Solving a PDE to estimate $E(X_t\mid B_t = x)$ where $X_t=\int_t^TB_s^2ds$

Question

Given that $B_t$ is standard Brownian Motion.I am supposed to find

$$ \mathop{\mathbb{E}_{B_t = x}} \int_{t}^{T}{B_s}^2ds$$

I used Ito's lemma and ended up with a PDE as follows

$$u_t + u_{xx} + x^2 = 0$$

with the condition

$$ u(T,x) = 0 $$

How would you go about solving the differential equation?

Answer 1

You seem to be asking for the value of $$u(t,x)=E\left(X_t\mid B_t=x\right)$$ for every $t$ in $(0,T)$ and every $x$, where $$X_t=\int_t^TB_s^2ds$$ Fix some $t$ in $(0,T)$. For every $s>t$, $$B_s=B_t+W_{s-t}$$ where $(W_u)$ is a standard Brownian motion independent of $B_t$. One knows that $E(W_u)=0$ and $E(W_u^2)=u$, hence the expansion $$X_t=\int_0^{T-t}(B_t^2+2B_tW_u+W_u^2)du$$ yields $$u(t,x)=\int_0^{T-t}(x^2+2x\cdot0+u)\,du$$ and finally, $$u(t,x)=(T-t)x^2+\tfrac12(T-t)^2$$