Method of characteristics for Burgers' equation with rectangular data
$begingroup$
Using the method of characteristics, find a solution to Burgers' equation
begin{cases}
u_t+left(frac{u^2}{2} right)_x =0 & text{in }mathbb{R}times(0,infty) \
qquad qquad , , u=g & text{on } mathbb{R} times{t=0}
end{cases}
with the initial conditions
$$g(x)=begin{cases}
0 & text{if }x < 0 \
1 & text{if }0 le x le 1 \
0 & text{if }x > 1
end{cases}$$
First, I realized that the equation $u_t+left(frac{u^2}{2} right)_x =0$ is equivalent to this form: $$u_t+uu_x =0$$
Then should I generally follow the method of solution as outlined in the answer of this page?
Note that this is not a duplicate question of that page. Rather I want to know if that page can be used for my problem, even though my ICs are different.
By the way, the solution printed in my book (PDE Evans, 2nd edition, page 142) is
$$u(x,t) =
begin{cases} 0 & text{if } x < 0 \
frac xt & text{if } 0 < x < t
\ 1 & text{if } t < x < 1 + frac t2
\ 0 & text{if } x > 1 + frac t2 tag{$0 le t le 2$}
end{cases}$$
pde hyperbolic-equations
$endgroup$
add a comment |
$begingroup$
Using the method of characteristics, find a solution to Burgers' equation
begin{cases}
u_t+left(frac{u^2}{2} right)_x =0 & text{in }mathbb{R}times(0,infty) \
qquad qquad , , u=g & text{on } mathbb{R} times{t=0}
end{cases}
with the initial conditions
$$g(x)=begin{cases}
0 & text{if }x < 0 \
1 & text{if }0 le x le 1 \
0 & text{if }x > 1
end{cases}$$
First, I realized that the equation $u_t+left(frac{u^2}{2} right)_x =0$ is equivalent to this form: $$u_t+uu_x =0$$
Then should I generally follow the method of solution as outlined in the answer of this page?
Note that this is not a duplicate question of that page. Rather I want to know if that page can be used for my problem, even though my ICs are different.
By the way, the solution printed in my book (PDE Evans, 2nd edition, page 142) is
$$u(x,t) =
begin{cases} 0 & text{if } x < 0 \
frac xt & text{if } 0 < x < t
\ 1 & text{if } t < x < 1 + frac t2
\ 0 & text{if } x > 1 + frac t2 tag{$0 le t le 2$}
end{cases}$$
pde hyperbolic-equations
$endgroup$
$begingroup$
It should be mentioned that the two forms $(frac{1}{2}u^2)_x$ and $uu_x$ are not equivalent. In general $f(u)$ being differentiable w.r.t. $x$ need not imply that $u$ is differentiable w.r.t. $x$. Consider for instance $u(x) = text{sign}(x)$
$endgroup$
– Hyperplane
Dec 4 '17 at 11:03
add a comment |
$begingroup$
Using the method of characteristics, find a solution to Burgers' equation
begin{cases}
u_t+left(frac{u^2}{2} right)_x =0 & text{in }mathbb{R}times(0,infty) \
qquad qquad , , u=g & text{on } mathbb{R} times{t=0}
end{cases}
with the initial conditions
$$g(x)=begin{cases}
0 & text{if }x < 0 \
1 & text{if }0 le x le 1 \
0 & text{if }x > 1
end{cases}$$
First, I realized that the equation $u_t+left(frac{u^2}{2} right)_x =0$ is equivalent to this form: $$u_t+uu_x =0$$
Then should I generally follow the method of solution as outlined in the answer of this page?
Note that this is not a duplicate question of that page. Rather I want to know if that page can be used for my problem, even though my ICs are different.
