## Weak solution and Galerkin method of parabolic equation

Consider the second order parabolic equation

$\displaystyle (1)\begin{cases} u_t+Lu=f\text{ in }\Omega_T\\ \quad\quad u=0 \text{ on } \partial \Omega\times [0,T]\\\quad\quad u(\cdot,0)=g\text{ on } \Omega\times\{t=0\}\end{cases}$

where $\displaystyle Lu=-(a^{ij}(x,t)u_{x_i})_{x_j}+b^i(x,t)u_{x_i}+c(x,t)u$ and $(a^{ij})$ is elliptic in $\overline{\Omega}_T$, $\Omega_T=\Omega\times (0,T]$.

For a weak solution $u$, at least we should have $u(\cdot,t)\in H^1_0(\Omega)$ for each $t\in [0,T]$. So it is natural to consider this function space

$\displaystyle L^p(0,T;X)$

which consists of all strongly measurable functions $u:[0,T]\to X$ with

$\displaystyle ||u||_{L^p(0,T;X)}:=\left(\int_0^T ||u||_X^pdt\right)^{1/p}\leq\infty$

with $1\leq p<\infty$ and

$\displaystyle ||u||_{L^\infty(0,T;X)}:=\text{ess}\,\sup_{t\in [0,T]}||u(\cdot,t)||_X\leq \infty$.

We will use $L^2(0,T;H^1_0(\Omega)$ in particular. In order to solve $(1)$, it is equivalent to solve

$\displaystyle (u', v)+(Lu,v)=(f,v)$ holds for $v\in H^1_0(\Omega)$, $0\leq t\leq T$

To make sense of $(u',v)$, we should require $u'=u_t\in L^2(0,T;H^{-1}(\Omega))$. So $(u',v)=\langle u',v\rangle$, the pairing of $H^{-1}(\Omega)$ and $H^1_0(\Omega)$.

As usual, we interpret $\displaystyle (Lu,v):=B(u,v;t)=\int_{\Omega} a^{ij}(x,t)u_{x_i}v_{x_j}+b^i(x,t)u_iv+c(x,t)uvdx$

Definition: $u$ is called the weak solution of $(1)$ if

$\displaystyle u\in L^2(0,T;H^1_0(\Omega)), u'\in L^2(0,T;H^{-1}(\Omega))$

and $u$ satisfies $\displaystyle u(0)=g$ and

$\displaystyle \langle u',v\rangle+B(u,v;t)=(f,v)$ for all $v\in H^1_0(\Omega)$

Galerkin method is trying to solve this in a finite space and let the dimension grows, these solutions will approximate a solution of $(1)$. As we all know, $(-\Delta)^{-1}$ is a compact self-adjoint operator, so the eigenfunctions of $(-\Delta)^{-1}$ form a basis of $H^1_0(\Omega)$. Suppose $w_1(x), w_2(x), \cdots,$ is such an orthogonal basis of $H^1_0(\Omega)$. Because $w_k$ are eigenfunction for different eigenvalues, they orthogonal with each other. Since $H^1_0(\Omega)$ is dense in $L^2(\Omega)$, we know $w_1(x), w_2(x), \cdots,$ actually form a basis of $L^2(\Omega)$. By normalizing them, we can require $w_1(x), w_2(x), \cdots$ are orthonormal basis of $L^2(\Omega)$.

So we can express $g=\sum g^kw_k$. We are finding the solution of the form

$u_m(x,t)=a^k_m(t)w_k(x)$ with $a_k(0)=g^k$

and wishing $u_m(x,t)$ will converge to a solution of $(1)$ as $m\to \infty$.

If $u_m$ is a weak solution of $(1)$, then

$\displaystyle \langle u_m',v\rangle+B(u_m,v ; t)=(f, v)$, $v\in \text{span}\{w_1, w_2, \cdots,w_m\}$

This indeed means

$a^k_m(t)'+ B_l^k a_m^l=f^k$

where $B_l^k=B(w_l,w_k;t)$, $f^k=(f,w_k)$.

This is a linear system of ODE with initial condition $a_k(0)=g^k$, so it has a unique solution for $0\leq t\leq T$, equivalently there exists $u_m$, for each $m$.

We need the following energy estimates

$\displaystyle \max_{0\leq t\leq T}||u_m(t)||_{L^2(\Omega)}+||u_m||_{L^2(0,T;H^1_0(\Omega))}+||u'||_{L^2(0,T;H^{-1}(\Omega))}\leq C(||f||_{L^2(0,T;L^2(\Omega))}+||g||_{L^2(\Omega)})$

Now we are going to let $m\to \infty$. From the energy estimate, we know $u_m$ is bounded in ${L^2(0,T;H^1_0(\Omega))}$ and $u'$ is bounded in $L^2(0,T;H^{-1}(\Omega))$, then

$u_m\rightharpoonup u$ weakly in ${L^2(0,T;H^1_0(\Omega))}$

$u_m'\rightharpoonup u$ weakly in ${L^2(0,T;H^{-1}(\Omega))}$

For any $v\in \text{span}\{w_1, w_2, \cdots,w_k\}$, if $m\geq k$, then

$\displaystyle \langle u'_m,v\rangle+B(u_m,v;t)=(f,v)$

Integrating with respect to $t$,

$\displaystyle \int^T_0 \langle u'_m,v\rangle+B(u_m,v;t)dt=\int^T_0 (f,v)dt$

sending $m\to \infty$, we have

$\displaystyle \int^T_0 \langle u',v\rangle+B(u,v;t)dt=\int^T_0 (f,v)dt$

Since such $v$ is dense in $L^2(0,T;H^1_0(\Omega))$, the above equality actually implies

$\displaystyle \langle u',v\rangle+B(u,v;t)=(f,v)$ a.e. $0\leq t\leq T$

If $u$ satisfies the initial condition, then $u$ is a weak solution of $(1)$. To verify $u(0)=g$, let us consider the equivalent form

$\displaystyle \langle u_m(T), v(T)\rangle- \langle u_m(0), v(0)\rangle-\int^T_0 \langle u_m,v'\rangle+B(u_m,v;t)dt=\int^T_0 (f,v)dt$

Let $v(T)=0$, we get

$\displaystyle -\langle u_m(0), v(0)\rangle-\int^T_0 \langle u_m,v'\rangle+B(u_m,v;t)dt=\int^T_0 (f,v)dt$

similarly for $u$ and such $v$, we have

$\displaystyle -\langle u(0), v(0)\rangle-\int^T_0 \langle u,v'\rangle+B(u,v;t)dt=\int^T_0 (f,v)dt$

Then we will have $\langle u_m(0), v(0)\rangle\to \langle u(0), v(0)\rangle$. Since $v(0)$ is arbitrary in $H^1_0(\Omega)$ and $u_m(0)\to g$ in $L^2$, then $u(0)=g$.

$\textbf{Remark:}$ refer to Evan’s book.