From: wolf Date: Sat, 17 Apr 2004 16:30:41 +0000 (+0000) Subject: Almost finish X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=6d4570a7d50ec9b61bfb2e12f5c8bf200b5623b8;p=dealii-svn.git Almost finish git-svn-id: https://svn.dealii.org/trunk@9041 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/doc/tutorial/chapter-2.step-by-step/step-15.data/img1.png b/deal.II/doc/tutorial/chapter-2.step-by-step/step-15.data/img1.png new file mode 100644 index 0000000000..e6db7d6b76 Binary files /dev/null and b/deal.II/doc/tutorial/chapter-2.step-by-step/step-15.data/img1.png differ diff --git a/deal.II/doc/tutorial/chapter-2.step-by-step/step-15.data/img10.png b/deal.II/doc/tutorial/chapter-2.step-by-step/step-15.data/img10.png new file mode 100644 index 0000000000..37e6f62972 Binary files /dev/null and b/deal.II/doc/tutorial/chapter-2.step-by-step/step-15.data/img10.png differ diff --git a/deal.II/doc/tutorial/chapter-2.step-by-step/step-15.data/img11.png 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Introduction

+ +

+Foreword +

+ +

+This program demonstrates a number of techniques that have not been shown in +previous example programs. In particular, it shows how to program for +one-dimensional problems, and some aspects of what to do with nonlinear +problems, in particular how to transfer the solution from one grid to the next +finer one. Apart from this, however, the program does not attempt to do much +more than to entertain those who sometimes like to play with maths. + +

+The application we chose is, as you will see, not even very well suited for +anything, since it is rather impossible to solve. When I started to write the +program, I was not aware of this, and it only turned out later that the +optimization problem we are looking at here is severely plagued by many, +likely even degenerate minima, and that we cannot really hope to find a global +one. What we do instead is to rather start the optimization from many initial +guesses (which is cheap since the problem is 1d), and hope that we can get a +reasonable best solution for some of them. While the whole thing, as an +application, is not very satisfactory, keep in mind that solving particular +applications is not the goal of the tutorial programs; rather, we would like +to demonstrate techniques of programming with deal.II, which is indeed the +focus here. + +

+ +

+The problem +

+ +

+Now for a description of the problem. In the book by Dacorogna on the +Calculus of Variations, I found the following statement, which confused me +tremendously at first (see Section 3.4.3, ``Lavrentiev Phenomenon'', very +slightly edited): + +

+

+Theorem 4.6: Let +

+

+
+$\displaystyle I(u)=\int_0^1 (x-u^3)^2 (u')^6\; dx.$ +

+

+ Let +

+

+
+$\displaystyle {\cal W}_1 = \{ u\in W^{1,\infty}(0,1) : u(0)=0, u(1)=1 \} +$ +

+

+

+
+$\displaystyle {\cal W}_2 = \{ u\in W^{1,1}(0,1) : u(0)=0, u(1)=1 \} +$ +

+

+

+

Then +

+

+
+$\displaystyle \inf_{u\in {\cal W}_1} I(u) \ge c_0 > 0 = \inf_{u\in {\cal W}_2} I(u). +$ +

+

+ Moreover the minimum of $ I(u)$ + over +$ {\cal W}_2$ + is attained by + +$ u(x)=x^{1/3}$ +. +
+

+

REMARKS. +
+

+

[...] +
+

+

ii) it is interesting to note that if one uses the usual finite element + methods (by taking piecewise affine functions, which are in +$ W^{1,\infty}$ +) + one will not be able to detect the minimum of some integrals such as the one + in the theorem. + +
+In other words: minimizing the energy functional over one space +( +$ W^{1,\infty}$ +) does not give the same value as when minimizing over a larger +space ($ W^{1,1}$ +). Furthermore, they give a rough estimate of the value of the +constant $ c_0$ +, which is +$ c_0=\tfrac{7^23^5}{2^{18}5^5}\approx 1.61\cdot +10^{-6}$ + (although by their calculation it is obvious that this estimate is +far too small, but the point of course is just to show that it is strictly +larger than zero). + +

