Question about the Dirac-delta function The 2019 Stack Overflow Developer Survey Results Are InDiffusion equation in an infinite mediumDirac Delta Function as Initial condition for 1D Diffusion PDE: ONE or TWO equations(conditions)?Solving Poisson's equation for $varrho(mathbfr) = sigma cosleft(frac2 piL xright) , delta(y)$Inverse fourier transform for the heat equationA specific partial differential equation using Fourier TransformFourier Transform Dirac DeltaUsing the Dirac Delta function in PDE'sRepresentation formula for harmonic function and Dirac Delta functionPartial Differential Equation using Fourier Sine TransformQuestions about the heat equation
Is "plugging out" electronic devices an American expression?
What is the accessibility of a package's `Private` context variables?
How to type this arrow in math mode?
What does Linus Torvalds mean when he says that Git "never ever" tracks a file?
Output the Arecibo Message
What is the meaning of Triage in Cybersec world?
Is bread bad for ducks?
What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?
What does ひと匙 mean in this manga and has it been used colloquially?
How are circuits which use complex ICs normally simulated?
Is this app Icon Browser Safe/Legit?
Did Scotland spend $250,000 for the slogan "Welcome to Scotland"?
Should I use my personal e-mail address, or my workplace one, when registering to external websites for work purposes?
What is the most effective way of iterating a std::vector and why?
Which Sci-Fi work first showed weapon of galactic-scale mass destruction?
Return to UK after having been refused entry years ago
The difference between dialogue marks
Why is the maximum length of OpenWrt’s root password 8 characters?
How to notate time signature switching consistently every measure
What tool would a Roman-age civilization have for the breaking of silver and other metals into dust?
Can someone be penalized for an "unlawful" act if no penalty is specified?
One word riddle: Vowel in the middle
Can one be advised by a professor who is very far away?
Can you compress metal and what would be the consequences?
Question about the Dirac-delta function
The 2019 Stack Overflow Developer Survey Results Are InDiffusion equation in an infinite mediumDirac Delta Function as Initial condition for 1D Diffusion PDE: ONE or TWO equations(conditions)?Solving Poisson's equation for $varrho(mathbfr) = sigma cosleft(frac2 piL xright) , delta(y)$Inverse fourier transform for the heat equationA specific partial differential equation using Fourier TransformFourier Transform Dirac DeltaUsing the Dirac Delta function in PDE'sRepresentation formula for harmonic function and Dirac Delta functionPartial Differential Equation using Fourier Sine TransformQuestions about the heat equation
$begingroup$
The equation$$ u_t = frac14 u_xx, ; ; ; (x,t) in Omega = (- infty,
infty) times (0, infty) $$
can be solved using Fourier transform and the solution is
$$ u(x,t) = frac1sqrt pi t intlimits_-infty^infty
g(y) mathrme^ - (x-y)^2/ t mathrmd y $$
What is the solution for initial condition $u(x,0) = delta(x+1)$?
Where $delta(x)$ is the Dirac delta function
thought:
We know $g(x) = u(x,0)$ is the initial condition. I know $delta(x+1) = infty$ when $x=-1$ and zero otherwise. So, if I plug this into my solution I get something of the form
$$ u(x,t) = frac1sqrtpi t * (infty ) * mathrme^ - (x+1)^2/ t $$
How to simplify such expression?
pde
edited Mar 23 at 18:41
Bernard
124k741117
asked Mar 23 at 18:39
JamesJames
2,636425
$endgroup$
add a comment |
$begingroup$
The equation$$ u_t = frac14 u_xx, ; ; ; (x,t) in Omega = (- infty,
infty) times (0, infty) $$
can be solved using Fourier transform and the solution is
$$ u(x,t) = frac1sqrt pi t intlimits_-infty^infty
g(y) mathrme^ - (x-y)^2/ t mathrmd y $$
What is the solution for initial condition $u(x,0) = delta(x+1)$?
