An inequality about linear PDE












6












$begingroup$


I am trying to solve the following problem:




Let $Omega$ be a bounded smooth domain in $mathbb{R}^n, ngeq 2$. Let $uin C^2(overline{Omega})$ be a solution of
$$left{begin{array}{ll}u_t-Delta u=f(x)& text{in } Omegatimes(0,infty)\ u=0&text{on }partial Omegatimes (0,infty) \ u=g(x) & text{on } Omegatimes{0}end{array}right.$$
Show that
$$max_{0leq tleq T} int_Omega u^2(x,t)dx+int_0^Tint_Omega|nabla u(x,t)|^2dx, dtleq Cleft(int_Omega g^2(x)dx+int_0^T|f(x)|^2 dx,dtright)$$
for some constant $C$ independent of $f, g$ and $u$.





My attempt:



Multiplying the first equation by $u$ and taking integration on $Omega$, we have
$$frac{1}{2}frac{d}{dt}||u||^2_{L^2}+int_Omega nabla ucdot nabla u=int_Omega fu.$$
(Here we also use the Green's identity and the boundary condition of $u$)



Taking integration on $[0,s]$ where $sleq T$. Then we have
$$frac{1}{2}||u(x,s)||^2_{L^2}+int_0^sint_Omega|nabla u(x,t)|^2dx, dt= frac{1}{2} int_Omega g^2(x)dx+int_0^sint_Omega fu dt$$
where
$$int_Omega fuleq ||f||_{L^2} ||u||_{L^2}leq epsilon ||u||_{L^2}^2 + C(epsilon ) ||f||_{L^2}$$



Then we have
$$(frac{1}{2}-sepsilon )||u(x,s)||^2_{L^2}+int_0^sint_Omega|nabla u(x,t)|^2dx, dt
leq frac{1}{2} int_Omega g^2(x)dx+int_0^s C(epsilon ) ||f||_{L^2} dt$$



Then I was stuck here. I don't know where can I derive the part
$$max_{0leq tleq T}||u(x,t)||.$$
Also, I am struggling how to make the two coefficients before the two terms on the left the same so that we can get the desired inequality.






real-analysis inequality pde







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Solved it. Tell me if someone needs the answer. The idea is right. Just need some subtle modification about $epsilon$.
    $endgroup$
    – Aolong Li
    Dec 14 '18 at 7:29






  • 2




    $begingroup$
    I for one would like to see your solution. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 14 '18 at 7:38








  • 2




    $begingroup$
    the usual convention is to post your own answer and accept it.
    $endgroup$
    – dezdichado
    Dec 14 '18 at 15:56










  • $begingroup$
    @dezdichado OK thanks! Will do!
    $endgroup$
    – Aolong Li
    Dec 14 '18 at 16:00
















6












$begingroup$


I am trying to solve the following problem:




Let $Omega$ be a bounded smooth domain in $mathbb{R}^n, ngeq 2$. Let $uin C^2(overline{Omega})$ be a solution of
$$left{begin{array}{ll}u_t-Delta u=f(x)& text{in } Omegatimes(0,infty)\ u=0&text{on }partial Omegatimes (0,infty) \ u=g(x) & text{on } Omegatimes{0}end{array}right.$$
Show that
$$max_{0leq tleq T} int_Omega u^2(x,t)dx+int_0^Tint_Omega|nabla u(x,t)|^2dx, dtleq Cleft(int_Omega g^2(x)dx+int_0^T|f(x)|^2 dx,dtright)$$
for some constant $C$ independent of $f, g$ and $u$.





My attempt:



Multiplying the first equation by $u$ and taking integration on $Omega$, we have
$$frac{1}{2}frac{d}{dt}||u||^2_{L^2}+int_Omega nabla ucdot nabla u=int_Omega fu.$$
(Here we also use the Green's identity and the boundary condition of $u$)



Taking integration on $[0,s]$ where $sleq T$. Then we have
$$frac{1}{2}||u(x,s)||^2_{L^2}+int_0^sint_Omega|nabla u(x,t)|^2dx, dt= frac{1}{2} int_Omega g^2(x)dx+int_0^sint_Omega fu dt$$
where
$$int_Omega fuleq ||f||_{L^2} ||u||_{L^2}leq epsilon ||u||_{L^2}^2 + C(epsilon ) ||f||_{L^2}$$



Then we have
$$(frac{1}{2}-sepsilon )||u(x,s)||^2_{L^2}+int_0^sint_Omega|nabla u(x,t)|^2dx, dt
leq frac{1}{2} int_Omega g^2(x)dx+int_0^s C(epsilon ) ||f||_{L^2} dt$$



