tag:blogger.com,1999:blog-22829559172760317642017-08-05T11:32:35.491+01:00M2AA1 Differential Equations 2016Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.comBlogger52125tag:blogger.com,1999:blog-2282955917276031764.post-6143963788962803142016-07-20T16:03:00.001+01:002016-08-02T22:10:41.351+01:00Preparation for the resit exam<div dir="ltr" style="text-align: left;" trbidi="on">Instructions on how to study for the resit exam in September 2016 are the same as for the May 2016 exam. </div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com2tag:blogger.com,1999:blog-2282955917276031764.post-18043401625047399502016-05-15T09:51:00.001+01:002016-05-15T09:51:03.547+01:00Picard's Theorem<div dir="ltr" style="text-align: left;" trbidi="on">One of the students has asked me: "In the past papers, the statement of the Picard Theorem is in a different form from the version given in the notes. Which one shall we use?"<br /><br />There are various versions of the Picard's Theorem, which although they may be formulated with different technical details, all essentially state the same thing, i.e. that unique solutions of ODEs can be found by means of Picard iteration. If you would be asked to state "Picard's Theorem" then of course I would be happy with any correct and sensible version of this Theorem. However, note that if I would be asking you to prove a certain result, your proof should of course be related to the Theorem that you provide, or a specific version of this Theorem (from the lectures) that I would ask you to prove.<br /><br />Past exam papers are useful to get an idea whether or not you are broadly well prepared for the exam. There is little point to learn answers to past exam papers by heart. Also in most cases, the provided "model answers" are not necessarily the unique correct formulation to an answer...<br /><br /><br />In general, let me assure you that I will give generous credit to answers to exam questions that demonstrate your understanding of the material that you are asked about, rather than split hairs over whether or not you provide exactly the answer that I would have given as a model...<br /></div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-69666169995492797652016-05-08T20:42:00.000+01:002016-05-08T20:42:44.780+01:00Persistence of transverse intersections<div dir="ltr" style="text-align: left;" trbidi="on">In Example 1.4.9 we discuss the persistence of transverse intersections as an application of the Impicit Function Theorem. Someone asked me about this in the second revision class. I will try to elucidate the final conclusion in this example from the notes.<br /><br />"Persistence" of the isolated intersection of two curves in \(\mathbb{R}^2\) in this example, means that if we "perturb" the curves slightly, then there remains to be a unique isolated intersection of these two curves near the original isolated intersection.<br /><br />We represent the two curves by differentiable functions that parametrize these curves: \(f,g:\mathbb{R}\to\mathbb{R}^2\). We assume the intersection to be at \(f(0)=g(0)\). We now consider a parametrized family of functions \(f_\lambda,g_\lambda:\mathbb{R}\to\mathbb{R}^2\), representing "perturbations of the original curves", where \(\lambda\) serves as the "small parameter" so that \(f_0=f\) and \(g_0=g\). We furthermore assume that the perturbations are such that the derivatives \(Df_\lambda(0)\) and \(Dg_\lambda(0)\) are continuous in \(\lambda\) near \(\lambda=0\).<br /><br />In the example it is proposed to consider the function \(h_\lambda(s,t):\mathbb{R}^2\to\mathbb{R}^2\) defined as \(h_\lambda(s,t):=f_\lambda(s)-g_\lambda(t).\) By construction \(h_0(0,0)=(0,0)\) and indeed the intersection points of the curves represented by \(f_\lambda\) and\(g_\lambda\) are given by \(h_\lambda^{-1}(0,0)\).<br /><br />It follows that \(Dh_\lambda(s,t)=(Df_\lambda(s),-Dg_\lambda(s))\), as in the notes. This two-by-two matrix is non-singular (ie has no zero eigenvalue, or - equivalently - is invertible) if and only if the two-dimensional vectors \(Df_\lambda(s)\) and \(Dg_\lambda(t)\) are not parallel (ie not real multiples of each other).<br /><br />We now use this to analyze the intersection at \(\lambda=0\): when \(Df_0(0)\) and \(Dg_0(0)\) (which are the tangent vectors to the respective curves at the intersection point) are not parallel, then \(Dh_0(0,0)\) is invertible and the intersection of the two curves at \(f_0(0)=g_0(0)\) is isolated (there is a neighbourhood of this point, where there is no other intersection).