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Eigenvalues and eigenfunctions of the Fredholm integral equation of the second kind

Homogeneous Fredholm Equation of Second KindSpectrum of eigenvalues and eigenfunctionsFinding eigenvalues and eigenfunctions for a BVPEigenvalues and Eigenfunctions of Integral EquationHow to find eigenvalues and eigenfunctions of this boundary value problem?Finding eigenvalues and eigenfunctions of this boundary value problemFind the eigenvalues and eigenfunctions for $y''+lambda y=0$ where $y'(1)=0$ and $y'(2)=0$eigenfunctions $y''+lambda y=0$ and $y^prime (0)=0$ , $y^prime (1)=0$The eigenvalues and the eigenvectors of $y^(4) = lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0$Eigenvalues and eigenvectors of the second derivative

1

1

$begingroup$

I'm trying to solve the following question:

Find the eigenfunctions and eigenvalues of the following integral operator:

beginequation
Ku(x) = int_0^pi(sin(x)sin(y)+alphacos(x)cos(y))u(y) dy.
endequation

My attempted solution is as follows:

Let $(lambda, u)$ be such an eigenpair. Then,

beginalign
lambda u(x) & = int_0^pi(sin(x)sin(y)+alphacos(x)cos(y))u(y)dy \
& = sin(x)int_0^pisin(y)u(y)dy + alpha cos(x)int_0^pisin(y)u(y)dy \
endalign

We convert the above equation into a differential equation. Differentiating twice, we get,

beginequation
lambda u^''(x) = - u(x).
endequation

As $lambda = 0$ is clearly not an eigenvalue (if not, then the only possible eigenfunction is $u = 0$ almost everywhere), we have the general solution is,

beginequation
u(x) = Asinbigg(fracxsqrtlambdabigg) + Bcosbigg(fracxsqrtlambdabigg).
endequation

We need two boundary conditions. From the eigenvalue equation above, we have,

beginalign
u(0) & = fracalphalambda int_0^pisin(y)u(y)dy \
u(pi) & = frac-alphalambda int_0^pisin(y)u(y)dy \
endalign

However, I don't how to proceed from here. The boundary conditions above are functions of $u$.

Please provide relevant hints. This is a homework problem, and I would like to finish it on my own.

functional-analysis eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 11 at 10:08

user82261

asked Mar 11 at 9:44

user82261user82261

26517

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You got $u(0)=-u(pi)$. One more boundary condition and you will be able to eliminate $A$ and $B$ in the general solution, concluding the exercise.
  $endgroup$
  – Giuseppe Negro
  Mar 11 at 10:14
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I'm trying to solve the following question:

Find the eigenfunctions and eigenvalues of the following integral operator:

beginequation
Ku(x) = int_0^pi(sin(x)sin(y)+alphacos(x)cos(y))u(y) dy.
endequation

My attempted solution is as follows:

Let $(lambda, u)$ be such an eigenpair. Then,

beginalign
lambda u(x) & = int_0^pi(sin(x)sin(y)+alphacos(x)cos(y))u(y)dy \
& = sin(x)int_0^pisin(y)u(y)dy + alpha cos(x)int_0^pisin(y)u(y)dy \
endalign

We convert the above equation into a differential equation. Differentiating twice, we get,

beginequation
lambda u^''(x) = - u(x).
endequation

As $lambda = 0$ is clearly not an eigenvalue (if not, then the only possible eigenfunction is $u = 0$ almost everywhere), we have the general solution is,

beginequation
u(x) = Asinbigg(fracxsqrtlambdabigg) + Bcosbigg(fracxsqrtlambdabigg).
endequation

We need two boundary conditions. From the eigenvalue equation above, we have,

beginalign
u(0) & = fracalphalambda int_0^pisin(y)u(y)dy \
u(pi) & = frac-alphalambda int_0^pisin(y)u(y)dy \
endalign

However, I don't how to proceed from here. The boundary conditions above are functions of $u$.

Please provide relevant hints. This is a homework problem, and I would like to finish it on my own.

functional-analysis eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 11 at 10:08

user82261

asked Mar 11 at 9:44

user82261user82261

26517

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You got $u(0)=-u(pi)$. One more boundary condition and you will be able to eliminate $A$ and $B$ in the general solution, concluding the exercise.
  $endgroup$
  – Giuseppe Negro
  Mar 11 at 10:14
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to solve the following question:

Find the eigenfunctions and eigenvalues of the following integral operator:

beginequation
Ku(x) = int_0^pi(sin(x)sin(y)+alphacos(x)cos(y))u(y) dy.
endequation

My attempted solution is as follows:

Let $(lambda, u)$ be such an eigenpair. Then,

beginalign
lambda u(x) & = int_0^pi(sin(x)sin(y)+alphacos(x)cos(y))u(y)dy \
& = sin(x)int_0^pisin(y)u(y)dy + alpha cos(x)int_0^pisin(y)u(y)dy \
endalign

