Fdtxjr

Should a narrator ever describe things based on a character's view instead of facts?

Has the laser at Magurele, Romania reached a tenth of the Sun's power?

What can I do if I am asked to learn different programming languages very frequently?

categorizing a variable turns it from insignificant to significant

"Marked down as someone wanting to sell shares." What does that mean?

Relations between homogeneous polynomials

Why does the frost depth increase when the surface temperature warms up?

Should I warn a new PhD Student?

Is divisi notation needed for brass or woodwind in an orchestra?

Trouble reading roman numeral notation with flats

Are hand made posters acceptable in Academia?

How to get directions in deep space?

How can a new country break out from a developed country without war?

Taking the numerator and the denominator

How to test the sharpness of a knife?

Why does a 97 / 92 key piano exist by Bosendorfer?

Sort with assumptions

Turning a hard to access nut?

Travelling in US for more than 90 days

How do I prevent inappropriate ads from appearing in my game?

Would this string work as string?

Is this saw blade faulty?

What properties make a magic weapon befit a Rogue more than a DEX-based Fighter?

What is this high flying aircraft over Pennsylvania?

Change of coordinates breaks the equality of mixed partial derivatives

Transforming hyperbolic PDE into normal formHow to construct a counterexample for the reduction of linear second order PDE in several variablesReducing heat equation into nondimensional formChange of variable, calculation of new partial derivative expressionGeneral Solution for $x^2u_xx-y^2u_yy=0$Change of Variables - independence of variablesUnderstanding how to obtain the substitution to reduce a PDE to canonical formTrying to understand hyperbolic canonical form transformationVariable transformation PDEWave equation in hyperbolic canonical form

1

1

$begingroup$

I'm trying to reduce the PDE

$$u_xx + 2 u_xy + u_yy = 0,$$
to its canonical form $u_eta eta = 0.$

According to the book that I'm using, $epsilon (x,y) = y-x$ and $eta (x,y) = x$ is a valid coordinate transformation, so lets use it.

After doing the algebra,

$$u_xx = -1[u_epsilon epsilon * (-1) + u_epsilon eta ] + [(-1)* u_eta epsilon + u_eta eta],$$

$$u_yy = u_epsilon epsilon$$

$$u_xy = (-1)[(-1)u_epsilon epsilon] + [u_eta epsilon],$$
but $$u_yx = (-1) u_epsilon epsilon + u_epsilon eta not = u_xy.$$

So, what is wrong in with this transformation ?, and why the mixed partials are not equal in $epsilon, eta$ coordinates ?

I mean, I plug everything into the PDE, by $u_xy$ does not transforms the PDE into its canonical form, but $u_yx $ does.

I checked my calculations, but I couldn't find an error either.

ordinary-differential-equations pde

share|cite|improve this question

asked Mar 13 at 18:20

onurcanbektasonurcanbektas

3,48411037

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Where dd the extra factor of $-1$ in $u_xy$ come from?
  $endgroup$
  – amd
  Mar 13 at 20:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amd You are right, I that should be a plus! .Thanks a lot for pointing out. if you post your comment as an answer, I will accept.
  $endgroup$
  – onurcanbektas
  Mar 14 at 4:28
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I'm trying to reduce the PDE

$$u_xx + 2 u_xy + u_yy = 0,$$
to its canonical form $u_eta eta = 0.$

According to the book that I'm using, $epsilon (x,y) = y-x$ and $eta (x,y) = x$ is a valid coordinate transformation, so lets use it.

After doing the algebra,

$$u_xx = -1[u_epsilon epsilon * (-1) + u_epsilon eta ] + [(-1)* u_eta epsilon + u_eta eta],$$

$$u_yy = u_epsilon epsilon$$

$$u_xy = (-1)[(-1)u_epsilon epsilon] + [u_eta epsilon],$$
but $$u_yx = (-1) u_epsilon epsilon + u_epsilon eta not = u_xy.$$

So, what is wrong in with this transformation ?, and why the mixed partials are not equal in $epsilon, eta$ coordinates ?

I mean, I plug everything into the PDE, by $u_xy$ does not transforms the PDE into its canonical form, but $u_yx $ does.

I checked my calculations, but I couldn't find an error either.

ordinary-differential-equations pde

share|cite|improve this question

asked Mar 13 at 18:20

onurcanbektasonurcanbektas

3,48411037

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Where dd the extra factor of $-1$ in $u_xy$ come from?
  $endgroup$
  – amd
  Mar 13 at 20:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amd You are right, I that should be a plus! .Thanks a lot for pointing out. if you post your comment as an answer, I will accept.
  $endgroup$
  – onurcanbektas
  Mar 14 at 4:28
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to reduce the PDE

$$u_xx + 2 u_xy + u_yy = 0,$$
to its canonical form $u_eta eta = 0.$

According to the book that I'm using, $epsilon (x,y) = y-x$ and $eta (x,y) = x$ is a valid coordinate transformation, so lets use it.

After doing the algebra,

$$u_xx = -1[u_epsilon epsilon * (-1) + u_epsilon eta ] + [(-1)* u_eta epsilon + u_eta eta],$$

$$u_yy = u_epsilon epsilon$$

$$u_xy = (-1)[(-1)u_epsilon epsilon] + [u_eta epsilon],$$
but $$u_yx = (-1) u_epsilon epsilon + u_epsilon eta not = u_xy.$$

So, what is wrong in with this transformation ?, and why the mixed partials are not equal in $epsilon, eta$ coordinates ?

