Ordinary Differential Equations: Graduate Level Problems .
Ordinary Differential Equations: Graduate Level Problemsand SolutionsIgor Yanovsky1
Ordinary Differential EquationsIgor Yanovsky, 20052Disclaimer: This handbook is intended to assist graduate students with qualifyingexamination preparation. Please be aware, however, that the handbook might contain,and almost certainly contains, typos as well as incorrect or inaccurate solutions. I cannot be made responsible for any inaccuracies contained in this handbook.
Ordinary Differential EquationsIgor Yanovsky, 20053Contents1 Preliminaries1.1 Gronwall Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.2 Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5662 Linear Systems2.1 Existence and Uniqueness . . . . . . . . . . . . . . . . . . . . . . .2.2 Fundamental Matrix . . . . . . . . . . . . . . . . . . . . . . . . . .2.2.1 Distinct Eigenvalues or Diagonalizable . . . . . . . . . . . .2.2.2 Arbitrary Matrix . . . . . . . . . . . . . . . . . . . . . . . .2.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.3 Asymptotic Behavior of Solutions of Linear Systems with Constantefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.4 Variation of Constants . . . . . . . . . . . . . . . . . . . . . . . . .2.5 Classification of Critical Points . . . . . . . . . . . . . . . . . . . .2.5.1 Phase Portrait . . . . . . . . . . . . . . . . . . . . . . . . .2.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.7 Stability and Asymptotic Stability . . . . . . . . . . . . . . . . . .2.8 Conditional Stability . . . . . . . . . . . . . . . . . . . . . . . . . .2.9 Asymptotic Equivalence . . . . . . . . . . . . . . . . . . . . . . . .2.9.1 Levinson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Co. . . . . . . . . . . . . . . . . . .7777781011121213232526263 Lyapunov’s Second Method3.1 Hamiltonian Form . . . . . . . . . . . .3.2 Lyapunov’s Theorems . . . . . . . . . .3.2.1 Stability (Autonomous Systems)3.3 Periodic Solutions . . . . . . . . . . . .3.4 Invariant Sets and Stability . . . . . . .3.5 Global Asymptotic Stability . . . . . . .3.6 Stability (Non-autonomous Systems) . .3.6.1 Examples . . . . . . . . . . . . .272729293538404141.4 Poincare-Bendixson Theory425 Sturm-Liouville Theory5.1 Sturm-Liouville Operator . . . . . . . . .5.2 Existence and Uniqueness for Initial-Value5.3 Existence of Eigenvalues . . . . . . . . . .5.4 Series of Eigenfunctions . . . . . . . . . .5.5 Lagrange’s Identity . . . . . . . . . . . . .5.6 Green’s Formula . . . . . . . . . . . . . .5.7 Self-Adjointness . . . . . . . . . . . . . . .5.8 Orthogonality of Eigenfunctions . . . . . .5.9 Real Eigenvalues . . . . . . . . . . . . . .5.10 Unique Eigenfunctions . . . . . . . . . . .5.11 Rayleigh Quotient . . . . . . . . . . . . .5.12 More Problems . . . . . . . . . . . . . . .48484848494949506667697072. . . . . .Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 Variational (V) and Minimization (M) Formulations.97