We convert the above equation into a differential equation. Differentiating twice, we get,

beginequation
lambda u^''(x) = - u(x).
endequation

As $lambda = 0$ is clearly not an eigenvalue (if not, then the only possible eigenfunction is $u = 0$ almost everywhere), we have the general solution is,

beginequation
u(x) = Asinbigg(fracxsqrtlambdabigg) + Bcosbigg(fracxsqrtlambdabigg).
endequation

We need two boundary conditions. From the eigenvalue equation above, we have,

beginalign
u(0) & = fracalphalambda int_0^pisin(y)u(y)dy \
u(pi) & = frac-alphalambda int_0^pisin(y)u(y)dy \
endalign

However, I don't how to proceed from here. The boundary conditions above are functions of $u$.

Please provide relevant hints. This is a homework problem, and I would like to finish it on my own.

functional-analysis eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 11 at 10:08

user82261

asked Mar 11 at 9:44

user82261user82261

26517

$endgroup$

share|cite|improve this question

edited Mar 11 at 10:08

user82261

asked Mar 11 at 9:44

user82261user82261

26517

share|cite|improve this question

edited Mar 11 at 10:08

user82261

asked Mar 11 at 9:44

user82261user82261

26517

asked Mar 11 at 9:44

user82261user82261

26517

asked Mar 11 at 9:44

user82261user82261

26517

1 Answer
1

active

oldest

votes

0

$begingroup$

There is a error, you have that

beginalign
u(0) & = fracalphalambda int_0^picos(y)u(y)dy \
u(pi) & = frac-alphalambda int_0^picos(y)u(y)dy \
endalign

So $u(0)=-u(pi)$. Another condition can be

beginalign
u(fracpi2) & = fracalphalambda int_0^pisin(y)u(y)dy \
u(frac3pi2) & = frac-alphalambda int_0^pisin(y)u(y)dy \
endalign

So $u(fracpi2)=-u(frac3pi2)$

But

$u(0)=Asin(frac0sqrtlambda)+Bcos(frac0sqrtlambda)=B$

and

$u(pi)= Asin(fracpisqrtlambda)+Bcos(fracpisqrtlambda)$

So you have that

$Asin(fracpisqrtlambda) =-B(1+cos(fracpisqrtlambda))$

To the other hand

$u(fracpi2)=Asin(fracpi2sqrtlambda)+Bcos(fracpi2sqrtlambda) $

while

$u(frac3pi2)= Asin(frac3pi2sqrtlambda) +Bcos(frac3pi2sqrtlambda) $

so

$A(sin(fracpi2sqrtlambda)+sin(frac3pi2sqrtlambda))=-B(cos(fracpi2sqrtlambda) + cos(3fracpi2sqrtlambda) $

share|cite|improve this answer

edited Mar 11 at 10:56

answered Mar 11 at 10:23

Federico FalluccaFederico Fallucca

2,270210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get rid of the factor of $sin$ in the equation for $u(pi)$. We haven't assumed $lambda$ is an integer.
  $endgroup$
  – user82261
  Mar 11 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ops, you’re right. Thanks you
  $endgroup$
  – Federico Fallucca
  Mar 11 at 10:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can't quite figure out how to deal with the case $A neq 0, B neq 0$. In that case, I have two equations in three unknowns, $A, B$ and $lambda$.
  $endgroup$
  – user82261
  Mar 11 at 10:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think I get you mean you mean. We can plug the general solution in the eigenvalue equation to make the above argument sufficient as well, and get the answer.
  $endgroup$
  – user82261
  Mar 11 at 10:58
  

  
  
  
  

add a comment | 

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0

$begingroup$

There is a error, you have that

beginalign
u(0) & = fracalphalambda int_0^picos(y)u(y)dy \
u(pi) & = frac-alphalambda int_0^picos(y)u(y)dy \
endalign

So $u(0)=-u(pi)$. Another condition can be

beginalign
u(fracpi2) & = fracalphalambda int_0^pisin(y)u(y)dy \
u(frac3pi2) & = frac-alphalambda int_0^pisin(y)u(y)dy \
endalign

So $u(fracpi2)=-u(frac3pi2)$

But

$u(0)=Asin(frac0sqrtlambda)+Bcos(frac0sqrtlambda)=B$

and

$u(pi)= Asin(fracpisqrtlambda)+Bcos(fracpisqrtlambda)$

So you have that

$Asin(fracpisqrtlambda) =-B(1+cos(fracpisqrtlambda))$

To the other hand

$u(fracpi2)=Asin(fracpi2sqrtlambda)+Bcos(fracpi2sqrtlambda) $

while

$u(frac3pi2)= Asin(frac3pi2sqrtlambda) +Bcos(frac3pi2sqrtlambda) $

so

$A(sin(fracpi2sqrtlambda)+sin(frac3pi2sqrtlambda))=-B(cos(fracpi2sqrtlambda) + cos(3fracpi2sqrtlambda) $