+While the theorem was not surprising, the remark stunned me at first. After +all, we know that we can approximate functions in $ W^{1,1}$ + to arbitrary +accuracy. Also, although it is true that finite element functions are in + +$ W^{1,\infty}$ +, this statement is not really accurate: if the function itself +is bounded pointwise by, say, a constant $ C$ +, then its gradient is bounded by +$ 2C/h$ +, and thus +$ \Vert u_h\Vert _{1,\infty} \le 2C/h$ +. That means that we should be +able to lift this limit just by mesh refinement. Finite element functions are +therefore only in +$ W^{1,\infty}$ + if one considers them on a fixed grid, not on +a sequence of successively finer grids. (Note, we can only lift the +boundedness in $ W^{1,1}$ + in the same way by considering functions that +oscillate at cell frequency; these, however, do not converge in any reasonable +measure.) + +

+So it took me a while to see where the problem lies. Here it is: While we are +able to approximate functions to arbitrary accuracies in Sobolev + norms, this does not necessarily also hold with respect to the functional +$ I(u)$ +! After all, this functional was made to show exactly these +pathologies. + +

+What happens in this case is actually not so difficult to understand. Let us +look at what happens if we plug the lowest-order (piecewise linear) +interpolant $ i_hu$ + of the optimal solution $ u=x^{1/3}$ + into the functional +$ I(u)$ +: on the leftmost cell, the left end of $ i_hu$ + is tagged to zero by the +boundary condition, and the right end has the value +$ i_hu(h)=u(h)=h^{1/3}$ +. So +let us only consider the contribution of this single cell to $ I(u)$ +: +
+

+ + + + + + + + + + + + + + + + + + +
$\displaystyle \int_0^h (x-(i_hu)^3)^2 ((i_hu)')^6 dx$$\displaystyle =$$\displaystyle \int_0^h (x-(h^{1/3}x)^3)^2 ((h^{1/3}/h)')^6 dx$ + 
 $\displaystyle =$$\displaystyle h^{-4} \int_0^h (x^2-2hx^4+h^2x^6) dx$ + 
 $\displaystyle =$$\displaystyle h^{-4} (h^3/3-2h^5/5+h^9/7)$ + 
 $\displaystyle =$$\displaystyle {\cal O}(h^{-1}).$ + 
+
+ +Ups, even the contribution of the first cell blows up under mesh refinement, +and we have not even summed up the contributions of the other cells! + +

+It turns out, that the other cells are not really problematic (since the +gradient is bounded there by a constant independent of $ h$ +), but we cannot +really avoid the trouble with the first cell: if instead of the interpolant we +choose some other finite element function that is closer on average to +$ x^{1/3}$ + than the interpolant above, then we have to increase the slope of +this function, since we have to obey the boundary condition at the left +end. But then we are hit by the weight $ (u')^6$ +. This weight is simply too +strong! + +

+On the other hand, the interpolation of the linear function +$ \varphi(x)=x$ + +connecting the boundary values has the finite energy +$ I(i_h\varphi)=1/10$ +, +independent of the mesh size. Thus, +$ i_hx^{1/3}$ + cannot be the minimizer of the +energy as +$ h\rightarrow 0$ +. This is also easy to see by noting that +the minimal value of $ I$ + cannot increase under mesh +refinement: if it is finite for some function on some mesh, then it must be +smaller or equal to that value on a finer mesh, since the original function is +still in the space spanned by the shape functions on the finer grid, as finite +element spaces are nested. However, the computation above shows that we should +not be surprised if the value of the functional does not converge to zero, but +rather some finite value. + +

+There is one more conclusion to be drawn from the blow-up lesson above: we +cannot expect the finite dimensional approximation to be close to the root +function at the left end of the domain, for any mesh we choose! Because, if it +would, then its energy would have to blow up. And we will see exactly this +in the results section below. + +

