I am given a temperature function $T(t,x)$ defined on $\Omega$, and an area $\Gamma \subset \partial \Omega$ (part of the boundary of $\Omega$) where heat can get out of the cup.
I know the following details:
- $\partial_t T = \Delta T$ in $(0,\infty) \times \Omega$
- $T = 0$ on $[0,\infty)\times \Gamma$
- $\nabla T \cdot n = 0$ on $[0,\infty) \times \partial \Omega \backslash\Gamma $
- $T = T_0$ on $0\times \Omega$.
I now want to show that
$$ \int_{\Omega} T(t,x)^2 dx = \int_{\Omega} T_0(x)^2 dx - 2 \int_0^t\int_{\Omega} |\nabla T(s,x)|^2dxds. $$
I don't know how, and I was given a vague hint: Multiply one of the four given identities (boundary conditions) by something and then integrate over $\Omega$, but I don't know what to multiply by.
Similar questions did this multiplying "trick" and then used integration by parts.
I thought about the following, but it doesn't lead anywhere I think:
$$\int_{\Omega} T(t,x) (T(t,x) - T_0(x)) = \int_{\Omega} T(t,x) \int_0^t \partial_t T(s,x)dsdx \\= \int_{\Omega} \int_0^t (\nabla\cdot\nabla T(s,x))\cdot T(t,x) ds dx \\= \int_{\partial \Omega \backslash \Gamma} \int_0^t \nabla T(s,x)\cdot T(t,x)\cdot n ds ds_x \\+ \int_{\Gamma} \int_0^t \nabla T(s,x)\cdot T(t,x)\cdot n ds ds_x,$$ which would be equal to zero, since $T$ is zero on $\Gamma$ and $\nabla T \cdot n$ is zero on $\partial \Omega \backslash \Gamma$.
I think the minus two might be coming from the identity:
$$ (a - b)^2 = a^2 - 2ab + b^2,$$
but I don't see how.
Here are some things that I did figure out:
$$ \int_{\Omega} T(t,x) dx = \int_{\Omega} T_0(x) dx,\quad \text{ for all } t>0 $$
$$T(t,x) = T_0(x) + \int_0^t \partial_t T(s,x)ds$$
$$ T(t,x) - T_0(x) = \int_0^t \partial_t T(s,x)ds =\int_0^t \Delta T(s,x)ds$$