it. Can you help me? Find the distribution 𝑢(𝑥, 𝑡) by writing the wave equation and boundary conditions for a rod (one dimension) of length L=1 unit, with both ends fixed and whose initial displacement is given by 𝑓(𝑥), whose initial velocity is equal to zero. (𝑐2 = 1, 𝑘= 0.01) 𝑓(𝑥) =ksin(3𝜋x)
Thanks Dr. Brunton. I have finished 3 of your playlist. Vector Calc-PDE is the forth one I am watching. You are very passionate about these math theories. Your explanation are very easy to follow. Just wanted to thank you for your effort. These playlists are going benefit many future generations to come.
That is awesome! Thanks for sharing!
This guy is incredible. he has helped me so much.Thank you so much
Fun fact about history of music and science. Equal temperament, the way we divide octaves in notes in multiple log_2(1/12) was rediscovered in mid1500 by Vincenzo Galilei. He's Galileo father
Whoa, that is super cool! I didn't know that
Thank you for the wonderful tie-in with the guitar near the end!
Thank you Dr.Brunton , I am watching your videos from Iran and wish all the best for you . Your lessons really helped me to pass the mathematics successfully . And i would like to watch your Data analyze videos ❤
Ce valeureux Professeur est génial, il a le don d'enseigner et de simplifier les concepts qu'on prenait parfois pour des citadelles impénétrables . Un grand Merci pour vous cher Monsieur . may God Bless you , I know it's hard, but. you have to publish more for the best of your thirsty and faithful audience, . Thanks,
Great..finally I have understood..why we choose -lambda square and how sin or cos vanishes
for the negative sign, similar to the heat equation video, diffision was negative bc it was the state returning to equillibrium (exuding heat to the env), similarly the string will be returning to equillibrium in a non-preturbed state (at rest)
at least kinda how i think of it, might help others with sign of lambdas
I like this way of thinking.
Excellent, truly. Thank you for posting.
Frequency: number of waves passing by a specific point per second. Period: time it takes for one wave cycle to complete. The relation between frequency (f) and time period (T) is given by f=1/T. Notice that (f) increases when L is shortened.
the step to eliminate the sin solution part is not clear. and the constant c is employed twice in 2 different uses-
But that's nitpicking. great lecture
Thanks for letting me know -- always good to know what could be more clear.
He removed the sin part because sin(c£t) when t= 0 is equal to zero.
Sin (0) = 0 . So we removed it .
Because according to initial condition when t= 0 , U(x,0) = f(x).
@ares9748 that just means that sin in G doesn't contribute to u at t=0. It still doesn't contradict the initial condition, so why remove it?
Maybe the sin term in the general solution for G(t) should not have been dropped off? the coefficient associated with that term will be determined by a 2nd initial condition, i.e., u"(x,0).
agreed!
agreed! +1
Yeah since it's second order we need two I.C's.
At 28:08 he assumed implicitly that dU/dt (x,0) =0 which means the initial velocity is zero. So that's an extra initial condition that was not mentioned at the beginning.
Hi Steve, or anyone, one more question at 28:05 please! You drop sin(c*lambda*t) in G(t) by arguing at Initial Condition when t=0 we have f(x) not zero. But that doesn't give you reason to drop sin(...). Unlike in F(x) you drop cos(lambda*x) because Boundary Condition requires zero so you drop all that's non-zero. And why your original assumption of separation of F(x)G(t) is valid? Any intuition?
Many thanks if anyone can advice!
I was having the same question, and I read @khaledqaraman 's comment, saying
"At 28:08 he assumed implicitly that dU/dt (x,0) =0 which means the initial velocity is zero. So that's an extra initial condition that was not mentioned at the beginning."
Now taking derivative of the cos you get sin so that is why you drop the sin in G(t). Hope this helps.
Thank you for the informative video.
Thank you!
this video is perfect🥰 thank you so much
18:00ish the solutions with sin cos or e, are all equivalents. No need to say only cos? A sin and cos are the same with a dephase angle. Its just a matter of taste really.
Really good analysis. Would love a 2D adaptation to emphasize interactions between indices :)
At 28:00, I don't think I follow why Steve ignored the sin() part of G just because the Initial condition is equal to zero. I think we need to solve for the coefficient of the sine part of G just like we did for F. Because both G and F have the form 'A*cos() + B*sin()' we, really need 4 givens (2 initial and 2 boundary conditions) instead of 3. I added my own, setting Ut (the time derivative of U at time 0) equal to 0 and then it followed that the coefficient of the sine part of G had to be zero to satisfy that. I think that is the right way to do it... What do you think?
I was having the same question, and I read @khaledqaraman 's comment, saying
"At 28:08 he assumed implicitly that dU/dt (x,0) =0 which means the initial velocity is zero. So that's an extra initial condition that was not mentioned at the beginning."
Now taking derivative of the cos you get sin so that is why you drop the sin in G(t). Hope this helps.
You are the best ever!
Hello, thank you for this incredible video! I am a high school student from Korea, and I’m trying to better understand wave equations. I’ve learned that for non-standing waves, the wave function can be expressed as y=Asin(kx−ωt+ϕ). Could you explain how this form of the wave function is derived?
