Showing that the Riemann invariant $frac{1}{2} (u^2+v^2) + int frac{c(p)^2}{p} dp $ is conserved along the...
$begingroup$
I need to show that the Riemann invariant $R = frac{1}{2} (u^2+v^2) + int frac{c(p)^2}{p} dp $ is conserved along the characteristic $dy/dx = v/u$. My system of equations are:
begin{aligned}
(pu)_x + (pv)_y &= 0 \
p(uu_x + vu_y) + c(p)^2p_x &= 0 \
p(uv_x +vv_y)+c(p)^2p_y &= 0
end{aligned}
The system is hyperbolic if $u^2 + v^2 > c^2$ where $c=c(p)$ (cf. this post). The left eigenvector corresponding to the eigenvalue $lambda = dy/dx = v/u$ is $(0,1,v/u)$. To find the Riemann invariant I do:
$$
begin{pmatrix}
0 & 1 & v/u
end{pmatrix}
begin{pmatrix}
u & p & 0 \
c^2 & pu & 0 \
0 & 0 & pu
end{pmatrix}
frac{d}{dx}begin{pmatrix} p \ u \ vend{pmatrix} = begin{pmatrix}
0 & 0 & 0
end{pmatrix}
$$
which gives me
$$
begin{pmatrix}
c^2 & pu & pv
end{pmatrix}
frac{d}{dx}begin{pmatrix} p \ u \ vend{pmatrix}= begin{pmatrix}
0 & 0 & 0
end{pmatrix}
$$
I am stuck on the final part as to how this implies that $frac{d}{dx}[R] =0$. I am struggling to see how $frac{d}{dx} int frac{c^2(p)}{p}dp = frac{c^2(p) {dp}/{dx}}{p}$ which would make $frac{d}{dx}[R] =0$.
pde systems-of-equations hyperbolic-equations
edited Dec 14 '18 at 15:22
Harry49
7,49431341
asked Dec 13 '18 at 13:56
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
I need to show that the Riemann invariant $R = frac{1}{2} (u^2+v^2) + int frac{c(p)^2}{p} dp $ is conserved along the characteristic $dy/dx = v/u$. My system of equations are:
begin{aligned}
(pu)_x + (pv)_y &= 0 \
p(uu_x + vu_y) + c(p)^2p_x &= 0 \
p(uv_x +vv_y)+c(p)^2p_y &= 0
end{aligned}
The system is hyperbolic if $u^2 + v^2 > c^2$ where $c=c(p)$ (cf. this post). The left eigenvector corresponding to the eigenvalue $lambda = dy/dx = v/u$ is $(0,1,v/u)$. To find the Riemann invariant I do:
$$
begin{pmatrix}
0 & 1 & v/u
end{pmatrix}
begin{pmatrix}
u & p & 0 \
c^2 & pu & 0 \
0 & 0 & pu
end{pmatrix}
frac{d}{dx}begin{pmatrix} p \ u \ vend{pmatrix} = begin{pmatrix}
0 & 0 & 0
end{pmatrix}
$$
which gives me
$$
begin{pmatrix}
c^2 & pu & pv
end{pmatrix}
frac{d}{dx}begin{pmatrix} p \ u \ vend{pmatrix}= begin{pmatrix}
0 & 0 & 0
end{pmatrix}
$$
I am stuck on the final part as to how this implies that $frac{d}{dx}[R] =0$. I am struggling to see how $frac{d}{dx} int frac{c^2(p)}{p}dp = frac{c^2(p) {dp}/{dx}}{p}$ which would make $frac{d}{dx}[R] =0$.