Then I was stuck here. I don't know where can I derive the part
$$max_{0leq tleq T}||u(x,t)||.$$
Also, I am struggling how to make the two coefficients before the two terms on the left the same so that we can get the desired inequality.






real-analysis inequality pde







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Solved it. Tell me if someone needs the answer. The idea is right. Just need some subtle modification about $epsilon$.
    $endgroup$
    – Aolong Li
    Dec 14 '18 at 7:29






  • 2




    $begingroup$
    I for one would like to see your solution. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 14 '18 at 7:38








  • 2




    $begingroup$
    the usual convention is to post your own answer and accept it.
    $endgroup$
    – dezdichado
    Dec 14 '18 at 15:56










  • $begingroup$
    @dezdichado OK thanks! Will do!
    $endgroup$
    – Aolong Li
    Dec 14 '18 at 16:00














6












6








6


2



$begingroup$


I am trying to solve the following problem:




Let $Omega$ be a bounded smooth domain in $mathbb{R}^n, ngeq 2$. Let $uin C^2(overline{Omega})$ be a solution of
$$left{begin{array}{ll}u_t-Delta u=f(x)& text{in } Omegatimes(0,infty)\ u=0&text{on }partial Omegatimes (0,infty) \ u=g(x) & text{on } Omegatimes{0}end{array}right.$$
Show that
$$max_{0leq tleq T} int_Omega u^2(x,t)dx+int_0^Tint_Omega|nabla u(x,t)|^2dx, dtleq Cleft(int_Omega g^2(x)dx+int_0^T|f(x)|^2 dx,dtright)$$
for some constant $C$ independent of $f, g$ and $u$.





My attempt:



Multiplying the first equation by $u$ and taking integration on $Omega$, we have
$$frac{1}{2}frac{d}{dt}||u||^2_{L^2}+int_Omega nabla ucdot nabla u=int_Omega fu.$$
(Here we also use the Green's identity and the boundary condition of $u$)



Taking integration on $[0,s]$ where $sleq T$. Then we have
$$frac{1}{2}||u(x,s)||^2_{L^2}+int_0^sint_Omega|nabla u(x,t)|^2dx, dt= frac{1}{2} int_Omega g^2(x)dx+int_0^sint_Omega fu dt$$
where
$$int_Omega fuleq ||f||_{L^2} ||u||_{L^2}leq epsilon ||u||_{L^2}^2 + C(epsilon ) ||f||_{L^2}$$



Then we have
$$(frac{1}{2}-sepsilon )||u(x,s)||^2_{L^2}+int_0^sint_Omega|nabla u(x,t)|^2dx, dt
leq frac{1}{2} int_Omega g^2(x)dx+int_0^s C(epsilon ) ||f||_{L^2} dt$$



Then I was stuck here. I don't know where can I derive the part
$$max_{0leq tleq T}||u(x,t)||.$$
Also, I am struggling how to make the two coefficients before the two terms on the left the same so that we can get the desired inequality.






real-analysis inequality pde







share|cite|improve this question









$endgroup$




I am trying to solve the following problem:




Let $Omega$ be a bounded smooth domain in $mathbb{R}^n, ngeq 2$. Let $uin C^2(overline{Omega})$ be a solution of
$$left{begin{array}{ll}u_t-Delta u=f(x)& text{in } Omegatimes(0,infty)\ u=0&text{on }partial Omegatimes (0,infty) \ u=g(x) & text{on } Omegatimes{0}end{array}right.$$
Show that
$$max_{0leq tleq T} int_Omega u^2(x,t)dx+int_0^Tint_Omega|nabla u(x,t)|^2dx, dtleq Cleft(int_Omega g^2(x)dx+int_0^T|f(x)|^2 dx,dtright)$$
for some constant $C$ independent of $f, g$ and $u$.





My attempt:



Multiplying the first equation by $u$ and taking integration on $Omega$, we have
$$frac{1}{2}frac{d}{dt}||u||^2_{L^2}+int_Omega nabla ucdot nabla u=int_Omega fu.$$
(Here we also use the Green's identity and the boundary condition of $u$)



Taking integration on $[0,s]$ where $sleq T$. Then we have
$$frac{1}{2}||u(x,s)||^2_{L^2}+int_0^sint_Omega|nabla u(x,t)|^2dx, dt= frac{1}{2} int_Omega g^2(x)dx+int_0^sint_Omega fu dt$$
where
$$int_Omega fuleq ||f||_{L^2} ||u||_{L^2}leq epsilon ||u||_{L^2}^2 + C(epsilon ) ||f||_{L^2}$$



