Lax entropy condition for convex flux and sign of shock speed
$begingroup$
Assume that $u$ is an entropy solution to $u_t+partial_x(A(u))=0$, where $A$ is a non-decreasing function on $mathbb{R}$ such that $A(0)=0$ and $$A'(u_ell)>gamma'(t)=frac{A(u_ell)-A(u_r)}{u_ell-u_r}>A'(u_r)$$ (namely the Lax' entropy condition, where $u_ell, u_r$ are of the usual meanings). Prove that the shocks travel from left to right if $A$ is convex.
I got frustrated after spending a long time on it (not sure if one is supposed to consider the shock speed $gamma'(t)$, which is always non-negative since $A$ is non-decreasing). I believe that I missed something here and that there is some very simple way to solve this problem. Can someone help me? Thanks.
pde hyperbolic-equations
edited Dec 20 '18 at 9:54
Harry49
6,96631238
asked Oct 3 '16 at 8:28
Liebster JugendtraumLiebster Jugendtraum
1689
$endgroup$
add a comment |
$begingroup$
Assume that $u$ is an entropy solution to $u_t+partial_x(A(u))=0$, where $A$ is a non-decreasing function on $mathbb{R}$ such that $A(0)=0$ and $$A'(u_ell)>gamma'(t)=frac{A(u_ell)-A(u_r)}{u_ell-u_r}>A'(u_r)$$ (namely the Lax' entropy condition, where $u_ell, u_r$ are of the usual meanings). Prove that the shocks travel from left to right if $A$ is convex.
I got frustrated after spending a long time on it (not sure if one is supposed to consider the shock speed $gamma'(t)$, which is always non-negative since $A$ is non-decreasing). I believe that I missed something here and that there is some very simple way to solve this problem. Can someone help me? Thanks.
pde hyperbolic-equations
edited Dec 20 '18 at 9:54
Harry49
6,96631238
asked Oct 3 '16 at 8:28
Liebster JugendtraumLiebster Jugendtraum
1689
$endgroup$
add a comment |
$begingroup$
Assume that $u$ is an entropy solution to $u_t+partial_x(A(u))=0$, where $A$ is a non-decreasing function on $mathbb{R}$ such that $A(0)=0$ and $$A'(u_ell)>gamma'(t)=frac{A(u_ell)-A(u_r)}{u_ell-u_r}>A'(u_r)$$ (namely the Lax' entropy condition, where $u_ell, u_r$ are of the usual meanings). Prove that the shocks travel from left to right if $A$ is convex.
I got frustrated after spending a long time on it (not sure if one is supposed to consider the shock speed $gamma'(t)$, which is always non-negative since $A$ is non-decreasing). I believe that I missed something here and that there is some very simple way to solve this problem. Can someone help me? Thanks.
pde hyperbolic-equations
edited Dec 20 '18 at 9:54
Harry49
6,96631238
asked Oct 3 '16 at 8:28
Liebster JugendtraumLiebster Jugendtraum
1689
$endgroup$
Assume that $u$ is an entropy solution to $u_t+partial_x(A(u))=0$, where $A$ is a non-decreasing function on $mathbb{R}$ such that $A(0)=0$ and $$A'(u_ell)>gamma'(t)=frac{A(u_ell)-A(u_r)}{u_ell-u_r}>A'(u_r)$$ (namely the Lax' entropy condition, where $u_ell, u_r$ are of the usual meanings). Prove that the shocks travel from left to right if $A$ is convex.
I got frustrated after spending a long time on it (not sure if one is supposed to consider the shock speed $gamma'(t)$, which is always non-negative since $A$ is non-decreasing). I believe that I missed something here and that there is some very simple way to solve this problem. Can someone help me? Thanks.
pde hyperbolic-equations
pde hyperbolic-equations
edited Dec 20 '18 at 9:54
Harry49
6,96631238
asked Oct 3 '16 at 8:28
Liebster JugendtraumLiebster Jugendtraum
1689
edited Dec 20 '18 at 9:54
Harry49
6,96631238
asked Oct 3 '16 at 8:28
Liebster JugendtraumLiebster Jugendtraum
1689
edited Dec 20 '18 at 9:54
Harry49
6,96631238
edited Dec 20 '18 at 9:54
Harry49
6,96631238
edited Dec 20 '18 at 9:54
Harry49
6,96631238
6,96631238
asked Oct 3 '16 at 8:28
Liebster JugendtraumLiebster Jugendtraum
1689
asked Oct 3 '16 at 8:28
Liebster JugendtraumLiebster Jugendtraum
1689
asked Oct 3 '16 at 8:28
Liebster JugendtraumLiebster Jugendtraum
1689
1689
add a comment |
add a comment |
1 Answer
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$begingroup$
The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
$$
gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
$$
so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
$$
A'(u_ell) > gamma'(t) > A'(u_ell)
$$
to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.
answered Dec 20 '18 at 10:22
Harry49Harry49
6,96631238
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
$$
gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
$$
so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
$$
A'(u_ell) > gamma'(t) > A'(u_ell)
$$
to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.
answered Dec 20 '18 at 10:22
Harry49Harry49
6,96631238
$endgroup$
add a comment |
$begingroup$
The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
$$
gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
$$
so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
$$
A'(u_ell) > gamma'(t) > A'(u_ell)
$$
to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.
answered Dec 20 '18 at 10:22
Harry49Harry49
6,96631238
$endgroup$
add a comment |
$begingroup$
The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
$$
gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
$$
so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
$$
A'(u_ell) > gamma'(t) > A'(u_ell)
$$
to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.
answered Dec 20 '18 at 10:22
Harry49Harry49
6,96631238
$endgroup$
The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
$$
gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
$$
so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
$$
A'(u_ell) > gamma'(t) > A'(u_ell)
$$
to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.
answered Dec 20 '18 at 10:22
Harry49Harry49
6,96631238
edited Dec 20 '18 at 13:30
answered Dec 20 '18 at 10:22
Harry49Harry49
6,96631238
answered Dec 20 '18 at 10:22
Harry49Harry49
6,96631238
answered Dec 20 '18 at 10:22
Harry49Harry49
6,96631238
6,96631238
add a comment |
add a comment |
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