Where $delta(x)$ is the Dirac delta function
thought:
We know $g(x) = u(x,0)$ is the initial condition. I know $delta(x+1) = infty$ when $x=-1$ and zero otherwise. So, if I plug this into my solution I get something of the form
$$ u(x,t) = frac1sqrtpi t * (infty ) * mathrme^ - (x+1)^2/ t $$
How to simplify such expression?
pde
edited Mar 23 at 18:41
Bernard
124k741117
asked Mar 23 at 18:39
JamesJames
2,636425
$endgroup$
1
$begingroup$
Forget the informal "$=infty$ when $x=0$ and $=0$ otherwise" description of the delta-function. The (or a) proper definition of the delta function is in how it acts under an integral, namely $int_-infty^infty f(x)delta(x-a)rm dx equiv f(a)$. This is all you need here.
$endgroup$
– Winther
Mar 23 at 19:25
add a comment |
$begingroup$
The equation$$ u_t = frac14 u_xx, ; ; ; (x,t) in Omega = (- infty,
infty) times (0, infty) $$
can be solved using Fourier transform and the solution is
$$ u(x,t) = frac1sqrt pi t intlimits_-infty^infty
g(y) mathrme^ - (x-y)^2/ t mathrmd y $$
What is the solution for initial condition $u(x,0) = delta(x+1)$?
Where $delta(x)$ is the Dirac delta function
thought:
We know $g(x) = u(x,0)$ is the initial condition. I know $delta(x+1) = infty$ when $x=-1$ and zero otherwise. So, if I plug this into my solution I get something of the form
$$ u(x,t) = frac1sqrtpi t * (infty ) * mathrme^ - (x+1)^2/ t $$
How to simplify such expression?
pde
edited Mar 23 at 18:41
Bernard
124k741117
asked Mar 23 at 18:39
JamesJames
2,636425
$endgroup$
The equation$$ u_t = frac14 u_xx, ; ; ; (x,t) in Omega = (- infty,
infty) times (0, infty) $$
can be solved using Fourier transform and the solution is
$$ u(x,t) = frac1sqrt pi t intlimits_-infty^infty
g(y) mathrme^ - (x-y)^2/ t mathrmd y $$
What is the solution for initial condition $u(x,0) = delta(x+1)$?
Where $delta(x)$ is the Dirac delta function
thought:
We know $g(x) = u(x,0)$ is the initial condition. I know $delta(x+1) = infty$ when $x=-1$ and zero otherwise. So, if I plug this into my solution I get something of the form
$$ u(x,t) = frac1sqrtpi t * (infty ) * mathrme^ - (x+1)^2/ t $$
How to simplify such expression?
pde
pde
edited Mar 23 at 18:41
Bernard
124k741117
asked Mar 23 at 18:39
JamesJames
2,636425
edited Mar 23 at 18:41
Bernard
124k741117
asked Mar 23 at 18:39
JamesJames
2,636425
edited Mar 23 at 18:41
Bernard
124k741117
edited Mar 23 at 18:41
Bernard
124k741117
edited Mar 23 at 18:41
Bernard
124k741117
124k741117
asked Mar 23 at 18:39
JamesJames
2,636425
asked Mar 23 at 18:39
JamesJames
2,636425
asked Mar 23 at 18:39
JamesJames
2,636425
2,636425
1
$begingroup$
Forget the informal "$=infty$ when $x=0$ and $=0$ otherwise" description of the delta-function. The (or a) proper definition of the delta function is in how it acts under an integral, namely $int_-infty^infty f(x)delta(x-a)rm dx equiv f(a)$. This is all you need here.
$endgroup$
– Winther
Mar 23 at 19:25
add a comment |
1
$begingroup$
Forget the informal "$=infty$ when $x=0$ and $=0$ otherwise" description of the delta-function. The (or a) proper definition of the delta function is in how it acts under an integral, namely $int_-infty^infty f(x)delta(x-a)rm dx equiv f(a)$. This is all you need here.
$endgroup$
– Winther
Mar 23 at 19:25
1
1
$begingroup$
Forget the informal "$=infty$ when $x=0$ and $=0$ otherwise" description of the delta-function. The (or a) proper definition of the delta function is in how it acts under an integral, namely $int_-infty^infty f(x)delta(x-a)rm dx equiv f(a)$. This is all you need here.
$endgroup$
– Winther
Mar 23 at 19:25
$begingroup$
Forget the informal "$=infty$ when $x=0$ and $=0$ otherwise" description of the delta-function. The (or a) proper definition of the delta function is in how it acts under an integral, namely $int_-infty^infty f(x)delta(x-a)rm dx equiv f(a)$. This is all you need here.
