# Tag Info

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Quasi-Newton methods construct an approximate Hessian for an arbitrary smooth objective function $f(x)$ using values of $\nabla f$ evaluated at the current and previous points. At each iteration of the method the quasi-Newton approximate Hessian is updated using the gradient evaluated at the latest iterate, $x^{(k)}$. These approximate Hessians aren't ...

12

Short answer If you only want second order accuracy and no embedded error estimation, chances are that you'll be happy with Strang splitting: half-step of reaction, full step of diffusion, half step of reaction. Long answer Reaction-diffusion, even with linear reaction, is famous for demonstrating splitting error. Indeed, it can be much worse, including "...

11

Your example is a pretty good indication that the two derivatives (with respect to $x$ and with respect to $u$) do not commute :) (In fact, they're very different beasts -- one is a Fréchet derivative, the other a weak derivative.) Rather, you should consider $a(u) = (\partial_x u)^2$ as the composition $a = f\circ g$, of two functions $$f: v(x)\mapsto v(... 10 According to Wolfram Alpha, x^5+3(x-1)=0 has no closed-form solution, so you can forget about a nice closed-form expression. :) I see nothing wrong with Newton's method; it should be quick and accurate, and with some analysis like the one you are sketching I think you can identify a safe starting point and prove global convergence. It might even be faster ... 9 I assume, that you have conducted a space discretization, so that you are about solving the (vector-valued) ODE$$ \dot u_h(t) = F_h(t,u_h(t)), \text{ on [0,T] }, u_h(0) = \alpha. $$via a numerical scheme \Phi that advances the approximation u_h^n at the current time instance t=t^n to the next value u_h^{n+1} at t=t^{n+1}:=t^n+\tau. Then your ... 8 For an unconstrained linear SPD problem, BFGS convergence matches unpreconditioned conjugate gradients. Differential operators appearing in structural mechanics problems are usually ill-conditioned with many large eigenvalues, making unpreconditioned iterative methods hopelessly slow. To deal with the troublesome spectrum, structural mechanics solvers ... 8 Yes: See Higham's book "Accuracy And Stability of Numerical Algorithms", second edition, chapter 25: Nonlinear Systems and Newton's Method. In particular, see the section on the "limiting residual" in terms of a condition number for the Jacobian. It may well be that since your system is ill-conditioned, that you quickly hit the limiting residual and your ... 7 For solving nonlinear partial equations, say -\Delta y + f(y) = 0, there are two common approaches: Fixed point iteration: Pick y^0 and for k=1,\dots solve$$ -\Delta y^{k+1} + f(y^k) = 0.$$(That is, you replace y by y^{k+1} in all linear terms and by y^k in all nonlinear terms.) Newton's method: Pick y^0 and for k=1,\dots solve$$ -\Delta \...

7

This is somehow unexpected, but my recent experience with solving a system of nonlinear equations is that treating them as the right hand side of a system of ordinary equations and then evolve the system with an ODE solver can be considerably faster than with the usual Newton-Raphson iteration. It sounds like you're doing some sort of pseudotransient ...

7

The issues you're running into now are not a failing of Newton-Raphson, but a question of coupling. You're doing iterated sequential coupling -- solving each equation sequentially and then iterating until (hopeful) convergence. No solver choice in place of NR is going to fix this lack of convergence, as long as you are doing iterated sequential coupling. ...

7

It's a bit easier to see if you write your equation in the a semi-discretised system of the form $u^{\prime}(t) = F(u(t))$ and with the application of the $\theta$-method and approximating $u^{\prime}(t) \approx (w^{n+1} - w^{n})/\tau$ this gives, $$w^{n+1} - w^{n} - (1-\theta) \tau F(w^n) - \theta\tau F(w^{n+1}) = 0$$ with unknown vector $w^{n+1}$ and ...

6

Off the top of my head, there are a number of things that may be going wrong. You might want to verify the following Are the function values or derivatives you are computing numerically stable? Is the rounding error in their evaluation smooth? You can verify this visually by plotting an extremely small interval around a known root such that the function ...

6

A few specific examples: Quasi-Newton methods avoid computing second derivatives by using an approximation to the Hessian which is updated at each iteration by a low rank (rank one or rank two) update. This makes factoring the Hessian (or equivalently keeping the inverse in product form) easy to do computationally. There are also limited memory Quasi-...

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If you need Jacobian matrix information for a numerical method, you should calculate the Jacobian matrix of the discretized form of the equations, since that will be consistent with the discretized equations you are solving.

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There are multiple ways of implementing Dirichlet boundary conditions when using the finite element method. For each unknown $i$ of the system belonging to the Dirichlet boundary, you can zero out row $i$ of the Jacobian and entry $i$ of the right-hand side. At that rate, you can reformulate the problem for just the unknowns in the interior, which reduces ...

