Relation between second derivatives and mixed derivatives

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

I'm studying the first chapter of Fourier Analysis -- An Introduction and unfortunately I don't understand some transitions.

There is an equation given (wave equation):
$$fracpartial^2upartial t^2 = fracpartial^2upartial x^2. tag1$$
Moreover we know that:
$$u(x,t) = F(x+t) + G(x-t)$$
is the solution of the equation $(1)$, where $F, G$ are twice differentiable functions.

We are to show that every solution takes this form.

Now we define new variables:

1. $xi = x+t$,


2. $eta = x-t$.

We define a new function:
$$v(xi, eta) = u(x,t).$$
Everything is quite obvious for the time being.

I don't understand the next step however. My book says, that: The change of variables formula shows that v satisfies:
$$fracpartial^2vpartial xi partial eta =0. tag2$$
I don't know why. I would appreciate any explanation.

My book also says that integrating $(2)$ twice will give us:
$$v(xi, eta) = F(xi) + G(eta).$$
Here again I don't know what the integration should look like.

Thanks for any help!

differential-equations derivatives wave-equation

share|cite|improve this question

edited Jul 18 at 20:29

hardmath

28.2k94592

asked Jul 18 at 20:16

Hendrra

925313

add a comment | 

up vote
2
down vote

favorite

I'm studying the first chapter of Fourier Analysis -- An Introduction and unfortunately I don't understand some transitions.

There is an equation given (wave equation):
$$fracpartial^2upartial t^2 = fracpartial^2upartial x^2. tag1$$
Moreover we know that:
$$u(x,t) = F(x+t) + G(x-t)$$
is the solution of the equation $(1)$, where $F, G$ are twice differentiable functions.

We are to show that every solution takes this form.

Now we define new variables:

1. $xi = x+t$,


2. $eta = x-t$.

We define a new function:
$$v(xi, eta) = u(x,t).$$
Everything is quite obvious for the time being.

I don't understand the next step however. My book says, that: The change of variables formula shows that v satisfies:
$$fracpartial^2vpartial xi partial eta =0. tag2$$
I don't know why. I would appreciate any explanation.

My book also says that integrating $(2)$ twice will give us:
$$v(xi, eta) = F(xi) + G(eta).$$
Here again I don't know what the integration should look like.

Thanks for any help!

differential-equations derivatives wave-equation

share|cite|improve this question

edited Jul 18 at 20:29

hardmath

28.2k94592

asked Jul 18 at 20:16

Hendrra

925313

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I'm studying the first chapter of Fourier Analysis -- An Introduction and unfortunately I don't understand some transitions.

There is an equation given (wave equation):
$$fracpartial^2upartial t^2 = fracpartial^2upartial x^2. tag1$$
Moreover we know that:
$$u(x,t) = F(x+t) + G(x-t)$$
is the solution of the equation $(1)$, where $F, G$ are twice differentiable functions.

We are to show that every solution takes this form.

Now we define new variables:

1. $xi = x+t$,


2. $eta = x-t$.

We define a new function:
$$v(xi, eta) = u(x,t).$$
Everything is quite obvious for the time being.

I don't understand the next step however. My book says, that: The change of variables formula shows that v satisfies:
$$fracpartial^2vpartial xi partial eta =0. tag2$$
I don't know why. I would appreciate any explanation.

My book also says that integrating $(2)$ twice will give us:
$$v(xi, eta) = F(xi) + G(eta).$$
Here again I don't know what the integration should look like.

Thanks for any help!

differential-equations derivatives wave-equation

share|cite|improve this question

edited Jul 18 at 20:29

hardmath

28.2k94592

asked Jul 18 at 20:16

Hendrra

925313

I'm studying the first chapter of Fourier Analysis -- An Introduction and unfortunately I don't understand some transitions.

There is an equation given (wave equation):
$$fracpartial^2upartial t^2 = fracpartial^2upartial x^2. tag1$$
Moreover we know that:
$$u(x,t) = F(x+t) + G(x-t)$$
is the solution of the equation $(1)$, where $F, G$ are twice differentiable functions.

We are to show that every solution takes this form.

Now we define new variables:

1. $xi = x+t$,


2. $eta = x-t$.

We define a new function:
$$v(xi, eta) = u(x,t).$$
Everything is quite obvious for the time being.

I don't understand the next step however. My book says, that: The change of variables formula shows that v satisfies:
$$fracpartial^2vpartial xi partial eta =0. tag2$$
I don't know why. I would appreciate any explanation.

My book also says that integrating $(2)$ twice will give us:
$$v(xi, eta) = F(xi) + G(eta).$$
Here again I don't know what the integration should look like.

Thanks for any help!

differential-equations derivatives wave-equation

share|cite|improve this question

edited Jul 18 at 20:29

hardmath

28.2k94592

asked Jul 18 at 20:16

Hendrra

925313

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 20:29

hardmath

28.2k94592

edited Jul 18 at 20:29

hardmath

28.2k94592

edited Jul 18 at 20:29

hardmath

28.2k94592

28.2k94592

asked Jul 18 at 20:16

Hendrra

925313

asked Jul 18 at 20:16

Hendrra

925313

asked Jul 18 at 20:16

Hendrra

925313

925313

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

That's because of this $$dfracpartial upartial x=dfracpartial upartial etadfracpartial etapartial x+dfracpartial upartial xidfracpartial xipartial x=dfracpartial upartial eta+dfracpartial upartial xi$$therefore$$dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x=dfracpartial^2 upartialeta^2+2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$similarly $$dfracpartialpartial tdfracpartial upartial t=dfracpartial^2 upartialeta^2-2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$therefore $$dfracpartial^2 upartial t^2=dfracpartial^2 upartial x^2$$ yields to $$dfracpartial^2 upartialxipartialeta=0$$

share|cite|improve this answer

edited Jul 18 at 22:05

answered Jul 18 at 21:15

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you dear @Isham...
  – Mostafa Ayaz
  Jul 18 at 22:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1 yw @mostafa .....
  – Isham
  Jul 18 at 22:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! I wonder however how does $dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x$ work?
  – Hendrra
  Jul 19 at 18:02
  