$endgroup$
– Winther
Mar 23 at 19:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In your statement, I think you mean that the function $g(y) = u(y,0)$ contains the initial conditions. Then to answer your question, you need to perform the integral with $g(y) = delta(y+1)$. This can be done quite easily since integration against a $delta$ simply evaluates the integrand at the support of the $delta$. So in your case, we would have $u(x,t) = frac1sqrtpi t e^-(x+1)^2/t$, and there is no need for the "infinity" that you write above. (Edit: I completely agree with the comment by Winther, you should probably not think of the delta function as being "equal to infinity" at the origin, just use the integral property that Winther stated as the defining property of the delta function.)
Edit to answer your comment: If we replace the $delta(x-1)$ by a Heaviside $H(x-1)$ then since
$$H(x-1) = cases0,,textfor x<1,\ 1,,textfor x>1,$$
we need to evaluate the integral
$$u(x,t)=frac1sqrtpi tint_1^inftye^-(x-y)^2/t ,mathrmdy.$$
I think you should be able to do this in terms of the error function.
answered Mar 23 at 19:21
bobcliffebobcliffe
1506
$endgroup$
$begingroup$
thanks for your asnwer. I have a question, how if we replace delta function with heaviside $H(x-1)$? How would solution look now?
$endgroup$
– James
Mar 23 at 19:28
$begingroup$
@JimmySabater I have added some extra comments for you :)
$endgroup$
– bobcliffe
Mar 24 at 16:16
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%2f3159684%2fquestion-about-the-dirac-delta-function%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In your statement, I think you mean that the function $g(y) = u(y,0)$ contains the initial conditions. Then to answer your question, you need to perform the integral with $g(y) = delta(y+1)$. This can be done quite easily since integration against a $delta$ simply evaluates the integrand at the support of the $delta$. So in your case, we would have $u(x,t) = frac1sqrtpi t e^-(x+1)^2/t$, and there is no need for the "infinity" that you write above. (Edit: I completely agree with the comment by Winther, you should probably not think of the delta function as being "equal to infinity" at the origin, just use the integral property that Winther stated as the defining property of the delta function.)
Edit to answer your comment: If we replace the $delta(x-1)$ by a Heaviside $H(x-1)$ then since
$$H(x-1) = cases0,,textfor x<1,\ 1,,textfor x>1,$$
we need to evaluate the integral
$$u(x,t)=frac1sqrtpi tint_1^inftye^-(x-y)^2/t ,mathrmdy.$$
I think you should be able to do this in terms of the error function.
answered Mar 23 at 19:21
bobcliffebobcliffe
1506
$endgroup$
$begingroup$
thanks for your asnwer. I have a question, how if we replace delta function with heaviside $H(x-1)$? How would solution look now?
$endgroup$
– James
Mar 23 at 19:28
$begingroup$
@JimmySabater I have added some extra comments for you :)
$endgroup$
– bobcliffe
Mar 24 at 16:16
add a comment |
$begingroup$
In your statement, I think you mean that the function $g(y) = u(y,0)$ contains the initial conditions. Then to answer your question, you need to perform the integral with $g(y) = delta(y+1)$. This can be done quite easily since integration against a $delta$ simply evaluates the integrand at the support of the $delta$. So in your case, we would have $u(x,t) = frac1sqrtpi t e^-(x+1)^2/t$, and there is no need for the "infinity" that you write above. (Edit: I completely agree with the comment by Winther, you should probably not think of the delta function as being "equal to infinity" at the origin, just use the integral property that Winther stated as the defining property of the delta function.)
Edit to answer your comment: If we replace the $delta(x-1)$ by a Heaviside $H(x-1)$ then since
$$H(x-1) = cases0,,textfor x<1,\ 1,,textfor x>1,$$
we need to evaluate the integral
$$u(x,t)=frac1sqrtpi tint_1^inftye^-(x-y)^2/t ,mathrmdy.$$
I think you should be able to do this in terms of the error function.
answered Mar 23 at 19:21
bobcliffebobcliffe
1506
$endgroup$
$begingroup$
thanks for your asnwer. I have a question, how if we replace delta function with heaviside $H(x-1)$? How would solution look now?