Ordinary Differential Equations7 Euler-Lagrange Equations7.1 Rudin-Osher-Fatemi . .7.1.1 Gradient Descent7.2 Chan-Vese . . . . . . . .7.3 Problems . . . . . . . .Igor Yanovsky, 2005.4.1031031041051068 Integral Equations1108.1 Relations Between Differential and Integral Equations . . . . . . . . . . 1108.2 Green’s Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1139 Miscellaneous11910 Dominant Balance12411 Perturbation Theory125
Ordinary Differential Equations1Igor Yanovsky, 20055PreliminariesCauchy-Peano. dudt f (t, u)u(t0 ) u0t0 t t1(1.1)f (t, u) is continuous in the rectangle R {(t, u) : t0 t t0 a, u u0 b}.b). Then u(t) with continuous first derivativeM max f (t, u) , and α min(a, MRs.t. it satisfies (1.1) for t0 t t0 α.Local Existence via Picard Iteration.f (t, u) is continuous in the rectangle R {(t, u) : t0 t t0 a, u u0 b}.Assume f is Lipschitz in u on R. f (t, u) f (t, v) L u v b). Then a unique u(t), with u, duM max f (t, u) , and α min(a, Mdt continuousRon [t0 , t0 β], β (0, α] s.t. it satisfies (1.1) for t0 t t0 β.Power Series.du f (t, u)dtu(0) u0u(t) 1 dj u(0)tjj! dtji.e.j 0d2 u(0) (ft fu f ) 0dt2Fixed Point Iteration. xn x kn x0 x k 1 xn 1 xn kn x1 x0 k 1 x xn lim xm xn kn (1 k k2 · · · ) x1 x0 m Picard Iteration. Approximates (1.1). Initial guess: u0 (t) u0 tun 1 (t) T un (t) u0 f (s, un (s))ds.t0Differential Inequality. v(t) piecewise continuous on t0 t t0 a.u(t) and dudt continuous on some interval. Ifdu v(t)u(t)dt t u(t) u(t0 )ev(s)dst0 Proof. Multiply both sides by e tt0 v(s)ds. Thenddt [e tt0v(s)dsu(t)] 0.kn x1 x0 1 k
Ordinary Differential Equations1.1Igor Yanovsky, 20056Gronwall InequalityGronwall Inequality. u(t), v(t) continuous on [t0 , t0 a]. v(t) 0, c 0. tu(t) c v(s)u(s)dst0 t u(t) c et0v(s)dst0 t t0 aProof. Multiply both sides by v(t): t u(t)v(t) v(t) c v(s)u(s)dst0Denote A(t) c hypothesis: tv(s)u(s)ds t0 tu(t) A(t) A(t0 )et0dAdt tv(s)ds v(t)A(t). By differential inequality andv(s)ds cet0.Error Estimates. f (t, u(t)) continuous on R {(t, u) : t t0 a, u u0 b}f (t, u(t)) Lipschitz in u: f (t, A) f (t, B) L A B u1 (t), u2 (t) are 1 , 2 approximate solutionsdu1 f (t, u1 (t)) R1 (t),dt R1 (t) 1du2 f (t, u2 (t)) R2 (t),dt u1 (t0 ) u2 (t0 ) δ R2 (t) 2 u1 (t) u2 (t) (δ a( 1 2 ))ea·Lt0 t t0 aGeneralized Gronwall Inequality. w(s), u(s) 0 tu(t) w(t) v(s)u(s)dst0 t u(t) w(t) tv(s)w(s) esv(x)dxdst0Improved Error Estimate (Fundamental Inequality). u1 (t) u2 (t) δeL(t t0 ) 1.2( 1 2 ) L(t t0)(e 1)LTrajectoriesLet K D compact. If for the trajectory Z {(t, z(t)) : α t β)} we have thatβ , then Z lies outside of K for all t sufficiently close to β.
