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.

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