<br /><br />Considering a small variation of \(\lambda\), we note that by application of the Implicit Function Theorem to \(h_0\), for sufficiently small \(\lambda\) there exist continuous functions \(s(\lambda)\) and \(t(\lambda)\) so that \((s(\lambda),t(\lambda))\) is the element of \(h_\lambda ^{-1}(0,0)\) near \((0,0)=(s(0),t(0))\). This unique "continuation" of the original solution \(0,0\) is of course also isolated since if \(Dh_0(0,0)\) is invertible then so is \(Dh_\lambda(s(\lambda),t(\lambda))\) by continuity of all the dependences; so the vectors \(Df_\lambda(s(\lambda))\) and \(Dg_\lambda(t(\lambda))\) will not be parallel for sufficiently small \(\lambda\).</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-54095503235395762992016-05-06T18:52:00.000+01:002016-05-08T19:31:36.597+01:002009 exam question 2<div dir="ltr" style="text-align: left;" trbidi="on">A student has asked me go in more detail of the model answers for parts 2009 (c)(iii) and (d)(ii)<br /><br />First let me recall some generalities about the Jordan Chevalley decomposition. In the (complex) Jordan form, the diagonal part of the matrix is semi-simple - as it obviously has a diagonal complex Jordan form - and the remaining off-diagonal part is nilpotent (as it is upper or lower triangular with zero diagonal; one easily verifies that taking powers of such matrices eventually always results in the 0 matrix). One also easily verifies in Jordan form that the diagonal part and the upper or lower triangular part of the matrix commute with each other. A very convenient fact is that the properties "semi-simple", "nilpotent" and "commutation" are intrinsic and do not depend on the choice of coordinates:<br /><br />If \(A^k=0\) then \((TAT^{-1})^k=0\) for any invertible matrix \(T\).<br />If \(A\) is complex diagonalizable, then so is \(TAT^{-1}\) for any invertible matrix \(T\).<br />If \(A\) and \(B\) commute, i.e. \(AB=BA\), then also \((TAT^{-1})(TBT^{-1})=(TBT^{-1})(TAT^{-1})\) for any invertible matrix \(T\).<br /><br />So we observe that we can prove the Jordan-Chevalley decomposition (and its uniqueness) directly from the Jordan normal form.<br /><br />We proved in an exercise that \(\exp(A+B)=\exp(A)\exp(B)\) if \(AB=BA\). So in particular, if \(A=A_s+A_n\) is the Jordan-Chevalley decomposition, then \(\exp(At)=\exp(A_st)\exp(A_nt)\). This is very useful since the first part \(\exp(A_st)\) depends only on the eigenvalues of \(A\), and thus contains terms depending only on \(e^{\lambda_i t}\) where \(\lambda_i\) denote eigenvalues of \(A\) (if eigenvalues are complex this leads to terms with dependencies \(e^{\Re(\lambda_i)t}\cos(\Im(\lambda_i) t)\) and \(e^{\Re(\lambda_i)t}\sin(\Im(\lambda_i) t))\), were \(\Re\) and \(\Im\) denote the real and imaginary parts, respectively).<br /><br />We know that sometimes we also have polynomial terms appearing in the expression \(\exp(At)\). These polynomials come from the second part \(\exp(A_nt)\) since \(\exp(A_n t)=\sum_{m=0}^{k-1}\frac{A_n^m}{m!} t^m\) (this follows from the fact that \(A_n^k=0\)).<br /><br />The question (c)(iii) is about the Jordan-Chevalley decomposition of \(\exp(At)\). The only thing to check is that we can write this as sum of a nilpotent and a semi-simple matrix which commute with each other. (The Jordan-Chevalley decomposition theorem than asserts that this decomposition is in fact unique.) <br /><br />The question contains the hint that \(\exp(A_s t)\) is semi-simple. We can see this by verifying that if \(TA_sT^{-1}\) is (complex) diagonal, then so is \(T\exp(A_st)T^{-1}=\exp(TA_sT^{-1}t)\).<br /><br />Let us check whether indeed the semi-simple part of \(\exp(At)\) is equal to \(\exp(A_st)\) (in the sense of the Jordan-Chevalley decomposition). We write \(\exp(At)=\exp(A_st)+N\) where \(N:=[\exp(At)-\exp(A_st)]\). Now we recall that \(\exp(At)=\exp(A_st)\exp(A_nt)\) so \(N=\exp(A_st)[\exp(A_nt)-1]\) and as these two factors commute we have \(N^k=\exp(A_s t k)[\exp(A_nt)-1] ^k\) and if \(A_n^k=0\) we also have \([\exp(A_n t)-1]^k=0\) since \([\exp(A_n t)-1]=p(A_n)\) is a polynomial in \(A_n\) with \(p(0)=0\). Thus \(N\) is nilpotent, and it is readily checked that \(N\) also commutes with \(\exp(A_s t)\). So \(\exp(A t)=\exp(A_s t)+N\) is the Jordan-Chevalley decomposition of \(\exp(A t)\) where \(\exp(A_s t)\) is the semi-simple part and \(N=\exp(A_s t)[\exp(A_n t)-1]\) is the nilpotent part.