General solution vs General Integral
$begingroup$
I have been reading some notes about Quasi Linear PDE and I stuck to the following issue:
Consider the PDE: $xz_x+yz_y=xe^{-z}$
then my notes read as:
"the general solution is: $G(x/y, e^z-x)=0$ . (1)
and so $$e^z-x=g(x/y)$$ , where $G,g$ are $C^1$ functions."
In my opinion this is not right and we are losing solutions that way.
REMARK: The only way I know we can do a similar thing is when the following PDE
$az_x+bz_y=0$ (2) is given, where $a,b$ are functions of $x,y$ and they do not vanish simultaneously. Then if $u=c$ is the general solution of the ODE:
$$frac{dx}{a}=frac{dy}{b}$$, then $z=f(u)$ , where $f$ is any $C^1$ function.
This last is taken from the book Introduction to PDE with applications by Zachamanoglou and Thoe, Example 2.2 page 62.
Is my cosideration correct?
ordinary-differential-equations pde
asked Dec 14 '18 at 17:04
dmtridmtri
1,5082521
$endgroup$
add a comment |
$begingroup$
I have been reading some notes about Quasi Linear PDE and I stuck to the following issue:
Consider the PDE: $xz_x+yz_y=xe^{-z}$
then my notes read as:
"the general solution is: $G(x/y, e^z-x)=0$ . (1)
and so $$e^z-x=g(x/y)$$ , where $G,g$ are $C^1$ functions."
In my opinion this is not right and we are losing solutions that way.
REMARK: The only way I know we can do a similar thing is when the following PDE
$az_x+bz_y=0$ (2) is given, where $a,b$ are functions of $x,y$ and they do not vanish simultaneously. Then if $u=c$ is the general solution of the ODE:
$$frac{dx}{a}=frac{dy}{b}$$, then $z=f(u)$ , where $f$ is any $C^1$ function.
This last is taken from the book Introduction to PDE with applications by Zachamanoglou and Thoe, Example 2.2 page 62.
Is my cosideration correct?
ordinary-differential-equations pde
asked Dec 14 '18 at 17:04
dmtridmtri
1,5082521
$endgroup$
add a comment |
$begingroup$
I have been reading some notes about Quasi Linear PDE and I stuck to the following issue:
Consider the PDE: $xz_x+yz_y=xe^{-z}$
then my notes read as:
"the general solution is: $G(x/y, e^z-x)=0$ . (1)
and so $$e^z-x=g(x/y)$$ , where $G,g$ are $C^1$ functions."
In my opinion this is not right and we are losing solutions that way.
REMARK: The only way I know we can do a similar thing is when the following PDE
$az_x+bz_y=0$ (2) is given, where $a,b$ are functions of $x,y$ and they do not vanish simultaneously. Then if $u=c$ is the general solution of the ODE:
$$frac{dx}{a}=frac{dy}{b}$$, then $z=f(u)$ , where $f$ is any $C^1$ function.
This last is taken from the book Introduction to PDE with applications by Zachamanoglou and Thoe, Example 2.2 page 62.
Is my cosideration correct?
ordinary-differential-equations pde
asked Dec 14 '18 at 17:04
dmtridmtri
1,5082521
$endgroup$
I have been reading some notes about Quasi Linear PDE and I stuck to the following issue:
Consider the PDE: $xz_x+yz_y=xe^{-z}$
then my notes read as:
"the general solution is: $G(x/y, e^z-x)=0$ . (1)
and so $$e^z-x=g(x/y)$$ , where $G,g$ are $C^1$ functions."
In my opinion this is not right and we are losing solutions that way.
REMARK: The only way I know we can do a similar thing is when the following PDE
$az_x+bz_y=0$ (2) is given, where $a,b$ are functions of $x,y$ and they do not vanish simultaneously. Then if $u=c$ is the general solution of the ODE:
$$frac{dx}{a}=frac{dy}{b}$$, then $z=f(u)$ , where $f$ is any $C^1$ function.
This last is taken from the book Introduction to PDE with applications by Zachamanoglou and Thoe, Example 2.2 page 62.
