If $u$ is a solution to the wave equation (Cauchy) then $|u(x,t)|le A/t$ for some $A$












1












$begingroup$



Let $u(x,t)$ be a solution for the Cauchy Problem



$$u_{tt}-Delta_xu = 0mbox{ in $mathbb{R}^3times mathbb{R}$}$$
$$u(x,0) = f(x)mbox{ in $mathbb{R}^3$}$$ $$u_t(x,0) = g(x) mbox{in
$mathbb{R}^3$}$$



where $f$ and $g$ are of class $C^3$ and $C^2$ respectively in
$mathbb{R}^3$ which are null in the complementar of a compact. Show
that there exists a constant $A$ such that



$$|u(x,t)|le A/t, xinmathbb{R}^3, tge 1$$



Find, also, an estimative for the constant $A$ in terms of $f$ and
$g$.




UPDATE:



I've found the solution enter image description here



but I need to understand why the intersection with the support is at most $4pi R^2$. As I understand, the support of the data is the support of $f$. The intersection of $S_t(x)$ with this support should be what?



I'm trying to imagine the complement of the ball $B(0,R)$ which contains the support of $f$ (is the support $g$ necessary?). I must take the intersection with $S_x(t)$ but I do not know what to do






real-analysis integration pde wave-equation







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Let $u(x,t)$ be a solution for the Cauchy Problem



    $$u_{tt}-Delta_xu = 0mbox{ in $mathbb{R}^3times mathbb{R}$}$$
    $$u(x,0) = f(x)mbox{ in $mathbb{R}^3$}$$ $$u_t(x,0) = g(x) mbox{in
    $mathbb{R}^3$}$$



    where $f$ and $g$ are of class $C^3$ and $C^2$ respectively in
    $mathbb{R}^3$ which are null in the complementar of a compact. Show
    that there exists a constant $A$ such that



    $$|u(x,t)|le A/t, xinmathbb{R}^3, tge 1$$



    Find, also, an estimative for the constant $A$ in terms of $f$ and
    $g$.




    UPDATE:



    I've found the solution enter image description here



    but I need to understand why the intersection with the support is at most $4pi R^2$. As I understand, the support of the data is the support of $f$. The intersection of $S_t(x)$ with this support should be what?



    I'm trying to imagine the complement of the ball $B(0,R)$ which contains the support of $f$ (is the support $g$ necessary?). I must take the intersection with $S_x(t)$ but I do not know what to do






    real-analysis integration pde wave-equation







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$



      Let $u(x,t)$ be a solution for the Cauchy Problem



      $$u_{tt}-Delta_xu = 0mbox{ in $mathbb{R}^3times mathbb{R}$}$$
      $$u(x,0) = f(x)mbox{ in $mathbb{R}^3$}$$ $$u_t(x,0) = g(x) mbox{in
      $mathbb{R}^3$}$$



      where $f$ and $g$ are of class $C^3$ and $C^2$ respectively in
      $mathbb{R}^3$ which are null in the complementar of a compact. Show
      that there exists a constant $A$ such that



      $$|u(x,t)|le A/t, xinmathbb{R}^3, tge 1$$



      Find, also, an estimative for the constant $A$ in terms of $f$ and
      $g$.




      UPDATE:



      I've found the solution enter image description here



      but I need to understand why the intersection with the support is at most $4pi R^2$. As I understand, the support of the data is the support of $f$. The intersection of $S_t(x)$ with this support should be what?



      I'm trying to imagine the complement of the ball $B(0,R)$ which contains the support of $f$ (is the support $g$ necessary?). I must take the intersection with $S_x(t)$ but I do not know what to do






      real-analysis integration pde wave-equation







      share|cite|improve this question











      $endgroup$





      Let $u(x,t)$ be a solution for the Cauchy Problem



      $$u_{tt}-Delta_xu = 0mbox{ in $mathbb{R}^3times mathbb{R}$}$$
      $$u(x,0) = f(x)mbox{ in $mathbb{R}^3$}$$ $$u_t(x,0) = g(x) mbox{in
      $mathbb{R}^3$}$$



      where $f$ and $g$ are of class $C^3$ and $C^2$ respectively in
      $mathbb{R}^3$ which are null in the complementar of a compact. Show
      that there exists a constant $A$ such that



      $$|u(x,t)|le A/t, xinmathbb{R}^3, tge 1$$



      Find, also, an estimative for the constant $A$ in terms of $f$ and
      $g$.




      UPDATE:



      I've found the solution enter image description here



      but I need to understand why the intersection with the support is at most $4pi R^2$. As I understand, the support of the data is the support of $f$. The intersection of $S_t(x)$ with this support should be what?



      I'm trying to imagine the complement of the ball $B(0,R)$ which contains the support of $f$ (is the support $g$ necessary?). I must take the intersection with $S_x(t)$ but I do not know what to do






      real-analysis integration pde wave-equation




      real-analysis integration pde wave-equation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 19:04







      Lucas Zanella

















      asked Nov 27 '18 at 19:33









      Lucas ZanellaLucas Zanella

      92411330




      92411330






















          1 Answer
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          +50







          $begingroup$

          We use the following fact.