Ordinary Differential Equations2Igor Yanovsky, 20057Linear Systems2.1Existence and UniquenessA(t), g(t) continuous, then can solvey A(t)y g(t)(2.1)y(t0 ) y0For uniqueness, need RHS to satisfy Lipshitz condition.2.2Fundamental MatrixA matrix whose columns are solutions of y A(t)y is called a solution matrix.A solution matrix whose columns are linearly independent is called a fundamentalmatrix.F (t) is a fundamental matrix if:1) F (t) is a solution matrix;2) det F (t) 0.Either det M (t) 0 t R, or det M (t) 0 t R.F (t)c is a solution of (2.1), where c is a column vector.If F (t) is a fundamental matrix, can use it to solve:y (t) A(t)y(t), y(t0 ) y0i.e. since F (t)c t0 F (t0 )c y0 c F 1 (t0 )y0 y(t) F (t)F (t0 ) 1 y02.2.1Distinct Eigenvalues or DiagonalizableF (t) [eλ1t v1 , . . . , eλnt vn ]2.2.2eAt F (t)CArbitrary Matrixi) Find generalized eigenspaces Xj {x : (A λj I)nj x 0};ii) Decompose initial vector η v1 · · · vk , vj Xj ,solve for v1 , . . . , vk in terms of components of ηy(t) k j 1λj tej 1 i n ti 0i!(A λj I)i vjiii) Plug in η e1 , . . . , en successively to get y1 (t), . . . , yn (t) columns of F (t).Note: y(0) η, F (0) I.(2.2)
Ordinary Differential Equations2.2.3Igor Yanovsky, 20058ExamplesExample 1. Show that the solutions of the following system of differential equationsremain bounded as t :u v uv u 1 1uu . The eigenvalues of A are λ1,2 12 23 i, so 1 0vvthe eigenvalues are distinct diagonalizable. Thus, F (t) [eλ1 t v1 , eλ2t v2 ] is a fundamental matrix. Since Re(λi) 12 0, the solutions to y Ay remain bounded ast .Proof. 1)2) u v u u u ,u u u 0,u u (u )2 u u 0,1 d1 d 2 222 dt ((u ) ) (u ) 2 dt (u ) 0,t11 2 2 22 ((u ) ) 2 (u ) t0 (u ) dt const,11 2 22 ((u ) ) 2 (u ) const, (u , u) is bounded. 1 Example 2. Let A be the matrix given by: A 20the generalized eigenspaces, and a fundamental matrix010for 32 . Find the eigenvalues,2the system y (t) Ay.Proof. det(A λI) (1 λ)2 (2 λ). The eigenvalues and their multiplicities:λ1 1, n1 2; λ2 2, n2 1. Determine subspaces X1 and X2 , (A λj I)nj x 0.(A 2I)x 0(A I)2 x 0To find X1 : 0 0 30 0 30 0 3x102 2 0 22 0 2x 0 0 8x2 0 .(A I) x x30 0 10 0 10 0 10 αdim X1 2. x3 0, x1 , x2 arbitrary X1 β , any α, β C .0To find X2 : 0 1 0 3 1 0 3x1(A 2I)x 2 1 2 x 0 1 8 x2 0 .x000 000 0 3 3dim X2 1. x3 γ, x1 3γ, x2 8γ X2 γ 8 , any γ C .1 X,v X,suchthatinitialvectorη is decomposed as η v1 v2 . Needtofindv1122 α3γη1 η2 β 8γ .η30γ η1 3η33η3 v1 η2 8η3 , v2 8η3 .0η3
Ordinary Differential Equations y(t) k j 1λj tej 1 i n ti 0i!Igor Yanovsky, 20059(A λj I)i vj eλ1 t (I t(A I))v1 eλ2 t v2 η1 3η33η3 et (I t(A I))v1 e2tv2 et (I t(A I)) η2 8η3 e2t 8η3 0η3 1 03tη1 3η33η3t 2t 2t 12t eη2 8η38η3 . e0 0 1 t0η3 η1Note: y(0) η η2 .η3 100 To find a fundamental matrix, putting η successively equal to0,1, 0 001in this formula, we obtain the three linearly independentsolutionsthatweuseas 101columns of the matrix. If η 0 , y1 (t) et 2t . If η 1 , y2 (t) 000 0et 1 .0 330t2t 6t 8 e8 . The fundamental matrix isIf η 0, y3 (t) e011 t e0 3et 3e2tF (t) eAt 2tet et ( 6t 8)et 8e2t 00e2tNote: At t 0, F (t) reduces to I.