By the way, the solution printed in my book (PDE Evans, 2nd edition, page 142) is
$$u(x,t) =
begin{cases} 0 & text{if } x < 0 \
frac xt & text{if } 0 < x < t
\ 1 & text{if } t < x < 1 + frac t2
\ 0 & text{if } x > 1 + frac t2 tag{$0 le t le 2$}
end{cases}$$
pde hyperbolic-equations
$endgroup$
Using the method of characteristics, find a solution to Burgers' equation
begin{cases}
u_t+left(frac{u^2}{2} right)_x =0 & text{in }mathbb{R}times(0,infty) \
qquad qquad , , u=g & text{on } mathbb{R} times{t=0}
end{cases}
with the initial conditions
$$g(x)=begin{cases}
0 & text{if }x < 0 \
1 & text{if }0 le x le 1 \
0 & text{if }x > 1
end{cases}$$
First, I realized that the equation $u_t+left(frac{u^2}{2} right)_x =0$ is equivalent to this form: $$u_t+uu_x =0$$
Then should I generally follow the method of solution as outlined in the answer of this page?
Note that this is not a duplicate question of that page. Rather I want to know if that page can be used for my problem, even though my ICs are different.
By the way, the solution printed in my book (PDE Evans, 2nd edition, page 142) is
$$u(x,t) =
begin{cases} 0 & text{if } x < 0 \
frac xt & text{if } 0 < x < t
\ 1 & text{if } t < x < 1 + frac t2
\ 0 & text{if } x > 1 + frac t2 tag{$0 le t le 2$}
end{cases}$$
pde hyperbolic-equations
pde hyperbolic-equations
edited Dec 10 '18 at 10:05
Harry49
6,18331132
6,18331132
asked Jun 22 '14 at 2:06
CookieCookie
8,725123682
8,725123682
$begingroup$
It should be mentioned that the two forms $(frac{1}{2}u^2)_x$ and $uu_x$ are not equivalent. In general $f(u)$ being differentiable w.r.t. $x$ need not imply that $u$ is differentiable w.r.t. $x$. Consider for instance $u(x) = text{sign}(x)$
$endgroup$
– Hyperplane
Dec 4 '17 at 11:03
add a comment |
$begingroup$
It should be mentioned that the two forms $(frac{1}{2}u^2)_x$ and $uu_x$ are not equivalent. In general $f(u)$ being differentiable w.r.t. $x$ need not imply that $u$ is differentiable w.r.t. $x$. Consider for instance $u(x) = text{sign}(x)$
$endgroup$
– Hyperplane
Dec 4 '17 at 11:03
$begingroup$
It should be mentioned that the two forms $(frac{1}{2}u^2)_x$ and $uu_x$ are not equivalent. In general $f(u)$ being differentiable w.r.t. $x$ need not imply that $u$ is differentiable w.r.t. $x$. Consider for instance $u(x) = text{sign}(x)$
$endgroup$
– Hyperplane
Dec 4 '17 at 11:03
$begingroup$
It should be mentioned that the two forms $(frac{1}{2}u^2)_x$ and $uu_x$ are not equivalent. In general $f(u)$ being differentiable w.r.t. $x$ need not imply that $u$ is differentiable w.r.t. $x$. Consider for instance $u(x) = text{sign}(x)$
$endgroup$
– Hyperplane
Dec 4 '17 at 11:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The variable $u$ is constant along the characteristic curves, which satisfy
begin{aligned}
x'(t) & = u(x(t),t) , ,
\
& = u(x(0),0) , .
end{aligned}
Thus, the latter are straight lines in the $x$-$t$ plane, determined by the initial data. Here, the initial data is piecewise constant, i.e. we solve Riemann problems. As displayed in the figure below,
- characteristics separate in the vicinity of $x=0$, and a rarefaction wave occurs;
- characteristics cross in the vicinity of $x=1$, and a shock-wave occurs.
The rarefaction wave is a continuous self-similar solution, deduced from the self-similarity Ansatz $u (x,t)=v (x/t)$. Indeed,
$$
partial_t
v(x/t) + v(x/t), partial_x v(x/t) = left(v(x/t) - frac{x}{t}right) frac{v'(x/t)}{t} , ,
$$
and thus, $v(x/t) = x/t$.
