Find weak solution to Riemann problem for conservation law The 2019 Stack Overflow Developer...
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Find weak solution to Riemann problem for conservation law
The 2019 Stack Overflow Developer Survey Results Are InMethod of characteristics for Burgers' equation with rectangular dataEntropy Solution of the Burgers' EquationBurgers' Equation Shock SolutionsRiemann problem of nonconvex scalar conservation lawsShock-wave solution for PDE $u_t+(u-1)u_x=2$Conservation law $A_t + (A^{3/2})_x = 0$ for flood water waveFinding weak solutions of conservation law $u_t + (u^4)_x = 0$Rarefaction solution to Riemann problem for $x/t=0$Weak Solution to Conservation Lawreversibility scalar conservation lawWhat is the use of the notion of consistency for Riemann solvers?Riemann problem of nonconvex scalar conservation lawsConservation law and entropy condition problemFind the weak solution of the conservation lawThe Rankine-Hugoniot jump conditions for conservation and balance lawsDefinition of weak solution of a PDE that is given in the nondivergent formWeak solution for Burgers' equationShock of Burgers equation $u_t+uu_x=0$ at $t=0$
$begingroup$
Find the weak solution of the following conservation law
$$ u_t + (u^2)_x = 0 $$
with the initial condition
$$ u(x,0) = leftlbrace
begin{aligned}
&u_l & &text{if } x < 0 ,\
&u_r & &text{if } x > 0.
end{aligned}right. $$
Consider both cases $u_l>u_r$ and $u_l<u_r$. Fin the solution at $x=0$ in each case.
Attempt
We have equation: $u_t + 2 u u_x = 0 $ and characteristics are given by $t' = 1 $ and $x' = 2u $ and $u' = 0$ and so $u = const$, $t = s$, $x = 2 u s + r $ so that
$$ x = 2 u(x,0) t + r $$
are characteristicts. so that
$$ x = begin{cases} 2 u_l t + r, ; ; r < 0 \ 2 u_r t + r, ; ; r > 0 end{cases} $$
so thereis a shock formation at $x=0$. Any help in how to continue this problem?
pde hyperbolic-equations
$endgroup$
add a comment |
$begingroup$
Find the weak solution of the following conservation law
$$ u_t + (u^2)_x = 0 $$
with the initial condition
$$ u(x,0) = leftlbrace
begin{aligned}
&u_l & &text{if } x < 0 ,\
&u_r & &text{if } x > 0.
end{aligned}right. $$
Consider both cases $u_l>u_r$ and $u_l<u_r$. Fin the solution at $x=0$ in each case.
Attempt
We have equation: $u_t + 2 u u_x = 0 $ and characteristics are given by $t' = 1 $ and $x' = 2u $ and $u' = 0$ and so $u = const$, $t = s$, $x = 2 u s + r $ so that
$$ x = 2 u(x,0) t + r $$
are characteristicts. so that
$$ x = begin{cases} 2 u_l t + r, ; ; r < 0 \ 2 u_r t + r, ; ; r > 0 end{cases} $$
so thereis a shock formation at $x=0$. Any help in how to continue this problem?
pde hyperbolic-equations
$endgroup$
add a comment |
$begingroup$
Find the weak solution of the following conservation law
$$ u_t + (u^2)_x = 0 $$
with the initial condition
$$ u(x,0) = leftlbrace
begin{aligned}
&u_l & &text{if } x < 0 ,\
&u_r & &text{if } x > 0.
end{aligned}right. $$
Consider both cases $u_l>u_r$ and $u_l<u_r$. Fin the solution at $x=0$ in each case.
