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











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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










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    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










    share|cite|improve this question


























      up vote
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      favorite
      1









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      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










      share|cite|improve this question
















      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






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      edited Dec 3 at 19:04

























      asked Nov 27 at 19:33









      Lucas Zanella

      92311330




      92311330






















          1 Answer
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          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























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










          • Two-dimensional Hausdorff measure. The surface area
            – Federico
            Dec 3 at 19:20










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










          • Read carefully. Where did I say that $S_x(t)subset B_R$?
            – Federico
            Dec 3 at 19:23










          • I thought $C$ was $S_x(t)$
            – Lucas Zanella
            Dec 3 at 19:23











          Your Answer





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          up vote
          2
          down vote



          accepted
          +50










          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























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










          • Two-dimensional Hausdorff measure. The surface area
            – Federico
            Dec 3 at 19:20










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










          • Read carefully. Where did I say that $S_x(t)subset B_R$?
            – Federico
            Dec 3 at 19:23










          • I thought $C$ was $S_x(t)$
            – Lucas Zanella
            Dec 3 at 19:23















          up vote
          2
          down vote



          accepted
          +50










          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























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










          • Two-dimensional Hausdorff measure. The surface area
            – Federico
            Dec 3 at 19:20










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










          • Read carefully. Where did I say that $S_x(t)subset B_R$?
            – Federico
            Dec 3 at 19:23










          • I thought $C$ was $S_x(t)$
            – Lucas Zanella
            Dec 3 at 19:23













          up vote
          2
          down vote



          accepted
          +50







          up vote
          2
          down vote



          accepted
          +50




          +50




          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














          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 at 19:25

























          answered Dec 3 at 19:18









          Federico

          4,068512




          4,068512












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










          • Two-dimensional Hausdorff measure. The surface area
            – Federico
            Dec 3 at 19:20










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










          • Read carefully. Where did I say that $S_x(t)subset B_R$?
            – Federico
            Dec 3 at 19:23










          • I thought $C$ was $S_x(t)$
            – Lucas Zanella
            Dec 3 at 19:23


















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










          • Two-dimensional Hausdorff measure. The surface area
            – Federico
            Dec 3 at 19:20










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










          • Read carefully. Where did I say that $S_x(t)subset B_R$?
            – Federico
            Dec 3 at 19:23










          • I thought $C$ was $S_x(t)$
            – Lucas Zanella
            Dec 3 at 19:23
















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




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












          Two-dimensional Hausdorff measure. The surface area
          – Federico
          Dec 3 at 19:20




          Two-dimensional Hausdorff measure. The surface area
          – Federico
          Dec 3 at 19:20












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




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












          Read carefully. Where did I say that $S_x(t)subset B_R$?
          – Federico
          Dec 3 at 19:23




          Read carefully. Where did I say that $S_x(t)subset B_R$?
          – Federico
          Dec 3 at 19:23












          I thought $C$ was $S_x(t)$
          – Lucas Zanella
          Dec 3 at 19:23




          I thought $C$ was $S_x(t)$
          – Lucas Zanella
          Dec 3 at 19:23


















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