Analysis III Serie 13 Musterlösung

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1 Ana-3 Hs 22 Analysis III Serie 3 Musterlösung Abgabe: Freitag, , Uhr, in der Vorlesung * Aufgabe Welche der folgenden Aussagen sind wahr und welche sind falsch? (Mit Begründung) (i) Sei A R 3 3 eine 3 3 Matrix mit spec A = { i,, i}, dann gibt es genau 3 verschiedene A-invariante R-Unterräume der Dimension. / False. Suppose we had 3 A-invarinat subspaces, i.e. R 3 = E E 2 E 3. Since A is linear, we thus have A Ei = λ i I Ei, with λ i R. Thus A has three real eigenvalues, which is impossible. / (ii) Sei f : R 2 R 2 ein stetiges Vektorfeld. Dann schneiden sich Lösungskurven der Differenzialgleichung ẋ = f (x) entweder niemals oder sie fallen ganz zusammen. / False. Consider f (x) = (3x 2/3, ), then for the initial value x = (, ), x is a solution as well as x(t) = (t 3, ). These curves intersect only at the origin. / (iii) Sei f : D R 2 ein Lipschitz-stetiges Vektorfeld auf der offenen Einheitskugel D = {x R 2 : x < } mit Lipschitz-Konstante K >. Dann besitzt das Anfangswertproblem ẋ = f (x), x() = x D stets eine globale Lösung. / False. Consider f (x) = 2x which is Lipschitz with L-constant 2 >. The general solution is given by t e 2t x. However, for any x which is not the origin, the solution curve leaves D in finite time and thus is not a global solution. / * Aufgabe 2 Sei A R n n eine n n Matrix. Zeige, dass das Iterationsverfahren von Picard für das Anfangswertproblem ẋ = Ax, x() = x 7. Juni 23 3:4 Serie 3 Seite von 8

2 Ana-3 Hs 22 die Lösung ergibt. R R n, t e ta x / The iteration scheme is given by u (t) = x and u m (t) = x + We claim that u m (t) = n m f (u m (s)) ds, m. n! (ta)n x, which is evidently true for m =. If we proceed by induction, then u m+ (t) = x + f (u m (s)) ds = x + A n! sn A n x ds n m = x + ( ) t s n ds A n+ x n! n m = x + n m = x + = n m n m+ n! n + An+ x t n+ (n + )! (ta)n+ x n! (ta)n x. So the claim is true, and u m (t) e ta x uniformly on any compact time interval as m. / * Aufgabe 3 (i) Zeige, dass auf R das Anfangswertproblem ẋ = x, x() =, überabzählbar viele Lösungen besitzt. / Clearly, x is a solution, and also x(t) = 2 as we compute s ds = 4 t2 sign t ẋ(t) = 2 t = x(t). In fact, any solution in between,, t a, x a (t) = 4 (t a)2, t > a, 4 (t + a)2, t < a, for a solves the equation. / 7. Juni 23 3:4 Serie 3 Seite 2 von 8

3 Ana-3 Hs 22 (ii) Zeige, dass auf R das Anfangswertproblem ẋ = x, x() =, lokal um t = eine eindeutige Lösung besitzt, diese aber nicht global eindeutig ist. / By reparametrizing the solutions found in (i), t y a (t) x a (t a), we obtain a family of solutions with y a () =, hence the solutions cannot be globally unique. However, f C on (, ), and hence locally lipschitz there. Consequently, by the existence and uniqueness theorem, there exists some ε > such that the solutions with initial value are unique as long as t ε. / Phase portrait of some solutions for x = (left) and x = (right). 7. Juni 23 3:4 Serie 3 Seite 3 von 8