Attempt
We have equation: $u_t + 2 u u_x = 0 $ and characteristics are given by $t' = 1 $ and $x' = 2u $ and $u' = 0$ and so $u = const$, $t = s$, $x = 2 u s + r $ so that
$$ x = 2 u(x,0) t + r $$
are characteristicts. so that
$$ x = begin{cases} 2 u_l t + r, ; ; r < 0 \ 2 u_r t + r, ; ; r > 0 end{cases} $$
so thereis a shock formation at $x=0$. Any help in how to continue this problem?
pde hyperbolic-equations
$endgroup$
Find the weak solution of the following conservation law
$$ u_t + (u^2)_x = 0 $$
with the initial condition
$$ u(x,0) = leftlbrace
begin{aligned}
&u_l & &text{if } x < 0 ,\
&u_r & &text{if } x > 0.
end{aligned}right. $$
Consider both cases $u_l>u_r$ and $u_l<u_r$. Fin the solution at $x=0$ in each case.
Attempt
We have equation: $u_t + 2 u u_x = 0 $ and characteristics are given by $t' = 1 $ and $x' = 2u $ and $u' = 0$ and so $u = const$, $t = s$, $x = 2 u s + r $ so that
$$ x = 2 u(x,0) t + r $$
are characteristicts. so that
$$ x = begin{cases} 2 u_l t + r, ; ; r < 0 \ 2 u_r t + r, ; ; r > 0 end{cases} $$
so thereis a shock formation at $x=0$. Any help in how to continue this problem?
pde hyperbolic-equations
pde hyperbolic-equations
edited Mar 25 at 7:52
Harry49
8,73431345
8,73431345
asked Mar 22 at 0:29
JamesJames
2,636425
2,636425
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is very similar to the Riemann problem of the inviscid Burgers' equation (see e.g. (1), (2), (3), (4) and related posts). For this type of problem, weak solutions are not unique. Thus, I guess that the problem statement asks for the entropy solution. I will provide a detailed general answer for the case of conservation laws $u_t + f(u)_x = 0$ with Riemann data $u(x<0,0) = u_l$ and $u(x>0,0) = u_r$, where the flux $f$ is smooth and either convex or concave. If the flux has inflection points, the more general solution is provided here.
In the case of convex or concave flux $f$, there are only two possible types of waves:
shock waves. If the solution is a shock wave with speed $s$,
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x < s t, \
&u_r & &text{if } st < x,
end{aligned}right.
$$
then the speed of shock must satisfy the Rankine-Hugoniot jump condition $s = frac{f(u_l)-f(u_r)}{u_l-u_r}$. Moreover, to be admissible, the shock wave must satisfy the Lax entropy condition $f'(u_l) > s > f'(u_r)$,
where $f'$ denotes the derivative of $f$.rarefaction waves. The are obtained from the self-similarity Ansatz $u(x,t) = v(xi)$ with $xi = x/t$, which leads to the identity $f'(v(xi)) = xi$. Since $f'$ is an increasing function, we can invert the previous equation to find $v(xi) = (f')^{-1}(xi)$. The final solution reads
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x leq f'(u_l) t, \
&(f')^{-1}(x/t) & &text{if } f'(u_l) t leq x leq f'(u_r) t, \
&u_r & &text{if } f'(u_r) t leq x,
end{aligned}right.
$$
where $(f')^{-1}$ denotes the reciprocal function of $f'$. One notes that this solution requires $f'(u_l) leq f'(u_r)$.
In the present case, the flux $f: u mapsto u^2$ is a smooth convex function, so that its derivative $f':umapsto 2u$ is increasing. Shock waves are obtained for $u_l geq u_r$ (cf. Lax entropy condition), and rarefaction waves are obtained for $u_l leq u_r$. In the first case, the shock speed deduced from the Rankine-Hugoniot condition reads $s = u_l + u_r$. The value of the solution at $x=0$ for positive times is $u_r$ if $s < 0$, and $u_l$ otherwise. In the second case, the reciprocal of the derivative is given by $(f')^{-1} : xi mapsto xi/2$. The value of the solution at $x=0$ for positive times is $u_r$ if $u_r < 0$, $u_l$ if $u_l > 0$, and $0$ otherwise (i.e., if $u_l < 0 < u_r $).