Is my cosideration correct?
ordinary-differential-equations pde
ordinary-differential-equations pde
asked Dec 14 '18 at 17:04
dmtridmtri
1,5082521
asked Dec 14 '18 at 17:04
dmtridmtri
1,5082521
edited Dec 15 '18 at 17:09
dmtri
asked Dec 14 '18 at 17:04
dmtridmtri
1,5082521
asked Dec 14 '18 at 17:04
dmtridmtri
1,5082521
asked Dec 14 '18 at 17:04
dmtridmtri
1,5082521
1,5082521
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$xu_x+yu_y=xe^{-u}$$
The Charpit-Lagrange equations are :
$$frac{dx}{x}=frac{dy}{y}=frac{du}{xe^{-u}}$$
First equation of characteristic curves , from $frac{dx}{x}=frac{dy}{y}$ :
$$frac{x}{y}=c_1$$
Second equation of characteristic curves , from $frac{dx}{x}=frac{du}{xe^{-u}}$:
$$e^u-x=c_2$$
The general solution of the PDE, expressed on the form of implicit equation is :
$$Phileft(frac{x}{y}:,:e^u-xright)=0$$
where $Phi$ is an arbitrary function of two variables.
Or, equivalently on explicit form :
$e^u-x=Fleft(frac{x}{y}right)$
$$u(x,y)=lnleft|x+Fleft(frac{x}{y}right)right|$$
where $F$ is an arbitrary function.
The function $F$ had to be determined according to some boundary condition. Since there is no such condition specified in the wording of the problem, further calculus is not possible.
Nevertheless, the above result is consistent with the solution (which is correct) noted in the OP question.
answered Dec 15 '18 at 10:39
JJacquelinJJacquelin
44k21853
$endgroup$
$begingroup$
Thanks for the explicit answer(+25), but what I am worried about is how do You derive from the general integral $Phi(u_1,u_2)=0$ the solution: $u_2=F(u_1)$ ....especially when $Phi_{u_2}=0$...ie you cannot solve for $u_2$. Are the general integral and general solution equivalent?
$endgroup$
– dmtri
Dec 15 '18 at 11:57
1
$begingroup$
Of course it is not always possible to solve $Phi(u_1,u_2)=0$ for an explicit form $u_2=F(u_1)$ in terms of standard elementary and/or special functions. So, the equivalence is purely symbolic. If it is not possible today one can imagine that a convenient brand-new special function will be standardised in the futur, then allowing to do it. In any case everybody can define convenient particular special functions (Which should be of no practical use until being standardized and become common in math literature. That is a different kettle of fish).
$endgroup$
– JJacquelin
Dec 15 '18 at 14:27
$begingroup$
Thanks, I liked that!!, especially the kettle and fish!
$endgroup$
– dmtri
Dec 15 '18 at 15:29
$begingroup$
I made also some corrections to my original post, especially in the REMARK area.
$endgroup$
– dmtri
Dec 15 '18 at 17:14
1
$begingroup$
OK I agree now. So my answer about your remark area is no longer valid. I delete it.
$endgroup$
– JJacquelin
Dec 16 '18 at 7:59
|
show 1 more 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%2f3039642%2fgeneral-solution-vs-general-integral%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
$begingroup$
$$xu_x+yu_y=xe^{-u}$$
The Charpit-Lagrange equations are :
$$frac{dx}{x}=frac{dy}{y}=frac{du}{xe^{-u}}$$
First equation of characteristic curves , from $frac{dx}{x}=frac{dy}{y}$ :
$$frac{x}{y}=c_1$$
Second equation of characteristic curves , from $frac{dx}{x}=frac{du}{xe^{-u}}$:
$$e^u-x=c_2$$
The general solution of the PDE, expressed on the form of implicit equation is :
$$Phileft(frac{x}{y}:,:e^u-xright)=0$$
where $Phi$ is an arbitrary function of two variables.
Or, equivalently on explicit form :
$e^u-x=Fleft(frac{x}{y}right)$
$$u(x,y)=lnleft|x+Fleft(frac{x}{y}right)right|$$
where $F$ is an arbitrary function.
The function $F$ had to be determined according to some boundary condition. Since there is no such condition specified in the wording of the problem, further calculus is not possible.