pde systems-of-equations hyperbolic-equations
edited Dec 14 '18 at 15:22
Harry49
7,49431341
asked Dec 13 '18 at 13:56
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
I need to show that the Riemann invariant $R = frac{1}{2} (u^2+v^2) + int frac{c(p)^2}{p} dp $ is conserved along the characteristic $dy/dx = v/u$. My system of equations are:
begin{aligned}
(pu)_x + (pv)_y &= 0 \
p(uu_x + vu_y) + c(p)^2p_x &= 0 \
p(uv_x +vv_y)+c(p)^2p_y &= 0
end{aligned}
The system is hyperbolic if $u^2 + v^2 > c^2$ where $c=c(p)$ (cf. this post). The left eigenvector corresponding to the eigenvalue $lambda = dy/dx = v/u$ is $(0,1,v/u)$. To find the Riemann invariant I do:
$$
begin{pmatrix}
0 & 1 & v/u
end{pmatrix}
begin{pmatrix}
u & p & 0 \
c^2 & pu & 0 \
0 & 0 & pu
end{pmatrix}
frac{d}{dx}begin{pmatrix} p \ u \ vend{pmatrix} = begin{pmatrix}
0 & 0 & 0
end{pmatrix}
$$
which gives me
$$
begin{pmatrix}
c^2 & pu & pv
end{pmatrix}
frac{d}{dx}begin{pmatrix} p \ u \ vend{pmatrix}= begin{pmatrix}
0 & 0 & 0
end{pmatrix}
$$
I am stuck on the final part as to how this implies that $frac{d}{dx}[R] =0$. I am struggling to see how $frac{d}{dx} int frac{c^2(p)}{p}dp = frac{c^2(p) {dp}/{dx}}{p}$ which would make $frac{d}{dx}[R] =0$.
pde systems-of-equations hyperbolic-equations
edited Dec 14 '18 at 15:22
Harry49
7,49431341
asked Dec 13 '18 at 13:56
pablo_mathscobarpablo_mathscobar
996
$endgroup$
I need to show that the Riemann invariant $R = frac{1}{2} (u^2+v^2) + int frac{c(p)^2}{p} dp $ is conserved along the characteristic $dy/dx = v/u$. My system of equations are:
begin{aligned}
(pu)_x + (pv)_y &= 0 \
p(uu_x + vu_y) + c(p)^2p_x &= 0 \
p(uv_x +vv_y)+c(p)^2p_y &= 0
end{aligned}
The system is hyperbolic if $u^2 + v^2 > c^2$ where $c=c(p)$ (cf. this post). The left eigenvector corresponding to the eigenvalue $lambda = dy/dx = v/u$ is $(0,1,v/u)$. To find the Riemann invariant I do:
$$
begin{pmatrix}
0 & 1 & v/u
end{pmatrix}
begin{pmatrix}
u & p & 0 \
c^2 & pu & 0 \
0 & 0 & pu
end{pmatrix}
frac{d}{dx}begin{pmatrix} p \ u \ vend{pmatrix} = begin{pmatrix}
0 & 0 & 0
end{pmatrix}
$$
which gives me
$$
begin{pmatrix}
c^2 & pu & pv
end{pmatrix}
frac{d}{dx}begin{pmatrix} p \ u \ vend{pmatrix}= begin{pmatrix}
0 & 0 & 0
end{pmatrix}
$$
I am stuck on the final part as to how this implies that $frac{d}{dx}[R] =0$. I am struggling to see how $frac{d}{dx} int frac{c^2(p)}{p}dp = frac{c^2(p) {dp}/{dx}}{p}$ which would make $frac{d}{dx}[R] =0$.
pde systems-of-equations hyperbolic-equations
pde systems-of-equations hyperbolic-equations
edited Dec 14 '18 at 15:22
Harry49
7,49431341
asked Dec 13 '18 at 13:56
pablo_mathscobarpablo_mathscobar
996
edited Dec 14 '18 at 15:22
Harry49
7,49431341
asked Dec 13 '18 at 13:56
pablo_mathscobarpablo_mathscobar
996
edited Dec 14 '18 at 15:22
Harry49
7,49431341
edited Dec 14 '18 at 15:22
Harry49
7,49431341
edited Dec 14 '18 at 15:22
Harry49
7,49431341
7,49431341
asked Dec 13 '18 at 13:56
pablo_mathscobarpablo_mathscobar
996
asked Dec 13 '18 at 13:56
pablo_mathscobarpablo_mathscobar
996
asked Dec 13 '18 at 13:56
pablo_mathscobarpablo_mathscobar
996
996
add a comment |
add a comment |
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