+ +

+What to do? +

+ +

+After this somewhat theoretical introduction, let us just once in our life +have fun with pure mathematics, and actually see what happens in this problem +when we run the finite element method on it. So here it goes: to find the +minimum of $ I(u)$ +, we have to find its stationary point. The condition for +this reads +

+
+ + + + +
$\displaystyle I'(u,\varphi) = \int_0^1 6 (x-u^3) (u')^5 \{ (x-u^3)\varphi' - u^2 u' \varphi\}\ dx,$ +   
+

+for all test functions $ \varphi$ + from the same space as that from which we +take $ u$ +, but with zero boundary conditions. If this space allows us to +integrate by parts, then we could associate this with a two point boundary +value problem +

+
+ + + + +
$\displaystyle -(x-u^3) u^2(u')^6 - \frac{d}{dx} \left\{(x-u^3)^2 (u')^5\right\} = 0, \qquad\qquad u(0)=0, \quad u(1)=1.$ +(1)
+

+Note that this equation degenerates wherever $ u^3=x$ +, which is at least the +case at +$ x\in\{0,1\}$ + due to the prescribed boundary values for $ u$ +, but +possibly at other places as well. However, for finite elements, we will want +to have the equation in weak form anyway. Since the equation is still +nonlinear, one may be tempted to we compute iterates + +$ u_{k+1}=u_k+\alpha_k\delta u_k$ + using a Newton method for updates +$ \delta +u_k$ +, like in +

+
+ + + + +
$\displaystyle I''(u_k,\delta u_k,\varphi) = -I'(u_k, \varphi).$ +   
+

+However, since +$ I''(u_k,\cdot,\cdot)$ + may be an indefinite operator (and, as +numerical experiments indicate, is in fact during typical computations), we +don't want to use this. Instead, we use a gradient method, for which we +compute updates according to the following scheme: +

+
+ + + + +
$\displaystyle \left<\delta u_k,\varphi\right> = -I'(u_k, \varphi).$ +(2)
+

+For the scalar product on the left hand side, there are multiple valid ways; +we choose the mesh dependent definition +$ \left<u,v\right> = \int_\Omega (uv + +h(x)^2 \nabla u\cdot \nabla v)\; dx$ +, where the weight $ h(x)^2$ +, i.e. using +the local mesh width, is chosen so that the definition is dimensionally +consistent. It also yields a matrix on the left hand side that is simple to +invert, as it is the sum of the well-conditioned mass matrix, and a Laplace +matrix times a factor that counters the growth of condition number of the +Laplace matrix. + +

+The step length $ \alpha_k$ + is then computed using a one-dimensional line search +finding +

+
+ + + + +
$\displaystyle \alpha_k = \arg\min_\alpha I(u_k+\alpha\delta u_k),$ +(3)
+

+or at least an approximation to this using a one-dimensional Newton method +which itself has a line search. The details of this can be found in the code. +We iterate the updates and line searches until the change in energy $ I(u_k)$ + +becomes too small to warrant any further iterations. + +

+The basic idea that you should get in all this is that we formulate the +optimization method in a function space, and will only discretize each step +separately. A number of subsequent steps will be done on the same mesh, before +we refine it and go on to do the same on the next finer mesh. + +

+As for mesh refinement, it is instructional to recall how residual based error +estimates like the one used in the Kelly et al. error estimator are usually +derived (the Kelly estimator is the one that we have used in most of the +previous example programs). In a similar way, by looking at the residual of +the strong form of the nonlinear equation we attempt here to solve, see +equation (1), we may be tempted to consider the following +expression for refinement of cell $ K$ +: +
+