The shock speed $s$ is given by the Rankine-Hugoniot jump condition:
$$
s = (1+0)/2, .
$$
As long as the rarefaction and the shock don't interact, the solution is
therefore
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq xleq t , , \
& 1 &&text{if }; t leq x < 1+ t/2 , ,\
& 0 &&text{if }; 1+t/2 < x , ,
end{aligned}
right.
$$
valid for times $t<t^*$ such that $t^* = 1 + t^*/2 = 2$. At the time $t^*$, both waves interact. The new shock speed is determined from the Rankine-Hugoniot condition
$$
x'(t) = (x(t)/t+0)/2 , ,
$$
with initial shock speed $x'(t^*) = s$.
Hence, the solution for $tgeq t^*$ is
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq x< sqrt{2t} , , \
& 0 &&text{if }; sqrt{2t} < x , .
end{aligned}right.
$$
$endgroup$
add a comment |
$begingroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$
$u(x,0)=begin{cases}0&text{if}~x<0\1&text{if}~0le xle1\0&text{if}~x>1end{cases}$ :
$therefore u=begin{cases}0&text{if}~x-ut<0\1&text{if}~0le x-utle1\0&text{if}~x-ut>1end{cases}=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1end{cases}$
Hence $u(x,t)=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1\c&text{otherwise}end{cases}$
$endgroup$
$begingroup$
How were you able to simply remove the $u$? e.g. $0 le x-ut le 1 rightarrow 0 le x-t le 1$...
$endgroup$
– Cookie
Jun 22 '14 at 5:58
1
$begingroup$
@legâteauaufromage I suppose it is due to the initial condition $u(x,0) = 1$ when $0 ≤ x ≤ 1$
$endgroup$
– BRabbit27
Oct 19 '14 at 13:14
$begingroup$
@Cookie Note that the method of characteristics (which gives $u=F(x−ut)$) can only be used as long as characteristics don't cross, which occurs already at $t=0$ here. See my answer for details.
$endgroup$
– Harry49
Dec 2 '17 at 15:24
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f843118%2fmethod-of-characteristics-for-burgers-equation-with-rectangular-data%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The variable $u$ is constant along the characteristic curves, which satisfy
begin{aligned}
x'(t) & = u(x(t),t) , ,
\
& = u(x(0),0) , .
end{aligned}
Thus, the latter are straight lines in the $x$-$t$ plane, determined by the initial data. Here, the initial data is piecewise constant, i.e. we solve Riemann problems. As displayed in the figure below,
- characteristics separate in the vicinity of $x=0$, and a rarefaction wave occurs;
- characteristics cross in the vicinity of $x=1$, and a shock-wave occurs.
The rarefaction wave is a continuous self-similar solution, deduced from the self-similarity Ansatz $u (x,t)=v (x/t)$. Indeed,
$$
partial_t
v(x/t) + v(x/t), partial_x v(x/t) = left(v(x/t) - frac{x}{t}right) frac{v'(x/t)}{t} , ,
$$
and thus, $v(x/t) = x/t$.
The shock speed $s$ is given by the Rankine-Hugoniot jump condition:
$$
s = (1+0)/2, .
$$
As long as the rarefaction and the shock don't interact, the solution is
therefore
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq xleq t , , \
& 1 &&text{if }; t leq x < 1+ t/2 , ,\
& 0 &&text{if }; 1+t/2 < x , ,
end{aligned}
right.
$$
valid for times $t<t^*$ such that $t^* = 1 + t^*/2 = 2$. At the time $t^*$, both waves interact. The new shock speed is determined from the Rankine-Hugoniot condition
$$
x'(t) = (x(t)/t+0)/2 , ,
$$
with initial shock speed $x'(t^*) = s$.
Hence, the solution for $tgeq t^*$ is
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq x< sqrt{2t} , , \
& 0 &&text{if }; sqrt{2t} < x , .
end{aligned}right.
$$
$endgroup$
add a comment |
$begingroup$
The variable $u$ is constant along the characteristic curves, which satisfy
begin{aligned}
x'(t) & = u(x(t),t) , ,
\
& = u(x(0),0) , .
end{aligned}
Thus, the latter are straight lines in the $x$-$t$ plane, determined by the initial data. Here, the initial data is piecewise constant, i.e. we solve Riemann problems. As displayed in the figure below,
- characteristics separate in the vicinity of $x=0$, and a rarefaction wave occurs;
- characteristics cross in the vicinity of $x=1$, and a shock-wave occurs.
