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Calculus of Variations is what happens when you stop optimizing values and start optimizing geometries. The unknown isn’t a single x...it’s a whole curve y(x). And the thing you’re minimizing usually isn’t a formula you can eyeball...it’s an integral that judges the entire shape. For our first lecture, we... show more
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![What Curve Beats a Straight Line? Take two fixed points A and B. A bead slides without friction under gravity. Among all possible curves connecting A to B, which one gives the smallest travel time? This is the Brachistochrone problem. The time functional is T[y] = ∫ √((1 + y′²)/(2gy)) dx where y is the vertical drop from the starting point. The winning curve is not a straight line. It is a cycloid x(θ) = a(θ - sin θ) y(θ) = a(1 - cos θ) The bead first falls steeply to gain speed, then uses that speed across the remaining distance. The path is longer than the straight segment, yet the travel time is smaller. The shortest path solves geometry. The fastest path solves variation. #CalculusOfVariations #Brachistochrone #Cycloid #ClassicalMechanics #MathematicalPhysics #Optimization #DifferentialEquations #PhysicsAnimation #Mathematics](https://image.24vids.com/tw-2070226540810428705/amplify_video_thumb/2070226188677644288/img/-_2rUKA6DxvJw6EX.jpg)
![Lecture 4 on Calculus of Variations You might wonder...If I’m optimizing a shape...a curve, a surface, a whole path, what does "take the derivative and set it to zero" even mean? Do I take the damn derivative with respect to a curve/surface? 🤔 In normal calculus the variable is a number x, so the reflex is clean...f′(x)=0. In calculus of variations the variable is a whole function...the geometry itself, like a curve y(x) (or a surface z(x,y)). So the derivative can’t be a single slope. It has to be a pointwise sensitivity, i.e. how the objective reacts to tiny local deformations. You’re holding a whole shape, like a curve y(x). Your objective isn’t f(x) anymore, it’s a functional J[y], and as we've seen with our first there examples, usually an integral that depends on the entire curve (often through y and y’). To talk about a “derivative”, you do the only thing that makes sense: you nudge the entire curve by a tiny amount and see how J changes. Pick a wiggle shape η(x). It’s not random...it’s any admissible deformation direction. Admissible just means it obeys the constraints. If the endpoints are fixed, you force η(0)=η(1)=0 so the wiggle doesn’t move the endpoints. Then scale that wiggle by a small number ε and define the perturbed curve yε(x)=y(x)+εη(x). Now treat ε like the usual scalar in a Taylor expansion. As ε→0, J[y+εη] expands as J[y+εη] = J[y] + ε · (first-order term depending linearly on η) + o(ε). So the difference is J[y+εη] - J[y] = ε · (linear functional of η) + o(ε). For the standard integral of a Lagrangian problems, that linear functional can be written as an inner product with some function of x: J[y+εη] - J[y] = ε ∫ (δJ/δy)(x) η(x) dx + o(ε). That’s the definition-level meaning of δJ/δy: it’s the unique pointwise sensitivity function that makes this identity true for every admissible η. If δJ/δy is positive at some x, then choosing η negative there decreases J; if δJ/δy is negative there, pushing y upward locally decreases J. It’s literally a map along the curve saying push this way to go downhill. Now translate “set the derivative to zero.” At a minimizer y*, the first-order change must vanish for every admissible wiggle: J[y*+εη] − J[y*] = o(ε) for all η. Plug in the expansion and the ε-term must be zero: ∫ (δJ/δy)(x) η(x) dx = 0 for all admissible η. Here’s the crucial logic step: the only way an integral against every test function η can be zero is if the integrand itself is zero (in the usual sense used in analysis). So you get δJ/δy = 0. For the common case J[y]=∫ L(x, y, y’) dx, you can compute δJ/δy explicitly and it becomes the Euler–Lagrange expression δJ/δy = ∂L/∂y − d/dx(∂L/∂y’). So if you name the Euler–Lagrange residual as “left-hand side” R(x) = ∂L/∂y − d/dx(∂L/∂y’), then “set the derivative to zero” is exactly R(x)=0. That’s why animation works so well. You don’t have to solve R=0 in one shot. You can evolve the curve in an artificial time τ by pushing it in the downhill direction: ∂y/∂τ = −R(y). Where the residual is large, the curve moves a lot; as the residual drains toward zero, the motion dies out and the curve settles into an extremal. In our animations, we start from an intentionally ugly curve/surface. Frame by frame the functional drops, the residual drains away, and the geometry relaxes into an extremal. #CalculusOfVariations #EulerLagrange #FunctionalDerivative #GradientFlow #Optimization #MathAnimation](https://image.24vids.com/tw-2005012934343467237/amplify_video_thumb/2005008764127076352/img/icWC3mGkeN2DfPJK.jpg)
![String Theory Lecture 1 A String Does Not Move Like a Point A point particle traces a line through spacetime. A string traces a surface. This is the first geometric shift in String Theory. Particle mechanics asks where one object is at time t, so its history is a curve. String Theory asks where every point of an extended object is at worldsheet time τ, so we need another coordinate telling us where we are along the string. For a point particle x(t) So, for one input of time we get a position in Spacetime. For a string Xᵘ(τ,σ) Here τ plays the role of time on the worldsheet, while σ labels position along the string. Freeze τ and vary σ, and you see the string at one instant. Let τ move, and that curve sweeps out a two-dimensional surface... the worldsheet. The same comparison appears in the action. For a relativistic point particle, the geometric action measures worldline length S = −m ∫ ds If we parameterize the path by t, the action has one integral, one parameter, and one tangent vector dxᵘ/dt For a string, the same idea grows by one dimension. The action measures area, not length. In Nambu-Goto form, S = −T ∫ dτ dσ √[−det hₐᵦ] Here T is the string tension. It plays a role similar to mass, but for an extended object. It weights the area of a surface rather than the length of a line. The particle action has ∫ dt because the history is one-dimensional. The string action has ∫ dτ dσ because the history is two-dimensional. We are no longer summing along a path, we are summing over a surface. The geometry changes for the same reason. For the particle, one derivative is enough dxᵘ/dt For the string, the geometry is built from two derivatives: ∂τXᵘ and ∂σXᵘ The first tells you how the string changes as worldsheet time flows. The second tells you how the embedding changes as you move along the string. Together they define the induced worldsheet metric hₐᵦ = ∂ₐXᵘ ∂ᵦXᵤ In plain terms, hₐᵦ measures tangent lengths and tangent angles on the worldsheet. From it, the area element is dA = dτ dσ √[−det hₐᵦ] This, the Nambu-Goto action is the direct analogue of the point-particle length action. The point particle extremizes length and the string extremizes area. For calculations, people usually switch to the Polyakov action: S = −(T/2) ∫ dτ dσ √[−γ] γᵃᵇ ∂ₐXᵘ ∂ᵦXᵤ This describes the same classical string dynamics, but the algebra is cleaner. After choosing conformal gauge, varying with respect to Xᵘ gives (∂²/∂τ² − ∂²/∂σ²) Xᵘ = 0 This is the first real dynamical payoff... a two-dimensional wave equation on the worldsheet. For a point particle, the equation of motion tells you how one position evolves along one path. For a string, it tells you how an entire curve evolves, with waves traveling along it. The term ∂²Xᵘ/∂τ² measures acceleration in worldsheet time, while ∂²Xᵘ/∂σ² measures curvature along the string. The time evolution is balanced by how the string bends along its own length. This is why strings have oscillation modes. A point particle has one trajectory. A string has many possible vibration patterns, each one a normal mode of the worldsheet wave equation. For a closed string, σ wraps around the loop Xᵘ(τ, σ + 2π) = Xᵘ(τ, σ) For an open string, one standard free-end condition is ∂σXᵘ = 0 at the endpoints. Solving the wave equation gives waves moving in opposite directions along the string Xᵘ(τ,σ) = Fᵘ(τ + σ) + Gᵘ(τ − σ) A function of τ + σ moves one way. A function of τ − σ moves the other. Therefore, a particle has a worldline, its action measures length, and its geometry uses one tangent. The string has a worldsheet, its action measures area, and its geometry uses two tangent directions. #StringTheory #TheoreticalPhysics #MathematicalPhysics #Physics #Spacetime](https://image.24vids.com/tw-2047925796963004514/amplify_video_thumb/2047925638485753856/img/d0IV19m4OAIfEf0I.jpg)
