Finally, an explanation that does not get rid of the higher order terms as quickly as we can. Doing that feels like an approximation while this is exactly true for all of the things we are working with.
I love this kind of video, going over mathematical concepts that we breezed past in engineering classes without proper explanations or derivations. Keep em coming!
oh man!! I was so confused by the multiplication by epsilon in the taylor expansion derivation from my course and I had just given up on understanding it. THANK YOU
Stick with it!! This is more advanced material than before. Videos are a little bit mathematical at first because it is setting up a framework that we will use. As soon as you see a few examples of this theory being used in practice, it will fall into place for you. It's a little like using Lagrange's equations, which made our lives very easy once we started applying it to mechanics problems, but we've never derived Lagrange's equations...until now. These first few videos on this playlist explain the theory behind Lagrange's equations and other energy methods. Once you see some practical examples of this being used (which will come in the next few videos), then, you'll return to these videos and they will make more sense. I am very much trying to get you guys quickly to the point where we are solving practical problems, but I can't get there without laying out some theory first. The math behind this theory is actually very beautiful, but for most people, Variational Calculus takes a little while to get used to. It took the world top scientists 160 years to get to Hamilton's Principle from Newtonian Mechanics and I'm trying to get you there in a few videos.
This is great. I just started the video and I'm holding my breath to see if Oliver Heaviside gets proper credit for inventing this operation. Now as I proceed I'm not seeing evidence that this delta operator is the the same as Dirac's delta operator.
Delta's occurs in many places in math and engineering. In this case, we are referring to the variational operator. It bears no relation other uses like the Dirac delta function or the Kronecker delta.
@عَدِيُّ to me it is not that much, since usually you don't have to deal with the uncertainty as you do see in quantum mechanics, however, you will have to do some real world solutions in which will be harder for some, due to the scale, but to me the larger values just make more leeway for "rookie mistakes", I personally do not think of it as confusing, but others may say otherwise to me it is a shame that modern physics, and maybe biology is the reason why most people hate high school and college, and i think that we should dive more deeply into physics just to show that it is very important to our everyday lives, most of the reasons for this is because of the intense math involved, but as Richard Feynman has done it, he makes sure that physics in general is easier by the Feynman technique for integrals, differentials and more, if only more people knew about those subjects, maybe the world may be different today.
@@universum-theuniverseexpla6565I have heard so many time about feyman technique...can you kindly briefly explain..I know there is a complete set of his lectures pdf available.
05:43 Suppose the value of I for the extremal path is a minimum, then the variation in I is δI = Ibar - I >0 for all paths very close to extremal path. For one particular path with tiny deviation δy we get a tiny positve value of δI. Now consider the path got by just changing the sign of δy. The integral only changes in its sign, but we must still get a tiny positive value for δI. To ensure this as the higher order terms approach 0 we must have that the integrand is 0.
Just some doubts: Firstly, when we say that in order for the variation of I to retain its sign in the vicinity of an extremal, everything multiplying "Dell y" should be zero. This is also considering that Dell y is arbitrary and can be negative or positive. I don't think I truly understand what it means for the variation of I to retain its sign, I understand that Dell y can be anything, and thus to extract a "stable" answer, the only possibility is to make it equal 0, which had no sign. If you could explain the meaning of the retention of the sign, it would confirm my understanding. Secondly, the way I understood it is, since the variation of I had to be infinitesimally small to achieve the perfect path, and this had to be true for any value of variation, it only makes sense that everything except the variation be zero. It is the only way to compute the path of zero variation, or, the optimum path. This was my intuition. Lastly, this is just a fault of my maths, but I don't understand how you converted the weak form to the strong form in both videos, thought that is my fault. Thank you for creating these videos and for your valuable time sir. (Please don't see this as a request to explain my every doubt, it is more of a note to myself for me to remember this stuff. A single hinting line in answer holds the same value as an explanation for me:)
At an extremal (min/max), the variation should maintain a consistent sign - if we're at a minimum, variations should stay above it; at a maximum, they should stay below. Since δy can be either positive or negative (it's arbitrary), the only way to ensure this consistency is if the coefficient multiplying δy equals zero. The weak to strong form conversion uses integration by parts to move derivatives from δy to other terms. The boundary terms disappear because we have fixed endpoints, leaving us with a cleaner expression where δy stands alone. I have a video that explains this in more detail (th-cam.com/video/4f1601w4syI/w-d-xo.html) Your intuition about the variation being infinitesimally small is exactly right! For the optimal path, the first variation must vanish for any arbitrary δy, which can only happen when the Euler-Lagrange equation equals zero. This is what leads us to our extremal solution. Hope this helps clarify the concepts! 🎯
Thanks for your videos, they help a lot. I've just started going through to try and refresh my knowledge in solving motion problems. If you consider making another simple 2DOF example, i would be interested to see a case with air resistance that is quadratic to velocity - for example a rock or rocket with initial velocity in x and y direction. Another thing i would be curious about is constrained motion. For example a trajectory is given by curve segments and a body is constrained to this trajectory by 2 contact points, while an external force acts on the body to move it. e.g. a long box on a slide with variable curvature, sliding down through gravity. Determining movement/rotation around the center of mass, and also if the body would stop due to friction in those 2 contact points.