<br /><br />In part d(ii), \(B\) is the projection to the eigenspace for eigenvalue \(-1\) with as kernel the generalised eigenspace for eigenvalue +1, and \(D=A_n\). As there is a Jordan block for eigenvalue +1 (and not for eigenvalue -1), the range of A_n is the eigenspace of A for eigenvalue +1 (check this by writing a 2-by-2 matrix with a Jordan block); the kernel of \(A_n\) is spanned by the eigenspaces of \(A\). Since the range of \(D\) lies inside the kernel of \(B\), and the range of \(B\) in the kernel of \(D\), it follows that \(DB=BD=0\).</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-46367215966678676012016-05-06T11:26:00.000+01:002016-05-06T11:26:22.676+01:002014 exam question 4 (v)<div dir="ltr" style="text-align: left;" trbidi="on">A student asked me about the model answer, which is very short and perhaps not so obviously correct.<br /><blockquote cite="mid:DB5PR06MB1623590ACCE3CA004BCD9C4CA7780@DB5PR06MB1623.eurprd06.prod.outlook.com" type="cite"> <div id="divtagdefaultwrapper" style="background-color: white; font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt;">'there exists \(x\) in \(A\) such that the \(\omega-\)limit set of \(x\) is not contained in the stable manifold of the singularity. Hence \(A\) contains a periodic orbit.' .</div></blockquote>The \(\omega\)-limit set of a point \(x\) are the points to which \(\phi^t(x)\) accumulates, ie points \(y\) so that there exists an increasing sequence \(t_n\) such that \(lim_{n\to\infty} \phi^{t_n}(x)=y\). <br />If an equilibrium \(y\) has a stable manifold, then the \(\omega\)-limit set of every point on this manifold equals \(y\) (as the point \(x\) converges to \(y\) under the flow). However, if we have a saddle point, there exists also an unstable manifold. If an initial point \(x\) does not lie on the stable manifold of an equilibrium, then by definition it does not converge to the equilibrium. It is a more subtle question whether it could accumulate to the equilibrium. I did not make this exam question, but the model answer if perhaps a bit too brief. It could namely be that a for point \(x\) that does not lie on the stable manifold of an equilibrium \(y\), we still have \(y\in\omega(x)\). For instance, there could be a heteroclinic or homoclinic cycle (consisting of equilibria and connecting orbits) to which \(x\) accumulates, and that contains the saddle equilibrium \(y\). In this exam question, there is only one equilibrium, so this implies that we only could have a homoclinic cycle (connecting orbit to one saddle), but this homoclinic cycle to a saddle would imply the existence of an equilibrium inside the area enclosed by the homoclinic loop (by PB arguments similar to the conclusion about the existence of an equilibrium inside the area enclosed by a periodic orbit) and as there is only one equilibrium in the system under consideration, this cannot be the case here. So there is no homoclinic cycle, there is only one equilibrium in \(A\) and orbits cannot leave \(A\) but also do not converge to the equilibrium. Then by PB they need to accumulate to a periodic solution (in \(A\)).<br /><br /><blockquote cite="mid:DB5PR06MB1623590ACCE3CA004BCD9C4CA7780@DB5PR06MB1623.eurprd06.prod.outlook.com" type="cite"> <div id="divtagdefaultwrapper" style="background-color: white; font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt;"><br /></div></blockquote></div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-64258252877933131152016-05-03T17:42:00.001+01:002016-05-03T17:42:15.132+01:00Questionnaire<div dir="ltr" style="text-align: left;" trbidi="on">In the final revision class I handed out a questionnaire to get a more detailed feedback about the course beyond SOLE. If you have not filled out and handed in the form to me yet, you can find an electronic copy of the questionnaire <a href="https://imperialcollegelondon.box.com/s/vvm035zeny325bpcr52axy0dprs1cliw" target="_blank">here</a>. Please fill it out and send it to me by e-mail or print it out and leave it in my pigeonhole. Your feedback is very much appreciated.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-82454356740447883922016-05-03T12:19:00.002+01:002016-05-03T12:19:23.389+01:00Second revision class<div dir="ltr" style="text-align: left;" trbidi="on">I discussed the application of Poincare-Bendixson theory to make sketches of phase portraits. The short note/summary I used can be found <a href="https://imperialcollegelondon.box.com/s/reile7c0hv7rnul2rxn7045ntrt376qp" target="_blank">here</a></div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-56776321803424447972016-04-30T13:13:00.000+01:002016-05-02T11:03:04.258+01:002013 exam question 2.