The rarefaction wave is a continuous self-similar solution, deduced from the self-similarity Ansatz $u (x,t)=v (x/t)$. Indeed,
$$
partial_t
v(x/t) + v(x/t), partial_x v(x/t) = left(v(x/t) - frac{x}{t}right) frac{v'(x/t)}{t} , ,
$$
and thus, $v(x/t) = x/t$.
The shock speed $s$ is given by the Rankine-Hugoniot jump condition:
$$
s = (1+0)/2, .
$$
As long as the rarefaction and the shock don't interact, the solution is
therefore
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq xleq t , , \
& 1 &&text{if }; t leq x < 1+ t/2 , ,\
& 0 &&text{if }; 1+t/2 < x , ,
end{aligned}
right.
$$
valid for times $t<t^*$ such that $t^* = 1 + t^*/2 = 2$. At the time $t^*$, both waves interact. The new shock speed is determined from the Rankine-Hugoniot condition
$$
x'(t) = (x(t)/t+0)/2 , ,
$$
with initial shock speed $x'(t^*) = s$.
Hence, the solution for $tgeq t^*$ is
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq x< sqrt{2t} , , \
& 0 &&text{if }; sqrt{2t} < x , .
end{aligned}right.
$$
$endgroup$
add a comment |
$begingroup$
The variable $u$ is constant along the characteristic curves, which satisfy
begin{aligned}
x'(t) & = u(x(t),t) , ,
\
& = u(x(0),0) , .
end{aligned}
Thus, the latter are straight lines in the $x$-$t$ plane, determined by the initial data. Here, the initial data is piecewise constant, i.e. we solve Riemann problems. As displayed in the figure below,
- characteristics separate in the vicinity of $x=0$, and a rarefaction wave occurs;
- characteristics cross in the vicinity of $x=1$, and a shock-wave occurs.
The rarefaction wave is a continuous self-similar solution, deduced from the self-similarity Ansatz $u (x,t)=v (x/t)$. Indeed,
$$
partial_t
v(x/t) + v(x/t), partial_x v(x/t) = left(v(x/t) - frac{x}{t}right) frac{v'(x/t)}{t} , ,
$$
and thus, $v(x/t) = x/t$.
The shock speed $s$ is given by the Rankine-Hugoniot jump condition:
$$
s = (1+0)/2, .
$$
As long as the rarefaction and the shock don't interact, the solution is
therefore
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq xleq t , , \
& 1 &&text{if }; t leq x < 1+ t/2 , ,\
& 0 &&text{if }; 1+t/2 < x , ,
end{aligned}
right.
$$
valid for times $t<t^*$ such that $t^* = 1 + t^*/2 = 2$. At the time $t^*$, both waves interact. The new shock speed is determined from the Rankine-Hugoniot condition
$$
x'(t) = (x(t)/t+0)/2 , ,
$$
with initial shock speed $x'(t^*) = s$.
Hence, the solution for $tgeq t^*$ is
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq x< sqrt{2t} , , \
& 0 &&text{if }; sqrt{2t} < x , .
end{aligned}right.
$$
$endgroup$
The variable $u$ is constant along the characteristic curves, which satisfy
begin{aligned}
x'(t) & = u(x(t),t) , ,
\
& = u(x(0),0) , .
end{aligned}
Thus, the latter are straight lines in the $x$-$t$ plane, determined by the initial data. Here, the initial data is piecewise constant, i.e. we solve Riemann problems. As displayed in the figure below,
- characteristics separate in the vicinity of $x=0$, and a rarefaction wave occurs;
- characteristics cross in the vicinity of $x=1$, and a shock-wave occurs.