Nevertheless, the above result is consistent with the solution (which is correct) noted in the OP question.
answered Dec 15 '18 at 10:39
JJacquelinJJacquelin
44k21853
$endgroup$
$begingroup$
Thanks for the explicit answer(+25), but what I am worried about is how do You derive from the general integral $Phi(u_1,u_2)=0$ the solution: $u_2=F(u_1)$ ....especially when $Phi_{u_2}=0$...ie you cannot solve for $u_2$. Are the general integral and general solution equivalent?
$endgroup$
– dmtri
Dec 15 '18 at 11:57
1
$begingroup$
Of course it is not always possible to solve $Phi(u_1,u_2)=0$ for an explicit form $u_2=F(u_1)$ in terms of standard elementary and/or special functions. So, the equivalence is purely symbolic. If it is not possible today one can imagine that a convenient brand-new special function will be standardised in the futur, then allowing to do it. In any case everybody can define convenient particular special functions (Which should be of no practical use until being standardized and become common in math literature. That is a different kettle of fish).
$endgroup$
– JJacquelin
Dec 15 '18 at 14:27
$begingroup$
Thanks, I liked that!!, especially the kettle and fish!
$endgroup$
– dmtri
Dec 15 '18 at 15:29
$begingroup$
I made also some corrections to my original post, especially in the REMARK area.
$endgroup$
– dmtri
Dec 15 '18 at 17:14
1
$begingroup$
OK I agree now. So my answer about your remark area is no longer valid. I delete it.
$endgroup$
– JJacquelin
Dec 16 '18 at 7:59
|
show 1 more comment
$begingroup$
$$xu_x+yu_y=xe^{-u}$$
The Charpit-Lagrange equations are :
$$frac{dx}{x}=frac{dy}{y}=frac{du}{xe^{-u}}$$
First equation of characteristic curves , from $frac{dx}{x}=frac{dy}{y}$ :
$$frac{x}{y}=c_1$$
Second equation of characteristic curves , from $frac{dx}{x}=frac{du}{xe^{-u}}$:
$$e^u-x=c_2$$
The general solution of the PDE, expressed on the form of implicit equation is :
$$Phileft(frac{x}{y}:,:e^u-xright)=0$$
where $Phi$ is an arbitrary function of two variables.
Or, equivalently on explicit form :
$e^u-x=Fleft(frac{x}{y}right)$
$$u(x,y)=lnleft|x+Fleft(frac{x}{y}right)right|$$
where $F$ is an arbitrary function.
The function $F$ had to be determined according to some boundary condition. Since there is no such condition specified in the wording of the problem, further calculus is not possible.
Nevertheless, the above result is consistent with the solution (which is correct) noted in the OP question.
answered Dec 15 '18 at 10:39
JJacquelinJJacquelin
44k21853
$endgroup$
$begingroup$
Thanks for the explicit answer(+25), but what I am worried about is how do You derive from the general integral $Phi(u_1,u_2)=0$ the solution: $u_2=F(u_1)$ ....especially when $Phi_{u_2}=0$...ie you cannot solve for $u_2$. Are the general integral and general solution equivalent?
$endgroup$
– dmtri
Dec 15 '18 at 11:57
1
$begingroup$
Of course it is not always possible to solve $Phi(u_1,u_2)=0$ for an explicit form $u_2=F(u_1)$ in terms of standard elementary and/or special functions. So, the equivalence is purely symbolic. If it is not possible today one can imagine that a convenient brand-new special function will be standardised in the futur, then allowing to do it. In any case everybody can define convenient particular special functions (Which should be of no practical use until being standardized and become common in math literature. That is a different kettle of fish).
$endgroup$
– JJacquelin
Dec 15 '18 at 14:27
$begingroup$
Thanks, I liked that!!, especially the kettle and fish!
$endgroup$
– dmtri
Dec 15 '18 at 15:29
$begingroup$
I made also some corrections to my original post, especially in the REMARK area.
$endgroup$
– dmtri
Dec 15 '18 at 17:14
1
$begingroup$
OK I agree now. So my answer about your remark area is no longer valid. I delete it.