$endgroup$
– James
Mar 23 at 19:28
$begingroup$
@JimmySabater I have added some extra comments for you :)
$endgroup$
– bobcliffe
Mar 24 at 16:16
add a comment |
$begingroup$
In your statement, I think you mean that the function $g(y) = u(y,0)$ contains the initial conditions. Then to answer your question, you need to perform the integral with $g(y) = delta(y+1)$. This can be done quite easily since integration against a $delta$ simply evaluates the integrand at the support of the $delta$. So in your case, we would have $u(x,t) = frac1sqrtpi t e^-(x+1)^2/t$, and there is no need for the "infinity" that you write above. (Edit: I completely agree with the comment by Winther, you should probably not think of the delta function as being "equal to infinity" at the origin, just use the integral property that Winther stated as the defining property of the delta function.)
Edit to answer your comment: If we replace the $delta(x-1)$ by a Heaviside $H(x-1)$ then since
$$H(x-1) = cases0,,textfor x<1,\ 1,,textfor x>1,$$
we need to evaluate the integral
$$u(x,t)=frac1sqrtpi tint_1^inftye^-(x-y)^2/t ,mathrmdy.$$
I think you should be able to do this in terms of the error function.
answered Mar 23 at 19:21
bobcliffebobcliffe
1506
$endgroup$
In your statement, I think you mean that the function $g(y) = u(y,0)$ contains the initial conditions. Then to answer your question, you need to perform the integral with $g(y) = delta(y+1)$. This can be done quite easily since integration against a $delta$ simply evaluates the integrand at the support of the $delta$. So in your case, we would have $u(x,t) = frac1sqrtpi t e^-(x+1)^2/t$, and there is no need for the "infinity" that you write above. (Edit: I completely agree with the comment by Winther, you should probably not think of the delta function as being "equal to infinity" at the origin, just use the integral property that Winther stated as the defining property of the delta function.)
Edit to answer your comment: If we replace the $delta(x-1)$ by a Heaviside $H(x-1)$ then since
$$H(x-1) = cases0,,textfor x<1,\ 1,,textfor x>1,$$
we need to evaluate the integral
$$u(x,t)=frac1sqrtpi tint_1^inftye^-(x-y)^2/t ,mathrmdy.$$
I think you should be able to do this in terms of the error function.
answered Mar 23 at 19:21
bobcliffebobcliffe
1506
edited Mar 24 at 16:15
answered Mar 23 at 19:21
bobcliffebobcliffe
1506
answered Mar 23 at 19:21
bobcliffebobcliffe
1506
answered Mar 23 at 19:21
bobcliffebobcliffe
1506
1506
$begingroup$
thanks for your asnwer. I have a question, how if we replace delta function with heaviside $H(x-1)$? How would solution look now?
$endgroup$
– James
Mar 23 at 19:28
$begingroup$
@JimmySabater I have added some extra comments for you :)
$endgroup$
– bobcliffe
Mar 24 at 16:16
add a comment |
$begingroup$
thanks for your asnwer. I have a question, how if we replace delta function with heaviside $H(x-1)$? How would solution look now?
$endgroup$
– James
Mar 23 at 19:28
$begingroup$
@JimmySabater I have added some extra comments for you :)
$endgroup$
– bobcliffe
Mar 24 at 16:16
$begingroup$
thanks for your asnwer. I have a question, how if we replace delta function with heaviside $H(x-1)$? How would solution look now?
$endgroup$
– James
Mar 23 at 19:28
$begingroup$
thanks for your asnwer. I have a question, how if we replace delta function with heaviside $H(x-1)$? How would solution look now?
$endgroup$
– James
Mar 23 at 19:28
$begingroup$
@JimmySabater I have added some extra comments for you :)
$endgroup$
– bobcliffe
Mar 24 at 16:16
$begingroup$
@JimmySabater I have added some extra comments for you :)
$endgroup$
– bobcliffe
Mar 24 at 16:16
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%2f3159684%2fquestion-about-the-dirac-delta-function%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
1
$begingroup$
Forget the informal "$=infty$ when $x=0$ and $=0$ otherwise" description of the delta-function. The (or a) proper definition of the delta function is in how it acts under an integral, namely $int_-infty^infty f(x)delta(x-a)rm dx equiv f(a)$. This is all you need here.
$endgroup$
– Winther
Mar 23 at 19:25