$endgroup$
$begingroup$
there is another bounty here: math.stackexchange.com/questions/3157673/…
$endgroup$
– Mikey Spivak
Mar 25 at 8:00
add a comment |
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$begingroup$
This is very similar to the Riemann problem of the inviscid Burgers' equation (see e.g. (1), (2), (3), (4) and related posts). For this type of problem, weak solutions are not unique. Thus, I guess that the problem statement asks for the entropy solution. I will provide a detailed general answer for the case of conservation laws $u_t + f(u)_x = 0$ with Riemann data $u(x<0,0) = u_l$ and $u(x>0,0) = u_r$, where the flux $f$ is smooth and either convex or concave. If the flux has inflection points, the more general solution is provided here.
In the case of convex or concave flux $f$, there are only two possible types of waves:
shock waves. If the solution is a shock wave with speed $s$,
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x < s t, \
&u_r & &text{if } st < x,
end{aligned}right.
$$
then the speed of shock must satisfy the Rankine-Hugoniot jump condition $s = frac{f(u_l)-f(u_r)}{u_l-u_r}$. Moreover, to be admissible, the shock wave must satisfy the Lax entropy condition $f'(u_l) > s > f'(u_r)$,
where $f'$ denotes the derivative of $f$.rarefaction waves. The are obtained from the self-similarity Ansatz $u(x,t) = v(xi)$ with $xi = x/t$, which leads to the identity $f'(v(xi)) = xi$. Since $f'$ is an increasing function, we can invert the previous equation to find $v(xi) = (f')^{-1}(xi)$. The final solution reads
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x leq f'(u_l) t, \
&(f')^{-1}(x/t) & &text{if } f'(u_l) t leq x leq f'(u_r) t, \
&u_r & &text{if } f'(u_r) t leq x,
end{aligned}right.
$$
where $(f')^{-1}$ denotes the reciprocal function of $f'$. One notes that this solution requires $f'(u_l) leq f'(u_r)$.
In the present case, the flux $f: u mapsto u^2$ is a smooth convex function, so that its derivative $f':umapsto 2u$ is increasing. Shock waves are obtained for $u_l geq u_r$ (cf. Lax entropy condition), and rarefaction waves are obtained for $u_l leq u_r$. In the first case, the shock speed deduced from the Rankine-Hugoniot condition reads $s = u_l + u_r$. The value of the solution at $x=0$ for positive times is $u_r$ if $s < 0$, and $u_l$ otherwise. In the second case, the reciprocal of the derivative is given by $(f')^{-1} : xi mapsto xi/2$. The value of the solution at $x=0$ for positive times is $u_r$ if $u_r < 0$, $u_l$ if $u_l > 0$, and $0$ otherwise (i.e., if $u_l < 0 < u_r $).
$endgroup$
$begingroup$
there is another bounty here: math.stackexchange.com/questions/3157673/…
$endgroup$
– Mikey Spivak
Mar 25 at 8:00
add a comment |
$begingroup$
This is very similar to the Riemann problem of the inviscid Burgers' equation (see e.g. (1), (2), (3), (4) and related posts). For this type of problem, weak solutions are not unique. Thus, I guess that the problem statement asks for the entropy solution. I will provide a detailed general answer for the case of conservation laws $u_t + f(u)_x = 0$ with Riemann data $u(x<0,0) = u_l$ and $u(x>0,0) = u_r$, where the flux $f$ is smooth and either convex or concave. If the flux has inflection points, the more general solution is provided here.
