$-y''+(1+x)y= lambda y$ for which value of $lambda$ the eqn has nonzero solution

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The problem is the following
$$begincases-y''+(1+x)y= λy\
y(0)=y(1)=0\ xin(0,1)endcases$$ For which value if $lambda$ the problem has nonzero solutions?



The answers given are



  1. for all $lambda<0$

  2. for all $lambdain[0,1]$.

  3. for some $lambdain(2,infty)$

  4. for a countable number of $lambda$'s






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  • What have you tried so far? Are there theorems in your course you can use, which might help you? For example something about eigenvalues of linear elliptic operators?
    – humanStampedist
    Aug 22 at 8:25










  • What about your work?
    – Cesareo
    Aug 22 at 8:31










  • Same question in math.stackexchange.com/questions/1838947/…, math.stackexchange.com/questions/2318378/…, similar question in math.stackexchange.com/questions/2686316/….
    – LutzL
    Aug 22 at 9:17














up vote
0
down vote

favorite












The problem is the following
$$begincases-y''+(1+x)y= λy\
y(0)=y(1)=0\ xin(0,1)endcases$$ For which value if $lambda$ the problem has nonzero solutions?



The answers given are



  1. for all $lambda<0$

  2. for all $lambdain[0,1]$.

  3. for some $lambdain(2,infty)$

  4. for a countable number of $lambda$'s






share|cite|improve this question






















  • What have you tried so far? Are there theorems in your course you can use, which might help you? For example something about eigenvalues of linear elliptic operators?
    – humanStampedist
    Aug 22 at 8:25










  • What about your work?
    – Cesareo
    Aug 22 at 8:31










  • Same question in math.stackexchange.com/questions/1838947/…, math.stackexchange.com/questions/2318378/…, similar question in math.stackexchange.com/questions/2686316/….
    – LutzL
    Aug 22 at 9:17












up vote
0
down vote

favorite









up vote
0
down vote

favorite











The problem is the following
$$begincases-y''+(1+x)y= λy\
y(0)=y(1)=0\ xin(0,1)endcases$$ For which value if $lambda$ the problem has nonzero solutions?



The answers given are



  1. for all $lambda<0$

  2. for all $lambdain[0,1]$.

  3. for some $lambdain(2,infty)$

  4. for a countable number of $lambda$'s






share|cite|improve this question














The problem is the following
$$begincases-y''+(1+x)y= λy\
y(0)=y(1)=0\ xin(0,1)endcases$$ For which value if $lambda$ the problem has nonzero solutions?



The answers given are



  1. for all $lambda<0$

  2. for all $lambdain[0,1]$.

  3. for some $lambdain(2,infty)$

  4. for a countable number of $lambda$'s








share|cite|improve this question













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edited Aug 22 at 8:42









Davide Morgante

2,480423




2,480423










asked Aug 22 at 8:01









user532616

12




12











  • What have you tried so far? Are there theorems in your course you can use, which might help you? For example something about eigenvalues of linear elliptic operators?
    – humanStampedist
    Aug 22 at 8:25










  • What about your work?
    – Cesareo
    Aug 22 at 8:31










  • Same question in math.stackexchange.com/questions/1838947/…, math.stackexchange.com/questions/2318378/…, similar question in math.stackexchange.com/questions/2686316/….
    – LutzL
    Aug 22 at 9:17
















  • What have you tried so far? Are there theorems in your course you can use, which might help you? For example something about eigenvalues of linear elliptic operators?
    – humanStampedist
    Aug 22 at 8:25










  • What about your work?
    – Cesareo
    Aug 22 at 8:31










  • Same question in math.stackexchange.com/questions/1838947/…, math.stackexchange.com/questions/2318378/…, similar question in math.stackexchange.com/questions/2686316/….
    – LutzL
    Aug 22 at 9:17















What have you tried so far? Are there theorems in your course you can use, which might help you? For example something about eigenvalues of linear elliptic operators?
– humanStampedist
Aug 22 at 8:25




What have you tried so far? Are there theorems in your course you can use, which might help you? For example something about eigenvalues of linear elliptic operators?
– humanStampedist
Aug 22 at 8:25












What about your work?
– Cesareo
Aug 22 at 8:31




What about your work?
– Cesareo
Aug 22 at 8:31












Same question in math.stackexchange.com/questions/1838947/…, math.stackexchange.com/questions/2318378/…, similar question in math.stackexchange.com/questions/2686316/….
– LutzL
Aug 22 at 9:17




