General solution od PDE of second order Announcing the arrival of Valued Associate #679: Cesar...
Karn the great creator - 'card from outside the game' in sealed
Why are vacuum tubes still used in amateur radios?
macOS: Name for app shortcut screen found by pinching with thumb and three fingers
Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?
How does the math work when buying airline miles?
If Windows 7 doesn't support WSL, then what is "Subsystem for UNIX-based Applications"?
Trademark violation for app?
Flight departed from the gate 5 min before scheduled departure time. Refund options
Misunderstanding of Sylow theory
Strange behavior of Object.defineProperty() in JavaScript
Putting class ranking in CV, but against dept guidelines
What is the meaning of 'breadth' in breadth first search?
What makes a man succeed?
Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?
Why are my pictures showing a dark band on one edge?
A term for a woman complaining about things/begging in a cute/childish way
Did any compiler fully use 80-bit floating point?
What is best way to wire a ceiling receptacle in this situation?
Is the IBM 5153 color display compatible with the Tandy 1000 16 color modes?
The Nth Gryphon Number
Are sorcerers unable to use the Careful Spell metamagic option on themselves?
How long can equipment go unused before powering up runs the risk of damage?
How to compare two different files line by line in unix?
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?
General solution od PDE of second order
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Monge-Ampere PDE is hyperbolic ($f(x) < 0$) - General Solution About finding the general solution of first-order totally nonlinear PDE with two independent variablesAnalytical solution to PDEGeneral integral of an PDESolutions to the PDE :$2Vfrac{partial I}{partial V}+2Wfrac{partial I}{partial W}=I$General solution of partial differential equation 2. orderGeneral solution of particular first-order nonlinear PDEQuasilinear 2nd order PDE, apply initial data to general solutionProblem with satisfying Boundary conditions for 1D heat PDEApplying infinite boundary conditions to PDE solution by the separation of variables
$begingroup$
I have following PDE equation:
$frac{partial^2T(s,x)}{partial x^2}=frac{s}{a}T(x,s)$.
Why general solution is found by form: $T=c_1 e^{sqrt{frac{s}{a}}x}+c_2 e^{-sqrt{frac{s}{a}}x}$. I know that it was from $lambda^2=frac{s}{a}$ but it is not clear for me where is last equation from( I don't understand proof and evaluation of this) ?
pde
asked Mar 25 at 18:02
EugenSEugenS
62
$endgroup$
add a comment |
$begingroup$
I have following PDE equation:
$frac{partial^2T(s,x)}{partial x^2}=frac{s}{a}T(x,s)$.
Why general solution is found by form: $T=c_1 e^{sqrt{frac{s}{a}}x}+c_2 e^{-sqrt{frac{s}{a}}x}$. I know that it was from $lambda^2=frac{s}{a}$ but it is not clear for me where is last equation from( I don't understand proof and evaluation of this) ?
pde
asked Mar 25 at 18:02
EugenSEugenS
62
$endgroup$
1
$begingroup$
Thiis is a rather long elaboration. Isn't that provided on your textbook or notes ?
$endgroup$
– Rebellos
Mar 25 at 18:13
$begingroup$
Unfortunately, it is not provided in my textbook, otherwise I shouldn't post my question:)
$endgroup$
– EugenS
Mar 26 at 2:38
add a comment |
$begingroup$
I have following PDE equation:
$frac{partial^2T(s,x)}{partial x^2}=frac{s}{a}T(x,s)$.
Why general solution is found by form: $T=c_1 e^{sqrt{frac{s}{a}}x}+c_2 e^{-sqrt{frac{s}{a}}x}$. I know that it was from $lambda^2=frac{s}{a}$ but it is not clear for me where is last equation from( I don't understand proof and evaluation of this) ?
pde
asked Mar 25 at 18:02
EugenSEugenS
62
$endgroup$
I have following PDE equation:
$frac{partial^2T(s,x)}{partial x^2}=frac{s}{a}T(x,s)$.
Why general solution is found by form: $T=c_1 e^{sqrt{frac{s}{a}}x}+c_2 e^{-sqrt{frac{s}{a}}x}$. I know that it was from $lambda^2=frac{s}{a}$ but it is not clear for me where is last equation from( I don't understand proof and evaluation of this) ?
pde
pde
asked Mar 25 at 18:02
EugenSEugenS
62
asked Mar 25 at 18:02
EugenSEugenS
62
asked Mar 25 at 18:02
EugenSEugenS
62
asked Mar 25 at 18:02
EugenSEugenS
62
asked Mar 25 at 18:02
EugenSEugenS
62
62
1
$begingroup$
Thiis is a rather long elaboration. Isn't that provided on your textbook or notes ?
$endgroup$
– Rebellos
Mar 25 at 18:13
$begingroup$
Unfortunately, it is not provided in my textbook, otherwise I shouldn't post my question:)
$endgroup$
– EugenS
Mar 26 at 2:38
add a comment |
1
$begingroup$
Thiis is a rather long elaboration. Isn't that provided on your textbook or notes ?
$endgroup$
– Rebellos
Mar 25 at 18:13
$begingroup$
Unfortunately, it is not provided in my textbook, otherwise I shouldn't post my question:)
$endgroup$
– EugenS
Mar 26 at 2:38
1
1
$begingroup$
Thiis is a rather long elaboration. Isn't that provided on your textbook or notes ?