Loosely speaking, in dropping the higher order terms of the Taylor Series, can we also say (besides invoking the boundary conditions) that we are essentially taking the limit (in analogy with the differential dx) of epsilon as it approaches 0 (the variation at the boundary)? That is, after we differentiate the TOTAL variation of I in finding an extremum.
I am so thankful for your explenations! Question 1: I have one question, I didnt get the background of : at 14:05 when we do the PI on the Right Term - why does the derivative falls apart from the ∂y‘ to ∂y ? Question 2: And why do we want it to do that ? And (a Bit previous) why do we change the sign from ∂(T) (for total diff. i understood that) to ∂(I) I ? And question 3: 15:51 why is everithing zero when we multiply it with del y ? I thought that this ∂y simply Shows is the sign and If its 0 .. so where do i Take the Information from that multiplied with it = 0 ? Thanks a lot you help so much !! 🎉
1. It's not the derivative ∂y‘, but rather the variation 𝛿y' that we are integrating. The formula for integration by parts is: int u dv = uv - int v du. So, in our problem, dv = 𝛿y' therefore integrating gives us v = 𝛿y. Then we plug this into the formula. 2. This is how we integrate by parts. Just plugging into the formula above. Perhaps try to review this technique. 3. The reason we say that the entire expression must be equal to zero is as follows... - If the value of the integral, I, is to be a maximum or minimum, then the integral must not change its sign for all possible variations of 𝛿y. - However 𝛿y is arbitrary (ie can be positive or negative), therefore the part that multiplies 𝛿y equal to 0. - We conclude from this that 𝛿I = 0 is the necessary condition to find an extremal. Hope this makes sense.
"in the vicinity of an extremum" So we are restricted to SMALL values of δy = εη in order to get the result? An infinitesimal variation of y makes the variation of the integral be zero if we are at an extremum?
There is no restriction on ε as such, but we are interested in the behavior near extremals, and in the vicinity of extremals, values of ε --> 0, so really, the infinitesimal values are the ones of interest. And, "yes" to your last question. That's exactly the definition of an extremal - it's a stationary point so a little nudge by a variational amount (in terms of any variable) will have no change on the value of the function (because the slope is zero).
@@Freeball99 Thanks. Also, why are we interested in retaining the sign of the integral? Seems like merely a more complicated way of saying that the variation of the integral must be zero. Edit: Actually, isn't it the variation of the integral and not the integral itself that must retain its sign? For example Wikipedia states so.
why del(y) and dI/depsilon at epsilon=0 are equivalent? I mean del(I)=dI/depsilon epsilon, at epsilon=0 so it is true without assuming dI/depsilon is zero
It doesn't matter what causes it to be zero. The fact that both of these give us zero demonstrates that setting the first variation δI = 0 is the equivalent of setting dI/dε (@ ε=0) to 0. We are just showing that we will find the extremal using either method.
This is very nice presentation. I especially like how organized you are in your presentation. And you even number each equation. But I don't think it's necessary to do the dI/de | e= 0. In fact, I still don't fully understand it's rational in the first derivation of E-L. Seems to me, once you have equation 17, it's not hard to see that both epsilon and eta are free choices and can approach zero, but never be zero, that the rest in brackets has to be zero to make delta I = zero every where. Thus the E-L equations. If you can pick e = 0 at anytime, why bring it in to be zero. It's hard to see how it's a free parameter, if you have to force it. I think I need to study the first video!!! Think of it like this. You know at extrema of f(x) that f'(x) = O. Did you set x or epsilon in f to zero? You just know you can make epsilon arbitrarily close to zero, but never zero ie df/dx = lim e approach 0 f(x + e) - f(x) / e. So df/dx has to be zero (not epsilon and not x, like wise epsilon and not eta). What I don't understand is how lagrange did this at age of 19!!!