$endgroup$
– JJacquelin
Dec 16 '18 at 7:59
|
show 1 more comment
$begingroup$
$$xu_x+yu_y=xe^{-u}$$
The Charpit-Lagrange equations are :
$$frac{dx}{x}=frac{dy}{y}=frac{du}{xe^{-u}}$$
First equation of characteristic curves , from $frac{dx}{x}=frac{dy}{y}$ :
$$frac{x}{y}=c_1$$
Second equation of characteristic curves , from $frac{dx}{x}=frac{du}{xe^{-u}}$:
$$e^u-x=c_2$$
The general solution of the PDE, expressed on the form of implicit equation is :
$$Phileft(frac{x}{y}:,:e^u-xright)=0$$
where $Phi$ is an arbitrary function of two variables.
Or, equivalently on explicit form :
$e^u-x=Fleft(frac{x}{y}right)$
$$u(x,y)=lnleft|x+Fleft(frac{x}{y}right)right|$$
where $F$ is an arbitrary function.
The function $F$ had to be determined according to some boundary condition. Since there is no such condition specified in the wording of the problem, further calculus is not possible.
Nevertheless, the above result is consistent with the solution (which is correct) noted in the OP question.
answered Dec 15 '18 at 10:39
JJacquelinJJacquelin
44k21853
$endgroup$
$$xu_x+yu_y=xe^{-u}$$
The Charpit-Lagrange equations are :
$$frac{dx}{x}=frac{dy}{y}=frac{du}{xe^{-u}}$$
First equation of characteristic curves , from $frac{dx}{x}=frac{dy}{y}$ :
$$frac{x}{y}=c_1$$
Second equation of characteristic curves , from $frac{dx}{x}=frac{du}{xe^{-u}}$:
$$e^u-x=c_2$$
The general solution of the PDE, expressed on the form of implicit equation is :
$$Phileft(frac{x}{y}:,:e^u-xright)=0$$
where $Phi$ is an arbitrary function of two variables.
Or, equivalently on explicit form :
$e^u-x=Fleft(frac{x}{y}right)$
$$u(x,y)=lnleft|x+Fleft(frac{x}{y}right)right|$$
where $F$ is an arbitrary function.
The function $F$ had to be determined according to some boundary condition. Since there is no such condition specified in the wording of the problem, further calculus is not possible.
Nevertheless, the above result is consistent with the solution (which is correct) noted in the OP question.
answered Dec 15 '18 at 10:39
JJacquelinJJacquelin
44k21853
edited Dec 16 '18 at 8:00
answered Dec 15 '18 at 10:39
JJacquelinJJacquelin
44k21853
answered Dec 15 '18 at 10:39
JJacquelinJJacquelin
44k21853
answered Dec 15 '18 at 10:39
JJacquelinJJacquelin
44k21853
44k21853
$begingroup$
Thanks for the explicit answer(+25), but what I am worried about is how do You derive from the general integral $Phi(u_1,u_2)=0$ the solution: $u_2=F(u_1)$ ....especially when $Phi_{u_2}=0$...ie you cannot solve for $u_2$. Are the general integral and general solution equivalent?
$endgroup$
– dmtri
Dec 15 '18 at 11:57
1
$begingroup$
Of course it is not always possible to solve $Phi(u_1,u_2)=0$ for an explicit form $u_2=F(u_1)$ in terms of standard elementary and/or special functions. So, the equivalence is purely symbolic. If it is not possible today one can imagine that a convenient brand-new special function will be standardised in the futur, then allowing to do it. In any case everybody can define convenient particular special functions (Which should be of no practical use until being standardized and become common in math literature. That is a different kettle of fish).
$endgroup$
– JJacquelin
Dec 15 '18 at 14:27
$begingroup$
Thanks, I liked that!!, especially the kettle and fish!
$endgroup$
– dmtri
Dec 15 '18 at 15:29
$begingroup$
I made also some corrections to my original post, especially in the REMARK area.
$endgroup$
– dmtri
Dec 15 '18 at 17:14
1
$begingroup$
OK I agree now. So my answer about your remark area is no longer valid. I delete it.
