Trefftz Discontinuous Galerkin Method for Friedrichs Systems with Linear Relaxation: Application to the P₁ Model

Lemma A.1.

The conditions (A.4) are

with λ=±εv⁢σa⁢σt.

Proof.

First, a necessary condition for (A.4) to admit a solution is det⁡(A1⁢λ-R)=0. Since

one deduces

With this choice for λ, the matrix A1⁢λ+R reads:

and one notices that

Thanks to the first and the second equations of (A.4) one has

From (A.6) one gets (A1⁢λ+R)⁢𝐪1=𝟎, therefore 𝐪3=𝟎. ∎

Proposition A.2.

The P1 model (A.1) admits the following four solutions:

Proof.

One notices Ker⁡(A1⁢λ+R)=Span⁡((σt,∓σa)T). Thus with 𝐰=(σt,∓σa)T and the relations (A.5) one gets

From the last equality of (A.5) one sees that -A1⁢𝐪1-𝐪2∈ℑ⁡(A1⁢λ+R) which implies

Since the matrices A1 and R are symmetric, Ker⁡((A1⁢λ+R)T)=Ker⁡(A1⁢λ+R)=Span⁡(𝐰). A necessary condition is then (-A1⁢𝐪1-𝐪2,𝐰)=0 which is equivalent to

Finally, let 𝐪0=(q01,q02)T. From the fourth equation of (A.5) one gets q01=1σa⁢(β⁢σa-σt2⁢σt⁢σa∓σt⁢q02). Thus one can choose 𝐪0 of the form

with γ∈ℝ. In summary, one has the following relations:

where β,γ∈ℝ. Because the solutions are of the form 𝐮⁢(x,t)=(𝐪0+x⁢𝐪1+t⁢𝐪2+x⁢t⁢𝐪3)⁢eλ⁢x, with λ=±εc⁢σa⁢σt, one finds the four basis functions (A.7). ∎

Lemma A.3.

The following four functions are linear combinations of the solutions (A.7):

Proof.

One defines the following linear combinations of the functions (A.7):

Then defining the four solutions

one gets the functions given in the statement of the lemma. ∎

Proposition A.4.

When σa→0 (σt→σsε2), the solutions given in Lemma A.3 tend to the following functions:

Proof.

One notices that

The limits of 𝐞~1⁢(x,t), 𝐞~2⁢(x,t) and 𝐞~3⁢(x,t) are simply obtained by using the expressions (A.8) in the functions given in Lemma A.3. The limit of the second component of 𝐞~4⁢(x,t) can be obtained in a similar way. It remains to study the first component of 𝐞~4⁢(x,t). One has

Because

one gets the expression of the limit of 𝐞~4⁢(x,t). ∎

Lemma B.1.

Assume that the hypotheses of Proposition 7.2 are satisfied. Then for all 0≤l≤n-2 one has the identity

Proof.

Let l∈ℕ, 0≤l≤n-2. For l1∈ℤ, -1≤l1≤l-1 we define the function

where we use the convention ∑p=ab=0 for a,b∈ℤ and b<a. First we show f⁢(l1)=f⁢(l1+1) for -1≤l1≤l-1. Because u is solution to equation (B.1) one notices

Now we consider the definition (B.2) of the function f and study the difference f⁢(l1+1)-f⁢(l1). Cancelling all elements which appear in both f⁢(l1) and f⁢(l1+1), one finds

Using equality (B.3) to reformulate the fifth term on the right-hand side, one gets

Ordering the terms with respect to the derivatives gives

Using the definitions (7.3) and (7.4), one finds

Therefore one has f⁢(l1+1)-f⁢(l1)=0 for all -1≤l1≤l-1. One deduces f⁢(-1)=f⁢(l) which can be written as

Noticing from (7.5) that αl+20⁢(𝐱)=βl+20⁢(𝐱) and αl+21⁢(𝐱)=βl+21⁢(𝐱), one incorporates the two corresponding terms in the second sum. So one finds equality (B.1). ∎

Proof of Proposition 7.2.

Start from the Taylor expansion (7.2). From definition (7.3) one has αnp⁢(𝐱)=Tnp⁢(𝐱) and αn-1p⁢(𝐱)=Tn-1p⁢(𝐱). Therefore

One can recursively use equality (B.1) from l=n-2 to l=0. More precisely, rearranging the first sum, one has

One can reformulate the terms between brackets using (B.1) with the index correspondence n-2=l. One finds

One can now recursively repeat this simple operation using equality (B.1) for l=n-3 to l=0. One finally gets formula (B.4) where the first line is written for n=2, the term [⋅] becomes a sum and the last term remains unchanged:

That is,

Noticing from (7.5) that α00⁢(𝐱)=β00⁢(𝐱),α10⁢(𝐱)=β10⁢(𝐱),α11⁢(𝐱)=β11⁢(𝐱), one finds the expression (7.6). ∎