Using the fact this equation is equidimensional. Calculate all associated eigenvalues.












1












$begingroup$


Let



$$x^2y''+xy'+lambda y=0,,,,,,y(1)=0,,,y(b)=0tag1$$



Using the fact this equation is equidimensional. Calculate all associated eigenvalues.



Is $lambda=0$ an eigenvalue?



Prove there exist infinite eigenvalues with one minimum, but there doesn't exist a maximum eigenvalue.



My attempt:



Note the equation $(1)$ is a Cauchy-Euler equation.



Suppose $y=x^r$ then $y'=r x^{r-1}$, $y''=r(r-1)x^{r-2}$.



Then



$$x^2r(r-1)x^{r-2}+xr x^{r-1}+r x^r=0implies(r^2-r)x^r=0iff r_1=0,,,, r_2=-1$$



Then the solution for the ODE is:
$$y(x)=C_1+C_2x^{-1}$$



Here I'm a little confused, to solve my problem. Can someone help me?






pde







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Don't use $x^lambda$, use $x^r$ where $r$ is to be determined (it won't be $lambda$).
    $endgroup$
    – Robert Israel
    Jan 2 at 18:54






  • 1




    $begingroup$
    Ok, i go to edit that @RobertIsrael
    $endgroup$
    – Bvss12
    Jan 2 at 18:59






  • 1




    $begingroup$
    Using $lambda$ for the power was the source of your confusion. The equation should be $r^2+lambda=0$
    $endgroup$
    – Dylan
    Jan 2 at 19:17










  • $begingroup$
    Ummm, yes you have reason. I go to solve that and update my question. @Dylan
    $endgroup$
    – Bvss12
    Jan 2 at 19:19
















1












$begingroup$


Let



$$x^2y''+xy'+lambda y=0,,,,,,y(1)=0,,,y(b)=0tag1$$



Using the fact this equation is equidimensional. Calculate all associated eigenvalues.



Is $lambda=0$ an eigenvalue?



Prove there exist infinite eigenvalues with one minimum, but there doesn't exist a maximum eigenvalue.



My attempt:



Note the equation $(1)$ is a Cauchy-Euler equation.



Suppose $y=x^r$ then $y'=r x^{r-1}$, $y''=r(r-1)x^{r-2}$.



Then



$$x^2r(r-1)x^{r-2}+xr x^{r-1}+r x^r=0implies(r^2-r)x^r=0iff r_1=0,,,, r_2=-1$$



Then the solution for the ODE is:
$$y(x)=C_1+C_2x^{-1}$$



Here I'm a little confused, to solve my problem. Can someone help me?






pde







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Don't use $x^lambda$, use $x^r$ where $r$ is to be determined (it won't be $lambda$).
    $endgroup$
    – Robert Israel
    Jan 2 at 18:54






  • 1




    $begingroup$
    Ok, i go to edit that @RobertIsrael
    $endgroup$
    – Bvss12
    Jan 2 at 18:59






  • 1




    $begingroup$
    Using $lambda$ for the power was the source of your confusion. The equation should be $r^2+lambda=0$
    $endgroup$
    – Dylan
    Jan 2 at 19:17










  • $begingroup$
    Ummm, yes you have reason. I go to solve that and update my question. @Dylan
    $endgroup$
    – Bvss12
    Jan 2 at 19:19














1












1








1





$begingroup$


Let



$$x^2y''+xy'+lambda y=0,,,,,,y(1)=0,,,y(b)=0tag1$$



Using the fact this equation is equidimensional. Calculate all associated eigenvalues.



Is $lambda=0$ an eigenvalue?



Prove there exist infinite eigenvalues with one minimum, but there doesn't exist a maximum eigenvalue.



My attempt:



Note the equation $(1)$ is a Cauchy-Euler equation.



Suppose $y=x^r$ then $y'=r x^{r-1}$, $y''=r(r-1)x^{r-2}$.



Then



$$x^2r(r-1)x^{r-2}+xr x^{r-1}+r x^r=0implies(r^2-r)x^r=0iff r_1=0,,,, r_2=-1$$



Then the solution for the ODE is:
$$y(x)=C_1+C_2x^{-1}$$



Here I'm a little confused, to solve my problem. Can someone help me?






pde







share|cite|improve this question











$endgroup$




Let



$$x^2y''+xy'+lambda y=0,,,,,,y(1)=0,,,y(b)=0tag1$$



Using the fact this equation is equidimensional. Calculate all associated eigenvalues.