Ordinary Differential Equations2.3Igor Yanovsky, 200510Asymptotic Behavior of Solutions of Linear Systems with Constant CoefficientsIf all λj of A are such that Re(λj ) 0, then every solution φ(t) of the system y Ayapproaches zero as t . φ(t) K̂e σt or eAt Ke σt .If, in addition, there are λj such that Re(λj ) 0 and are simple, then eAt K, andhence every solution of y Ay is bounded.Also, see the section on Stability and Asymptotic Stability.Proof. λ1 , λ2, . . . , λk are eigenvalues and n1 , n2 , . . . , nk are their corresponding multiplicities. Consider (2.2), i.e. the solution y satisfying y(0) η istAy(t) e η k λj tej 1 i n tj 1i 0i!(A λj I)i vj .Subdivide the right hand side of equality above into two summations, i.e.:1) λj , s.t. nj 1, Re(λj ) 0;2) λj , s.t. nj 2, Re(λj ) 0.k y(t) j 1 eλj t vj (nj 1) Re(λj ) 0 y(t) k eλj t I t(A λj I) · · · j 1 eλj t I vj j 1 k Re(λj ) 0 σt Ke c max(ck, K) const indep of tRe(λj ) 0k σt vj Kej 1 σ max(Re(λj ), Re(λj ) 0) σt ck max vj Kej(nj 2) tnj 1(A λj I)nj 1 vj .(nj 1)! e σtmax vj K. j 0 as t indep of t
Ordinary Differential Equations2.4Igor Yanovsky, 200511Variation of ConstantsDerivation: Variation of constants is a method to determine a solution of y A(t)y g(t), provided we know a fundamental matrix for the homogeneous system y A(t)y.Let F be a fundamental matrix. Look for solution of the form ψ(t) F (t)v(t), wherev is a vector to be determined. (Note that if v is a constant vector, then ψ satisfiesthe homogeneous system and thus for the present purpose v(t) c is ruled out.)Substituting ψ(t) F (t)v(t) into y A(t)y g(t), we getψ (t) F (t)v(t) F (t)v (t) A(t)F (t)v(t) g(t)Since F is a fundamental matrix of the homogeneous system, F (t) A(t)F (t). Thus,F (t)v (t) g(t),v (t) F 1 (t)g(t), tF 1 (s)g(s)ds.v(t) t0 tTherefore, ψ(t) F (t)F 1 (s)g(s)ds.t0Variation of Constants Formula: Every solution y of y A(t)y g(t) has theform: ty(t) φh (t) ψp(t) F (t)c F (t)F 1 (s)g(s)dst0where ψp is the solution satisfying initial condition ψp(t0 ) 0 and φh (t) is that solutionof the homogeneous system satisfying the same initial condition at t0 as y, φh (t0 ) y0 .F (t) eAt is the fundamental matrix of y Ay with F (0) I. Therefore, everysolution of y Ay has the form y(t) eAt c for a suitably chosen constant vector c.(t t0 )Ay(t) e ty0 e(t s)A g(s)dst0That is, to find the general solution of (2.1), use (2.2) to get a fundamental matrixF (t). t tThen, add e(t s)A g(s)ds F (t) F 1 (s)g(s)ds to F (t)c.t0t0
Ordinary Differential Equations2.5Igor Yanovsky, 200512Classification of Critical Pointsy Ay. Change of variable y T z, where T is nonsingular constant matrix (to bedetermined). z T 1 AT zThe solution is passing through (c1 , c2 ) at t 0.1) λ1 , λ2 are real.z λ1 00 λ2zc1 eλ1 tc2 eλ2 ta) λ2 λ1 0 z2 (t) c(z1 (t))p, p 1 Improper Node (tilted toward z2 -axis)Improper Node (tilted toward z2 -axis)b) λ2 λ1 0 z2 (t) c(z1 (t))p, p 1Proper Nodec) λ2 λ1 , A diagonalizable z2 cz1Saddle Pointd) λ2 0 λ1 z1 (t) c(z2 (t))p, p 0λ 1z2) λ2 λ1 , A non-diagonalizable, z 0 λeλt teλtc1c1 c2 t z Improper Node eλtλt0 ec2c2σ νz z3) λ1,2 σ iν. ν σc1 cos(νt) c2 sin(νt)Spiral Point z eσt c1 sin(νt) c2 cos(νt) z 2.5.1Phase PortraitLocate stationary points by setting:dudt f (u, v) 0dvdt g(u, v) 0(u0 , v0 ) is a stationary point. In order to classify a stationary point, need to findeigenvalues of a linearized that point. system at f f u v g g u vdet(J (u0 ,v0 ) J(f (u, v), g(u, v)) Find λj ’s such that.λI) 0.
