Solve inviscid Burgers' equation with shock [closed]
$begingroup$
Consider the initial value problem for the Burgers' equation:
begin{equation}
begin{cases}u_t+uu_x=0\u(x,0)=phi(x)end{cases}
end{equation}
where
$$phi(x)=
begin{cases}
2 & text{if }xleq0\
2-x & text{if }0leq xleq2\
0 & text{if }x geq 2.
end{cases}
$$
Find $u(x,t)$ for all $t > 0$.
Show that the entropy condition is satisfied along the shock curve.
Could someone help me out with this problem.
I know that the method of characteristics gives $u=phi(x-ut)$:
$$
u(x,t) = leftlbracebegin{aligned}
&2 &&text{if}quad xleq 2t\
&tfrac{2-x}{1-t} &&text{if}quad 2tleq x leq 2\
&0 &&text{if}quad xgeq 2
end{aligned}right.
$$
pde hyperbolic-equations
asked Dec 22 '18 at 12:19
F.OF.O
738
$endgroup$
closed as off-topic by user10354138, Paul Frost, Alex Provost, José Carlos Santos, amWhy Dec 22 '18 at 17:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Paul Frost, Alex Provost, José Carlos Santos, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the initial value problem for the Burgers' equation:
begin{equation}
begin{cases}u_t+uu_x=0\u(x,0)=phi(x)end{cases}
end{equation}
where
$$phi(x)=
begin{cases}
2 & text{if }xleq0\
2-x & text{if }0leq xleq2\
0 & text{if }x geq 2.
end{cases}
$$
Find $u(x,t)$ for all $t > 0$.
Show that the entropy condition is satisfied along the shock curve.
Could someone help me out with this problem.
I know that the method of characteristics gives $u=phi(x-ut)$:
$$
u(x,t) = leftlbracebegin{aligned}
&2 &&text{if}quad xleq 2t\
&tfrac{2-x}{1-t} &&text{if}quad 2tleq x leq 2\
&0 &&text{if}quad xgeq 2
end{aligned}right.
$$
pde hyperbolic-equations
asked Dec 22 '18 at 12:19
F.OF.O
738
$endgroup$
closed as off-topic by user10354138, Paul Frost, Alex Provost, José Carlos Santos, amWhy Dec 22 '18 at 17:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Paul Frost, Alex Provost, José Carlos Santos, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the initial value problem for the Burgers' equation:
begin{equation}
begin{cases}u_t+uu_x=0\u(x,0)=phi(x)end{cases}
end{equation}
where
$$phi(x)=
begin{cases}
2 & text{if }xleq0\
2-x & text{if }0leq xleq2\
0 & text{if }x geq 2.
end{cases}
$$
Find $u(x,t)$ for all $t > 0$.
Show that the entropy condition is satisfied along the shock curve.
Could someone help me out with this problem.
I know that the method of characteristics gives $u=phi(x-ut)$:
$$
u(x,t) = leftlbracebegin{aligned}
&2 &&text{if}quad xleq 2t\
&tfrac{2-x}{1-t} &&text{if}quad 2tleq x leq 2\
&0 &&text{if}quad xgeq 2
end{aligned}right.
$$
pde hyperbolic-equations
asked Dec 22 '18 at 12:19
F.OF.O
738
$endgroup$
Consider the initial value problem for the Burgers' equation:
begin{equation}
begin{cases}u_t+uu_x=0\u(x,0)=phi(x)end{cases}
end{equation}
where
$$phi(x)=
begin{cases}
2 & text{if }xleq0\
2-x & text{if }0leq xleq2\
0 & text{if }x geq 2.
end{cases}
$$
Find $u(x,t)$ for all $t > 0$.
Show that the entropy condition is satisfied along the shock curve.
Could someone help me out with this problem.
I know that the method of characteristics gives $u=phi(x-ut)$:
$$
u(x,t) = leftlbracebegin{aligned}
&2 &&text{if}quad xleq 2t\
&tfrac{2-x}{1-t} &&text{if}quad 2tleq x leq 2\
&0 &&text{if}quad xgeq 2
end{aligned}right.