          • If $Csubsetmathbb R^3$ is a convex set contained in a ball $B_R$, then $mathcal H^2(partial C)leq mathcal H^2(partial B_R)=4pi R^2$.


          With your notation, let $E_x(t)$ be the ball bounded by $S_x(t)$ and let $B_R$ the ball that contains $mathrm{supp}(f)$ and $mathrm{supp}(g)$. Then $E_x(t)cap B_R$ is convex and contained in $B_R$. Its surface is composed of two pieces, one of which is $partial E_x(t)cap B_R = S_x(t)cap B_R$. Therefore
          $$
          mathcal H^2bigl(S_x(t)cap B_Rbigr)
          leq mathcal H^2bigl(partial(E_x(t)cap B_R)bigr)
          leq mathcal H^2(partial B_R)=4pi R^2 .
          $$



          Edit



          Obviously, I applied the initial fact with $C=E_x(t)cap B_R$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What is $mathcal H^2$?
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:19










          • $begingroup$
            Two-dimensional Hausdorff measure. The surface area
            $endgroup$
            – Federico
            Dec 3 '18 at 19:20










          • $begingroup$
            Why is $C$, that is, $S_t(x)$ in my example, contained in $B_R$? I thought it was not necessairy contained
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:22










          • $begingroup$
            Read carefully. Where did I say that $S_x(t)subset B_R$?
            $endgroup$
            – Federico
            Dec 3 '18 at 19:23










          • $begingroup$
            I thought $C$ was $S_x(t)$
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:23











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2





          +50







          $begingroup$

          We use the following fact.




          • If $Csubsetmathbb R^3$ is a convex set contained in a ball $B_R$, then $mathcal H^2(partial C)leq mathcal H^2(partial B_R)=4pi R^2$.


          With your notation, let $E_x(t)$ be the ball bounded by $S_x(t)$ and let $B_R$ the ball that contains $mathrm{supp}(f)$ and $mathrm{supp}(g)$. Then $E_x(t)cap B_R$ is convex and contained in $B_R$. Its surface is composed of two pieces, one of which is $partial E_x(t)cap B_R = S_x(t)cap B_R$. Therefore
          $$
          mathcal H^2bigl(S_x(t)cap B_Rbigr)
          leq mathcal H^2bigl(partial(E_x(t)cap B_R)bigr)
          leq mathcal H^2(partial B_R)=4pi R^2 .
          $$



          Edit



          Obviously, I applied the initial fact with $C=E_x(t)cap B_R$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What is $mathcal H^2$?
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:19










          • $begingroup$
            Two-dimensional Hausdorff measure. The surface area
            $endgroup$
            – Federico
            Dec 3 '18 at 19:20










          • $begingroup$
            Why is $C$, that is, $S_t(x)$ in my example, contained in $B_R$? I thought it was not necessairy contained
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:22










          • $begingroup$
            Read carefully. Where did I say that $S_x(t)subset B_R$?
            $endgroup$
            – Federico
            Dec 3 '18 at 19:23










          • $begingroup$
            I thought $C$ was $S_x(t)$
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:23
















          2





          +50







          $begingroup$

          We use the following fact.




          • If $Csubsetmathbb R^3$ is a convex set contained in a ball $B_R$, then $mathcal H^2(partial C)leq mathcal H^2(partial B_R)=4pi R^2$.


          With your notation, let $E_x(t)$ be the ball bounded by $S_x(t)$ and let $B_R$ the ball that contains $mathrm{supp}(f)$ and $mathrm{supp}(g)$. Then $E_x(t)cap B_R$ is convex and contained in $B_R$. Its surface is composed of two pieces, one of which is $partial E_x(t)cap B_R = S_x(t)cap B_R$. Therefore
          $$
          mathcal H^2bigl(S_x(t)cap B_Rbigr)
          leq mathcal H^2bigl(partial(E_x(t)cap B_R)bigr)
          leq mathcal H^2(partial B_R)=4pi R^2 .
          $$



          Edit



          Obviously, I applied the initial fact with $C=E_x(t)cap B_R$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What is $mathcal H^2$?
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:19










          • $begingroup$
            Two-dimensional Hausdorff measure. The surface area
            $endgroup$
            – Federico
            Dec 3 '18 at 19:20










          • $begingroup$
            Why is $C$, that is, $S_t(x)$ in my example, contained in $B_R$? I thought it was not necessairy contained
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:22










          • $begingroup$
            Read carefully. Where did I say that $S_x(t)subset B_R$?
            $endgroup$
            – Federico
            Dec 3 '18 at 19:23










          • $begingroup$
            I thought $C$ was $S_x(t)$
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:23














          2





          +50







          2





          +50



          2




          +50



          $begingroup$

          We use the following fact.