6

The approximation error is $\mathcal{O}(\varepsilon)$, where $\varepsilon$ is the parameter in the finite difference calculation. It will be good to know how this approximation affects the number of CG iterations compared to having the complete Hessian? Are there theoretical/empirical understandings about this? Yes. In short, there are problems where CG ...

6

This is sometimes known as Buzano's inequality (http://www.jstor.org/stable/2159168). In general, if $\|x\|=1$, $P=xx^{\top}$ is a projection operator, so a simple application of Cauchy-Schwarz leads to $$2|b^{\top}xx^{\top}a|-|b^{\top}a| \leq |b^{\top}(2P-I)a| \leq \|a\|\,\|b\|,$$ which gives Buzano's inequality $$|b^{\top}xx^{\top}a| \leq \frac{\|a\|\,\|... 6 A low rank update to an n by n matrix A is an update of the form A=A+UU^{T} where U is matrix with n rows but very few columns (typically just one or two.) If the matrix UU^{T} has k columns, then its rank is at most k, so this is a low-rank update to the A matrix. Low rank updates are important in computational linear algebra and ... 6 The reason is that GMRES can only be used for solving linear equations, i.e. equations of the form Ax=b, where A is some matrix and x,b are vectors. What GMRES does, essentially, is it approximates multiplication by the matrix A^{-1} using a matrix polynomial of A. In this case (I assume) f(y^{n+1},t) is not necessarily linear in the vector y^{... 5 You would use backward euler method to solve a differential equation of the form u_t =f(u,t) where f is not necessarily a linear function in u. When f is non-linear, then the backward euler method results in a set of non-linear equations that need to be solved for each time step. Ergo, Newton-raphson can be used to solve it. For example, take$$\...

5

Your observed quadratic convergence indicates that the Jacobian is likely correct. Have you looked at the solutions for your under-resolved configurations? Galerkin optimality uses the operator norm, which contains only the symmetric part, thus the solution of the discrete system could be quite different from the projection of the exact solution. This ...

5

I'm not sure if this is the answer, but consider the case as $\alpha \rightarrow \infty$. In this case you're trying to solve $\iint_{\Omega} \cosh(P) dxdy = 0$ And since cosh looks like this: You can see that integrating it over any domain will never give you zero. If the diffusion of $P$ ($\nabla^2 P$) is not large enough to counteract this limit then ...

5

I suspect the main problem is the magnitude of the values. If you divide through by $\cosh$ to make all the numbers smaller, then ApproxFun doesn't seem to have a problem finding all the roots. function lhs(ε, α) A, B = sqrt((α-1)/(2ε)), sqrt((α+1)/(2ε)) return sqrt(α*α-1)*(cos(A) - 1/cosh(B)) - sin(A)*tanh(B) end u = Fun(x -> lhs(1e-3, x), [1.0,...

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I don't know much about tracking implicit surfaces, so I'm just going to start with the optimization problem and go from there. The optimization problem is, at the core, nonlinear least squares, and can therefore be solved efficiently by the Gauss-Newton method, wherein at each step one linearizes the problem at the current guess, solves the linear least ...

5

To calculate the reaction forces at a node, Abaqus (or any structural FE code) simply sums the internal forces for all elements attached to that node. The reaction forces are the negative of that sum. For an Abaqus user element, the internal forces for the element are returned from subroutine UEL in the RHS array. The returned stiffness matrix (Jacobian), ...

5

The Netwon(-Raphson) method helps you finding the root(s) of a function $f(x)$, that is, those values of $x_0$ for which $f(x_0) = 0$. In your case, the unknown value is $c$, so we start by writing down the problem in terms of a function $f(c)$ that is zero when $c$ takes the "correct" value (which we call $c_0$). Since this looks like homework, I don't ...

5

Not sure where you get your equation for $\epsilon$, but ultimately your approximation for the Jacobian matvec operation is a finite difference approximation to the directed derivative of $F(\cdot)$. This means that you will want as accurate of an approximation to this directed derivative as possible while being sure to avoid numerical round-off issues. For ...

4

Using a line search or trust region method along with Newton's method typically does the trick of finding the zero reliably. Check any advanced numerical methods or optimization book.

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Can you bound your zero? If you can come up with a good heuristic for the approximate location of the zero and then compute bounds for it over an interval that is not too large, the performance of many methods is improved. Here's some Matlab code I use to perform basic root bracketing for monotonic functions. In terms of interval methods, @Pedro's ...

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(1): This depends on the behavior of derivatives of the Jacobian (sic!) in the null space of the Jacobian at the solution. In practice, nobody calculates these derivatives, and i didn't even bother to remember the precise conditions. (2) works, though convergence is only linear. To get superlinear convergence (at least in the majority of cases), one may ...

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