In the case of convex or concave flux $f$, there are only two possible types of waves:
shock waves. If the solution is a shock wave with speed $s$,
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x < s t, \
&u_r & &text{if } st < x,
end{aligned}right.
$$
then the speed of shock must satisfy the Rankine-Hugoniot jump condition $s = frac{f(u_l)-f(u_r)}{u_l-u_r}$. Moreover, to be admissible, the shock wave must satisfy the Lax entropy condition $f'(u_l) > s > f'(u_r)$,
where $f'$ denotes the derivative of $f$.rarefaction waves. The are obtained from the self-similarity Ansatz $u(x,t) = v(xi)$ with $xi = x/t$, which leads to the identity $f'(v(xi)) = xi$. Since $f'$ is an increasing function, we can invert the previous equation to find $v(xi) = (f')^{-1}(xi)$. The final solution reads
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x leq f'(u_l) t, \
&(f')^{-1}(x/t) & &text{if } f'(u_l) t leq x leq f'(u_r) t, \
&u_r & &text{if } f'(u_r) t leq x,
end{aligned}right.
$$
where $(f')^{-1}$ denotes the reciprocal function of $f'$. One notes that this solution requires $f'(u_l) leq f'(u_r)$.
In the present case, the flux $f: u mapsto u^2$ is a smooth convex function, so that its derivative $f':umapsto 2u$ is increasing. Shock waves are obtained for $u_l geq u_r$ (cf. Lax entropy condition), and rarefaction waves are obtained for $u_l leq u_r$. In the first case, the shock speed deduced from the Rankine-Hugoniot condition reads $s = u_l + u_r$. The value of the solution at $x=0$ for positive times is $u_r$ if $s < 0$, and $u_l$ otherwise. In the second case, the reciprocal of the derivative is given by $(f')^{-1} : xi mapsto xi/2$. The value of the solution at $x=0$ for positive times is $u_r$ if $u_r < 0$, $u_l$ if $u_l > 0$, and $0$ otherwise (i.e., if $u_l < 0 < u_r $).
$endgroup$
$begingroup$
there is another bounty here: math.stackexchange.com/questions/3157673/…
$endgroup$
– Mikey Spivak
Mar 25 at 8:00
add a comment |
$begingroup$
This is very similar to the Riemann problem of the inviscid Burgers' equation (see e.g. (1), (2), (3), (4) and related posts). For this type of problem, weak solutions are not unique. Thus, I guess that the problem statement asks for the entropy solution. I will provide a detailed general answer for the case of conservation laws $u_t + f(u)_x = 0$ with Riemann data $u(x<0,0) = u_l$ and $u(x>0,0) = u_r$, where the flux $f$ is smooth and either convex or concave. If the flux has inflection points, the more general solution is provided here.
In the case of convex or concave flux $f$, there are only two possible types of waves:
shock waves. If the solution is a shock wave with speed $s$,
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x < s t, \
&u_r & &text{if } st < x,
end{aligned}right.
$$
then the speed of shock must satisfy the Rankine-Hugoniot jump condition $s = frac{f(u_l)-f(u_r)}{u_l-u_r}$. Moreover, to be admissible, the shock wave must satisfy the Lax entropy condition $f'(u_l) > s > f'(u_r)$,
where $f'$ denotes the derivative of $f$.rarefaction waves. The are obtained from the self-similarity Ansatz $u(x,t) = v(xi)$ with $xi = x/t$, which leads to the identity $f'(v(xi)) = xi$. Since $f'$ is an increasing function, we can invert the previous equation to find $v(xi) = (f')^{-1}(xi)$. The final solution reads
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x leq f'(u_l) t, \
&(f')^{-1}(x/t) & &text{if } f'(u_l) t leq x leq f'(u_r) t, \
&u_r & &text{if } f'(u_r) t leq x,
end{aligned}right.
$$
where $(f')^{-1}$ denotes the reciprocal function of $f'$. One notes that this solution requires $f'(u_l) leq f'(u_r)$.