          • If $Csubsetmathbb R^3$ is a convex set contained in a ball $B_R$, then $mathcal H^2(partial C)leq mathcal H^2(partial B_R)=4pi R^2$.


          With your notation, let $E_x(t)$ be the ball bounded by $S_x(t)$ and let $B_R$ the ball that contains $mathrm{supp}(f)$ and $mathrm{supp}(g)$. Then $E_x(t)cap B_R$ is convex and contained in $B_R$. Its surface is composed of two pieces, one of which is $partial E_x(t)cap B_R = S_x(t)cap B_R$. Therefore
          $$
          mathcal H^2bigl(S_x(t)cap B_Rbigr)
          leq mathcal H^2bigl(partial(E_x(t)cap B_R)bigr)
          leq mathcal H^2(partial B_R)=4pi R^2 .
          $$



          Edit



          Obviously, I applied the initial fact with $C=E_x(t)cap B_R$.






          share|cite|improve this answer











          $endgroup$



          We use the following fact.




          • If $Csubsetmathbb R^3$ is a convex set contained in a ball $B_R$, then $mathcal H^2(partial C)leq mathcal H^2(partial B_R)=4pi R^2$.


          With your notation, let $E_x(t)$ be the ball bounded by $S_x(t)$ and let $B_R$ the ball that contains $mathrm{supp}(f)$ and $mathrm{supp}(g)$. Then $E_x(t)cap B_R$ is convex and contained in $B_R$. Its surface is composed of two pieces, one of which is $partial E_x(t)cap B_R = S_x(t)cap B_R$. Therefore
          $$
          mathcal H^2bigl(S_x(t)cap B_Rbigr)
          leq mathcal H^2bigl(partial(E_x(t)cap B_R)bigr)
          leq mathcal H^2(partial B_R)=4pi R^2 .
          $$



          Edit



          Obviously, I applied the initial fact with $C=E_x(t)cap B_R$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 19:25

























          answered Dec 3 '18 at 19:18









          FedericoFederico

          5,004514




          5,004514












          • $begingroup$
            What is $mathcal H^2$?
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:19










          • $begingroup$
            Two-dimensional Hausdorff measure. The surface area
            $endgroup$
            – Federico
            Dec 3 '18 at 19:20










          • $begingroup$
            Why is $C$, that is, $S_t(x)$ in my example, contained in $B_R$? I thought it was not necessairy contained
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:22










          • $begingroup$
            Read carefully. Where did I say that $S_x(t)subset B_R$?
            $endgroup$
            – Federico
            Dec 3 '18 at 19:23










          • $begingroup$
            I thought $C$ was $S_x(t)$
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:23


















          • $begingroup$
            What is $mathcal H^2$?
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:19










          • $begingroup$
            Two-dimensional Hausdorff measure. The surface area
            $endgroup$
            – Federico
            Dec 3 '18 at 19:20










          • $begingroup$
            Why is $C$, that is, $S_t(x)$ in my example, contained in $B_R$? I thought it was not necessairy contained
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:22










          • $begingroup$
            Read carefully. Where did I say that $S_x(t)subset B_R$?
            $endgroup$
            – Federico
            Dec 3 '18 at 19:23










          • $begingroup$
            I thought $C$ was $S_x(t)$
            $endgroup$
            – Lucas Zanella
            Dec 3 '18 at 19:23
















          $begingroup$
          What is $mathcal H^2$?
          $endgroup$
          – Lucas Zanella
          Dec 3 '18 at 19:19




          $begingroup$
          What is $mathcal H^2$?
          $endgroup$
          – Lucas Zanella
          Dec 3 '18 at 19:19












          $begingroup$
          Two-dimensional Hausdorff measure. The surface area
          $endgroup$
          – Federico
          Dec 3 '18 at 19:20




          $begingroup$
          Two-dimensional Hausdorff measure. The surface area
          $endgroup$
          – Federico
          Dec 3 '18 at 19:20












          $begingroup$
          Why is $C$, that is, $S_t(x)$ in my example, contained in $B_R$? I thought it was not necessairy contained
          $endgroup$
          – Lucas Zanella
          Dec 3 '18 at 19:22




          $begingroup$
          Why is $C$, that is, $S_t(x)$ in my example, contained in $B_R$? I thought it was not necessairy contained
          $endgroup$
          – Lucas Zanella
          Dec 3 '18 at 19:22












          $begingroup$
          Read carefully. Where did I say that $S_x(t)subset B_R$?
          $endgroup$
          – Federico
          Dec 3 '18 at 19:23




          $begingroup$
          Read carefully. Where did I say that $S_x(t)subset B_R$?
          $endgroup$
          – Federico
          Dec 3 '18 at 19:23












          $begingroup$
          I thought $C$ was $S_x(t)$
          $endgroup$
          – Lucas Zanella
          Dec 3 '18 at 19:23




          $begingroup$
          I thought $C$ was $S_x(t)$
          $endgroup$
          – Lucas Zanella
          Dec 3 '18 at 19:23


















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