Ordinary Differential Equations2.6Igor Yanovsky, 200513ProblemsProblem (F’92, #4). Consider the autonomous differential equationvxx v v 3 v0 0in which v0 is a constant.4, this equation has 3 stationary points and classify their type.a) Show that for v02 27b) For v0 0, draw the phase plane for this equation.Proof. a) We havev v v 3 v0 0.In order to find and analyze the stationary points of an ODE above, we write it as afirst-order system.y1 v,y2 v .y1 v y2 0,y2 v v v 3 v0 y13 y1 v0 0.The function f (y1 ) y13 y1 y1 (y12 1) has zeros y1 0, y1 1, y1 1.See the figure.It’s derivative f (y1 ) 3y12 1 has zeros y1 13 , y1 13 .22, f ( 13 ) 3 .At these points, f ( 13 ) 3 33 If v0 0, y2 is exactly this function f (y1 ), with 3 zeros.2,v0 only raises or lowers this function. If v0 3 3i.e. v02 427 ,the system would have 3 stationary points:Stationary points: (p1 , 0), (p2 , 0), (p3 , 0),with p1 p2 p3 .y1 y2y2 y13 f (y1 , y2 ), y1 v0 g(y1 , y2 ).In order to classify a stationary point, need to find eigenvalues of a linearized systemat that point. f f01 y1 y2. J(f (y1 , y2 ), g(y1, y2 )) g g3y12 1 0 y y1 For (y1 , y2 ) (pi, 0) : λ1det(J (pi ,0) λI) 23pi 1 λλ 3p2i 1.2 λ2 3p2 1 0.iAt y1 p1 13 , λ 0 λ . (p1 ,0) is Saddle Point.At 13 y1 p2 13 , λ C, Re(λ ) 0. (p2 ,0) is Stable ConcentricCircles.At y1 p3 13 , λ 0 λ . (p3 ,0) is Saddle Point.
Ordinary Differential EquationsIgor Yanovsky, 2005b) For v0 0,y1 y2 0,y2 y13 y1 0.Stationary points: ( 1, 0), (0, 0), (1, 0). J(f (y1 , y2 ), g(y1, y2 )) For (y1 , y2 ) (0, 0) : λ 1det(J (0,0) λI) 1 λ013y12 1 0 . λ2 1 0.λ i.(0,0) is Stable Concentric Circles (Center). For (y1 , y2 ) ( 1, 0) : λ 1 λ2 2 0.det(J ( 1,0) λI) 2 λ λ 2.(-1,0) and (1,0) are Saddle Points.14
Ordinary Differential EquationsIgor Yanovsky, 200515Problem (F’89, #2). Let V (x, y) x2 (x 1)2 y 2 . Consider the dynamical systemdx V ,dt x Vdy .dt ya) Find the critical points of this system and determine their linear stability.b) Show that V decreases along any solution of the system.c) Use (b) to prove that if z0 (x0 , y0 ) is an isolated minimum of V then z0 is anasymptotically stable equilibrium.Proof. a) We havex 4x3 6x2 2xy 2y. x x(4x2 6x 2) 0y 2y 0.Stationary points: (0, 0), J(f (y1 , y2 ), g(y1, y2 )) 1 f x g x2, 0 , (1, 0). f y g y 12x2 12x 2 00 2 For (x, y) (0, 0) : 2 λ0det(J (0,0) λI) 0 2 λ ( 2 λ)( 2 λ) 0.y Ay, λ1 λ2 0, A diagonalizable.(0,0) is Stable ProperNode.1 For (x, y) 2 , 0 :1 λ0det(J ( 1 ,0) λI) 20 2 λ (1 λ)( 2 λ) 0.λλ2 1. λ1 0 λ2 .1 2,1is Unstable Saddle Point.2 ,0 For (x, y) (1, 0) : 2 λ0det(J (1,0) λI) 0 2 λ ( 2 λ)( 2 λ) 0.y Ay, λ1 λ2 0, A diagonalizable.(1,0) is Stable Proper Node. .