$endgroup$
– JJacquelin
Dec 16 '18 at 7:59
|
show 1 more comment
$begingroup$
Thanks for the explicit answer(+25), but what I am worried about is how do You derive from the general integral $Phi(u_1,u_2)=0$ the solution: $u_2=F(u_1)$ ....especially when $Phi_{u_2}=0$...ie you cannot solve for $u_2$. Are the general integral and general solution equivalent?
$endgroup$
– dmtri
Dec 15 '18 at 11:57
1
$begingroup$
Of course it is not always possible to solve $Phi(u_1,u_2)=0$ for an explicit form $u_2=F(u_1)$ in terms of standard elementary and/or special functions. So, the equivalence is purely symbolic. If it is not possible today one can imagine that a convenient brand-new special function will be standardised in the futur, then allowing to do it. In any case everybody can define convenient particular special functions (Which should be of no practical use until being standardized and become common in math literature. That is a different kettle of fish).
$endgroup$
– JJacquelin
Dec 15 '18 at 14:27
$begingroup$
Thanks, I liked that!!, especially the kettle and fish!
$endgroup$
– dmtri
Dec 15 '18 at 15:29
$begingroup$
I made also some corrections to my original post, especially in the REMARK area.
$endgroup$
– dmtri
Dec 15 '18 at 17:14
1
$begingroup$
OK I agree now. So my answer about your remark area is no longer valid. I delete it.
$endgroup$
– JJacquelin
Dec 16 '18 at 7:59
$begingroup$
Thanks for the explicit answer(+25), but what I am worried about is how do You derive from the general integral $Phi(u_1,u_2)=0$ the solution: $u_2=F(u_1)$ ....especially when $Phi_{u_2}=0$...ie you cannot solve for $u_2$. Are the general integral and general solution equivalent?
$endgroup$
– dmtri
Dec 15 '18 at 11:57
$begingroup$
Thanks for the explicit answer(+25), but what I am worried about is how do You derive from the general integral $Phi(u_1,u_2)=0$ the solution: $u_2=F(u_1)$ ....especially when $Phi_{u_2}=0$...ie you cannot solve for $u_2$. Are the general integral and general solution equivalent?
$endgroup$
– dmtri
Dec 15 '18 at 11:57
1
1
$begingroup$
Of course it is not always possible to solve $Phi(u_1,u_2)=0$ for an explicit form $u_2=F(u_1)$ in terms of standard elementary and/or special functions. So, the equivalence is purely symbolic. If it is not possible today one can imagine that a convenient brand-new special function will be standardised in the futur, then allowing to do it. In any case everybody can define convenient particular special functions (Which should be of no practical use until being standardized and become common in math literature. That is a different kettle of fish).
$endgroup$
– JJacquelin
Dec 15 '18 at 14:27
$begingroup$
Of course it is not always possible to solve $Phi(u_1,u_2)=0$ for an explicit form $u_2=F(u_1)$ in terms of standard elementary and/or special functions. So, the equivalence is purely symbolic. If it is not possible today one can imagine that a convenient brand-new special function will be standardised in the futur, then allowing to do it. In any case everybody can define convenient particular special functions (Which should be of no practical use until being standardized and become common in math literature. That is a different kettle of fish).
$endgroup$
– JJacquelin
Dec 15 '18 at 14:27
$begingroup$
Thanks, I liked that!!, especially the kettle and fish!
$endgroup$
– dmtri
Dec 15 '18 at 15:29
$begingroup$
Thanks, I liked that!!, especially the kettle and fish!
$endgroup$
– dmtri
Dec 15 '18 at 15:29
$begingroup$
I made also some corrections to my original post, especially in the REMARK area.
$endgroup$
– dmtri
Dec 15 '18 at 17:14
$begingroup$
I made also some corrections to my original post, especially in the REMARK area.
$endgroup$
– dmtri
Dec 15 '18 at 17:14
1
1
$begingroup$
OK I agree now. So my answer about your remark area is no longer valid. I delete it.
$endgroup$
– JJacquelin
Dec 16 '18 at 7:59
$begingroup$
OK I agree now. So my answer about your remark area is no longer valid. I delete it.
$endgroup$
– JJacquelin
Dec 16 '18 at 7:59
|
show 1 more 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%2f3039642%2fgeneral-solution-vs-general-integral%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