Then we have
$$(frac{1}{2}-sepsilon )||u(x,s)||^2_{L^2}+int_0^sint_Omega|nabla u(x,t)|^2dx, dt
leq frac{1}{2} int_Omega g^2(x)dx+int_0^s C(epsilon ) ||f||_{L^2} dt$$



Then I was stuck here. I don't know where can I derive the part
$$max_{0leq tleq T}||u(x,t)||.$$
Also, I am struggling how to make the two coefficients before the two terms on the left the same so that we can get the desired inequality.






real-analysis inequality pde




real-analysis inequality pde






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 7:18









Aolong LiAolong Li

880615




880615








  • 1




    $begingroup$
    Solved it. Tell me if someone needs the answer. The idea is right. Just need some subtle modification about $epsilon$.
    $endgroup$
    – Aolong Li
    Dec 14 '18 at 7:29






  • 2




    $begingroup$
    I for one would like to see your solution. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 14 '18 at 7:38








  • 2




    $begingroup$
    the usual convention is to post your own answer and accept it.
    $endgroup$
    – dezdichado
    Dec 14 '18 at 15:56










  • $begingroup$
    @dezdichado OK thanks! Will do!
    $endgroup$
    – Aolong Li
    Dec 14 '18 at 16:00














  • 1




    $begingroup$
    Solved it. Tell me if someone needs the answer. The idea is right. Just need some subtle modification about $epsilon$.
    $endgroup$
    – Aolong Li
    Dec 14 '18 at 7:29






  • 2




    $begingroup$
    I for one would like to see your solution. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 14 '18 at 7:38








  • 2




    $begingroup$
    the usual convention is to post your own answer and accept it.
    $endgroup$
    – dezdichado
    Dec 14 '18 at 15:56










  • $begingroup$
    @dezdichado OK thanks! Will do!
    $endgroup$
    – Aolong Li
    Dec 14 '18 at 16:00








1




1




$begingroup$
Solved it. Tell me if someone needs the answer. The idea is right. Just need some subtle modification about $epsilon$.
$endgroup$
– Aolong Li
Dec 14 '18 at 7:29




$begingroup$
Solved it. Tell me if someone needs the answer. The idea is right. Just need some subtle modification about $epsilon$.
$endgroup$
– Aolong Li
Dec 14 '18 at 7:29




2




2




$begingroup$
I for one would like to see your solution. Cheers!
$endgroup$
– Robert Lewis
Dec 14 '18 at 7:38






$begingroup$
I for one would like to see your solution. Cheers!
$endgroup$
– Robert Lewis
Dec 14 '18 at 7:38






2




2




$begingroup$
the usual convention is to post your own answer and accept it.
$endgroup$
– dezdichado
Dec 14 '18 at 15:56




$begingroup$
the usual convention is to post your own answer and accept it.
$endgroup$
– dezdichado
Dec 14 '18 at 15:56












$begingroup$
@dezdichado OK thanks! Will do!
$endgroup$
– Aolong Li
Dec 14 '18 at 16:00




$begingroup$
@dezdichado OK thanks! Will do!
$endgroup$
– Aolong Li
Dec 14 '18 at 16:00










1 Answer
1






active

oldest

votes


















1












$begingroup$

What I did above tells us that
$$||u(x,s)||^2_{L^2}+frac{1}{frac{1}{2}-sepsilon}int_0^sint_Omega|nabla u(x,t)|^2dx, dt
leq frac{1}{frac{1}{2}-sepsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^s C(epsilon ) ||f||_{L^2} dt right) $$
for any $s>0$.



Now, pick up an appropriate $epsilon>0$ such that
$$frac{1}{2}-epsilon>0$$
and
$$int_0^Tint_Omega |nabla u|^2 dx dtleq frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt$$
where $t_0$ is the point where $||u(x,t)||^2_{L^2}$ obtain the maximum. Then we have
$$begin{eqnarray}||u(x,t_0)||^2_{L^2}+int_0^Tint_Omega |nabla u|^2 dx dt &leq& ||u(x,t_0)||^2_{L^2} + frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt\
&leq& frac{1}{frac{1}{2}-epsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)\
&leq& C'left( int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)
end{eqnarray}$$
for some constant $C'$.






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%2f3039058%2fan-inequality-about-linear-pde%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












    $begingroup$

    What I did above tells us that
    $$||u(x,s)||^2_{L^2}+frac{1}{frac{1}{2}-sepsilon}int_0^sint_Omega|nabla u(x,t)|^2dx, dt
    leq frac{1}{frac{1}{2}-sepsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^s C(epsilon ) ||f||_{L^2} dt right) $$
    for any $s>0$.



