$d^2u/dx^2 - d^2u/dy^2 = f(x,y)$ $rightarrow$ $d^2u/dξdη =...












0














$$frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = f(x,y) implies frac{partial^2u}{partial ξpartial η} = frac{1}{4}fleft(frac{1}{2}(ξ+η),frac{1}{2}(η-ξ)right)$$



Setting $ξ = x − y, η = x + y$ I get $x = frac{1}{2}(ξ+η)$ and $y=frac{1}{2}(η-ξ)$



I cant seem to show that $frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = 4times frac{partial^2u}{partial ξpartial η}$



My working is:



$$partial u/partial x = frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$



$$partial u/partial x = -frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$



$$frac{partial^2u}{partial x^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}+ frac{1}{2}frac{partial^2u}{partial ξpartial η}$$



$$frac{partial^2u}{partial y^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}- frac{1}{2}frac{partial^2u}{partial ξpartial η}$$



Where am I going wrong?
Could someone please help?






linear-algebra pde partial-derivative







share|cite|improve this question
























  • Are the $d$s in your question intended to be $partial$?
    – B. Mehta
    Nov 26 '18 at 19:08










  • Yes they are, i dont know how to get the partial symbol
    – pablo_mathscobar
    Nov 26 '18 at 19:08










  • As you can see in my edit, you can use partial to get that symbol.
    – B. Mehta
    Nov 26 '18 at 19:27
















0














$$frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = f(x,y) implies frac{partial^2u}{partial ξpartial η} = frac{1}{4}fleft(frac{1}{2}(ξ+η),frac{1}{2}(η-ξ)right)$$



Setting $ξ = x − y, η = x + y$ I get $x = frac{1}{2}(ξ+η)$ and $y=frac{1}{2}(η-ξ)$



I cant seem to show that $frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = 4times frac{partial^2u}{partial ξpartial η}$



My working is:



$$partial u/partial x = frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$



$$partial u/partial x = -frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$



$$frac{partial^2u}{partial x^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}+ frac{1}{2}frac{partial^2u}{partial ξpartial η}$$



$$frac{partial^2u}{partial y^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}- frac{1}{2}frac{partial^2u}{partial ξpartial η}$$



Where am I going wrong?
Could someone please help?






linear-algebra pde partial-derivative







share|cite|improve this question
























  • Are the $d$s in your question intended to be $partial$?
    – B. Mehta
    Nov 26 '18 at 19:08










  • Yes they are, i dont know how to get the partial symbol
    – pablo_mathscobar
    Nov 26 '18 at 19:08










  • As you can see in my edit, you can use partial to get that symbol.
    – B. Mehta
    Nov 26 '18 at 19:27














0












0








0







$$frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = f(x,y) implies frac{partial^2u}{partial ξpartial η} = frac{1}{4}fleft(frac{1}{2}(ξ+η),frac{1}{2}(η-ξ)right)$$



Setting $ξ = x − y, η = x + y$ I get $x = frac{1}{2}(ξ+η)$ and $y=frac{1}{2}(η-ξ)$



I cant seem to show that $frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = 4times frac{partial^2u}{partial ξpartial η}$



My working is:



$$partial u/partial x = frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$



$$partial u/partial x = -frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$



$$frac{partial^2u}{partial x^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}+ frac{1}{2}frac{partial^2u}{partial ξpartial η}$$



$$frac{partial^2u}{partial y^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}- frac{1}{2}frac{partial^2u}{partial ξpartial η}$$



Where am I going wrong?
Could someone please help?






linear-algebra pde partial-derivative







share|cite|improve this question















$$frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = f(x,y) implies frac{partial^2u}{partial ξpartial η} = frac{1}{4}fleft(frac{1}{2}(ξ+η),frac{1}{2}(η-ξ)right)$$



Setting $ξ = x − y, η = x + y$ I get $x = frac{1}{2}(ξ+η)$ and $y=frac{1}{2}(η-ξ)$



I cant seem to show that $frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = 4times frac{partial^2u}{partial ξpartial η}$



My working is:



$$partial u/partial x = frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$



$$partial u/partial x = -frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$



$$frac{partial^2u}{partial x^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}+ frac{1}{2}frac{partial^2u}{partial ξpartial η}$$



$$frac{partial^2u}{partial y^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}- frac{1}{2}frac{partial^2u}{partial ξpartial η}$$



Where am I going wrong?
Could someone please help?






linear-algebra pde partial-derivative




linear-algebra pde partial-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 19:14









B. Mehta

11.8k22144




11.8k22144










asked Nov 26 '18 at 19:04









pablo_mathscobar

836




836












  • Are the $d$s in your question intended to be $partial$?
    – B. Mehta
    Nov 26 '18 at 19:08










  • Yes they are, i dont know how to get the partial symbol
    – pablo_mathscobar
    Nov 26 '18 at 19:08










  • As you can see in my edit, you can use partial to get that symbol.
    – B. Mehta
    Nov 26 '18 at 19:27


















  • Are the $d$s in your question intended to be $partial$?
    – B. Mehta
    Nov 26 '18 at 19:08










  • Yes they are, i dont know how to get the partial symbol
    – pablo_mathscobar
    Nov 26 '18 at 19:08










  • As you can see in my edit, you can use partial to get that symbol.
    – B. Mehta
    Nov 26 '18 at 19:27
















Are the $d$s in your question intended to be $partial$?
– B. Mehta
Nov 26 '18 at 19:08




Are the $d$s in your question intended to be $partial$?
– B. Mehta
Nov 26 '18 at 19:08












Yes they are, i dont know how to get the partial symbol
– pablo_mathscobar
Nov 26 '18 at 19:08




Yes they are, i dont know how to get the partial symbol
– pablo_mathscobar
Nov 26 '18 at 19:08












As you can see in my edit, you can use partial to get that symbol.
– B. Mehta
Nov 26 '18 at 19:27




As you can see in my edit, you can use partial to get that symbol.
– B. Mehta
Nov 26 '18 at 19:27










1 Answer
1






active

oldest

votes


















0














You applied the chain rule incorrectly.



Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.






share|cite|improve this answer





















  • thank you, i was making a stupid mistake
    – pablo_mathscobar
    Nov 26 '18 at 19:17











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014765%2fd2u-dx2-d2u-dy2-fx-y-rightarrow-d2u-d%25ce%25bed%25ce%25b7-frac14f-frac1%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









0














You applied the chain rule incorrectly.



Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.






share|cite|improve this answer





















  • thank you, i was making a stupid mistake
    – pablo_mathscobar
    Nov 26 '18 at 19:17
















0














You applied the chain rule incorrectly.



Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.






share|cite|improve this answer





















  • thank you, i was making a stupid mistake
    – pablo_mathscobar
    Nov 26 '18 at 19:17














0












0








0






You applied the chain rule incorrectly.



Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.






share|cite|improve this answer












You applied the chain rule incorrectly.



Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 19:13









MisterRiemann

5,7791624




5,7791624












  • thank you, i was making a stupid mistake
    – pablo_mathscobar
    Nov 26 '18 at 19:17


















  • thank you, i was making a stupid mistake
    – pablo_mathscobar
    Nov 26 '18 at 19:17
















thank you, i was making a stupid mistake
– pablo_mathscobar
Nov 26 '18 at 19:17




thank you, i was making a stupid mistake
– pablo_mathscobar
Nov 26 '18 at 19:17


















draft saved

draft discarded




















































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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014765%2fd2u-dx2-d2u-dy2-fx-y-rightarrow-d2u-d%25ce%25bed%25ce%25b7-frac14f-frac1%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more