  
  
  
  

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2855957%2frelation-between-second-derivatives-and-mixed-derivatives%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

That's because of this $$dfracpartial upartial x=dfracpartial upartial etadfracpartial etapartial x+dfracpartial upartial xidfracpartial xipartial x=dfracpartial upartial eta+dfracpartial upartial xi$$therefore$$dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x=dfracpartial^2 upartialeta^2+2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$similarly $$dfracpartialpartial tdfracpartial upartial t=dfracpartial^2 upartialeta^2-2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$therefore $$dfracpartial^2 upartial t^2=dfracpartial^2 upartial x^2$$ yields to $$dfracpartial^2 upartialxipartialeta=0$$

share|cite|improve this answer

edited Jul 18 at 22:05

answered Jul 18 at 21:15

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you dear @Isham...
  – Mostafa Ayaz
  Jul 18 at 22:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1 yw @mostafa .....
  – Isham
  Jul 18 at 22:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! I wonder however how does $dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x$ work?
  – Hendrra
  Jul 19 at 18:02
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

That's because of this $$dfracpartial upartial x=dfracpartial upartial etadfracpartial etapartial x+dfracpartial upartial xidfracpartial xipartial x=dfracpartial upartial eta+dfracpartial upartial xi$$therefore$$dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x=dfracpartial^2 upartialeta^2+2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$similarly $$dfracpartialpartial tdfracpartial upartial t=dfracpartial^2 upartialeta^2-2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$therefore $$dfracpartial^2 upartial t^2=dfracpartial^2 upartial x^2$$ yields to $$dfracpartial^2 upartialxipartialeta=0$$

share|cite|improve this answer

edited Jul 18 at 22:05

answered Jul 18 at 21:15

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you dear @Isham...
  – Mostafa Ayaz
  Jul 18 at 22:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1 yw @mostafa .....
  – Isham
  Jul 18 at 22:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! I wonder however how does $dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x$ work?
  – Hendrra
  Jul 19 at 18:02
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

That's because of this $$dfracpartial upartial x=dfracpartial upartial etadfracpartial etapartial x+dfracpartial upartial xidfracpartial xipartial x=dfracpartial upartial eta+dfracpartial upartial xi$$therefore$$dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x=dfracpartial^2 upartialeta^2+2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$similarly $$dfracpartialpartial tdfracpartial upartial t=dfracpartial^2 upartialeta^2-2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$therefore $$dfracpartial^2 upartial t^2=dfracpartial^2 upartial x^2$$ yields to $$dfracpartial^2 upartialxipartialeta=0$$

share|cite|improve this answer

edited Jul 18 at 22:05

answered Jul 18 at 21:15

Mostafa Ayaz

8,6023630

That's because of this $$dfracpartial upartial x=dfracpartial upartial etadfracpartial etapartial x+dfracpartial upartial xidfracpartial xipartial x=dfracpartial upartial eta+dfracpartial upartial xi$$therefore$$dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x=dfracpartial^2 upartialeta^2+2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$similarly $$dfracpartialpartial tdfracpartial upartial t=dfracpartial^2 upartialeta^2-2dfracpartial^2 upartialxipartialeta+dfracpartial^2 upartialxi^2$$therefore $$dfracpartial^2 upartial t^2=dfracpartial^2 upartial x^2$$ yields to $$dfracpartial^2 upartialxipartialeta=0$$

share|cite|improve this answer

edited Jul 18 at 22:05

answered Jul 18 at 21:15

Mostafa Ayaz

8,6023630

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 22:05

edited Jul 18 at 22:05

edited Jul 18 at 22:05

answered Jul 18 at 21:15

Mostafa Ayaz

8,6023630

answered Jul 18 at 21:15

Mostafa Ayaz

8,6023630

answered Jul 18 at 21:15

Mostafa Ayaz

8,6023630

8,6023630

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you dear @Isham...
  – Mostafa Ayaz
  Jul 18 at 22:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1 yw @mostafa .....
  – Isham
  Jul 18 at 22:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! I wonder however how does $dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x$ work?
  – Hendrra
  Jul 19 at 18:02
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you dear @Isham...
  – Mostafa Ayaz
  Jul 18 at 22:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1 yw @mostafa .....
  – Isham
  Jul 18 at 22:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! I wonder however how does $dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x$ work?
  – Hendrra
  Jul 19 at 18:02
  

  
  
  
  

1

1

Thank you dear @Isham...
– Mostafa Ayaz
Jul 18 at 22:06

Thank you dear @Isham...
– Mostafa Ayaz
Jul 18 at 22:06

1

1

+1 yw @mostafa .....
– Isham
Jul 18 at 22:06

+1 yw @mostafa .....
– Isham
Jul 18 at 22:06

Thank you! I wonder however how does $dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x$ work?
– Hendrra
Jul 19 at 18:02

Thank you! I wonder however how does $dfracpartialpartial xdfracpartial upartial x=dfracpartialpartial etadfracpartial upartial x+dfracpartialpartial xidfracpartial upartial x$ work?
– Hendrra
Jul 19 at 18:02

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2855957%2frelation-between-second-derivatives-and-mixed-derivatives%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Name Email

Name Email

Name

Email

Name Email

Name

Email