In the present case, the flux $f: u mapsto u^2$ is a smooth convex function, so that its derivative $f':umapsto 2u$ is increasing. Shock waves are obtained for $u_l geq u_r$ (cf. Lax entropy condition), and rarefaction waves are obtained for $u_l leq u_r$. In the first case, the shock speed deduced from the Rankine-Hugoniot condition reads $s = u_l + u_r$. The value of the solution at $x=0$ for positive times is $u_r$ if $s < 0$, and $u_l$ otherwise. In the second case, the reciprocal of the derivative is given by $(f')^{-1} : xi mapsto xi/2$. The value of the solution at $x=0$ for positive times is $u_r$ if $u_r < 0$, $u_l$ if $u_l > 0$, and $0$ otherwise (i.e., if $u_l < 0 < u_r $).
$endgroup$
This is very similar to the Riemann problem of the inviscid Burgers' equation (see e.g. (1), (2), (3), (4) and related posts). For this type of problem, weak solutions are not unique. Thus, I guess that the problem statement asks for the entropy solution. I will provide a detailed general answer for the case of conservation laws $u_t + f(u)_x = 0$ with Riemann data $u(x<0,0) = u_l$ and $u(x>0,0) = u_r$, where the flux $f$ is smooth and either convex or concave. If the flux has inflection points, the more general solution is provided here.
In the case of convex or concave flux $f$, there are only two possible types of waves:
shock waves. If the solution is a shock wave with speed $s$,
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x < s t, \
&u_r & &text{if } st < x,
end{aligned}right.
$$
then the speed of shock must satisfy the Rankine-Hugoniot jump condition $s = frac{f(u_l)-f(u_r)}{u_l-u_r}$. Moreover, to be admissible, the shock wave must satisfy the Lax entropy condition $f'(u_l) > s > f'(u_r)$,
where $f'$ denotes the derivative of $f$.rarefaction waves. The are obtained from the self-similarity Ansatz $u(x,t) = v(xi)$ with $xi = x/t$, which leads to the identity $f'(v(xi)) = xi$. Since $f'$ is an increasing function, we can invert the previous equation to find $v(xi) = (f')^{-1}(xi)$. The final solution reads
$$
u(x,t) = leftlbracebegin{aligned}
&u_l & &text{if } x leq f'(u_l) t, \
&(f')^{-1}(x/t) & &text{if } f'(u_l) t leq x leq f'(u_r) t, \
&u_r & &text{if } f'(u_r) t leq x,
end{aligned}right.
$$
where $(f')^{-1}$ denotes the reciprocal function of $f'$. One notes that this solution requires $f'(u_l) leq f'(u_r)$.
In the present case, the flux $f: u mapsto u^2$ is a smooth convex function, so that its derivative $f':umapsto 2u$ is increasing. Shock waves are obtained for $u_l geq u_r$ (cf. Lax entropy condition), and rarefaction waves are obtained for $u_l leq u_r$. In the first case, the shock speed deduced from the Rankine-Hugoniot condition reads $s = u_l + u_r$. The value of the solution at $x=0$ for positive times is $u_r$ if $s < 0$, and $u_l$ otherwise. In the second case, the reciprocal of the derivative is given by $(f')^{-1} : xi mapsto xi/2$. The value of the solution at $x=0$ for positive times is $u_r$ if $u_r < 0$, $u_l$ if $u_l > 0$, and $0$ otherwise (i.e., if $u_l < 0 < u_r $).
edited Mar 27 at 17:19
answered Mar 25 at 7:50
Harry49Harry49
8,73431345
8,73431345
$begingroup$
there is another bounty here: math.stackexchange.com/questions/3157673/…
$endgroup$
– Mikey Spivak
Mar 25 at 8:00
add a comment |
$begingroup$
there is another bounty here: math.stackexchange.com/questions/3157673/…
$endgroup$
– Mikey Spivak
Mar 25 at 8:00
$begingroup$
there is another bounty here: math.stackexchange.com/questions/3157673/…
$endgroup$
– Mikey Spivak
Mar 25 at 8:00
$begingroup$
there is another bounty here: math.stackexchange.com/questions/3157673/…
$endgroup$
– Mikey Spivak
Mar 25 at 8:00
add a comment |
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