<div dir="ltr" style="text-align: left;" trbidi="on">A student has asked me about this exam question (I paraphrase):<br /><ul style="text-align: left;"><li>In part a, how is the stable manifold determined? <br /><br />The stable manifold of the equilibrium 0 of a linear ODE is precisely equal to the union of all (generalized) eigenspaces for eigenvalues that have negative real part. Namely, on these (generalized) eigenspaces, all solution curves converge (exponentially fast) to the equilibrium 0, whereas on the other eigenspaces solutions that start outside the equilbrium never converge to 0. These properties all follow from the explicit solutions of the linear ODE restricted to generalized eigenspaces. In the example at hand the matrix A has eigenvalues 1 and -1 and the eigenvector for -1 is (1,-1). Hence the stable manifold is the line through 0 and (1,-1). </li><li>In part d, why does the model answer use the Euler-Lagrange equation rather than the conservation of the Hamiltonian?<br /><br />Either route is possible. I will show here how we can obtain the answer using the conservation of the Hamiltonian. The Hamiltonain is given by \(H=y'f_{y'}[y]-f[y]=(2y)^2+(y')^2\). The level sets \(H=c\) are ellipses in the \(y-y'\) plane if \(c>0\). We observe that \(c\geq0\). \(H=0\) corresponds to an equilibrium. So let \(c=d^2\). Then we have \(y'=\pm\sqrt{d^2-(2y)^2}\) which can be solved by separation of variables<br />\[\int dt =\int \sqrt{d^2-(2y)^2}^{-1}.\] Then \[ t+T=\frac{1}{2}\tan^{-1}\left(\frac{2y}{d^2-4y^2}\right),\] where \(T\) is constant. After some algebraic manipulations, one can rewrite this as \(y=\pm d \cos(2(t+T))\). The boundary condition \(y(0)=0\) yields \(T=\pm\frac{\pi}{4}\) so that \(y=\pm d \sin(2t)\) and \(y(1)= 1\) implies that \(y(t)=\frac{\sin(2t)}{\sin(2)}\), in accordance with the model answer.<br /><br />Clearly in this case, the calculations using the Hamiltonian appear more involved than using the Euler-Lagrange equation. Which route to the answer is most efficient will depend on the example. My advice would be when approaching such a calculation, is to try one route and if it looks tedious, quickly try the other one as well to see if it makes a difference.</li></ul></div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com2tag:blogger.com,1999:blog-2282955917276031764.post-57802626437075775072016-04-30T11:35:00.001+01:002016-04-30T11:35:31.164+01:00Bifurcation points<div dir="ltr" style="text-align: left;" trbidi="on">A few students have asked me what they need to know about bifurcation points. Of course we only touched upon this topic superficially (and a more detailed analysis is well beyond the scope of M2AA1). If an equilibrium point is hyperbolic (no eigenvalues of the Jacobian are on the imaginary axis), then the flow near the equilibrium is determined by its linearized flow (Hartman Grobman theorem) and the flow does not essentially change under sufficiently small perturbations, so hyperbolicity is a counter indicator for (local) bifurcation. If an equilibrium is not hyperbolic, then small perturbations to the vector field may lead to substantial changes of the flow near the equilibrium, which would amount to a (local) bifurcation. We have not really discussed the precise analysis of the flow at a non-hyperbolic equilibrium point, so this is not something you need to master. However, if we have a parameter in our problem, and at one value of the parameter we have a non-hyperbolic equilibrium, we can often induce from the local behaviour near the hyperbolic equilibrium/equilibria before and/or after this parameter value, what may have happened at the bifurcation point. Nothing beyond this superficial level of analysis will be expected or required at the exam. <br /></div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-85439008336640493072016-04-30T11:19:00.000+01:002016-04-30T11:19:15.542+01:00Phase portrait sketch in case of a Lyapunov function that is a conserved quantity<div dir="ltr" style="text-align: left;" trbidi="on">A student asked me:"I was looking at the past exam paper from 2010, and in question 03. (a) (iii) when drawing the phase plane, was wondering how we are supposed to conclude that it is a periodic orbit? I'm not completely sure how to proceed after finding the nullclines and the direction of the solution curve."