4 Ana-3 Hs 22 * Aufgabe 4 Sei I R ein endliches abgeschlossenes Intervall um und B : I R n n stetig. Zeige, dass für jeden Punkt x R n das Anfangswertproblem ẋ(t) = B(t)x(t), x() = x, eine eindeutig bestimmte Lösung besitzt, die auf ganz I definiert ist. Hinweis: Verwende das Picard Iterationsverfahren aus der Vorlesung für die lokale Existenz von Lösungen und iteriere wiederum dieses Verfahren um globale Existenz zu erhalten. / Local existence. Choose T > such that [, T ] I and consider the Banach space E C([, T ], R n ) equipped with the sup norm [,T ]. We make the ansatz u (t) = x and u m (t) = x + B(s)u m (s) ds, m. Then for any m and t I, u m+ (t) u m (t) = B(s)(u m (s) u m (s)) ds, and thus in particular, u m+ u m [,T ] T B I u m u m [,T ] (T B I ) m u u [,T ]. If we choose T /(2 B I ), then q = T B I /2, and hence for m n +, u m u n [,T ] m k=n+ m k=n+ u k+ u k [,T ] q k u u [,T ] qn q u u [,T ]. So (u m ) is a Cauchy sequence in E which converges to some u E. Since the convergence is uniformly on [, T ], we further have u(t) = lim u m+ (t) = x + lim B(s)u m (s) ds = x + B(s)u(s) ds, m m thus u satisfies the integral equation, i.e. u C ([, T ], R n ) is a solution to the initial value problem. Global existence. Note that T = /(2 B I ) does not depend on the initial value x. Hence we can repeat the previous step for the initial value problem ẋ(t) = B(t + T )x(t), x() = u(t ), and obtain a solution v which satisfies v(t) = u(t ) + B(s + T )v(s) ds, t T, t + T I. 7. Juni 23 3:4 Serie 3 Seite 4 von 8

5 Ana-3 Hs 22 If we extend u onto the interval [, 2T ] I by u(t + T ) = v(t), then u is continuous on [, 2T ] I and for any t T with t + T I, u(t + T ) = u(t ) + T = x + = x + T +t B(s + T )v(s) ds B(s)u(s) ds + B(s)u(s) ds. B(s + T )u(s + T ) ds Therefore, u C ([, 2T ] I, R n ) is a solution of the original initial value problem defined on [, 2T ] I. Repeating this step, and proceeding analogously for t <, we obtain a global solution u C (I, R n ). Uniqueness. Suppose u and v are two solutions defined on some interval J I around t with u(t ) = v(t ) = x. Then by the integral equation u(t) v(t) = B(s)(u(s) v(s)) ds. t Let Q = max u(t) v(t), t J then by continuity this maximum is attained at some t. So Q = u(t ) v(t ) = B(s)(u(s) v(s)) ds B I t t Q. t By eventually shrinking J we have B I t t /2, so u and v coincide on J. Since t is arbitrary we conclude that u v independent of the size of J. / * Aufgabe 5 Sei f : R n R n ein Lipschitz stetiges Vektorfeld, f (x) f (y) K x y, x, y R n. Zeige, dass für jedes x R n das Anfangswertproblem ẋ = f (x), x() = x, eine eindeutig bestimmte globale Lösung R R n, t x(t) besitzt. Hinweis: Statt des Picard Iterationsverfahrens kann auch der Banachsche Fixpunktsatz verwendet werden um lokale Existenz und Eindeutigkeit zu erhalten. Iteriere anschliessend eines dieser Verfahren, um globale Existenz zu erhalten. / We present two possible solutions: A. Since f is Lipschitz, solutions exists locally, are unique, and satisfy the integral equation x(t) = x + f (x(s)) ds. We may rewrite the latter as x(t) = x + f ()t + ( f (x(s)) f () ) ds, 7. Juni 23 3:4 Serie 3 Seite 5 von 8