![Lecture 2 of our Physics-Informed Neural Networks mini-series. In Lecture 1 we made the idea visible...a neural network isn’t predicting a PDE solution, it is the candidate function uᵩ(x,t), and the PDE residual rᵩ(x,t) is the leash that keeps it honest. Now the natural question follows: How can a neural network be punished for breaking a PDE when nobody ever handed it the true solution, and the equation itself contains derivatives like uᵩₜₜ and uᵩₓₓ? Here’s the satisfying answer: A PINN doesn’t need the true answer to be corrected. It only needs a way to measure how wrong it is according to the PDE! The network outputs uᵩ(x,t). A software called "autodiff" is used to compute the derivatives (uᵩₓ, uᵩₜ, uᵩₓₓ, …) exactly by applying the chain rule through the network. Those derivatives get dropped into the PDE to produce rᵩ(x,t). If rᵩ is big at some point, the loss spikes there, and gradient descent pushes the parameters so that rᵩ shrinks. The math breakdown We want a function u(x,t) that satisfies a PDE on a domain Ω. In this lecture we keep a concrete nonlinear example in mind, the damped sine-Gordon equation uₜₜ(x,t) + γ uₜ(x,t) − c² uₓₓ(x,t) + sin(u(x,t)) = 0. A PINN replaces the unknown function u with a neural network uᵩ(x,t), where ᵩ means all the network parameters (weights and biases). Now we build the physics residual by plugging uᵩ into the PDE rᵩ(x,t) = uᵩₜₜ(x,t) + γ uᵩₜ(x,t) − c² uᵩₓₓ(x,t) + sin(uᵩ(x,t)). If uᵩ were a true solution, rᵩ would be 0 everywhere. So we sample points (xⱼ,tⱼ) inside the domain. These are collocation points. At each one we evaluate rᵩ, and we define a physics loss L_phys(ᵩ) = meanⱼ |rᵩ(xⱼ,tⱼ)|². This is the punishment mechanism. (Punish just means: if |rᵩ| is big, L_phys is big; training updates ᵩ to make L_phys smaller. Reward means the loss drops, so those parameter changes are kept.) The key question was where the derivatives come from. Since uᵩ is built out of differentiable operations, we can compute uᵩₜ(x,t), uᵩₜₜ(x,t), uᵩₓ(x,t), uᵩₓₓ(x,t), at any input (x,t) we choose. Imagine a simple differentiable model written as a sum of nonlinear features uᵩ(x,t) = Σₖ vₖ σ( wₖx x + wₖt t + bₖ ) + b₀. Then the derivatives are just chain rule uᵩₓ(x,t) = Σₖ vₖ σ′(·) wₖx uᵩₓₓ(x,t) = Σₖ vₖ σ″(·) (wₖx)² uᵩₜ(x,t) = Σₖ vₖ σ′(·) wₖt uᵩₜₜ(x,t) = Σₖ vₖ σ″(·) (wₖt)². So rᵩ(x,t) is an explicit computable number at every (x,t). For the damped sine-Gordon example, it’s the same story, just with one extra nonlinear term: rᵩ(x,t) = [uᵩₜₜ(x,t) + γ uᵩₜ(x,t) − c² uᵩₓₓ(x,t)] + sin(uᵩ(x,t)). A real PINN is a deeper composition of these same building blocks, but it’s still just a chain rule, and autodiff is the machinery that does that bookkeeping reliably for big graphs. Then we train by gradient descent on the total loss. Even if we use only physics for the moment, the update is conceptually just ᵩ ← ᵩ − η ∇ᵩ L_phys(ᵩ), with learning rate η. In practice we also include initial/boundary conditions or data, because PDEs aren’t uniquely determined without them L(ᵩ) = L_data(ᵩ) + λ L_phys(ᵩ) + L_bc/ic(ᵩ), where L_bc/ic(ᵩ) enforces things like uᵩ(x,0) ≈ u₀(x) and uᵩₜ(x,0) ≈ v₀(x), or boundary conditions at x = ±L. So Lecture 2’s punchline is simple: the PDE becomes a training signal. We keep differentiating uᵩ, measuring rᵩ, and updating ᵩ until the residual goes quiet across Ω. #PINNs #PhysicsInformedNeuralNetworks #ScientificMachineLearning #AutoDiff #Backpropagation #PDE #DifferentialEquations #Optimization #MachineLearning #AppliedMath #ComputationalPhysics](https://image.24vids.com/tw-2009748938266685482/amplify_video_thumb/2009748619386101760/img/q0u_ZxmvYMtDl2eQ.jpg)