Mixing
Solve initial value problem $u_t+x^2u_x=0$ by method of characteristics
$begingroup$
Related Question:
- Using method of characteristics to solve transport equation
Solve the initial value problem using the method of characteristics
$$u_t+x^2u_x=0 ~~~~-infty<x<infty,~~~ t> 0$$
$$
u(x,0) = left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
Here is what I have done so far.
We are given that $u_t+x^2u_x=0$. Therefore, we know that this is a type of transport equation in which we can write the general form as $A(x,t)u_t+B(x,t)u_x=0$. Hence, we have
$$frac{dt}{A}=frac{dx}{B}$$
as our characteristic equations. In our example,
$$frac{dt}{1}=frac{dx}{x^2}$$
So,
$$ frac{dx}{x^2}=dt~~Rightarrow~~~ -frac{1}{x} = t+C ~~Rightarrow~~~ C=-t-frac{1}{x}$$
Therefore, $u(x,t)$ is constant along the curves where $C=-t-frac{1}{x}$. Thus, the general solution is
$$u(x,t)=fBig(-t-frac{1}{x}Big)$$
where f is an arbitrary differentiable function. We now need to apply our initial data. We are given that
$$
u(x,0) = left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
Therefore,
$$u(x,0)=fBig(-frac{1}{x}Big)=left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
However, I'm not sure what to do with the piecewise-defined initial conditions after this step. We would have $f(-frac{1}{x})=0$ or $f(-frac{1}{x})=1$ which would indicate that f is a constant function.
pde
asked Jan 13 at 18:20
Axion004Axion004
374313
$endgroup$
add a comment |
$begingroup$
Related Question:
- Using method of characteristics to solve transport equation
Solve the initial value problem using the method of characteristics
$$u_t+x^2u_x=0 ~~~~-infty<x<infty,~~~ t> 0$$
$$
u(x,0) = left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
Here is what I have done so far.
We are given that $u_t+x^2u_x=0$. Therefore, we know that this is a type of transport equation in which we can write the general form as $A(x,t)u_t+B(x,t)u_x=0$. Hence, we have
$$frac{dt}{A}=frac{dx}{B}$$
as our characteristic equations. In our example,
$$frac{dt}{1}=frac{dx}{x^2}$$
So,
$$ frac{dx}{x^2}=dt~~Rightarrow~~~ -frac{1}{x} = t+C ~~Rightarrow~~~ C=-t-frac{1}{x}$$
Therefore, $u(x,t)$ is constant along the curves where $C=-t-frac{1}{x}$. Thus, the general solution is
$$u(x,t)=fBig(-t-frac{1}{x}Big)$$
where f is an arbitrary differentiable function. We now need to apply our initial data. We are given that
$$
u(x,0) = left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
Therefore,
$$u(x,0)=fBig(-frac{1}{x}Big)=left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
However, I'm not sure what to do with the piecewise-defined initial conditions after this step. We would have $f(-frac{1}{x})=0$ or $f(-frac{1}{x})=1$ which would indicate that f is a constant function.
pde
asked Jan 13 at 18:20
Axion004Axion004
374313
$endgroup$
$begingroup$
You have a shock forming due to your initial condition. You'll need to do a fair bit more work to get the solution. There are plenty of problems on the site providing a step by step solution on how to solve these types of problems.
$endgroup$
– Mattos
Jan 14 at 5:08
add a comment |
$begingroup$
Related Question:
- Using method of characteristics to solve transport equation
Solve the initial value problem using the method of characteristics
$$u_t+x^2u_x=0 ~~~~-infty<x<infty,~~~ t> 0$$
$$
u(x,0) = left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
Here is what I have done so far.
