Bourbaki

Exercise 8, § 2

Functions of a Real Variable, IV, 205

Bourbaki marks this one as harder than the rest.

Let GG be a complete normed algebra with unit e\mathbf{e} over the field R\mathbf{R}.

a) Let t↦a(t)t \mapsto \mathbf{a}(t) be a regulated function on an interval J⊂RJ \subset \mathbf{R} with values in GG. Show that the integral u\mathbf{u} of the linear equation dx/dt=a(t)xd\mathbf{x}/dt = \mathbf{a}(t)\mathbf{x} which takes the value e\mathbf{e} at a point t0∈Jt_0 \in J is invertible on JJ, and that its inverse is the solution of the equation dx/dt=−xa(t)d\mathbf{x}/dt = -\mathbf{x}\mathbf{a}(t) (if v\mathbf{v} is the integral of this last equation which takes the value e\mathbf{e} at the point t0t_0, consider the linear equations satisfied by uv\mathbf{uv} and vu\mathbf{vu}). Deduce that, for every x0∈G\mathbf{x}_0 \in G, the integral of dx/dt=a(t)xd\mathbf{x}/dt = \mathbf{a}(t)\mathbf{x} which takes the value x0\mathbf{x}_0 at the point t0t_0 is equal to u(t)x0\mathbf{u}(t)\mathbf{x}_0.

b) Let a(t),b(t)\mathbf{a}(t), \mathbf{b}(t) be two regulated functions on JJ, and u\mathbf{u} and v\mathbf{v} the integrals of the equations dx/dt=a(t)x,dx/dt=xb(t)d\mathbf{x}/dt = \mathbf{a}(t)\mathbf{x}, d\mathbf{x}/dt = \mathbf{x}\mathbf{b}(t) taking the value e\mathbf{e} at the point t0t_0. Show that the integral of the equation dx/dt=a(t)x+xb(t)d\mathbf{x}/dt = \mathbf{a}(t)\mathbf{x} + \mathbf{x}\mathbf{b}(t) which takes the value x0\mathbf{x}_0 at the point t0t_0 is equal to u(t)x0v(t)\mathbf{u}(t)\mathbf{x}_0 \mathbf{v}(t).

c) Let a(t),b(t),c(t),d(t)\mathbf{a}(t), \mathbf{b}(t), \mathbf{c}(t), \mathbf{d}(t) be four regulated functions on JJ, and (u,v)(\mathbf{u}, \mathbf{v}) a solution of the system of two linear equations

dxdt=a(t)x+b(t)y,dydt=c(t)x+d(t)y. \frac{d\mathbf{x}}{dt} = \mathbf{a}(t)\mathbf{x} + \mathbf{b}(t)\mathbf{y}, \qquad \frac{d\mathbf{y}}{dt} = \mathbf{c}(t)\mathbf{x} + \mathbf{d}(t)\mathbf{y}.

Show that, if vv is invertible on JJ, then w=uv−1w = uv^{-1} is an integral of the equation dz/dt=b(t)+a(t)z−zd(t)−zc(t)zd\mathbf{z}/dt = \mathbf{b}(t) + \mathbf{a}(t)\mathbf{z} - \mathbf{z}\mathbf{d}(t) - \mathbf{z}\mathbf{c}(t)\mathbf{z} (the "Ricatti equation"); and conversely. Deduce that every integral of this last equation on JJ, taking the value x0x_0 at the point t0t_0, is of the form (A(t)x0+B(t)e)(C(t)x0+D(t)e)−1(A(t)x_0 + B(t)e)(C(t)x_0 + D(t)e)^{-1}, where A(t)A(t), B(t)B(t), C(t)C(t) and D(t)D(t) are continuous maps from JJ into L(E)\mathscr{L}(E) satisfying the identity U.(xy)=(U.x)yU.(xy) = (U.x)y.

d) Let w1w_1 be an integral of dz/dt=b(t)+a(t)z−zd(t)−zc(t)zd\mathbf{z}/dt = \mathbf{b}(t) + \mathbf{a}(t)\mathbf{z} - \mathbf{z}\mathbf{d}(t) - \mathbf{z}\mathbf{c}(t)\mathbf{z} on JJ; show that if ww is another integral of this equation such that w−w1w - w_1 is invertible on JJ then w−w1w - w_1 can be expressed in terms of integrals of the equations dx/dt=−(d+cw1)xd\mathbf{x}/dt = -(d + cw_1)\mathbf{x} and dx/dt=x(a−w1c)d\mathbf{x}/dt = \mathbf{x}(\mathbf{a} - w_1\mathbf{c}) (consider the equation satisfied by (w−w1)−1(w - w_1)^{-1}, and use bb) ).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.