I might have misunderstood something, so I would really appreciate your explanation. Also, I’m not very confident in my English, so I apologize if my question is unclear. Thank you so much! :D
I hope you delve a bit into seismology too :)
Art.
why didn't you use the second initial condition u'(x,0)=g(x)?
Amazing. Thanks
Sir, why are they called eigen values and eigenfunction. Kindly explain. Your small effort will be a great help to me.thanks
I would recommend you to refer Linear algebra to understand that point. Once you understand eigenvalues it will be easy to understand eigenfunction. It is a bit tough but very beautiful.
But how is it that we take the constant as -lambda^2
Professor please show me that when a unit mass as a wave propagate and transfer energy to the mass energy is kept constant. I can find particle velocity and shear strain for a shear wave and the displacement at a particular point for any time t but I don’t get the total energy of at the point does not main the same value. As shear strain is directly related to the particle velocity, is it that I have to consider either particle velocity or shear strain plus displacement related velocity in the perpendicular direction of displacement. Please help me.
Newton wanted to apply music theory to his prism spectrum. He could "see" 6 colours. Red orange yellow green blue and the darker blue that he called Violet. But diatonic scale A-G is 7 notes. So he invented "indigo" to appear between blue and violet. Musical string analogy achieved 👍
"resonates" - very good. Comedy aside, great video.
Amazing, Than you
This is Oh My God, breath taking, amazing lecture!!! NEVER saw a professor explain this clear!!! Hi Steve, why you call lambda square Eigenvalue? How does this relate to matrix Eigenvalue? Thank you so much again for such vivid elegant explanation of wave equation video!
If you think of a differential operator D, applying to a function and setting a eigenvalue problem is:
D(y) = a*y
where "a" is a scalar and "y" is a real-value function. Solving for "y" gives y=e^(ax), so you can see that e^(ax) is an eigenvector or "eigenfunction", meanwhile "a" is it's eigen value.
In this case, the eigenvalues are infinitly many because it's a partial differential equation, meaning that it's has infinite solution. In a normal ODE, has finite many of them, so there is finite quantity of solutions.
@@demr04 WOW! clear! Really appreciate it Daniel!
@@Tyokok your welcome :)
@@demr04 this is real fun stuff
@@Tyokok yeah agree 🤓
Hi Steve, the last video you posted was the separation of variables one. I believe you skipped a video.
If you go to the "Vector Calculus and PDEs" playlist, they should all be there in order.
it turned out to be that we got a vector space with orthonormal basis of infinite dimension that has infinite amount eigenfunctions and their corresponding eigenvalues… just like in quantum physics
wouldn't g(t) have the cos term dropped rather than the sin?
TIL: fingers on guitar strings are high-pass filters
Would lambda be the eigen vectors and Bn be the eigen values? When I imagine an infinite sum of frequencies forming a solution, I think of each frequency as the eigen vector and Bn is the correct weight. I may be confusing eigen vectors for Fourier basis functions...
A linear combination of eigenvector don't need to be weigthed by its eigenvalues. In this case, the sines are eigenvector or "eigenfunction" of the differential operator, lambdas are the eigenvalues, and the Bs are the unique weights that can form the initial distribution with the fourier series.
12:35 any specific proof of why it's equal to constant?
hey hello it not need proof that space cant equal time at there like 5x is not equal to t or 5t or something it just can be if they equal a constant
The only function which can acept non related argument is the constant function, because the other case is for any f, g: R to R such that f(x) = g(y), means that x = f^-1(g(y)) or viceversa, which can't be because x and y are not related by any function.
how do they make these videos? does the prof just write backwards???
indeed this makes it even more next level. The explanation is in one direction but the writings are backwards
I think he write normally and then use video editing to flip it so it looks normal to us lol
I think they actually write on a glass panel, with the camera infront of him, that's why they make the background black and the pen colors bright.
Id die of embarsmemt having someone record me playing a guitar lol.😅
I wish i have your knowledge
Keep watching and you will!
Fouier Transform or Series?
I thought the same
what is Cn?
30:12
@@lt4376 thanks!
Me costó entender que "buzzcard" se refería a "buscar".
خسارة كون جاء يفهم بلعربي
The solution of wave eq. is too ugly here and it presented in a weak way. There are far better and cleaner ways of defining the solution analytically! Such a pity!
well, suggest one then
What is the ugly or weak?
@@demr04 The way of presenting the solution in comparison with others who did the same. Up to this point, almost everything was smooth and pretty. I think he needs to improve it.
this is a brilliant presentation by a master teacher. He has put so much work into it and then gives it to the community for free. He deserves our respect
Can you solve this question? I couldn't solve it. Can you help me?
Find the distribution 𝑢(𝑥, 𝑡) by writing the wave equation and boundary conditions for a rod (one dimension) of length L=1 unit, with both ends fixed and whose initial displacement is given by 𝑓(𝑥), whose initial velocity is equal to zero. (𝑐2 = 1, 𝑘= 0.01)
𝑓(𝑥) =ksin(3𝜋x)
"Harmonics of the planets" is real -- "Kirkwood Gaps" (en.wikipedia.org/wiki/Kirkwood_gap) :)