Same question in math.stackexchange.com/questions/1838947/…, math.stackexchange.com/questions/2318378/…, similar question in math.stackexchange.com/questions/2686316/….
– LutzL
Aug 22 at 9:17










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Just a partial answer: Multiply your equation with $y$ and integrate. Then you obtain
$$-int_0^1 y'' ydx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx$$
Now use integration by parts on the first integral to get
$$-[y'y]_0^1 + int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx,$$
which simplifies by the boundary values to
$$int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx.$$
$xin[0,1]$ further yields
$$int_0^1y^2, dxleq lambda int_0^1 y^2dx,$$
which exclude option 1) and 2). For $lambda=1$ you need go through the argument again and proceed by contradiction.



Existence can be done in a more general context. Look up Gilbarg/Trudinger's book on elliptic partial differential equations of second order. It should be Theorem 8.37.
This is a reference to the book:
https://www.springer.com/de/book/9783540411604






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    1 Answer
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    active

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    1 Answer
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    active

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    up vote
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    Just a partial answer: Multiply your equation with $y$ and integrate. Then you obtain
    $$-int_0^1 y'' ydx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx$$
    Now use integration by parts on the first integral to get
    $$-[y'y]_0^1 + int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx,$$
    which simplifies by the boundary values to
    $$int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx.$$
    $xin[0,1]$ further yields
    $$int_0^1y^2, dxleq lambda int_0^1 y^2dx,$$
    which exclude option 1) and 2). For $lambda=1$ you need go through the argument again and proceed by contradiction.



    Existence can be done in a more general context. Look up Gilbarg/Trudinger's book on elliptic partial differential equations of second order. It should be Theorem 8.37.
    This is a reference to the book:
    https://www.springer.com/de/book/9783540411604






    share|cite|improve this answer
























      up vote
      1
      down vote













      Just a partial answer: Multiply your equation with $y$ and integrate. Then you obtain
      $$-int_0^1 y'' ydx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx$$
      Now use integration by parts on the first integral to get
      $$-[y'y]_0^1 + int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx,$$
      which simplifies by the boundary values to
      $$int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx.$$
      $xin[0,1]$ further yields
      $$int_0^1y^2, dxleq lambda int_0^1 y^2dx,$$
      which exclude option 1) and 2). For $lambda=1$ you need go through the argument again and proceed by contradiction.



      Existence can be done in a more general context. Look up Gilbarg/Trudinger's book on elliptic partial differential equations of second order. It should be Theorem 8.37.
      This is a reference to the book:
      https://www.springer.com/de/book/9783540411604






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
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        Just a partial answer: Multiply your equation with $y$ and integrate. Then you obtain
        $$-int_0^1 y'' ydx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx$$
        Now use integration by parts on the first integral to get
        $$-[y'y]_0^1 + int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx,$$
        which simplifies by the boundary values to
        $$int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx.$$
        $xin[0,1]$ further yields
        $$int_0^1y^2, dxleq lambda int_0^1 y^2dx,$$
        which exclude option 1) and 2). For $lambda=1$ you need go through the argument again and proceed by contradiction.



        Existence can be done in a more general context. Look up Gilbarg/Trudinger's book on elliptic partial differential equations of second order. It should be Theorem 8.37.
        This is a reference to the book:
        https://www.springer.com/de/book/9783540411604






        share|cite|improve this answer












        Just a partial answer: Multiply your equation with $y$ and integrate. Then you obtain
        $$-int_0^1 y'' ydx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx$$
        Now use integration by parts on the first integral to get
        $$-[y'y]_0^1 + int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx,$$
        which simplifies by the boundary values to
        $$int_0^1(y')^2dx + int_0^1(1+x)y^2dx = lambda int_0^1y^2dx.$$
        $xin[0,1]$ further yields
        $$int_0^1y^2, dxleq lambda int_0^1 y^2dx,$$
        which exclude option 1) and 2). For $lambda=1$ you need go through the argument again and proceed by contradiction.



        Existence can be done in a more general context. Look up Gilbarg/Trudinger's book on elliptic partial differential equations of second order. It should be Theorem 8.37.
        This is a reference to the book:
        https://www.springer.com/de/book/9783540411604







        share|cite|improve this answer












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        answered Aug 22 at 8:45









        humanStampedist

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