$$
pde hyperbolic-equations
pde hyperbolic-equations
asked Dec 22 '18 at 12:19
F.OF.O
738
asked Dec 22 '18 at 12:19
F.OF.O
738
edited Feb 21 at 14:20
F.O
asked Dec 22 '18 at 12:19
F.OF.O
738
asked Dec 22 '18 at 12:19
F.OF.O
738
asked Dec 22 '18 at 12:19
F.OF.O
738
738
closed as off-topic by user10354138, Paul Frost, Alex Provost, José Carlos Santos, amWhy Dec 22 '18 at 17:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Paul Frost, Alex Provost, José Carlos Santos, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user10354138, Paul Frost, Alex Provost, José Carlos Santos, amWhy Dec 22 '18 at 17:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Paul Frost, Alex Provost, José Carlos Santos, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The solution obtained with the method of characteristics looks correct. One can observe that this solution is only valid for $t< 1$. At $t=1$, all characteristic curves with initial data in $0leq xleq 2$ intersect at the abscissa $x=2$. Thus, the method of characteristics breaks down ($t=1$ is called the breaking time). A shock wave is generated with data $u_L=2$ on the left and $u_R=0$ on the right. According to the Rankine-Hugoniot condition, the shock speed is $s=frac{1}{2}(0+2)=1$. Since $u_L>s>u_R$, the Lax entropy condition is satisfied. Finally, the entropy solution for $t>1$ reads
$$
u(x,t) = leftlbracebegin{aligned}
&2 & &text{if}quad x<1+t\
&0 & &text{if}quad x>1+t
end{aligned}right.
$$
answered Dec 22 '18 at 15:56
Harry49Harry49
7,49431341
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The solution obtained with the method of characteristics looks correct. One can observe that this solution is only valid for $t< 1$. At $t=1$, all characteristic curves with initial data in $0leq xleq 2$ intersect at the abscissa $x=2$. Thus, the method of characteristics breaks down ($t=1$ is called the breaking time). A shock wave is generated with data $u_L=2$ on the left and $u_R=0$ on the right. According to the Rankine-Hugoniot condition, the shock speed is $s=frac{1}{2}(0+2)=1$. Since $u_L>s>u_R$, the Lax entropy condition is satisfied. Finally, the entropy solution for $t>1$ reads
$$
u(x,t) = leftlbracebegin{aligned}
&2 & &text{if}quad x<1+t\
&0 & &text{if}quad x>1+t
end{aligned}right.
$$
answered Dec 22 '18 at 15:56
Harry49Harry49
7,49431341
$endgroup$
add a comment |
$begingroup$
The solution obtained with the method of characteristics looks correct. One can observe that this solution is only valid for $t< 1$. At $t=1$, all characteristic curves with initial data in $0leq xleq 2$ intersect at the abscissa $x=2$. Thus, the method of characteristics breaks down ($t=1$ is called the breaking time). A shock wave is generated with data $u_L=2$ on the left and $u_R=0$ on the right. According to the Rankine-Hugoniot condition, the shock speed is $s=frac{1}{2}(0+2)=1$. Since $u_L>s>u_R$, the Lax entropy condition is satisfied. Finally, the entropy solution for $t>1$ reads
$$
u(x,t) = leftlbracebegin{aligned}
&2 & &text{if}quad x<1+t\
&0 & &text{if}quad x>1+t
end{aligned}right.
$$
answered Dec 22 '18 at 15:56
Harry49Harry49
7,49431341
$endgroup$
add a comment |
$begingroup$
The solution obtained with the method of characteristics looks correct. One can observe that this solution is only valid for $t< 1$. At $t=1$, all characteristic curves with initial data in $0leq xleq 2$ intersect at the abscissa $x=2$. Thus, the method of characteristics breaks down ($t=1$ is called the breaking time). A shock wave is generated with data $u_L=2$ on the left and $u_R=0$ on the right. According to the Rankine-Hugoniot condition, the shock speed is $s=frac{1}{2}(0+2)=1$. Since $u_L>s>u_R$, the Lax entropy condition is satisfied. Finally, the entropy solution for $t>1$ reads
$$
u(x,t) = leftlbracebegin{aligned}
&2 & &text{if}quad x<1+t\
&0 & &text{if}quad x>1+t
end{aligned}right.
$$
answered Dec 22 '18 at 15:56
Harry49Harry49
7,49431341
$endgroup$
The solution obtained with the method of characteristics looks correct. One can observe that this solution is only valid for $t< 1$. At $t=1$, all characteristic curves with initial data in $0leq xleq 2$ intersect at the abscissa $x=2$. Thus, the method of characteristics breaks down ($t=1$ is called the breaking time). A shock wave is generated with data $u_L=2$ on the left and $u_R=0$ on the right. According to the Rankine-Hugoniot condition, the shock speed is $s=frac{1}{2}(0+2)=1$. Since $u_L>s>u_R$, the Lax entropy condition is satisfied. Finally, the entropy solution for $t>1$ reads
$$
u(x,t) = leftlbracebegin{aligned}
&2 & &text{if}quad x<1+t\
&0 & &text{if}quad x>1+t
end{aligned}right.
$$
answered Dec 22 '18 at 15:56
Harry49Harry49
7,49431341
answered Dec 22 '18 at 15:56
Harry49Harry49
7,49431341
answered Dec 22 '18 at 15:56
Harry49Harry49
7,49431341
answered Dec 22 '18 at 15:56
Harry49Harry49
7,49431341
7,49431341
add a comment |
add a comment |