The rarefaction wave is a continuous self-similar solution, deduced from the self-similarity Ansatz $u (x,t)=v (x/t)$. Indeed,
$$
partial_t
v(x/t) + v(x/t), partial_x v(x/t) = left(v(x/t) - frac{x}{t}right) frac{v'(x/t)}{t} , ,
$$
and thus, $v(x/t) = x/t$.
The shock speed $s$ is given by the Rankine-Hugoniot jump condition:
$$
s = (1+0)/2, .
$$
As long as the rarefaction and the shock don't interact, the solution is
therefore
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq xleq t , , \
& 1 &&text{if }; t leq x < 1+ t/2 , ,\
& 0 &&text{if }; 1+t/2 < x , ,
end{aligned}
right.
$$
valid for times $t<t^*$ such that $t^* = 1 + t^*/2 = 2$. At the time $t^*$, both waves interact. The new shock speed is determined from the Rankine-Hugoniot condition
$$
x'(t) = (x(t)/t+0)/2 , ,
$$
with initial shock speed $x'(t^*) = s$.
Hence, the solution for $tgeq t^*$ is
$$
u(x,t) =
leftlbrace
begin{aligned}
& 0 &&text{if }; xleq 0 , , \
& x/t &&text{if }; 0leq x< sqrt{2t} , , \
& 0 &&text{if }; sqrt{2t} < x , .
end{aligned}right.
$$
edited Oct 2 '17 at 12:13
answered Oct 1 '17 at 12:58
Harry49Harry49
6,18331132
6,18331132
add a comment |
add a comment |
$begingroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$
$u(x,0)=begin{cases}0&text{if}~x<0\1&text{if}~0le xle1\0&text{if}~x>1end{cases}$ :
$therefore u=begin{cases}0&text{if}~x-ut<0\1&text{if}~0le x-utle1\0&text{if}~x-ut>1end{cases}=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1end{cases}$
Hence $u(x,t)=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1\c&text{otherwise}end{cases}$
$endgroup$
$begingroup$
How were you able to simply remove the $u$? e.g. $0 le x-ut le 1 rightarrow 0 le x-t le 1$...
$endgroup$
– Cookie
Jun 22 '14 at 5:58
1
$begingroup$
@legâteauaufromage I suppose it is due to the initial condition $u(x,0) = 1$ when $0 ≤ x ≤ 1$
$endgroup$
– BRabbit27
Oct 19 '14 at 13:14
$begingroup$
@Cookie Note that the method of characteristics (which gives $u=F(x−ut)$) can only be used as long as characteristics don't cross, which occurs already at $t=0$ here. See my answer for details.
$endgroup$
– Harry49
Dec 2 '17 at 15:24
add a comment |
$begingroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$
$u(x,0)=begin{cases}0&text{if}~x<0\1&text{if}~0le xle1\0&text{if}~x>1end{cases}$ :
$therefore u=begin{cases}0&text{if}~x-ut<0\1&text{if}~0le x-utle1\0&text{if}~x-ut>1end{cases}=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1end{cases}$
Hence $u(x,t)=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1\c&text{otherwise}end{cases}$
$endgroup$
$begingroup$
How were you able to simply remove the $u$? e.g. $0 le x-ut le 1 rightarrow 0 le x-t le 1$...
$endgroup$
– Cookie
Jun 22 '14 at 5:58
1
$begingroup$
@legâteauaufromage I suppose it is due to the initial condition $u(x,0) = 1$ when $0 ≤ x ≤ 1$
$endgroup$
– BRabbit27
Oct 19 '14 at 13:14
$begingroup$
@Cookie Note that the method of characteristics (which gives $u=F(x−ut)$) can only be used as long as characteristics don't cross, which occurs already at $t=0$ here. See my answer for details.
