2xy 0 y = ln y 0. d(p 2 x p) =0. p = 1 ± p 1+Cx. 1 ± 1+Cx 1+Cx ln

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1 () Consider the di erential equation Find all solutions and the discriminant xy y = ln y Solution: Let F (x, y, p) = xp y ln p, where p = y Let us consider the surface M = {(x, y, p): F (x, y, p) =} in the space of -jets Solutions of the di erential equation correspond to curves in M whose tangent vectors lie on the contact planes dy pdx= So, we need to find the integral curves of the following system: F (x, y, p) = =) xp y ln p = df = =) pdx dy (x ) dp = p dy pdx= =) dy pdx= Using dy = pdx, the second equation becomes pdx+(x ) dx = Multiplying by p gives p That is, Thus, p x p dx +(xp ) dx = d(p x p) = p C, where C is a constant Solving for p gives 4 p = ± p +Cx x Using the first equation, y =xp ln p, we find y =± p p ± +Cx +Cx ln x The criminant is the set of points in the surface M = {(x, y, p): F (x, y, p) =} such = That is, the set of points in the -jet space such xp y = ln p x p = The second equation gives x = Plugging this into the first equation gives y = ln p p Thus, the criminant can be described by the parametric curve in the -jet space given by x(t) = t y(t) = z(t) =t The discriminant is the projection of the criminant onto the xy-plane via (x, y, p) 7! (x, y) This results in the parametric curve in the xy-plane given by ln t x(t) = t y(t) = ln t One can also describe this curve as the graph of y = + ln(x)

2 () Find a curve on the plane whose tangent lines form with the coordinate axis triangles of area a Remark: The following two solutions are ultimately equivalent Solution : The area of the triangle formed by the coordinate axes and the line Y = mx + b is A = b The tangent line to a curve Y = f(x) at a point (x, y) is given by Y = m f (x)x +(f(x) xf (x)) Thus, we require that or equivalently, (f(x) (f(x) xf (x)) f (x) Letting y = f(x) and p = f (x), we have (y =a, xf (x)) =(a) f (x) px) =(a) p, and hence y px = ±a p p Let F (x, y, p) := y px ± a p p Solutions to the di erential equation correspond to curves in M = {(x, y, p): F (x, y, p) =} whose tangent vectors lie on the contact planes dy pdx= Thus, we have =df = pdx+ dy +( x ± p a sgn(p)) dx p =( x ± p a sgn(p)) dx p This implies that x = ± p a sgn(p), so that x = a a, so p = In other words, f (x) =± a, p p x x so that f(x) =± a x Solution : We will use the fact that every curve y = f(x) is the envelope of its family of tangent lines Note first (cf Arnold: page ) that the envelope of a family of lines {y = px g(p)} pr is the curve y = f(x), where f is the Legendre transform of g Second, note that the family of lines {y = px ± a p p } pr form triangles of area a with the coordinate axes Thus, the required curve is the Legendre transform of g(p) =±a p p To calculate the Legendre transform of g, we (px ± ap p ) =, which x = p a, so p = a, so p = ± a Thus, p x x f(x) =px ± a p p = ± a x a x = a x

3 (3) Prove that the rank of any skew-symmetric bilinear form is even Solution (Sketch): Show that the eigenvalues of any skew-symmetric bilinear form are either zero or pure imaginary, and that the eigenvalues come in complex-conjugate pairs Thus, the number of non-zero eigenvalues ie, the rank of the bilinear form is even Solution : Let! be a skew-symmetric bilinear form on a finite-dimensional vector space V The result will follow from the following structure theorem Theorem: There exists a basis for V, denoted {u,,u k,e,,e m,f,,f m }, for which Such a basis is called a symplectic basis!(u i,v)=,!(e i,e j )=!(f i,f j )=!(e i,f j )= ij 8v V Proof of Theorem: Let U := {u V :!(u, v) =, 8v V } denote the nullspace of! Let {u,,u k } be a basis of U Write V = U W We will (inductively) decompose W into a direct sum of -dimensional subspaces If W =, we re done, so assume W 6= Let e W, e 6= Since e W, there exists f W with!(e,f ) 6= By rescaling, we may assume that!(e,f )= Let W := span{e,f } Let W! := {w W :!(w, v) =, 8v W } denote the symplectic complement to W in W, so that W = W W! Let e W!, e 6= Then there exists f W! with!(e,f ) 6= By rescaling, we may assume that!(e,f )= Repeat this process This process eventually terminates because dim V < Thus, we have a direct sum decomposition V = U span{e,f } span{e m,f m } With respect to a symplectic basis for!, the skew-symmetric bilinear form! can be written as Id!(u, v) =u T A v Id m m That is, with respect to a symplectic basis, the form! is represented by the matrix Id m m which has rank m In other words, rank(!) =m is even A,