The variation of I at the extremal will be zero. Consequently, in the vicinity of the extremal, the value of I will not change and so its signs stays the same. This is similar to differential calculus where the derivative/slope is zero at the extremal and changes sign in the vicinity. So, the statement is correct that I (rather than the variation of I) should retain its sign.
probably obvious but, isn't Y' completely determined by Y? Why would it appear seperately in the Taylor expansion? I mean, you can not fix Y and change Y', so what does partial derivative by Y' mean? In other words, we could also think of F as F(x,Y,Y',Y''), then would the total integral include a partial derivative by Y''? What is my confusion about? Thankful for any answers
Initially, we do treat y and its derivatives as independent parameters when taking the variation. The relationship between y and it’s derivatives is then accounted for when we integrate by parts and the combine terms to form the Euler-Lagrange equation. This enforces the relationship between y and its derivatives.
In essence I just applied the formula for the Taylor series of a function but replaced the function with a functional. Then, instead of taking the derivative, I took the variation (bearing in mind that we only vary the dependent variables so the variation of x is zero).
@@Freeball99 Hello, also asked this as a seperate comment but seeing that this is discussed here too: What does it mean to vary Y'. Afaiunderstand, Y' is completely decided by Y. what is the meaning of the partial derivative of Y' then? Thanks.
The varied path (y_bar) is the sum of the optimal path (y) plus a variation to that path (𝛿y). So y_bar(x) = y(x)+ 𝛿y(x). The variation is modeled as 𝛿y(x) = ε·η(x) where η(x) is a shape function and ε is the magnitude. NOTE: ε (which is considered to be small) is NOT a function of x. The shape function η(x) is completely arbitrary EXCEPT that it must satisfy the geometric boundary conditions, namely that η(x1) = 0 and η(x2) = 0 (ie the varied path must start and end at the same locations as the actual path so the variation at these points is zero).
hello, when we are doing integrate by parts in equation 13 (14:30 time) . I want to whether we are integrating only the second part of the equation or the whole integral. waiting for a reply.
Sorry for the delayed response, but I missed this until now. Yes, it's just the second term that is integrated by parts - since we need to remove the derivative on δy'.
In the region of a stationary point, a variational "nudge" away from this point should not change the value of the function. The idea is that if the value doesn't change, then its sign cannot change.
@@Freeball99 But isn't it true that what changes is just the variation of the functional not the value of the functional itself. I don't see why this variation cannot be changing signs. Could you please elaborate a bit? BTW, I am extremely grateful for these videos. I have been always shaky in the variational calculus part in my study of solid mechanics, and the mathematics of it seemed intimidating. Thanks a lot! Keep up the good work
so the delta of y' is not independent of the delta of y. why is it not important to have an independent delta of y' in such a general formulation? appreciate if anyone can provide an answer....
Not sure if I completely understand the question. Hopefully this covers it. The starting point is that we are dealing with path minimization type problems. In general, we are dealing with a functional that depends on an independent coordinate, x, and the dependent coordinate, y(x) and its derivatives (up to any order). We initially (in a previous video) treat y and its derivatives as independent coordinates, which allows us to derive the weak form of the governing equations. We then use the fact that y and its derivatives are not independent when we integrate by parts to arrive at the strong form.
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Finally, an explanation that does not get rid of the higher order terms as quickly as we can. Doing that feels like an approximation while this is exactly true for all of the things we are working with.
The clarity of though and logic the author is using is so so good.
This channel is honestly one of the best I've found when it comes to this kind of advanced calculus
I love this kind of video, going over mathematical concepts that we breezed past in engineering classes without proper explanations or derivations. Keep em coming!
This video is just brilliant. Explained very nicely.
oh man!! I was so confused by the multiplication by epsilon in the taylor expansion derivation from my course and I had just given up on understanding it. THANK YOU
Thank you for these videos. I've been struggling with the Calculus of Variations, and this has helped both mathematically and conceptually.
The lectures are very informative. Hats off to the facilitator and it would be better to have lecture notes in pdf.
The notes are available in PDF format from the download link in the video description.
You’re a very gifted teacher, the way you word things is concise and it all flows exceptionally well. 10/10 ❤ u dude no homo tho
i am lost in this now.... going to watch pervious video and this video again and again..