<br> In the 2010 exam question 3(a)(iii), we have a Lyapunov function \(V\) with \(\frac{d}{dt}V=0\). This means that the level sets \(V_C:=\{x~|V(x)=C\}\) are flow-invariant and thus that every solution curve must lie inside one particular level set. As the level sets \(V_C\) with \(C>0\) in this example are closed curves that do not contain equilibria, necessarily these level sets must be periodic solutions. The level set \(V_0\) is the unique equilibrium \((x,y)=(1,1)\).<br> A similar situation was encountered in Test 2, question 2(ii). </div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-83904986671597403512016-04-27T22:27:00.000+01:002016-04-27T22:27:40.897+01:00Lyapunov functions and phase portraits<div dir="ltr" style="text-align: left;" trbidi="on">A student asked me: "I have a question regarding Q3 on the 2008 M2AA1 paper. In the question you are asked to sketch the phase portrait of the system for various parameters. The question gives you two pictures of the lyapunov function for these parameters. I was wondering how one might use a lyapunov function to deduce the phase portrait as the question suggests. I have checked the answers for that question which don't give much detail, however the portraits look remarkable similar to the lyapunov functions. "<br /><br />Let \(V(x)\) be a Lyapunov function, ie \(\frac{d}{dt}V(x)\leq 0\). Then \(V(x(t)\) must not increase when \(t\) increases. It is also insightful to note that \(\frac{d}{dt}V(x)=\nabla V(x) \cdot\dot{x}\leq 0\). Recall that \(\nabla V(x)\) is the normal to level set \(\{\tilde{x}~|~V(\tilde{x})=V(x)\}\) at the point \(x\). Thus \( \nabla V(x)\cdot \dot{x}\leq 0\) indeed means that the vector \(\dot{x}\) does not point in the direction of higher level sets of \(V\)and thus that if \(t\geq\tau\) then \(V(x(t))\leq V(x(\tau))\). </div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-5548184276607457352016-04-25T21:48:00.003+01:002016-04-25T21:49:10.510+01:00Sketching phase portraits<div dir="ltr" style="text-align: left;" trbidi="on">A student asked me:<br />"With regards to Problem Sheet 7, question 2 (the first question on the sheet), after sketching the nullclines and determining the direction of flow at the nullclines, as well as the nature of the equilibrium point at (0.5,0.5) - stable equilibrium, how should I attempt to sketch the phase portrait, i.e. the diagram on the right in the answers. "<br /><br />It may be instructive to go through the steps of how to try sketching a phase portrait:<br /><br />Locally near equilibria:<br />1. Find the equilibria. (Depending on the type of the equations, this may be easy or impossible. In most exercises this is easy.)<br />2. Calculate the linearization of the vector field (ie Jacobian) at these equilibria and deduce , where possible (ie when hyperbolic), what the phase portrait should look like near these equilibria.<br /><br />More globally:<br />1. Where relevant or possible, determine a bounded invariant set to which all solutions are attracted (most of our examples have such a region).<br />2. Try drawing nullclines where the vector field (and thus the tangent to solution curves) is horizonal and vertical. These nullclines may be helpful. In simple examples, often the nullclines can be computed explicitly.<br />3. Where there are saddles, it may be useful to try sketching where stable and unstable manifolds may end up.<br />4. Determine possibilities for the \(\omega\)-limit sets, in view of the Poincare-Bendixson theory.<br /><br />All-in-all, this is what you should be doing when sketching a phase portrait. It often not possible to get all the properties of the flow this way, so that there still may be some unknowns. In particular, it is often hard to rule out the existence of a periodic solution around an attracting or repelling equilibrium point (unless it lies at the border of a forward invariant set, in which case no periodic solutions can encircle it).<br /><br />If at the exam, you think there is some ambiguity or unknown property of the phase portrait, you should just write that. "Sketching" a phase portrait is precisely that: provide those features which you are certain of and discuss which additional features may or may not be present, on the basis of the theory.