4 (4) Let us view C n as R n with the operation of multiplication by i Prove that SO(n) \ Sp(n; R) =GL(n; C) \ SO(n) =GL(n; C) \ Sp(n; R) =U(n) Notation: Let g denote the standard inner product, let! denote the standard symplectic form, and let h denote the standard Hermitian form We let z,w C n denote z =(x,,x n,y,,y n )= X x k + iy k w =(u,,u n,v,,v n )= X u k + iv k Solution (Sketch): Recall from basic linear algebra that Let J denote the matrix SO(n) ={A GL(n; R): A T A = Id and det R (A) =} U(n) ={A GL(n; C): A A =Id} Idn J = Id n We observe that the matrix J corresponds to multiplication by i That is, x x Jz = J n Idn x y = n Id n y = B C B C y n y n We observe also that J T = is, From these facts, it follows that x y y n x x n = i X x k + iy k = iz C A J is the matrix representation of! in the standard basis That!(z,w) =z T J T w GL(n; C) ={A GL(n; R): AJ = JA} Sp(n; R) ={A GL(n; R): A T JA = J} Using these four descriptions of the matrix groups SO(n), U(n), GL(n; C), Sp(n; R), the first three desired intersections are now fairly immediate For example: If A GL(n; C) \ O(n), then JA = AJ and A T A = Id, so A T JA = A T AJ = J, so A Sp(n; R) Thus, O(n) \ GL(n; C) Sp(n; R) Analogous arguments hold for the other two If A U(n), then A GL(n; C) and g(az, Az) =h(az, Az) =h(z,z) =g(z,z), so g O(n), so A Sp(n; R) by the preceding paragraph, and so A lies in any of these pairwise intersections Conversely, if A GL(n; C) \ SO(n), then h(az, Az) = g(az, Az) = g(z,z) = h(z,z), hence (by polarization) h(az, Aw) =h(z,w), so A U(n)

5 (4) Let us view C n as R n with the operation of multiplication by i Prove that SO(n) \ Sp(n; R) =GL(n; C) \ SO(n) =GL(n; C) \ Sp(n; R) =U(n) Notation: Let g denote the standard inner product, let! denote the standard symplectic form, and let h denote the standard Hermitian form We let z,w C n denote z =(x,,x n,y,,y n )= X x k + iy k w =(u,,u n,v,,v n )= X u k + iv k Solution (Sketch): Let us observe first that h(z,w) = X z k w k = X (x k + iy k )(u k iv k ) = X (x k u k + y k v k )+i(y k v k x k u k ) = g(z,w) i!(z,w) () From this, we observe secondly that!(z,w)+ig(z,w) =ih(z,w) =h(iz, w) =g(iz, w) i!(iz, w), and so by equating real parts, we obtain g(iz, w) =!(z, w) () Using equations () and (), the result is now straightforward to check If A SO(n) \ Sp(n; R), then h(az, Aw) =g(az, Aw) i!(az, Aw) =g(z,w) i!(z,w) =h(z,w), so A U(n) Thus, SO(n) \ Sp(n; R) U(n) If A U(n), then g(az, Aw) i!(az, Aw) = h(az, Aw) = h(z,w) = g(z, w) i!(z,w), so equating real and imaginary parts shows that A SO(n) \ Sp(n; R) Thus, U(n) SO(n) \ Sp(n; R) This shows that U(n) =SO(n) \ Sp(n; R) If A GL(n; C) \ SO(n), then!(az, Aw) = g(iaz, Aw) = g(a(iz), Aw) = g(iz, w) =!(z,w), so! Sp(n; R) Thus, GL(n; C) \ SO(n) U(n) If A GL(n; C) \ Sp(n; R), then g(az, Aw) =!(iaz, Aw) =!(A(iz), Aw) =!(iz, w) =g(z,w), so A SO(n) Thus, GL(n; C) \ Sp(n; R) U(n) Finally, if A U(n) = SO(n) \ Sp(n; R), then g(iaz, Aw) =!(Az, Aw) =!(z, w) =g(iz, w) =g(a(iz), Aw), so g(iaz Aiz, Aw) = Since g is non-degenerate, this forces A(iz) = i(az), so A GL(n; C)