Stick with it!! This is more advanced material than before. Videos are a little bit mathematical at first because it is setting up a framework that we will use. As soon as you see a few examples of this theory being used in practice, it will fall into place for you. It's a little like using Lagrange's equations, which made our lives very easy once we started applying it to mechanics problems, but we've never derived Lagrange's equations...until now. These first few videos on this playlist explain the theory behind Lagrange's equations and other energy methods. Once you see some practical examples of this being used (which will come in the next few videos), then, you'll return to these videos and they will make more sense.
I am very much trying to get you guys quickly to the point where we are solving practical problems, but I can't get there without laying out some theory first. The math behind this theory is actually very beautiful, but for most people, Variational Calculus takes a little while to get used to. It took the world top scientists 160 years to get to Hamilton's Principle from Newtonian Mechanics and I'm trying to get you there in a few videos.
took me some time but i get it now.
Thank you for excellent explanation.
Ur the GOAT 👑
Quite enlightening! Thanks!
This is great. I just started the video and I'm holding my breath to see if Oliver Heaviside gets proper credit for inventing this operation. Now as I proceed I'm not seeing evidence that this delta operator is the the same as Dirac's delta operator.
Delta's occurs in many places in math and engineering. In this case, we are referring to the variational operator. It bears no relation other uses like the Dirac delta function or the Kronecker delta.
very nice video and very helpfull
these videos are amazing, it really helps with my MIT class on quantum mechanics
@عَدِيُّ to me it is not that much, since usually you don't have to deal with the uncertainty as you do see in quantum mechanics, however, you will have to do some real world solutions in which will be harder for some, due to the scale, but to me the larger values just make more leeway for "rookie mistakes", I personally do not think of it as confusing, but others may say otherwise to me it is a shame that modern physics, and maybe biology is the reason why most people hate high school and college, and i think that we should dive more deeply into physics just to show that it is very important to our everyday lives, most of the reasons for this is because of the intense math involved, but as Richard Feynman has done it, he makes sure that physics in general is easier by the Feynman technique for integrals, differentials and more, if only more people knew about those subjects, maybe the world may be different today.
@@universum-theuniverseexpla6565 experimental physics mogs theoretical physics.
@@universum-theuniverseexpla6565I have heard so many time about feyman technique...can you kindly briefly explain..I know there is a complete set of his lectures pdf available.
Good Variations
05:43 Suppose the value of I for the extremal path is a minimum, then the variation in I is δI = Ibar - I >0 for all paths very close to extremal path. For one particular path with tiny deviation δy we get a tiny positve value of δI. Now consider the path got by just changing the sign of δy. The integral only changes in its sign, but we must still get a tiny positive value for δI. To ensure this as the higher order terms approach 0 we must have that the integrand is 0.
🏀
Great you just cleared my confusion, thanks
Just some doubts:
Firstly, when we say that in order for the variation of I to retain its sign in the vicinity of an extremal, everything multiplying "Dell y" should be zero. This is also considering that Dell y is arbitrary and can be negative or positive. I don't think I truly understand what it means for the variation of I to retain its sign, I understand that Dell y can be anything, and thus to extract a "stable" answer, the only possibility is to make it equal 0, which had no sign. If you could explain the meaning of the retention of the sign, it would confirm my understanding. Secondly, the way I understood it is, since the variation of I had to be infinitesimally small to achieve the perfect path, and this had to be true for any value of variation, it only makes sense that everything except the variation be zero. It is the only way to compute the path of zero variation, or, the optimum path. This was my intuition.
Lastly, this is just a fault of my maths, but I don't understand how you converted the weak form to the strong form in both videos, thought that is my fault.
Thank you for creating these videos and for your valuable time sir.
(Please don't see this as a request to explain my every doubt, it is more of a note to myself for me to remember this stuff. A single hinting line in answer holds the same value as an explanation for me:)
At an extremal (min/max), the variation should maintain a consistent sign - if we're at a minimum, variations should stay above it; at a maximum, they should stay below. Since δy can be either positive or negative (it's arbitrary), the only way to ensure this consistency is if the coefficient multiplying δy equals zero.
The weak to strong form conversion uses integration by parts to move derivatives from δy to other terms. The boundary terms disappear because we have fixed endpoints, leaving us with a cleaner expression where δy stands alone. I have a video that explains this in more detail (th-cam.com/video/4f1601w4syI/w-d-xo.html)
Your intuition about the variation being infinitesimally small is exactly right! For the optimal path, the first variation must vanish for any arbitrary δy, which can only happen when the Euler-Lagrange equation equals zero. This is what leads us to our extremal solution.