share|cite|improve this answer

edited Mar 11 at 10:56

answered Mar 11 at 10:23

Federico FalluccaFederico Fallucca

2,270210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get rid of the factor of $sin$ in the equation for $u(pi)$. We haven't assumed $lambda$ is an integer.
  $endgroup$
  – user82261
  Mar 11 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ops, you’re right. Thanks you
  $endgroup$
  – Federico Fallucca
  Mar 11 at 10:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can't quite figure out how to deal with the case $A neq 0, B neq 0$. In that case, I have two equations in three unknowns, $A, B$ and $lambda$.
  $endgroup$
  – user82261
  Mar 11 at 10:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think I get you mean you mean. We can plug the general solution in the eigenvalue equation to make the above argument sufficient as well, and get the answer.
  $endgroup$
  – user82261
  Mar 11 at 10:58
  

  
  
  
  

add a comment | 

0

$begingroup$

There is a error, you have that

beginalign
u(0) & = fracalphalambda int_0^picos(y)u(y)dy \
u(pi) & = frac-alphalambda int_0^picos(y)u(y)dy \
endalign

So $u(0)=-u(pi)$. Another condition can be

beginalign
u(fracpi2) & = fracalphalambda int_0^pisin(y)u(y)dy \
u(frac3pi2) & = frac-alphalambda int_0^pisin(y)u(y)dy \
endalign

So $u(fracpi2)=-u(frac3pi2)$

But

$u(0)=Asin(frac0sqrtlambda)+Bcos(frac0sqrtlambda)=B$

and

$u(pi)= Asin(fracpisqrtlambda)+Bcos(fracpisqrtlambda)$

So you have that

$Asin(fracpisqrtlambda) =-B(1+cos(fracpisqrtlambda))$

To the other hand

$u(fracpi2)=Asin(fracpi2sqrtlambda)+Bcos(fracpi2sqrtlambda) $

while

$u(frac3pi2)= Asin(frac3pi2sqrtlambda) +Bcos(frac3pi2sqrtlambda) $

so

$A(sin(fracpi2sqrtlambda)+sin(frac3pi2sqrtlambda))=-B(cos(fracpi2sqrtlambda) + cos(3fracpi2sqrtlambda) $

share|cite|improve this answer

edited Mar 11 at 10:56

answered Mar 11 at 10:23

Federico FalluccaFederico Fallucca

2,270210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get rid of the factor of $sin$ in the equation for $u(pi)$. We haven't assumed $lambda$ is an integer.
  $endgroup$
  – user82261
  Mar 11 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ops, you’re right. Thanks you
  $endgroup$
  – Federico Fallucca
  Mar 11 at 10:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can't quite figure out how to deal with the case $A neq 0, B neq 0$. In that case, I have two equations in three unknowns, $A, B$ and $lambda$.
  $endgroup$
  – user82261
  Mar 11 at 10:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think I get you mean you mean. We can plug the general solution in the eigenvalue equation to make the above argument sufficient as well, and get the answer.
  $endgroup$
  – user82261
  Mar 11 at 10:58
  

  
  
  
  

add a comment | 

$begingroup$

There is a error, you have that

beginalign
u(0) & = fracalphalambda int_0^picos(y)u(y)dy \
u(pi) & = frac-alphalambda int_0^picos(y)u(y)dy \
endalign

So $u(0)=-u(pi)$. Another condition can be

beginalign
u(fracpi2) & = fracalphalambda int_0^pisin(y)u(y)dy \
u(frac3pi2) & = frac-alphalambda int_0^pisin(y)u(y)dy \
endalign

So $u(fracpi2)=-u(frac3pi2)$

But

$u(0)=Asin(frac0sqrtlambda)+Bcos(frac0sqrtlambda)=B$

and

$u(pi)= Asin(fracpisqrtlambda)+Bcos(fracpisqrtlambda)$

So you have that

$Asin(fracpisqrtlambda) =-B(1+cos(fracpisqrtlambda))$

To the other hand

$u(fracpi2)=Asin(fracpi2sqrtlambda)+Bcos(fracpi2sqrtlambda) $

while

$u(frac3pi2)= Asin(frac3pi2sqrtlambda) +Bcos(frac3pi2sqrtlambda) $

so

$A(sin(fracpi2sqrtlambda)+sin(frac3pi2sqrtlambda))=-B(cos(fracpi2sqrtlambda) + cos(3fracpi2sqrtlambda) $

share|cite|improve this answer

edited Mar 11 at 10:56

answered Mar 11 at 10:23

Federico FalluccaFederico Fallucca

2,270210

$endgroup$

share|cite|improve this answer

edited Mar 11 at 10:56

answered Mar 11 at 10:23

Federico FalluccaFederico Fallucca

2,270210

answered Mar 11 at 10:23

Federico FalluccaFederico Fallucca

2,270210

answered Mar 11 at 10:23

Federico FalluccaFederico Fallucca

2,270210