Ordinary Differential EquationsIgor Yanovsky, 200516b) Show that V decreases along any solution of the system.dV Vxxt Vy yt Vx( Vx ) Vy ( Vy ) Vx2 Vy2 0.dtc) Use (b) to prove that if z0 (x0 , y0 ) is an isolated minimum of V then z0 is anasymptotically stable equilibrium.Lyapunov Theorem: If V (y) that is positive definite and for which V (y) is negativedefinite in a neighborhood of 0, then the zero solution is asymptotically stable.Let W (x, y) V (x, y) V (x0 , y0 ). Then, W (x0 , y0 ) 0.dVW (x, y) 0 in a neighborhood around (x0 , y0 ), and dWdt (x, y) 0 by (b). ( dt (x, y) 0and dVdt (x0 , y0 ) 0).(x0 , y0 ) is asymptotically stable.
Ordinary Differential EquationsIgor Yanovsky, 200517Problem (S’98, #1). Consider the undamped pendulum, whose equation isd2 p g sin p 0.dt2la) Describe all possible motions using a phase plane analysis.b) Derive an integral expression for the period of oscillation at a fixed energy E,and find the period at small E to first order.c) Show that there exists a critical energy for which the motion is not periodic.Proof. a) We havey1 py2 p .y1 p y2 0ggy2 p sin p sin y1 0.llStationary points: (nπ, 0).y1 y2 f (y1 , y2 ),gy2 sin y1 g(y1 , y2 ).l J(f1 (y1 , y2 ), f2 (y1 , y2 )) f1 y1 f2 y1 f1 y2 f2 y2 01 gl cos y1 0 . For (y1 , y2 ) (nπ, 0), n-even: λ 1 λ2 g 0.det(J (nπ,0) λI) gl l λ i g C,g 0, (nπ,0), n-even, are Stable Centers.lλ g R, g 0. (nπ,0), n-even, are Unstable Saddle Points.l For (y1 , y2 ) (nπ, 0), n-odd: λ 1 λ2 g 0.det(J (nπ,0) λI) gl λl g R,g 0, (nπ,0), n-odd, are Unstable Saddle Points.lλ i g C,g 0, (nπ,0), n-odd, are Stable Centers.l
Ordinary Differential EquationsIgor Yanovsky, 200518
Ordinary Differential EquationsIgor Yanovsky, 200519b) We havegp sin plg p p p sin pl1 d 2 g d(p ) (cos p)2 dtl dt1 2 g(p ) cos p2lE 0, 0, 0, Ẽ.1 2 g(p ) (1 cos p).2lSince we assume that p is small, we could replace sin p by p, and perform similarcalculations:gp p 0,lg p p p p 0,l1 d 2 1g d(p ) (p)2 0,2 dt2 l dt1 2 1g 2(p ) p E1 ,22lg(p )2 p2 E constant.lThus,p2(p )2 lEEg 1,which is an ellipse with radii E on p -axis, andlEgon p-axis.We derive an Integral Expression for the Period of oscillation at a fixed energy E.Note that at maximum amplitude (maximum displacement), p 0.Define p pmax to be the maximum displacement:g1E (p )2 (1 cos p),2l#2g2E (1 cos p),p L T T44Tp dt ,dt 42g002E L (1 cos p) % pmax T4p dpdt.T 4T 4002E 2g(1 cosp)2E 2gLL (1 cos p)Making change of variables: ξ p(t), dξ p (t)dt, we obtain pmaxdξ.T (pmax ) 402E 2g(1 cosξ)L
Ordinary Differential EquationsIgor Yanovsky, 200520Problem (F’94, #7).The weakly nonlinear approximation to the pendulum equation (ẍ sin x) is1ẍ x x3 .6(2.3)a) Draw the phase plane for (2.3).b) Prove that (2.3) has periodic solutions x(t) in the neighborhood of x 0.c) For such periodic solutions, define the amplitude as a maxt x(t). Find an integralformula for the period T of a periodic solution as a function of the amplitud
