Two dimensional Cauchy problems

[user40191]

Let y(t) = (y1(t), y2(t))^T and
A(t) = (a(t) b(t)
c(t) d(t)).

A(t) is a 2x2 matrix with a,b,c,d all polynomials in t. Consider the two dimensional Cauchy problem y'(t) = A(t)y(t), y(0)=y0.
Show that a solution exists for all t>=0.
Give a general condition on the A(t) which ensures global existence.

Please could you help me with this question - I don't know what to do. I need to use Picard somewhere I think but I don't know how to go about it.

 

[HallsofIvy]

Okay, you know the name "Picard" so I presume you have seen the proof- probably for one dimension. Do exactly the same thing: dy/dt= Ay. Convert that to an integral equation by integrating both sides: [itex]y= A\int_0^x y(t) dt[/itex]. Think of that as a "linear transformation", [itex]T(y)= A\int_0^x y(t)dt[/itex] and, just as in the regular "Picard theorem", show that, under restrictions on the domain, that is a "contraction mapping".

 

[Unredeemed]

Okay, you know the name "Picard" so I presume you have seen the proof- probably for one dimension. Do exactly the same thing: dy/dt= Ay. Convert that to an integral equation by integrating both sides: [itex]y= A\int_0^x y(t) dt[/itex]. Think of that as a "linear transformation", [itex]T(y)= A\int_0^x y(t)dt[/itex] and, just as in the regular "Picard theorem", show that, under restrictions on the domain, that is a "contraction mapping".

Can you take the matrix out of the integral even though its entries are polynomials in t?

 

[HallsofIvy]

You are right. I ignored the fact that entries are functions of t. It should be [tex]y= \int_0^x A(t)y(t)dt[/tex] and [tex]T(y)= \int_0^x A(t)y(t)dt[/tex].

 

[vanhees71]

For a general initial condition [itex]\vec{y}(0)=\vec{y}_0[/itex] the integral equation reads
[tex]\vec{y}(t)=\vec{y}_0+\int_0^t \mathrm{d} t' \hat{A}(t') \vec{y}(t').[/tex]
Now you can solve the problem easily by iteration. It's important to keep in mind that in general the matrices [itex]\hat{A}(t)[/itex] do not commute at different times!

 

[Unredeemed]

You are right. I ignored the fact that entries are functions of t. It should be [tex]y= \int_0^x A(t)y(t)dt[/tex] and [tex]T(y)= \int_0^x A(t)y(t)dt[/tex].

Would you mind going into a little more depth about how you'd modify the Picard's theorem proof here? I'm a little confused as the version I've seen is really long.