<br /><br />For the specific problem in question: it is completely reasonable to conclude that around the attracting equilibrium (0.5,0.5), the Poincare-bendixson theorem leave open the possibility of an \(\omega\)-limit set that is a periodic solution encircling this equilibrium. The argument I give in the model answer is correct, but not so easily verifiable (so you are not supposed to discover such a subtle property in an exam question, for instance).<br /><br />I hope this helps.<br /><br /><br /></div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com2tag:blogger.com,1999:blog-2282955917276031764.post-51000515477604249372016-04-23T20:27:00.001+01:002016-04-23T20:27:13.915+01:00Revision classes<div dir="ltr" style="text-align: left;" trbidi="on">Just to let you know that after consultation with the class rep, it has been decided that the scheduled revision classes of Tuesday 26 April 12:00-13:00 and Tuesday 3 May 12:00-13:00 in Clore, will take the form of problem classes. Graduate Teaching Assistants and I will be available for questions.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com1tag:blogger.com,1999:blog-2282955917276031764.post-80719366131242626082016-04-23T20:18:00.001+01:002016-04-26T14:16:46.000+01:00Guidance on past exam papers<div dir="ltr" style="text-align: left;" trbidi="on">Past summer M2AA1 exam papers can be found <a href="https://www.imperial.ac.uk/natural-sciences/departments/mathematics/study/students/undergraduate/pastexampapers/" target="_blank">here</a> (the course first featured in 2007-2008). I provide some guidance here on the relevance of the questions on the past papers for the current exam.<br /><br />2008: all questions<br />2009: all questions apart from Q3(c,d,e); for the model answers to Q3 (which do not seem to appear on the web), see <a href="https://imperialcollegelondon.box.com/s/urj1eo11uu0f72v9qnspm237cv9ql3wv" target="_blank">this link</a><br />2010: all questions<br />2011: not relevant<br />2012: not relevant<br />2013: all questions apart from Q3.<br />2014: all questions apart from Q1b(ii) and Q3 <br />2015: all questions apart from Q1(d) and Q3<br /><br />I was the setter (and lecturer of the course) in 2008, 2009 and 2010. It may be no surprise that the exam questions in these years are more representative of my style, than those of the other years.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-12497791716806868132016-04-23T18:12:00.004+01:002016-04-23T18:12:54.306+01:00Coordinate systems for calculus of variations <div dir="ltr" style="text-align: left;" trbidi="on">Your class rep wrote me "A student asked me about calculus of variations, if we have to be able to define coordinate systems ourselves? For example in Problem sheet 8 Question 8, students might struggle to get started on it."<br /><br />This question addresses a common type of anxiety that some of you may feel. It is unjustified.<br /><br />Problem sheet exercises are not necessarily model exam questions. When I list question 8 as "important" this means that I think it is important you understand how this problem is solved, and that it is a good exercise. Of course, in this question a "tricky" part is how to define the coordinates. If I would use this example as the base for an exam question, I would find it most important that you can show how to apply the Euler Lagrange equation. Most likely, as getting the coordinates right is perhaps a little tricky, I would most likely suggest you to use certain coordinates, since without a reasonable choice, you would not be able to get anywhere.<br /><br />As I told many of you before: it is my "problem" to set an exam that tests how well you understand the material. So I would aim to avoid exam questions to depend on "tricky" bits that are not at the core of the course material. <br /><br />So in answer to the question whether "[do] we have to be able to define coordinate systems ourselves?", the answer is "I would likely avoid obstacles, so if the choice of coordinates is not obvious, I would likely provide you with a suggestion."<br /><br />I hope this helps.