Hope this helps clarify the concepts! 🎯
Thanks for your videos, they help a lot. I've just started going through to try and refresh my knowledge in solving motion problems.
If you consider making another simple 2DOF example, i would be interested to see a case with air resistance that is quadratic to velocity - for example a rock or rocket with initial velocity in x and y direction.
Another thing i would be curious about is constrained motion. For example a trajectory is given by curve segments and a body is constrained to this trajectory by 2 contact points, while an external force acts on the body to move it. e.g. a long box on a slide with variable curvature, sliding down through gravity. Determining movement/rotation around the center of mass, and also if the body would stop due to friction in those 2 contact points.
Thanks for you suggestion. Might take me a little while to get there, but I will definitely tackle a constraint problem soon.
Thank you!!!
Wonderful! Thank yoy so much
Loosely speaking, in dropping the higher order terms of the Taylor Series, can we also say (besides invoking the boundary conditions) that we are essentially taking the limit (in analogy with the differential dx) of epsilon as it approaches 0 (the variation at the boundary)? That is, after we differentiate the TOTAL variation of I in finding an extremum.
Yes, I think that's correct.
thank you very much
Brilliant
I am so thankful for your explenations!
Question 1: I have one question, I didnt get the background of : at 14:05 when we do the PI on the Right Term - why does the derivative falls apart from the ∂y‘ to ∂y ?
Question 2: And why do we want it to do that ? And (a Bit previous) why do we change the sign from ∂(T) (for total diff. i understood that) to ∂(I) I ?
And question 3: 15:51 why is everithing zero when we multiply it with del y ? I thought that this ∂y simply Shows is the sign and If its 0 .. so where do i Take the Information from that multiplied with it = 0 ?
Thanks a lot you help so much !! 🎉
1. It's not the derivative ∂y‘, but rather the variation 𝛿y' that we are integrating.
The formula for integration by parts is:
int u dv = uv - int v du.
So, in our problem, dv = 𝛿y' therefore integrating gives us v = 𝛿y. Then we plug this into the formula.
2. This is how we integrate by parts. Just plugging into the formula above. Perhaps try to review this technique.
3. The reason we say that the entire expression must be equal to zero is as follows...
- If the value of the integral, I, is to be a maximum or minimum, then the integral must not change its sign for all possible variations of 𝛿y.
- However 𝛿y is arbitrary (ie can be positive or negative), therefore the part that multiplies 𝛿y equal to 0.
- We conclude from this that 𝛿I = 0 is the necessary condition to find an extremal.
Hope this makes sense.
It looks at the beginning that you have a lot of linearity but I'm not hearing any mention of that property.
I discuss linearity around the 10:45 mark.
Sir please make video on configuration space and phase space
"in the vicinity of an extremum"
So we are restricted to SMALL values of δy = εη in order to get the result? An infinitesimal variation of y makes the variation of the integral be zero if we are at an extremum?
There is no restriction on ε as such, but we are interested in the behavior near extremals, and in the vicinity of extremals, values of ε --> 0, so really, the infinitesimal values are the ones of interest.
And, "yes" to your last question. That's exactly the definition of an extremal - it's a stationary point so a little nudge by a variational amount (in terms of any variable) will have no change on the value of the function (because the slope is zero).
@@Freeball99 Thanks. Also, why are we interested in retaining the sign of the integral? Seems like merely a more complicated way of saying that the variation of the integral must be zero.
Edit: Actually, isn't it the variation of the integral and not the integral itself that must retain its sign? For example Wikipedia states so.
why del(y) and dI/depsilon at epsilon=0 are equivalent? I mean del(I)=dI/depsilon epsilon, at epsilon=0 so it is true without assuming dI/depsilon is zero
It doesn't matter what causes it to be zero. The fact that both of these give us zero demonstrates that setting the first variation δI = 0 is the equivalent of setting dI/dε (@ ε=0) to 0. We are just showing that we will find the extremal using either method.
Thanks a lot!!!
This is very nice presentation. I especially like how organized you are in your presentation. And you even number each equation. But I don't think it's necessary to do the dI/de | e= 0. In fact, I still don't fully understand it's rational in the first derivation of E-L.
Seems to me, once you have equation 17, it's not hard to see that both epsilon and eta are free choices and can approach zero, but never be zero, that the rest in brackets has to be zero to make delta I = zero every where. Thus the E-L equations.
If you can pick e = 0 at anytime, why bring it in to be zero. It's hard to see how it's a free parameter, if you have to force it. I think I need to study the first video!!!
