Wave Equation, is this problem well posed?











up vote
2
down vote

favorite
1












I am solving $$u_{tt}=4u_{xx}$$ $$u(x,0)=x^2-x$$ $$u_t(x,0)=cos x$$ $$u(0,t)=u(1,t)=0$$



I am trying to solve with the method of separable factors. So I am trying to find a equation like $u(x,t)=X(x)cdot T(t)$. But, I am finding a problem, because if the solution is like this, then $u_t(x,0)=X(x)cdot T'(0) = cos xRightarrow X(x) = frac{cos x}{T'(0)}$.



But if $X(x)$ is this way, then $X(0)=frac{cos 0}{T'(0)}neq 0$.



Am I solving this right or the problem is not well posed?






pde wave-equation







share|cite|improve this question


















  • 3




    The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
    – Winther
    Nov 22 at 20:11

















up vote
2
down vote

favorite
1












I am solving $$u_{tt}=4u_{xx}$$ $$u(x,0)=x^2-x$$ $$u_t(x,0)=cos x$$ $$u(0,t)=u(1,t)=0$$



I am trying to solve with the method of separable factors. So I am trying to find a equation like $u(x,t)=X(x)cdot T(t)$. But, I am finding a problem, because if the solution is like this, then $u_t(x,0)=X(x)cdot T'(0) = cos xRightarrow X(x) = frac{cos x}{T'(0)}$.



But if $X(x)$ is this way, then $X(0)=frac{cos 0}{T'(0)}neq 0$.



Am I solving this right or the problem is not well posed?






pde wave-equation







share|cite|improve this question


















  • 3




    The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
    – Winther
    Nov 22 at 20:11















up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I am solving $$u_{tt}=4u_{xx}$$ $$u(x,0)=x^2-x$$ $$u_t(x,0)=cos x$$ $$u(0,t)=u(1,t)=0$$



I am trying to solve with the method of separable factors. So I am trying to find a equation like $u(x,t)=X(x)cdot T(t)$. But, I am finding a problem, because if the solution is like this, then $u_t(x,0)=X(x)cdot T'(0) = cos xRightarrow X(x) = frac{cos x}{T'(0)}$.



But if $X(x)$ is this way, then $X(0)=frac{cos 0}{T'(0)}neq 0$.



Am I solving this right or the problem is not well posed?






pde wave-equation







share|cite|improve this question













I am solving $$u_{tt}=4u_{xx}$$ $$u(x,0)=x^2-x$$ $$u_t(x,0)=cos x$$ $$u(0,t)=u(1,t)=0$$



I am trying to solve with the method of separable factors. So I am trying to find a equation like $u(x,t)=X(x)cdot T(t)$. But, I am finding a problem, because if the solution is like this, then $u_t(x,0)=X(x)cdot T'(0) = cos xRightarrow X(x) = frac{cos x}{T'(0)}$.



But if $X(x)$ is this way, then $X(0)=frac{cos 0}{T'(0)}neq 0$.



Am I solving this right or the problem is not well posed?






pde wave-equation




pde wave-equation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 20:00









Matheus Popst de Campos

504




504








  • 3




    The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
    – Winther
    Nov 22 at 20:11
















  • 3




    The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
    – Winther
    Nov 22 at 20:11










3




3




The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
– Winther
Nov 22 at 20:11






The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
– Winther
Nov 22 at 20:11












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
$$
frac{T''}{T}=lambda = frac{X''}{X}
$$



leading to



$$
T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
$$



The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
$$
X_n(t)=C_nsin(npi x) \
T_n(t)=A_ncos(npi t)+B_nsin(npi t).
$$
The general solution (after combining constants) is
$$
u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
$$



The constants are determined by the initial conditions:
$$
x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
$$
So,
$$
A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
$$






share|cite|improve this answer





















    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: 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%2f3009581%2fwave-equation-is-this-problem-well-posed%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








    up vote
    0
    down vote



    accepted










    The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
    $$
    frac{T''}{T}=lambda = frac{X''}{X}
    $$



    leading to



    $$
    T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
    X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
    $$



    The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
    $$
    X_n(t)=C_nsin(npi x) \
    T_n(t)=A_ncos(npi t)+B_nsin(npi t).
    $$
    The general solution (after combining constants) is
    $$
    u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
    $$



    The constants are determined by the initial conditions:
    $$
    x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
    cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
    $$
    So,
    $$
    A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
    B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
      $$
      frac{T''}{T}=lambda = frac{X''}{X}
      $$



      leading to



      $$
      T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
      X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
      $$



      The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
      $$
      X_n(t)=C_nsin(npi x) \
      T_n(t)=A_ncos(npi t)+B_nsin(npi t).
      $$
      The general solution (after combining constants) is
      $$
      u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
      $$



      The constants are determined by the initial conditions:
      $$
      x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
      cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
      $$
      So,
      $$
      A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
      B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
      $$






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
        $$
        frac{T''}{T}=lambda = frac{X''}{X}
        $$



        leading to



        $$
        T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
        X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
        $$



        The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
        $$
        X_n(t)=C_nsin(npi x) \
        T_n(t)=A_ncos(npi t)+B_nsin(npi t).
        $$
        The general solution (after combining constants) is
        $$
        u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
        $$



        The constants are determined by the initial conditions:
        $$
        x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
        cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
        $$
        So,
        $$
        A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
        B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
        $$






        share|cite|improve this answer












        The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
        $$
        frac{T''}{T}=lambda = frac{X''}{X}
        $$



        leading to



        $$
        T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
        X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
        $$



        The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
        $$
        X_n(t)=C_nsin(npi x) \
        T_n(t)=A_ncos(npi t)+B_nsin(npi t).
        $$
        The general solution (after combining constants) is
        $$
        u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
        $$



        The constants are determined by the initial conditions:
        $$
        x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
        cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
        $$
        So,
        $$
        A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
        B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 at 4:31









        DisintegratingByParts

        58.3k42579




        58.3k42579






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3009581%2fwave-equation-is-this-problem-well-posed%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

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more