6 Ana-3 Hs 22 and use the lipschitz estimate to obtain x(t) ( x + T f () ) + K x(s) ds, for any t [, T ] for which the solution exists. Thus with Gronwall s inequality x(t) ( x + T f () )e K t. Consequently, any solution is bounded for finite time, and since R 2 is unbounded any solution exists for all time. B. For T > consider the Banach space E C([ T, T ], R n ) equipped with the sup norm [ T,T ]. Any solution of the initial value problem satisfies the integral equation u(t) = x + f (u(s)) ds. We thus consider the operator ( Λ: E E, u t x + f (u(s)) ds Clearly, Λu E since f and u are continuous on [ T, T ]. Moreover, (Λu Λv)(t) = f (u(s)) f (v(s)) ds, and hence Λu(t) Λv(t) R n f (u(s)) f (v(s)) R n ds K u(s) v(s) R n ds K t u v [ T,T ]. Taking the sup on the left hand side gives Λu Λv [ T,T ] KT u v [ T,T ], thus if T = /(2K), then Λ is a contraction on E and admits a unique fixed point u. Since any other solution ũ on [ T, T ] of the initial value problem also gives rise to a fixed point, we conclude ũ = u, hence uniqueness follows. ) Consequently, for any initial data x, the initial value problem has a unique solution defined on [ T, T ] where T = /(2K). Since T does not depend on the initial data x we may proceed as in (i), and extend the solution to [ 2T, 2T ], and further up to any time. /. * Aufgabe 6 7. Juni 23 3:4 Serie 3 Seite 6 von 8

7 Ana-3 Hs 22 (i) Transformiere die Differenzialgleichung im R 2, ẋ = y + x( x 2 y 2 ), ẏ = x + y( x 2 y 2 ), in Polarkoordinaten x = r cos ϕ, y = r sin ϕ. / First note that ẋ = ṙ cos ϕ ϕr sin ϕ, ẏ = ṙ sin ϕ + ϕr cos ϕ. Thus the equation may be written as ṙ cos ϕ ϕr sin ϕ = r sin ϕ + r cos ϕ( r 2 ), ṙ sin ϕ + ϕr cos ϕ = r cos ϕ + r sin ϕ( r 2 ). Multiplying the first row with cos ϕ and the second with sin ϕ we obtain ṙ cos 2 ϕ ϕr sin ϕ cos ϕ = r sin ϕ cos ϕ + r cos 2 ϕ( r 2 ), ṙ sin 2 ϕ + ϕr sin ϕ cos ϕ = r sin ϕ cos ϕ + r sin 2 ϕ( r 2 ), and summing both rows yields ṙ = r ( r 2 ). On the other hand, multiplying the first row with sin ϕ and the second with cos ϕ gives ṙ sin ϕ cos ϕ ϕr sin 2 ϕ = r sin 2 ϕ + r sin ϕ cos ϕ( r 2 ), ṙ sin ϕ cos ϕ + ϕr cos 2 ϕ = r cos 2 ϕ + r sin ϕ cos ϕ( r 2 ). Substracting the rows from each other gives 2 ϕr = 2r, and hence for all r >, ϕ =. / (ii) Bestimme die Lösungen mit den Anfangswerten (r, ϕ ) = (, ) und (r, ϕ ) = (, ). / Note that r and r are stationary solutions of ṙ = r ( r 2 ). Furthermore, ϕ(t) = ϕ t solves ϕ =. Consequently, the solutions are t (, t), t (, t). / (iii) Beschreibe diese beiden Lösungskurven in Polarkoordinaten und kartesischen Koordinaten. 7. Juni 23 3:4 Serie 3 Seite 7 von 8

8 Ana-3 Hs 22 / In polar coordinates the solutions are straight lines parallel to the ϕ-axis. In cartesian coordinates, they correspond to the stationary solution at (, ), (x(t), y(t)) = (, ) and a circle of radius, respectively, (x(t), y(t)) = (cos( t), sin( t)) = (cos t, sin t). / 7. Juni 23 3:4 Serie 3 Seite 8 von 8