    Now, pick up an appropriate $epsilon>0$ such that
    $$frac{1}{2}-epsilon>0$$
    and
    $$int_0^Tint_Omega |nabla u|^2 dx dtleq frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt$$
    where $t_0$ is the point where $||u(x,t)||^2_{L^2}$ obtain the maximum. Then we have
    $$begin{eqnarray}||u(x,t_0)||^2_{L^2}+int_0^Tint_Omega |nabla u|^2 dx dt &leq& ||u(x,t_0)||^2_{L^2} + frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt\
    &leq& frac{1}{frac{1}{2}-epsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)\
    &leq& C'left( int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)
    end{eqnarray}$$
    for some constant $C'$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      What I did above tells us that
      $$||u(x,s)||^2_{L^2}+frac{1}{frac{1}{2}-sepsilon}int_0^sint_Omega|nabla u(x,t)|^2dx, dt
      leq frac{1}{frac{1}{2}-sepsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^s C(epsilon ) ||f||_{L^2} dt right) $$
      for any $s>0$.



      Now, pick up an appropriate $epsilon>0$ such that
      $$frac{1}{2}-epsilon>0$$
      and
      $$int_0^Tint_Omega |nabla u|^2 dx dtleq frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt$$
      where $t_0$ is the point where $||u(x,t)||^2_{L^2}$ obtain the maximum. Then we have
      $$begin{eqnarray}||u(x,t_0)||^2_{L^2}+int_0^Tint_Omega |nabla u|^2 dx dt &leq& ||u(x,t_0)||^2_{L^2} + frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt\
      &leq& frac{1}{frac{1}{2}-epsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)\
      &leq& C'left( int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)
      end{eqnarray}$$
      for some constant $C'$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        What I did above tells us that
        $$||u(x,s)||^2_{L^2}+frac{1}{frac{1}{2}-sepsilon}int_0^sint_Omega|nabla u(x,t)|^2dx, dt
        leq frac{1}{frac{1}{2}-sepsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^s C(epsilon ) ||f||_{L^2} dt right) $$
        for any $s>0$.



        Now, pick up an appropriate $epsilon>0$ such that
        $$frac{1}{2}-epsilon>0$$
        and
        $$int_0^Tint_Omega |nabla u|^2 dx dtleq frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt$$
        where $t_0$ is the point where $||u(x,t)||^2_{L^2}$ obtain the maximum. Then we have
        $$begin{eqnarray}||u(x,t_0)||^2_{L^2}+int_0^Tint_Omega |nabla u|^2 dx dt &leq& ||u(x,t_0)||^2_{L^2} + frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt\
        &leq& frac{1}{frac{1}{2}-epsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)\
        &leq& C'left( int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)
        end{eqnarray}$$
        for some constant $C'$.






        share|cite|improve this answer











        $endgroup$



        What I did above tells us that
        $$||u(x,s)||^2_{L^2}+frac{1}{frac{1}{2}-sepsilon}int_0^sint_Omega|nabla u(x,t)|^2dx, dt
        leq frac{1}{frac{1}{2}-sepsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^s C(epsilon ) ||f||_{L^2} dt right) $$
        for any $s>0$.



        Now, pick up an appropriate $epsilon>0$ such that
        $$frac{1}{2}-epsilon>0$$
        and
        $$int_0^Tint_Omega |nabla u|^2 dx dtleq frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt$$
        where $t_0$ is the point where $||u(x,t)||^2_{L^2}$ obtain the maximum. Then we have
        $$begin{eqnarray}||u(x,t_0)||^2_{L^2}+int_0^Tint_Omega |nabla u|^2 dx dt &leq& ||u(x,t_0)||^2_{L^2} + frac{1}{frac{1}{2}-epsilon}int_0^{t_0}int_Omega |nabla u|^2 dx dt\
        &leq& frac{1}{frac{1}{2}-epsilon}left(frac{1}{2} int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)\
        &leq& C'left( int_Omega g^2(x)dx+int_0^T C(epsilon ) ||f||_{L^2} dtright)
        end{eqnarray}$$
        for some constant $C'$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 21 '18 at 7:22

























        answered Dec 21 '18 at 4:42









        Aolong LiAolong Li

        880615




        880615






























            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%2f3039058%2fan-inequality-about-linear-pde%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Brian Clough

            Image
            Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
            Read more

            Cáceres

            Image
            Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
            Read more