I am looking to compute the following using Ito's formula.
$$u(t,\beta_t) = \mathbb{E}(\int_t^T\beta_s^2ds|\beta_t)$$
Knowing the properties of brownian motion, it is rather easy to show that the above is equivalent to $\frac{1}{2}(T^2-t^2)$; however, i'm looking to apply Ito's formula to come up with a similar result. Given that $u$ is a martingale, it follows from Ito's formula that $u$ satisfies the homogenous heat equation:
$$u_t = -\frac{1}{2}u_{xx}$$ Though I am struggling to see how the solution aligns with what I found using the easier approach.
Side note:
My boundary conditions: $$u(T,x) = 0$$ $$u(0,0) = \mathbb{E}(\int_0^Tds) = T $$ Though I could be off here, as the expectation is confusing me
Edit:
My approach to finding $\frac{1}{2}(T^2-t^2)$ through knowledge of B.M.:
(1) By the tower property, using the fact that $\beta_t\in F_t$ $$u(t, \beta_t) = \mathbb{E}(\mathbb{E}(\int_t^T\beta_s^2ds|F_t)|\beta_t)$$
(2)Then given the integral is not within $F_t$, we have $$u(t,\beta_t) = \mathbb{E}(\mathbb{E}(\int_t^T\beta_s^2ds)|\beta_t)$$
(3)
$$u(t,\beta_t) = \mathbb{E}((\int_t^T\mathbb{E}(\beta_s^2)ds|\beta_t)$$
(4) Lastly,
$$u(t,\beta_t) = \mathbb{E}(T-t|\beta_t) = \frac{1}{2}(T^2-t^2)$$
Before going into the details of my solution, I'd like to point out how the first comment was right: your computations are wrong and indeed that expectation is $1/2(T^2-t^2)$. We can see it this way:
$$ \mathbb{E}[I] = \int_{t}^{T}\mathbb{E}[W_{s}^{2}] ds = \int_{t}^{T}s ds = \frac{1}{2}(T^2 - t^2) $$
I'm not sure whether by "using Ito" you meant to do what I'm about to, but here's what I did.
Let consider the function $f(t,W_t) = tW_{t}^{2}$ and apply Ito:
$$ TW_{T}^{2} - tW_{t}^{2} = \int_{t}^{T}(W_{s}^{2} + s) ds + \int_{t}^{T} 2s W_{s}^{2}dW_s = I + \frac{1}{2}(T^2-t^2) + 2\int_{t}^{T}sW_{s}^{2}dW_s $$
By taking the expectation, and noticing that the stochastic integral vanishes and $\mathbb{E}[W_{s}^{2}] = s$ we have once again the same result as before.
Another function we could have used is $f(W_t)=W_{t}^{4}$