$endgroup$
– Rebellos
Mar 25 at 18:13
$begingroup$
Thiis is a rather long elaboration. Isn't that provided on your textbook or notes ?
$endgroup$
– Rebellos
Mar 25 at 18:13
$begingroup$
Unfortunately, it is not provided in my textbook, otherwise I shouldn't post my question:)
$endgroup$
– EugenS
Mar 26 at 2:38
$begingroup$
Unfortunately, it is not provided in my textbook, otherwise I shouldn't post my question:)
$endgroup$
– EugenS
Mar 26 at 2:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Thank you All!
It is derived from Euler's assumption about form of partial solutions:
$$T = e^{lambda(s) x}.$$
Second partial derivative of $T$ in this case will be:$$frac{partial T(x,s)}{partial x^2 } = lambda^2 e^{lambda x} = lambda^2 T$$
Last expression equals $lambda^2 T = frac{s}{a}T$, from which we have that $$lambda_{1,2} = sqrt{frac{s}{a}}$$
Genereal solution is the sum of product of partial solutions and some coefficients: $$T = C_1 T_1+C_2 T_2 = C_1 e^{sqrt{frac{s}{a}} x}+C_2 e^{-sqrt{frac{s}{a}} x}$$
answered Mar 28 at 7:27
EugenSEugenS
62
$endgroup$
add a comment |
Your Answer
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%2f3162127%2fgeneral-solution-od-pde-of-second-order%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
$begingroup$
Thank you All!
It is derived from Euler's assumption about form of partial solutions:
$$T = e^{lambda(s) x}.$$
Second partial derivative of $T$ in this case will be:$$frac{partial T(x,s)}{partial x^2 } = lambda^2 e^{lambda x} = lambda^2 T$$
Last expression equals $lambda^2 T = frac{s}{a}T$, from which we have that $$lambda_{1,2} = sqrt{frac{s}{a}}$$
Genereal solution is the sum of product of partial solutions and some coefficients: $$T = C_1 T_1+C_2 T_2 = C_1 e^{sqrt{frac{s}{a}} x}+C_2 e^{-sqrt{frac{s}{a}} x}$$
answered Mar 28 at 7:27
EugenSEugenS
62
$endgroup$
add a comment |
$begingroup$
Thank you All!
It is derived from Euler's assumption about form of partial solutions:
$$T = e^{lambda(s) x}.$$
Second partial derivative of $T$ in this case will be:$$frac{partial T(x,s)}{partial x^2 } = lambda^2 e^{lambda x} = lambda^2 T$$
Last expression equals $lambda^2 T = frac{s}{a}T$, from which we have that $$lambda_{1,2} = sqrt{frac{s}{a}}$$
Genereal solution is the sum of product of partial solutions and some coefficients: $$T = C_1 T_1+C_2 T_2 = C_1 e^{sqrt{frac{s}{a}} x}+C_2 e^{-sqrt{frac{s}{a}} x}$$
answered Mar 28 at 7:27
EugenSEugenS
62
$endgroup$
add a comment |
$begingroup$
Thank you All!
It is derived from Euler's assumption about form of partial solutions:
$$T = e^{lambda(s) x}.$$
Second partial derivative of $T$ in this case will be:$$frac{partial T(x,s)}{partial x^2 } = lambda^2 e^{lambda x} = lambda^2 T$$
Last expression equals $lambda^2 T = frac{s}{a}T$, from which we have that $$lambda_{1,2} = sqrt{frac{s}{a}}$$
Genereal solution is the sum of product of partial solutions and some coefficients: $$T = C_1 T_1+C_2 T_2 = C_1 e^{sqrt{frac{s}{a}} x}+C_2 e^{-sqrt{frac{s}{a}} x}$$
answered Mar 28 at 7:27
EugenSEugenS
62
$endgroup$
Thank you All!
It is derived from Euler's assumption about form of partial solutions:
$$T = e^{lambda(s) x}.$$
Second partial derivative of $T$ in this case will be:$$frac{partial T(x,s)}{partial x^2 } = lambda^2 e^{lambda x} = lambda^2 T$$
Last expression equals $lambda^2 T = frac{s}{a}T$, from which we have that $$lambda_{1,2} = sqrt{frac{s}{a}}$$
Genereal solution is the sum of product of partial solutions and some coefficients: $$T = C_1 T_1+C_2 T_2 = C_1 e^{sqrt{frac{s}{a}} x}+C_2 e^{-sqrt{frac{s}{a}} x}$$
answered Mar 28 at 7:27
EugenSEugenS
62
answered Mar 28 at 7:27
EugenSEugenS
62
answered Mar 28 at 7:27
EugenSEugenS
62
answered Mar 28 at 7:27
EugenSEugenS
62
62
add a comment |
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%2f3162127%2fgeneral-solution-od-pde-of-second-order%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
$begingroup$
Thiis is a rather long elaboration. Isn't that provided on your textbook or notes ?
$endgroup$
– Rebellos
Mar 25 at 18:13
$begingroup$
Unfortunately, it is not provided in my textbook, otherwise I shouldn't post my question:)
$endgroup$
– EugenS
Mar 26 at 2:38