Think of it like this. You know at extrema of f(x) that f'(x) = O. Did you set x or epsilon in f to zero? You just know you can make epsilon arbitrarily close to zero, but never zero ie df/dx = lim e approach 0 f(x + e) - f(x) / e. So df/dx has to be zero (not epsilon and not x, like wise epsilon and not eta).
What I don't understand is how lagrange did this at age of 19!!!
Small question here: At 15:29 , shouldn't that be like- In the vicinity of extremal, 'variation of I' (not I) should retain its sign?
The variation of I at the extremal will be zero. Consequently, in the vicinity of the extremal, the value of I will not change and so its signs stays the same. This is similar to differential calculus where the derivative/slope is zero at the extremal and changes sign in the vicinity. So, the statement is correct that I (rather than the variation of I) should retain its sign.
probably obvious but, isn't Y' completely determined by Y? Why would it appear seperately in the Taylor expansion?
I mean, you can not fix Y and change Y', so what does partial derivative by Y' mean?
In other words, we could also think of F as F(x,Y,Y',Y''), then would the total integral include a partial derivative by Y''?
What is my confusion about?
Thankful for any answers
Initially, we do treat y and its derivatives as independent parameters when taking the variation. The relationship between y and it’s derivatives is then accounted for when we integrate by parts and the combine terms to form the Euler-Lagrange equation. This enforces the relationship between y and its derivatives.
How did you exactly calculate the taylor polynom of the functional?
In essence I just applied the formula for the Taylor series of a function but replaced the function with a functional. Then, instead of taking the derivative, I took the variation (bearing in mind that we only vary the dependent variables so the variation of x is zero).
@@Freeball99 Hello, also asked this as a seperate comment but seeing that this is discussed here too:
What does it mean to vary Y'. Afaiunderstand, Y' is completely decided by Y. what is the meaning of the partial derivative of Y' then?
Thanks.
I didn't get the difference between the varied path ( Y bar) and eta? Can you please elaborate more.
Thank you :)
The varied path (y_bar) is the sum of the optimal path (y) plus a variation to that path (𝛿y). So y_bar(x) = y(x)+ 𝛿y(x).
The variation is modeled as 𝛿y(x) = ε·η(x) where η(x) is a shape function and ε is the magnitude. NOTE: ε (which is considered to be small) is NOT a function of x.
The shape function η(x) is completely arbitrary EXCEPT that it must satisfy the geometric boundary conditions, namely that η(x1) = 0 and η(x2) = 0 (ie the varied path must start and end at the same locations as the actual path so the variation at these points is zero).
hello, when we are doing integrate by parts in equation 13 (14:30 time) . I want to whether we are integrating only the second part of the equation or the whole integral. waiting for a reply.
Sorry for the delayed response, but I missed this until now. Yes, it's just the second term that is integrated by parts - since we need to remove the derivative on δy'.
@@Freeball99 thankyou and great work you are doing
what does "I must retain sign" sentence meaning? I did not get the crux of this one.
In the region of a stationary point, a variational "nudge" away from this point should not change the value of the function. The idea is that if the value doesn't change, then its sign cannot change.
@@Freeball99 But isn't it true that what changes is just the variation of the functional not the value of the functional itself. I don't see why this variation cannot be changing signs. Could you please elaborate a bit?
BTW, I am extremely grateful for these videos. I have been always shaky in the variational calculus part in my study of solid mechanics, and the mathematics of it seemed intimidating. Thanks a lot! Keep up the good work
so the delta of y' is not independent of the delta of y. why is it not important to have an independent delta of y' in such a general formulation?
appreciate if anyone can provide an answer....
Not sure if I completely understand the question. Hopefully this covers it.
The starting point is that we are dealing with path minimization type problems. In general, we are dealing with a functional that depends on an independent coordinate, x, and the dependent coordinate, y(x) and its derivatives (up to any order).
We initially (in a previous video) treat y and its derivatives as independent coordinates, which allows us to derive the weak form of the governing equations. We then use the fact that y and its derivatives are not independent when we integrate by parts to arrive at the strong form.
which sources for explain Specifically the variations calculus??
Solid Mechanics: A Variational Approach by Dym & Shames
You are God
What the hill.. There is no God except Allah.
What application were you using?
The app is called "Paper" by WeTransfer. Running on an iPad Pro 13 inch and using an Apple Pencil. Quicktime is used to record the screen.
👏👏👏👍