We are given that $u_t+x^2u_x=0$. Therefore, we know that this is a type of transport equation in which we can write the general form as $A(x,t)u_t+B(x,t)u_x=0$. Hence, we have
$$frac{dt}{A}=frac{dx}{B}$$
as our characteristic equations. In our example,
$$frac{dt}{1}=frac{dx}{x^2}$$
So,
$$ frac{dx}{x^2}=dt~~Rightarrow~~~ -frac{1}{x} = t+C ~~Rightarrow~~~ C=-t-frac{1}{x}$$
Therefore, $u(x,t)$ is constant along the curves where $C=-t-frac{1}{x}$. Thus, the general solution is
$$u(x,t)=fBig(-t-frac{1}{x}Big)$$
where f is an arbitrary differentiable function. We now need to apply our initial data. We are given that
$$
u(x,0) = left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
Therefore,
$$u(x,0)=fBig(-frac{1}{x}Big)=left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
However, I'm not sure what to do with the piecewise-defined initial conditions after this step. We would have $f(-frac{1}{x})=0$ or $f(-frac{1}{x})=1$ which would indicate that f is a constant function.
pde
asked Jan 13 at 18:20
Axion004Axion004
374313
$endgroup$
Related Question:
- Using method of characteristics to solve transport equation
Solve the initial value problem using the method of characteristics
$$u_t+x^2u_x=0 ~~~~-infty<x<infty,~~~ t> 0$$
$$
u(x,0) = left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
Here is what I have done so far.
We are given that $u_t+x^2u_x=0$. Therefore, we know that this is a type of transport equation in which we can write the general form as $A(x,t)u_t+B(x,t)u_x=0$. Hence, we have
$$frac{dt}{A}=frac{dx}{B}$$
as our characteristic equations. In our example,
$$frac{dt}{1}=frac{dx}{x^2}$$
So,
$$ frac{dx}{x^2}=dt~~Rightarrow~~~ -frac{1}{x} = t+C ~~Rightarrow~~~ C=-t-frac{1}{x}$$
Therefore, $u(x,t)$ is constant along the curves where $C=-t-frac{1}{x}$. Thus, the general solution is
$$u(x,t)=fBig(-t-frac{1}{x}Big)$$
where f is an arbitrary differentiable function. We now need to apply our initial data. We are given that
$$
u(x,0) = left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
Therefore,
$$u(x,0)=fBig(-frac{1}{x}Big)=left{
begin{array}{ll}
1, & quad 0leq x leq 1 \
0, & quad otherwise
end{array}
right.
$$
However, I'm not sure what to do with the piecewise-defined initial conditions after this step. We would have $f(-frac{1}{x})=0$ or $f(-frac{1}{x})=1$ which would indicate that f is a constant function.
pde
pde
asked Jan 13 at 18:20
Axion004Axion004
374313
asked Jan 13 at 18:20
Axion004Axion004
374313
asked Jan 13 at 18:20
Axion004Axion004
374313
asked Jan 13 at 18:20
Axion004Axion004
374313
asked Jan 13 at 18:20
Axion004Axion004
374313
374313
$begingroup$
You have a shock forming due to your initial condition. You'll need to do a fair bit more work to get the solution. There are plenty of problems on the site providing a step by step solution on how to solve these types of problems.
$endgroup$
– Mattos
Jan 14 at 5:08
add a comment |
$begingroup$
You have a shock forming due to your initial condition. You'll need to do a fair bit more work to get the solution. There are plenty of problems on the site providing a step by step solution on how to solve these types of problems.
$endgroup$
– Mattos
Jan 14 at 5:08
$begingroup$
You have a shock forming due to your initial condition. You'll need to do a fair bit more work to get the solution. There are plenty of problems on the site providing a step by step solution on how to solve these types of problems.
$endgroup$
– Mattos
Jan 14 at 5:08
$begingroup$
You have a shock forming due to your initial condition. You'll need to do a fair bit more work to get the solution. There are plenty of problems on the site providing a step by step solution on how to solve these types of problems.
$endgroup$
– Mattos
Jan 14 at 5:08
add a comment |
0
active
oldest
votes
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%2f3072351%2fsolve-initial-value-problem-u-tx2u-x-0-by-method-of-characteristics%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3072351%2fsolve-initial-value-problem-u-tx2u-x-0-by-method-of-characteristics%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
$begingroup$
You have a shock forming due to your initial condition. You'll need to do a fair bit more work to get the solution. There are plenty of problems on the site providing a step by step solution on how to solve these types of problems.
$endgroup$
– Mattos
Jan 14 at 5:08