An initial- and boundary-value problem for Burgers' equation with no solution
up vote
3
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Prove that there is no solution to the following Cauchy problem:
$$begin{align}
u_t+uu_x& =0&quad x&in(-1,1), tgt0 label{1}tag{1}\
u(x,0)&=x&quad x&in[-1,1] label{2}tag{2}\
u(-1,t)&=-1& quad t&geq0 label{3}tag{3}\
u(1,t)&=1&quad t&geq0 label{4}tag{4}
end{align}$$
My attempt: Using the method of characteristics, I found the classical solution
$$
u=frac{x}{1+t} quad tgt{-1}
$$
which satisfies conditions eqref{1} and eqref{2}, but not eqref{3} and eqref{4}.
But how can I show that there is also no weak solution to this problem?
Thanks in advance.
pde hyperbolic-equations
edited Dec 5 at 13:17
Harry49
5,87421030
asked Dec 4 at 19:53
dmtri
1,3001521
add a comment |
up vote
3
down vote
favorite
Prove that there is no solution to the following Cauchy problem:
$$begin{align}
u_t+uu_x& =0&quad x&in(-1,1), tgt0 label{1}tag{1}\
u(x,0)&=x&quad x&in[-1,1] label{2}tag{2}\
u(-1,t)&=-1& quad t&geq0 label{3}tag{3}\
u(1,t)&=1&quad t&geq0 label{4}tag{4}
end{align}$$
My attempt: Using the method of characteristics, I found the classical solution
$$
u=frac{x}{1+t} quad tgt{-1}
$$
which satisfies conditions eqref{1} and eqref{2}, but not eqref{3} and eqref{4}.
But how can I show that there is also no weak solution to this problem?
Thanks in advance.
pde hyperbolic-equations
edited Dec 5 at 13:17
Harry49
5,87421030
asked Dec 4 at 19:53
dmtri
1,3001521
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Prove that there is no solution to the following Cauchy problem:
$$begin{align}
u_t+uu_x& =0&quad x&in(-1,1), tgt0 label{1}tag{1}\
u(x,0)&=x&quad x&in[-1,1] label{2}tag{2}\
u(-1,t)&=-1& quad t&geq0 label{3}tag{3}\
u(1,t)&=1&quad t&geq0 label{4}tag{4}
end{align}$$
My attempt: Using the method of characteristics, I found the classical solution
$$
u=frac{x}{1+t} quad tgt{-1}
$$
which satisfies conditions eqref{1} and eqref{2}, but not eqref{3} and eqref{4}.
But how can I show that there is also no weak solution to this problem?
Thanks in advance.
pde hyperbolic-equations
edited Dec 5 at 13:17
Harry49
5,87421030
asked Dec 4 at 19:53
dmtri
1,3001521
Prove that there is no solution to the following Cauchy problem:
$$begin{align}
u_t+uu_x& =0&quad x&in(-1,1), tgt0 label{1}tag{1}\
u(x,0)&=x&quad x&in[-1,1] label{2}tag{2}\
u(-1,t)&=-1& quad t&geq0 label{3}tag{3}\
u(1,t)&=1&quad t&geq0 label{4}tag{4}
end{align}$$
My attempt: Using the method of characteristics, I found the classical solution
$$
u=frac{x}{1+t} quad tgt{-1}
$$
which satisfies conditions eqref{1} and eqref{2}, but not eqref{3} and eqref{4}.
But how can I show that there is also no weak solution to this problem?
Thanks in advance.
pde hyperbolic-equations
pde hyperbolic-equations
edited Dec 5 at 13:17
Harry49
5,87421030
asked Dec 4 at 19:53
dmtri
1,3001521
edited Dec 5 at 13:17
Harry49
5,87421030
asked Dec 4 at 19:53
dmtri
1,3001521
edited Dec 5 at 13:17
Harry49
5,87421030
edited Dec 5 at 13:17
Harry49
5,87421030
edited Dec 5 at 13:17
Harry49
5,87421030
5,87421030
asked Dec 4 at 19:53
dmtri
1,3001521
asked Dec 4 at 19:53
dmtri
1,3001521
asked Dec 4 at 19:53
dmtri
1,3001521
1,3001521
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:
The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.
*The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.
**The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.
answered Dec 5 at 13:16
Harry49
5,87421030
Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
– dmtri
Dec 5 at 17:27
one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
– dmtri
Dec 5 at 18:06
I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
– dmtri
2 days ago
1
@dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
– Harry49
2 days ago
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
up vote
1
down vote
accepted
In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:
The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.
*The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.
**The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.
answered Dec 5 at 13:16
Harry49
5,87421030
Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
– dmtri
Dec 5 at 17:27
one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
– dmtri
Dec 5 at 18:06
I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
– dmtri
2 days ago
1
@dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
– Harry49
2 days ago
add a comment |
up vote
1
down vote
accepted
In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:
The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.
*The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.
**The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.
answered Dec 5 at 13:16
Harry49
5,87421030
Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
– dmtri
Dec 5 at 17:27
one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
– dmtri
Dec 5 at 18:06
I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
– dmtri
2 days ago
1
@dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
– Harry49
2 days ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:
The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.
*The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.
**The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.
answered Dec 5 at 13:16
Harry49
5,87421030
In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:
The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.
*The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.
**The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.
answered Dec 5 at 13:16
Harry49
5,87421030
edited 2 days ago
answered Dec 5 at 13:16
Harry49
5,87421030
answered Dec 5 at 13:16
Harry49
5,87421030
answered Dec 5 at 13:16
Harry49
5,87421030
5,87421030
Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
– dmtri
Dec 5 at 17:27
one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
– dmtri
Dec 5 at 18:06
I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
– dmtri
2 days ago
1
@dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
– Harry49
2 days ago
add a comment |
Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
– dmtri
Dec 5 at 17:27
one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
– dmtri
Dec 5 at 18:06
I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
– dmtri
2 days ago
1
@dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
– Harry49
2 days ago
Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
– dmtri
Dec 5 at 17:27
Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
– dmtri
Dec 5 at 17:27
one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
– dmtri
Dec 5 at 18:06
one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
– dmtri
Dec 5 at 18:06
I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
– dmtri
2 days ago
I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
– dmtri
2 days ago
1
1
@dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
– Harry49
2 days ago
@dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
– Harry49
2 days ago
add a comment |
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