# A Problem concerning the energy of a Galerkin Method

I'm reading an article from this website:

article

My question is about several formulations in that paper:

The paper is about a wave equation and the use of a Galerkin method to discretize in space.

(1) Page 4

why does the author use the fraction subscript? A convention?

(2) Page 5

Why $(\hat{q_h}v^-)_{j+\frac{1}{2}}$? Since this term is derived from the boundery of $I_j$, why not $q_h^-$ directly?

(3) Most important, the energy estimate, page 6

see (2.8) which is derived from (2.4). Why did the author simply differentiate over q and u, instead of using the chain rule?

Any help would be appreciated.

• I'm not sure I understand your third question, when you ask about using the chain rule. Does the following help? My interpretation of the setup is that in 2.4 the function $q_h$ is intended to be a function of time and space, but the test function $w$ is a spatial function only, and the desire is that $2.4$ holds for all test functions, over a time interval. Differentiating $2.4$ with respect to time therefore will only give time derivatives to $q$ and $v$. – Jason Knapp Jul 24 '14 at 13:10

If you have $u_{tt}=q_x$ and apply the projection you get $\int u_{tt} v dx-\int q_x v dx=0$. Now apply partial integration to $\int_{I_j} q_xv dx$ yielding $\int_{I_j} q_x v dx = - \int_{I_j}qv_x dx + [qv]_{\partial I_j}$ where $\partial I_j$ is the boundary of the interval, i.e. $j-1/2$ and $j+1/2$. The fact, that $[qv]_{\partial I_j}$ is written as $\hat qv^+-\hat qv^-$ means exactly this.
I agree with Jason Knapp. As you can see by the $dx$ of the integral, you project your residual onto the spatial domain $V_k$ which has no time dependency. Thus you can first take the derivative w.r.t time and set make a concrete choice of $w=q_h$.