+ + + + + + + + + +
$\displaystyle \eta_K^2$$\displaystyle =$$\displaystyle h^2 \left\Vert +(x-u_h^3) (u_h')^4 \left\{ u_h^2 (u_h')^2 + 5(x-u_h^3)u_h'' + 2u_h'(1-3u_h^2u_h') \right\} +\right\Vert^2_K +\notag$ +(4)
  $\displaystyle + +h \left\vert (x-u_h^3)^2 [(u_h')^5] \right\vert^2_{\partial K},$ +(5)
+
+ +where $ [\cdot]$ + is the jump of a quantity across an intercell boundary, and + +$ \vert\cdot\vert _{\partial K}$ + is the sum of the quantity evaluated at the two end +points of a cell. Note that in the evaluation of the jump, we have made use of +the fact that $ x-u_h^3$ + is a continuous quantity, and can therefore be taken +out of the jump operator. + +

+All these details actually matter - while writing the program I have played +around with many settings and different versions of the code, and the result +is that if you don't have a good line search, good stopping criteria, the +right metric (scalar product) for the gradient method, good initial values, +and a good refinement criterion, then the nonlinear solver gets stuck quite +readily for this highly nonlinear problem. Initially, I was hardly able to +find solutions for which the energy dropped below 0.005, while the energy +after the final iteration of the program as it is is usually around 0.0003, +and occasionally down to less than 3e-5. + +

+However, this is not enough. In the program, we start the solver on the coarse +mesh many times, with randomly perturbed starting values, and while it +converges it yields a different solution, with a different energy every +time. One can therefore not say that the solver converges to a certain energy, +and we can't answer the question what the smallest value of $ I(u)$ + might be in + +$ W^{1,\infty}$ +. This is unsatisfactory, but maybe to be expected for such a +contrived and pathological problem. Consider it an example in programming with +deal.II then, and not an example in solving this particular problem. + +

+ +

+Implementation +

+ +

+The program implements all the steps mentioned above, and we will discuss them +in the commented code below. In general, however, note that formulating the +Newton method in function spaces, and only discretizing afterwards has +consequences: we have to linearize around $ u_k$ + when we want to compute + +$ \delta +u_k$ +, and we have to sum up these two functions afterwards. However, +they may be living on different grids, if we have refined the grid before this +step, so we will have to present a way to actually get a function from one +grid to another. The SolutionTransfer class will help us here. On the +other hand, discretizing every nonlinear step separately has the advantage +that we can do the initial steps, when we are still far away from the +solution, on a coarse mesh, and only go on to more expensive computations when +we home in on an solution. We will use a +very simplistic strategy for when we refine the mesh (every fifth nonlinear +step), though. Realistic programs solving nonlinear problems will have to be more +clever in this respect, but it suffices for the purposes of this program. + +

+We will show some of the things that are really simple in 1d (but sometimes +different from what we are used to in 2d or 3d). Apart from this, the program +does not contain much new stuff, but if it explains a few of the techniques +that are available for nonlinear problems and in particular 1d problems, then +this is not so bad, after all. + +

+Note: As shown below, the program starts the nonlinear solver from 10 different +initial values, and outputs the results. This is not actually too many, but we +did so to keep run-time short (around 1:30 minutes on my laptop). If you want to +increase the number of realizations, you may want to switch to optimized mode +(by setting the ``debug-mode'' flag in the Makefile to ``off''), and increase +the number of realizations to a larger value. On the same machine as above, I +can compute 100 realizations in optimized mode in about 2 minutes. For +this particular program, the difference between debug and optimized mode is +thus about a factor of 7-8, which can be explained by the fact that we ask the +compiler to do optimizations on the code only in the latter mode, but in most +part due to the fact that in optimized mode all the ``Assert'' checks are +thrown out that make sure that function arguments are correct, and that check +the internal consistency of the library. The library contains several +thousands of these checks, and they significantly slow down debug +computations, but we feel that the benefit of finding programming errors +earlier and including where the problem exactly appeared to be of significantly +greater value than faster run-time. After all, all production runs of programs +should be done in optimized mode anyway. + +

+A slowdown of a factor of 7-8 is unusual, however. For 2d and 3d applications, +a typical value is around 4. +