Ordinary Differential Equations Igor Yanovsky, 2005 7 2LinearSystems 2.1 Existence and Uniqueness A(t),g(t) continuous, then can solve y A(t)y g(t) (2.1) y(t 0) y 0 For uniqueness, need RHS to satisfy Lipshitz condition. 2.2 Fundamental Matrix A matrix whose columns are solutions of y A(t)y is called a solution matrix.
Ordinary Di erential Equations by Morris Tenenbaum and Harry Pollard. Elementary Di erential Equations and Boundary Value Problems by William E. Boyce and Richard C. DiPrima. Introduction to Ordinary Di erential Equations by Shepley L. Ross. Di
69 of this semi-mechanistic approach, which we denote as Universal Di erential 70 Equations (UDEs) for universal approximators in di erential equations, are dif- 71 ferential equation models where part of the di erential equation contains an 72 embedded UA, such as a neural network, Chebyshev expansion, or a random 73 forest. 74 As a motivating example, the universal ordinary di erential .
1.2. Rewrite the linear 2 2 system of di erential equations (x0 y y0 3x y 4et as a linear second-order di erential equation. 1.3. Using the change of variables x u 2v; y 3u 4v show that the linear 2 2 system of di erential equations 8 : du dt 5u 8v dv dt u 2v can be rewritten as a linear second-order di erential equation. 5
Ordinary di erential equations This chapter contains three papers which are on the integer-order ordinary di erential equations for boundary value problem. 1.1 The symmetric positive solutions of four-point boundary value problems for nonlin-ear second-order di erential equations Abstract: In this paper, we are concerned with the existence of .
General theory of di erential equations of rst order 45 4.1. Slope elds (or direction elds) 45 4.1.1. Autonomous rst order di erential equations. 49 4.2. Existence and uniqueness of solutions for initial value problems 53 4.3. The method of successive approximations 59 4.4. Numerical methods for Di erential equations 62
Unit #15 - Di erential Equations Some problems and solutions selected or adapted from Hughes-Hallett Calculus. Basic Di erential Equations 1.Show that y x sin(x) ˇsatis es the initial value problem dy dx 1 cosx To verify anything is a solution to an equation, we sub it in and verify that the left and right hand sides are equal after
Parameter Estimation for Systems of Ordinary Di erential Equations Jonathan Calver Doctor of Philosophy Graduate Department of Computer Science University of Toronto 2019 We consider the formulation and solution of the inverse problem that arises when t-ting systems of ordinary di erential equations (ODEs) to observed data. This parameter
Di erential Equations The subject of ordinary di erential equations encompasses such a large eld that you can make a profession of it. There are however a small number of techniques in the subject that you have to know. These are the ones that come up so often in physical systems that you need both the skills to use them
