Uniqueness for ode coming from parabolic pde

[cliowa]

Hey all,
I was working a little on parabolic pde, and came across this (comes up in regularity theory). Consider a Hilbert triple [itex]V\subset H\subset V^*[/itex] (continuous embeddings) and a linear operator [itex]A(t)[/itex] from V to V*, where t ranges in some interval [0,T]. Now let [itex]w\in H^1(0,T;V^*)\cap L^2(0,T;V)[/itex] solve

[tex]w'=A(t)w-\int_0^t A'(\tau)w(\tau) d\tau, \quad w(0)=0 [/tex].

I want to show that this implies w=0. How could I do that?

I tried multiplying by w and integrating by parts, which results in

[tex] 1/2 (w,w) +\int_0^t a(s,w(s),w(s))ds= -\int_0^t (\int_0^s A(\tau)w(\tau d\tau,w(s))ds,[/tex]

where a(s,w(s),w(s)) is the induced quadratic form satisfying [itex] a(s,w(s),w(s))\geq \alpha ¦w(s)¦_V-\beta ¦w(s)¦_H[/itex] for constants >0, uniformly in t. How does this help me?

Best regards...cliowa

 

[jvicens]

Dear cliowa,

Thank you for sharing your findings on parabolic PDEs. It is an interesting problem and I will try to provide some guidance on how to approach it.

Firstly, let's recap the given problem. We have a Hilbert triple V\subset H\subset V^* and a linear operator A(t) from V to V*, where t ranges in some interval [0,T]. We also have a function w\in H^1(0,T;V^*)\cap L^2(0,T;V) that solves the parabolic PDE:

w'=A(t)w-\int_0^t A'(\tau)w(\tau) d\tau, \quad w(0)=0 .

To show that w=0, we can use the fact that w satisfies the following equation:

w'=A(t)w-\int_0^t A'(\tau)w(\tau) d\tau.

We can rearrange this equation to get:

w'-A(t)w=-\int_0^t A'(\tau)w(\tau) d\tau.

Now, let's multiply both sides by w and integrate from 0 to t:

\int_0^t w'w dt-\int_0^t A(t)ww dt=-\int_0^t (\int_0^t A'(\tau)w(\tau) d\tau)w dt.

Using integration by parts on the first integral, we get:

w(t)^2/2-\int_0^t w(\tau)w'(\tau) d\tau-\int_0^t A(t)ww dt=-\int_0^t (\int_0^t A'(\tau)w(\tau) d\tau)w dt.

Now, let's focus on the second integral on the left-hand side. We can use the Cauchy-Schwarz inequality to get:

-\int_0^t A(t)ww dt\leq \frac{1}{2}(w(t)^2+w(t)^2)-\frac{1}{2}\int_0^t A(t)^2w^2 dt.

Substituting this into the previous equation, we get:

w(t)^2/2-\int_0^t w(\tau)w'(\tau) d\tau-\frac{1}{2}\int