Krdytkyu

I am trying to show the following:
Let u solve the homogeneous heat equation in the cylinder $Omega$ x $(0, infty)$ with vanishing dirichlet data and initial condition g. Multiple the PDE by $tu_t$ and apply the energy method to show that $int_{Omega}| nabla u(x,t)|^2dx < frac{1}{t}int_{Omega}|g(x)|^2dx$.

I'm not sure how I apply the energy method once I have multiplied the PDE through by $tu_t$.

Hint

One dimensional Heat equation ($nabla u=u_x$)

$$u_t = Delta u=cu_{xx}$$

Energy method, define
$$E(t)=int_{Omega} u^2 dx $$

Due to vanishing dirichlet BC you get
$$E'(t)=int_{Omega} 2 u u_t dx =int_{Omega} 2 u c u_{xx} dx =
2c u u_xBig|_{partialOmega}-2cint_{Omega} u_x^2 dx =-2cint_{Omega} u_x^2 dx =quad
-2cint_{Omega}|nabla u|^2 dx le0
$$
You have also to explain here why you allowed to differentiate under the integral.

Since $E'le0$, $E$ is decreasing and, together with

$$E(0)=int_{Omega} g^2 dx $$
we got
$$E(t)le int_{Omega} g^2 dx, quad forall tge0$$

So you got
$$int_{Omega}|nabla u|^2 dx= -frac{1}{2c}E'(t)=frac{1}{2c}left|E'(t)right|$$

I'm actually fail to see why it should be
$< frac{1}{t}int_{Omega}|g(x)|^2dx$

Not quite sure how to use the hint, but here are two possible solutions (compared with Michael Medvinky's post, I take $c=1$).

1) Spectral method: Let
$$
-Delta=sum_{k=1}^inftylambda_k langlephi_k,cdotranglephi_k
$$
be the spectral decomposition of the Laplacian with Dirichlet boundary conditions. The solution of the heat equation is given by
$$
u(t,x)=sum_{k=1}^infty e^{-tlambda_k}langle phi_k,granglephi_k(x).
$$
Thus
$$
int_Omega lvert nabla u(t,x)rvert^2,dx=-int_Omega u(t,x)Delta u(t,x),dx=sum_{k=1}^infty lambda_k e^{-2tlambda_k}lvert langlephi_k,granglervert^2.
$$
Since $lambda e^{-2tlambda}leq (2et)^{-1}$ for $lambdageq 0$, we have
$$
int_Omega nabla u(t,x)rvert^2,dxleq frac{1}{2et}sum_{k=1}^infty lvertlanglephi_k,granglervert^2=frac{1}{2et}int_Omega g(x)^2,dx.
$$
If $gneq 0$, we get the strict inequality claimed in the question (of course it's simply not true for $g=0$).

2) Energy method: Let
$$E(t)=frac 1 2int_Omega u(t,x)^2,dx.$$
As shown in Michael Medvinky's post, the derivative of $E$ is given by
$$
E'(t)=-int_Omega lvert nabla u(t,x)rvert^2,dx.
$$
The second derivative of $E$ is
$$
E''(t)=frac{d}{dt}int_Omega u(t,x)Delta u(t,x),dx=int_Omega (Delta u(t,x))^2+u(t,x)Delta^2 u(t,x),dx=2int_Omega (Delta u(t,x))^2,dxgeq 0.
$$
Hence $E'(t)$ is increasing and we get
$$
frac1 2int_Omega g(x)^2,dx=E(0)=E(t)-int_0^t E'(s),dsgeq E(t)-t E'(t)geq -tE'(t)=tint_Omegalvert nabla u(t,x)rvert^2,dx.
$$
Again, you get the desired strict inequality from the question unless $g=0$.