6 (5) Prove that the plane fields given by equations dz ydx= and dz (xdy ydx)=are di eomorphic, but the plane fields dz (xdy ydx) = and dz (xdy+ydx) = are not Solution: Let = dz ydxand = dz (xdy ydx) Consider the normal vector fields # @z # @z Let : R 3! R 3 be a di eomorphism Note that = if and only if ( ) T ( # )= # This fact suggests a way of finding Namely, by observing that y y x A x A, we are led to search for bijections : R 3! R 3 with A One possibility is (x, y, z) =(x, y, z + xy) It is easy to check that is bijective and is invertible, so that is indeed a di eomorphism One can also verify that = To see that the plane fields dz (xdy ydx) = and dz (xdy+ ydx) = are not di emorphic, observe that the first is not integrable, whereas the second one is

7 (6) Consider the = u xy Solve the Cauchy problem for the initial data u(,y)=+y Solution : Let A denote the vector field in R given by A If (t) =(x(t),y(t)) is an integral curve of A, then (t) =A (t), so that x (t) =x(t) =) x(t) =x e t y (t) =y(t) =) y(t) =y e t Thus, the flow of A is given by t (x, y) =(xe t,ye t ) Let S = {(,s): s R} denote the initial hypersurface in R The flowout of A along the line S R is defined by One can check the following properties of We remark that (x, y) = (ln(x/), y/x) (t, s) := t (,s)=(e t, se t ) (which are generally true of flowouts): )=A Pulling back our Cauchy problem to the (t, = bu set, bu(,s)=+s, results in the new problem where bu := u = u Regarding s as fixed, we may view this problem as an ODE for bu(,s), namely bu = bu se t Multiplying by the integrating factor e (e t bu) = se t Integrating yields e t bu = se t + h(s) for some function h(s) Invoking the initial condition bu(,s)=+s shows that bu(t, s) = se t +(s +) e t Pulling back to the (x, y)-plane via, we find x u(x, y) =bu ln, y = x x +y xy + y x

8 (6) Consider the = u xy Solve the Cauchy problem for the initial data u(,y)=+y Solution : Let R 3 = R R have coordinates (x, y; z) Let be the characteristic vector field in R 3, ie: If (t) =(x(t),y(t),z(t)) is an integral curve of, then (t) = (t), so that x (t) =x(t) =) x(t) =x e t y (t) =y(t) =) y(t) =y e t z (t) =z(t) x(t)y(t) =) z(t) = x y e t +(z + x y )e t Thus, the flow of is given by t (x, y, z) =(xe t,ye t, xye t +(z + xy)e t ) Let S = {(,s): s R} be denote the initial hypersurface in R Let ': S! R denote '(,s)=+s Then Graph(') can be parametrized as ((,s), '(,s)) = (,s;+s ) The flowout of along the curve Graph(') R 3 is defined by (t, s) := t (, s, +s )=(e t, se t, se t +(s +) e t ) A solution to our Cauchy problem is the function u whose graph is the flowout above ie: Image( ) = Graph(u) Thus, setting we find that (e t, se t, se t +(s +) e t )=(x, y, u(x, y)), x =e t y = se t, t = ln(x/) s =y/x, so that x u(x, y) =bu ln, y = x x +y xy + y x