Is $lambda=0$ an eigenvalue?



Prove there exist infinite eigenvalues with one minimum, but there doesn't exist a maximum eigenvalue.



My attempt:



Note the equation $(1)$ is a Cauchy-Euler equation.



Suppose $y=x^r$ then $y'=r x^{r-1}$, $y''=r(r-1)x^{r-2}$.



Then



$$x^2r(r-1)x^{r-2}+xr x^{r-1}+r x^r=0implies(r^2-r)x^r=0iff r_1=0,,,, r_2=-1$$



Then the solution for the ODE is:
$$y(x)=C_1+C_2x^{-1}$$



Here I'm a little confused, to solve my problem. Can someone help me?






pde




pde






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 19:03









Bernard

119k639112




119k639112










asked Jan 2 at 18:44









Bvss12Bvss12

1,785618




1,785618








  • 1




    $begingroup$
    Don't use $x^lambda$, use $x^r$ where $r$ is to be determined (it won't be $lambda$).
    $endgroup$
    – Robert Israel
    Jan 2 at 18:54






  • 1




    $begingroup$
    Ok, i go to edit that @RobertIsrael
    $endgroup$
    – Bvss12
    Jan 2 at 18:59






  • 1




    $begingroup$
    Using $lambda$ for the power was the source of your confusion. The equation should be $r^2+lambda=0$
    $endgroup$
    – Dylan
    Jan 2 at 19:17










  • $begingroup$
    Ummm, yes you have reason. I go to solve that and update my question. @Dylan
    $endgroup$
    – Bvss12
    Jan 2 at 19:19














  • 1




    $begingroup$
    Don't use $x^lambda$, use $x^r$ where $r$ is to be determined (it won't be $lambda$).
    $endgroup$
    – Robert Israel
    Jan 2 at 18:54






  • 1




    $begingroup$
    Ok, i go to edit that @RobertIsrael
    $endgroup$
    – Bvss12
    Jan 2 at 18:59






  • 1




    $begingroup$
    Using $lambda$ for the power was the source of your confusion. The equation should be $r^2+lambda=0$
    $endgroup$
    – Dylan
    Jan 2 at 19:17










  • $begingroup$
    Ummm, yes you have reason. I go to solve that and update my question. @Dylan
    $endgroup$
    – Bvss12
    Jan 2 at 19:19








1




1




$begingroup$
Don't use $x^lambda$, use $x^r$ where $r$ is to be determined (it won't be $lambda$).
$endgroup$
– Robert Israel
Jan 2 at 18:54




$begingroup$
Don't use $x^lambda$, use $x^r$ where $r$ is to be determined (it won't be $lambda$).
$endgroup$
– Robert Israel
Jan 2 at 18:54




1




1




$begingroup$
Ok, i go to edit that @RobertIsrael
$endgroup$
– Bvss12
Jan 2 at 18:59




$begingroup$
Ok, i go to edit that @RobertIsrael
$endgroup$
– Bvss12
Jan 2 at 18:59




1




1




$begingroup$
Using $lambda$ for the power was the source of your confusion. The equation should be $r^2+lambda=0$
$endgroup$
– Dylan
Jan 2 at 19:17




$begingroup$
Using $lambda$ for the power was the source of your confusion. The equation should be $r^2+lambda=0$
$endgroup$
– Dylan
Jan 2 at 19:17












$begingroup$
Ummm, yes you have reason. I go to solve that and update my question. @Dylan
$endgroup$
– Bvss12
Jan 2 at 19:19




$begingroup$
Ummm, yes you have reason. I go to solve that and update my question. @Dylan
$endgroup$
– Bvss12
Jan 2 at 19:19