I mean, I plug everything into the PDE, by $u_xy$ does not transforms the PDE into its canonical form, but $u_yx $ does.

I checked my calculations, but I couldn't find an error either.

ordinary-differential-equations pde

share|cite|improve this question

asked Mar 13 at 18:20

onurcanbektasonurcanbektas

3,48411037

$endgroup$

share|cite|improve this question

asked Mar 13 at 18:20

onurcanbektasonurcanbektas

3,48411037

share|cite|improve this question

asked Mar 13 at 18:20

onurcanbektasonurcanbektas

3,48411037

asked Mar 13 at 18:20

onurcanbektasonurcanbektas

3,48411037

asked Mar 13 at 18:20

onurcanbektasonurcanbektas

3,48411037

1 Answer
1

active

oldest

votes

2

$begingroup$

It appears that you’ve got a stray factor of $-1$ in your expression for $u_xy$.

We have $$partial over partial x = partialepsilon over partial xpartial over partialepsilon + partialeta over partial xpartial over partialeta = -partial over partialepsilon + partial over partialeta \ partial over partial y = partialepsilon over partial ypartial over partialepsilon+partialeta over partial ypartial over partialeta = partial over partialepsilon$$ so, assuming equality of mixed partials with respect to $epsilon$ and $eta$, $$partial^2 over partial x , partial y = left(-partial over partialepsilon + partial over partialetaright)partial over partialepsilon = -partial^2 over partialepsilon^2+partial^2 over partialeta,partialepsilon = -partial^2 over partialepsilon^2+partial^2 over partialepsilon,partialeta = partial over partialepsilonleft(-partial over partialepsilon + partial over partialetaright) = partial^2 over partial y , partial x.$$

share|cite|improve this answer

answered Mar 14 at 6:51

amdamd

31.1k21051

$endgroup$

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146971%2fchange-of-coordinates-breaks-the-equality-of-mixed-partial-derivatives%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

It appears that you’ve got a stray factor of $-1$ in your expression for $u_xy$.

We have $$partial over partial x = partialepsilon over partial xpartial over partialepsilon + partialeta over partial xpartial over partialeta = -partial over partialepsilon + partial over partialeta \ partial over partial y = partialepsilon over partial ypartial over partialepsilon+partialeta over partial ypartial over partialeta = partial over partialepsilon$$ so, assuming equality of mixed partials with respect to $epsilon$ and $eta$, $$partial^2 over partial x , partial y = left(-partial over partialepsilon + partial over partialetaright)partial over partialepsilon = -partial^2 over partialepsilon^2+partial^2 over partialeta,partialepsilon = -partial^2 over partialepsilon^2+partial^2 over partialepsilon,partialeta = partial over partialepsilonleft(-partial over partialepsilon + partial over partialetaright) = partial^2 over partial y , partial x.$$

share|cite|improve this answer

answered Mar 14 at 6:51

amdamd

31.1k21051

$endgroup$

add a comment | 

2

$begingroup$

It appears that you’ve got a stray factor of $-1$ in your expression for $u_xy$.

We have $$partial over partial x = partialepsilon over partial xpartial over partialepsilon + partialeta over partial xpartial over partialeta = -partial over partialepsilon + partial over partialeta \ partial over partial y = partialepsilon over partial ypartial over partialepsilon+partialeta over partial ypartial over partialeta = partial over partialepsilon$$ so, assuming equality of mixed partials with respect to $epsilon$ and $eta$, $$partial^2 over partial x , partial y = left(-partial over partialepsilon + partial over partialetaright)partial over partialepsilon = -partial^2 over partialepsilon^2+partial^2 over partialeta,partialepsilon = -partial^2 over partialepsilon^2+partial^2 over partialepsilon,partialeta = partial over partialepsilonleft(-partial over partialepsilon + partial over partialetaright) = partial^2 over partial y , partial x.$$

share|cite|improve this answer

answered Mar 14 at 6:51

amdamd

31.1k21051

$endgroup$

add a comment | 

$begingroup$

It appears that you’ve got a stray factor of $-1$ in your expression for $u_xy$.

We have $$partial over partial x = partialepsilon over partial xpartial over partialepsilon + partialeta over partial xpartial over partialeta = -partial over partialepsilon + partial over partialeta \ partial over partial y = partialepsilon over partial ypartial over partialepsilon+partialeta over partial ypartial over partialeta = partial over partialepsilon$$ so, assuming equality of mixed partials with respect to $epsilon$ and $eta$, $$partial^2 over partial x , partial y = left(-partial over partialepsilon + partial over partialetaright)partial over partialepsilon = -partial^2 over partialepsilon^2+partial^2 over partialeta,partialepsilon = -partial^2 over partialepsilon^2+partial^2 over partialepsilon,partialeta = partial over partialepsilonleft(-partial over partialepsilon + partial over partialetaright) = partial^2 over partial y , partial x.$$

share|cite|improve this answer

answered Mar 14 at 6:51

amdamd

31.1k21051

$endgroup$

share|cite|improve this answer

answered Mar 14 at 6:51

amdamd

31.1k21051

answered Mar 14 at 6:51

amdamd

31.1k21051

answered Mar 14 at 6:51

amdamd

31.1k21051