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-38137511661575093912016-04-15T00:39:00.001+01:002016-04-15T00:39:14.645+01:00Lecture notes Chapter 5 and Chapter 6<div dir="ltr" style="text-align: left;" trbidi="on"><div class="post-title entry-title" itemprop="name"><span style="font-weight: normal;">Chapters 5 and 6 of the lectures notes correspond to Chapters 10 and 11 of <i>Differential equations, dynamical systems and an introduction to chaos </i>by M.W. Hirsch, S. Smale and R.L. Devaney. Please be aware that the entire book can be consulted digitally from the Imperial library website (where these chapters can also be downloaded; just search for the authors on the <a href="http://imp-primo.hosted.exlibrisgroup.com/primo_library/libweb/action/search.do?vid=ICL_VU1&reset_config=true" target="_blank">Library search</a> and you find a link to the digitcal copy). Because of copy-right issues I unfortunately cannot link on this webpage directly to these chapters. I understand that the class reps also sent around instructions how this material can be downloaded and consulted from the library website.</span></div><div class="post-title entry-title" itemprop="name" style="text-align: left;"><br /></div></div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-83941841900447109392016-04-09T19:54:00.002+01:002016-04-10T01:35:55.103+01:00Chapter 7 lecture notes<div dir="ltr" style="text-align: left;" trbidi="on">I have just posted typed up lecture notes of Chapter 7 on the Calculus of Variations. I apologize for the delay in finishing this. I also updated the instructions on "how to study for the exam" concerning this chapter of the notes. </div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-31044211423272634362016-03-29T23:24:00.002+01:002016-03-30T01:20:40.524+01:00Problem sheet 8 and answers<div dir="ltr" style="text-align: left;" trbidi="on">I have now posted model answers for problem sheet 8. Please note that questions 7 and 9 are not relevant for the exam (I updated the exam preparation note). I added a boundary condition to simplify question 6 somewhat.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-11252987420858864462016-03-22T12:23:00.000+00:002016-03-22T12:36:55.169+00:00Notes on synchronisation<div dir="ltr" style="text-align: left;" trbidi="on">My last lecture was loosely based on some <a href="http://http//arxiv.org/pdf/1112.2297.pdf" target="_blank">lecture notes</a> by <a href="http://www.icmc.usp.br/Portal/Pessoas/Detalhes.php?id=6938324" target="_blank">Dr Tiago Pereira</a>. This lecture connects my first lecture (where I mainly motivated the study of differential equations) to the material covered in this course. None of this material is examinable, but I hope it is interesting in context.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-62952298053898748392016-03-22T10:36:00.005+00:002016-03-22T10:36:49.670+00:00Problem class tuesday 22/3, 12-1pm, in Blacket 1004<div dir="ltr" style="text-align: left;" trbidi="on"><div class="MsoNormal">Room 340 is unfortunately not available. Blackett 1004 is on Level 10 of the Blackett building.</div><div class="MsoNormal">Come out of the lifts and turn left. Prof Turaev will discuss starred questions from the final problem sheet 8.</div></div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-12860073433316270792016-03-20T17:44:00.001+00:002016-03-20T17:44:57.459+00:00How to study for the 2016 exam?<div dir="ltr" style="text-align: left;" trbidi="on">Please see the notes that I posted in the left-hand side margin.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-66965196941978883282016-03-17T08:31:00.001+00:002016-03-17T08:31:29.390+00:00Final problem sheet nr 8 about calculus of variations.<div dir="ltr" style="text-align: left;" trbidi="on">This problem sheet has now been posted in the left-hand side margin.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-72166731502696662802016-03-10T20:59:00.002+00:002016-03-10T20:59:57.430+00:00Second test model answers<div dir="ltr" style="text-align: left;" trbidi="on">These have now been posted in the left-hand side margin.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-69286910703353998842016-03-10T08:01:00.002+00:002016-03-10T08:04:15.429+00:00Update on solutions to problem sheet 4<div dir="ltr" style="text-align: left;" trbidi="on">I corrected a typo in the solution to question 1(a) and reformulated the solution to question 5(a).</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0tag:blogger.com,1999:blog-2282955917276031764.post-90366583904765627432016-03-09T10:11:00.001+00:002016-03-09T10:11:57.813+00:00Additional office hour on wednesday 9 March 12-2pm in room 144<div dir="ltr" style="text-align: left;" trbidi="on">Questions will be answered about the material for the class test of tomorrow.</div>Jeroen Lambhttp://www.blogger.com/profile/14337866366830203975noreply@blogger.com0