1 Answer
1






active

oldest

votes


















1












$begingroup$

Considering



$$
left(x^2frac{d^2}{dt^2}+xfrac{d}{dt}+lambdaright)y = 0
$$



after the substitution $y = x^r$ we get



$$
x^rleft(r(r-1)+r +lambdaright)=0
$$



then



$$
r = pm i sqrtlambda
$$



then for $lambda > 0$ we have



$$
y = C_1sin(sqrtlambda ln x) + C_2cos(sqrtlambda ln x)
$$



and now from the boundary conditions



$$
y(1) = C_2 = 0\
y(b) = C_1sin(sqrtlambda ln b) = 0
$$



then we conclude



$$
sqrtlambdaln b = k piRightarrow lambda_ k = frac{pi^2}{ln^2 b}k^2
$$






share|cite|improve this answer









$endgroup$













    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%2f3059816%2fusing-the-fact-this-equation-is-equidimensional-calculate-all-associated-eigenv%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









    1












    $begingroup$

    Considering



    $$
    left(x^2frac{d^2}{dt^2}+xfrac{d}{dt}+lambdaright)y = 0
    $$



    after the substitution $y = x^r$ we get



    $$
    x^rleft(r(r-1)+r +lambdaright)=0
    $$



    then



    $$
    r = pm i sqrtlambda
    $$



    then for $lambda > 0$ we have



    $$
    y = C_1sin(sqrtlambda ln x) + C_2cos(sqrtlambda ln x)
    $$



    and now from the boundary conditions



    $$
    y(1) = C_2 = 0\
    y(b) = C_1sin(sqrtlambda ln b) = 0
    $$



    then we conclude



    $$
    sqrtlambdaln b = k piRightarrow lambda_ k = frac{pi^2}{ln^2 b}k^2
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Considering



      $$
      left(x^2frac{d^2}{dt^2}+xfrac{d}{dt}+lambdaright)y = 0
      $$



      after the substitution $y = x^r$ we get



      $$
      x^rleft(r(r-1)+r +lambdaright)=0
      $$



      then



      $$
      r = pm i sqrtlambda
      $$



      then for $lambda > 0$ we have



      $$
      y = C_1sin(sqrtlambda ln x) + C_2cos(sqrtlambda ln x)
      $$



      and now from the boundary conditions



      $$
      y(1) = C_2 = 0\
      y(b) = C_1sin(sqrtlambda ln b) = 0
      $$



      then we conclude



      $$
      sqrtlambdaln b = k piRightarrow lambda_ k = frac{pi^2}{ln^2 b}k^2
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Considering



        $$
        left(x^2frac{d^2}{dt^2}+xfrac{d}{dt}+lambdaright)y = 0
        $$



        after the substitution $y = x^r$ we get



        $$
        x^rleft(r(r-1)+r +lambdaright)=0
        $$



        then



        $$
        r = pm i sqrtlambda
        $$



        then for $lambda > 0$ we have



        $$
        y = C_1sin(sqrtlambda ln x) + C_2cos(sqrtlambda ln x)
        $$



        and now from the boundary conditions



        $$
        y(1) = C_2 = 0\
        y(b) = C_1sin(sqrtlambda ln b) = 0
        $$



        then we conclude



        $$
        sqrtlambdaln b = k piRightarrow lambda_ k = frac{pi^2}{ln^2 b}k^2
        $$






        share|cite|improve this answer









        $endgroup$



        Considering



        $$
        left(x^2frac{d^2}{dt^2}+xfrac{d}{dt}+lambdaright)y = 0
        $$



        after the substitution $y = x^r$ we get



        $$
        x^rleft(r(r-1)+r +lambdaright)=0
        $$



        then



        $$
        r = pm i sqrtlambda
        $$



        then for $lambda > 0$ we have



        $$
        y = C_1sin(sqrtlambda ln x) + C_2cos(sqrtlambda ln x)
        $$



        and now from the boundary conditions



        $$
        y(1) = C_2 = 0\
        y(b) = C_1sin(sqrtlambda ln b) = 0
        $$



        then we conclude



        $$
        sqrtlambdaln b = k piRightarrow lambda_ k = frac{pi^2}{ln^2 b}k^2
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 19:23









        CesareoCesareo

        8,5043516




        8,5043516






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059816%2fusing-the-fact-this-equation-is-equidimensional-calculate-all-associated-eigenv%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

            MongoDB - Not Authorized To Execute Command

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
            Read more

            How to fix TextFormField cause rebuild widget in Flutter

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
            Read more

            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

            Image
            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
            Read more