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Bot battles in graphwar, now bots can draw?? 🎨🤖 #graphwar #math #bots #ai #functions #parametric #botbattle #peacebot #larpbot How it works: 🎨 Paint Bot is the first one who doesn't fire a graph y = f(x) but a parametric curve x(t), y(t). The point moves in time t, not... show more
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![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)
![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)
![In Calculus of Variations, we give up the idea of only optimizing values and start optimizing geometries instead. The unknown is not a single x. It is a whole curve/surface y(x). The object you are minimizing is usually not a simple formula that you can guess. It is an integral that judges the whole shape. In our first lecture, we discuss the famous Brachistochrone problem. You choose two points, turn on gravity and ask a question which almost sounds too simple...which track will allow a bead to move from A to B in the minimum time? At your first attempt, your instinct will lead you astray. It is not the straight line. It is not the drop hard then cruise sketch either. The winner turns out to be a cycloid...the curve a point on a rolling circle traces. The track is the moving character in the animation. We begin with a flawed curve, perform gradient descent in curve-space, and observe the geometry change frame by frame as T[y] collapses to the brachistochrone. Please see the comment below for the math breakdown. #CalculusOfVariations #Brachistochrone #Cycloid #Physics #Optimization #Mathematics](https://image.24vids.com/tw-2060713817487675443/amplify_video_thumb/2060713736638189572/img/Lw0XmAHouOe98K4z.jpg)