$endgroup$
– Harry49
Dec 2 '17 at 15:24
add a comment |
$begingroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$
$u(x,0)=begin{cases}0&text{if}~x<0\1&text{if}~0le xle1\0&text{if}~x>1end{cases}$ :
$therefore u=begin{cases}0&text{if}~x-ut<0\1&text{if}~0le x-utle1\0&text{if}~x-ut>1end{cases}=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1end{cases}$
Hence $u(x,t)=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1\c&text{otherwise}end{cases}$
$endgroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$
$u(x,0)=begin{cases}0&text{if}~x<0\1&text{if}~0le xle1\0&text{if}~x>1end{cases}$ :
$therefore u=begin{cases}0&text{if}~x-ut<0\1&text{if}~0le x-utle1\0&text{if}~x-ut>1end{cases}=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1end{cases}$
Hence $u(x,t)=begin{cases}0&text{if}~x<0\1&text{if}~0le x-tle1\0&text{if}~x>1\c&text{otherwise}end{cases}$
answered Jun 22 '14 at 5:00
doraemonpauldoraemonpaul
12.6k31660
12.6k31660
$begingroup$
How were you able to simply remove the $u$? e.g. $0 le x-ut le 1 rightarrow 0 le x-t le 1$...
$endgroup$
– Cookie
Jun 22 '14 at 5:58
1
$begingroup$
@legâteauaufromage I suppose it is due to the initial condition $u(x,0) = 1$ when $0 ≤ x ≤ 1$
$endgroup$
– BRabbit27
Oct 19 '14 at 13:14
$begingroup$
@Cookie Note that the method of characteristics (which gives $u=F(x−ut)$) can only be used as long as characteristics don't cross, which occurs already at $t=0$ here. See my answer for details.
$endgroup$
– Harry49
Dec 2 '17 at 15:24
add a comment |
$begingroup$
How were you able to simply remove the $u$? e.g. $0 le x-ut le 1 rightarrow 0 le x-t le 1$...
$endgroup$
– Cookie
Jun 22 '14 at 5:58
1
$begingroup$
@legâteauaufromage I suppose it is due to the initial condition $u(x,0) = 1$ when $0 ≤ x ≤ 1$
$endgroup$
– BRabbit27
Oct 19 '14 at 13:14
$begingroup$
@Cookie Note that the method of characteristics (which gives $u=F(x−ut)$) can only be used as long as characteristics don't cross, which occurs already at $t=0$ here. See my answer for details.
$endgroup$
– Harry49
Dec 2 '17 at 15:24
$begingroup$
How were you able to simply remove the $u$? e.g. $0 le x-ut le 1 rightarrow 0 le x-t le 1$...
$endgroup$
– Cookie
Jun 22 '14 at 5:58
$begingroup$
How were you able to simply remove the $u$? e.g. $0 le x-ut le 1 rightarrow 0 le x-t le 1$...
$endgroup$
– Cookie
Jun 22 '14 at 5:58
1
1
$begingroup$
@legâteauaufromage I suppose it is due to the initial condition $u(x,0) = 1$ when $0 ≤ x ≤ 1$
$endgroup$
– BRabbit27
Oct 19 '14 at 13:14
$begingroup$
@legâteauaufromage I suppose it is due to the initial condition $u(x,0) = 1$ when $0 ≤ x ≤ 1$
$endgroup$
– BRabbit27
Oct 19 '14 at 13:14
$begingroup$
@Cookie Note that the method of characteristics (which gives $u=F(x−ut)$) can only be used as long as characteristics don't cross, which occurs already at $t=0$ here. See my answer for details.
$endgroup$
– Harry49
Dec 2 '17 at 15:24
$begingroup$
@Cookie Note that the method of characteristics (which gives $u=F(x−ut)$) can only be used as long as characteristics don't cross, which occurs already at $t=0$ here. See my answer for details.
$endgroup$
– Harry49
Dec 2 '17 at 15:24
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f843118%2fmethod-of-characteristics-for-burgers-equation-with-rectangular-data%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
It should be mentioned that the two forms $(frac{1}{2}u^2)_x$ and $uu_x$ are not equivalent. In general $f(u)$ being differentiable w.r.t. $x$ need not imply that $u$ is differentiable w.r.t. $x$. Consider for instance $u(x) = text{sign}(x)$
$endgroup$
– Hyperplane
Dec 4 '17 at 11:03