Proving the existence of a solution of the heat equation using semigroup methods

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I am trying to solve the following problem:




Prove the existence of a solution of the heat equation
$$
u_t=-Delta u +f(t), u(0)=u_0,
$$
with Dirichlet Boundary conditions. Identify the spaces that the data and solution should occupy.




This question is a specific version of question 12.9 in this book http://uxmym1.iimas.unam.mx/ramon/docs/RenRog.pdf



I am looking to use semigroup methods. In particular I am trying to verify the conditions of Lumer-Phillips Theorem, Theorem 12.22 in the book.



Let my operator $A=Delta$. Then if I can show that $A$ satisfies the conditions of Lumer-Phillips I will have that $A$ is the infinitesimal generator of a $C_0$-semigroup and so a solution will exist.



My Hilbert Space is $L^2(Omega)$, and I have that $D(A)=H^2(Omega)cap H^1_0(Omega)$, which is dense in $L^2(Omega)$. So the first condition of Lumer-Phillips is satisfied.



I still need to show the following:
$$
(2) text Re(x,Ax)leq w (x,x), text for some w, text for every xin D(A),
$$
$$
(3) text There exists a lambda_0>w text such that A-lambda_0 I text is onto.
$$
Here $(bullet,bullet)$ denotes the $L^2(Omega)$ inner product, i.e. $(u,v)=int_Omega u(x)overlinev(x) dx$.



I'm not sure how to show (2) or (3) and could use some help.



Thanks!







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    up vote
    0
    down vote

    favorite
    1












    I am trying to solve the following problem:




    Prove the existence of a solution of the heat equation
    $$
    u_t=-Delta u +f(t), u(0)=u_0,
    $$
    with Dirichlet Boundary conditions. Identify the spaces that the data and solution should occupy.




    This question is a specific version of question 12.9 in this book http://uxmym1.iimas.unam.mx/ramon/docs/RenRog.pdf



    I am looking to use semigroup methods. In particular I am trying to verify the conditions of Lumer-Phillips Theorem, Theorem 12.22 in the book.



    Let my operator $A=Delta$. Then if I can show that $A$ satisfies the conditions of Lumer-Phillips I will have that $A$ is the infinitesimal generator of a $C_0$-semigroup and so a solution will exist.



    My Hilbert Space is $L^2(Omega)$, and I have that $D(A)=H^2(Omega)cap H^1_0(Omega)$, which is dense in $L^2(Omega)$. So the first condition of Lumer-Phillips is satisfied.



    I still need to show the following:
    $$
    (2) text Re(x,Ax)leq w (x,x), text for some w, text for every xin D(A),
    $$
    $$
    (3) text There exists a lambda_0>w text such that A-lambda_0 I text is onto.
    $$
    Here $(bullet,bullet)$ denotes the $L^2(Omega)$ inner product, i.e. $(u,v)=int_Omega u(x)overlinev(x) dx$.



    I'm not sure how to show (2) or (3) and could use some help.



    Thanks!







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite
      1









      up vote
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      down vote

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      1





      I am trying to solve the following problem:




      Prove the existence of a solution of the heat equation
      $$
      u_t=-Delta u +f(t), u(0)=u_0,
      $$
      with Dirichlet Boundary conditions. Identify the spaces that the data and solution should occupy.




      This question is a specific version of question 12.9 in this book http://uxmym1.iimas.unam.mx/ramon/docs/RenRog.pdf



      I am looking to use semigroup methods. In particular I am trying to verify the conditions of Lumer-Phillips Theorem, Theorem 12.22 in the book.



      Let my operator $A=Delta$. Then if I can show that $A$ satisfies the conditions of Lumer-Phillips I will have that $A$ is the infinitesimal generator of a $C_0$-semigroup and so a solution will exist.



      My Hilbert Space is $L^2(Omega)$, and I have that $D(A)=H^2(Omega)cap H^1_0(Omega)$, which is dense in $L^2(Omega)$. So the first condition of Lumer-Phillips is satisfied.



      I still need to show the following:
      $$
      (2) text Re(x,Ax)leq w (x,x), text for some w, text for every xin D(A),
      $$
      $$
      (3) text There exists a lambda_0>w text such that A-lambda_0 I text is onto.
      $$
      Here $(bullet,bullet)$ denotes the $L^2(Omega)$ inner product, i.e. $(u,v)=int_Omega u(x)overlinev(x) dx$.



      I'm not sure how to show (2) or (3) and could use some help.



      Thanks!







      share|cite|improve this question














      I am trying to solve the following problem:




      Prove the existence of a solution of the heat equation
      $$
      u_t=-Delta u +f(t), u(0)=u_0,
      $$
      with Dirichlet Boundary conditions. Identify the spaces that the data and solution should occupy.




      This question is a specific version of question 12.9 in this book http://uxmym1.iimas.unam.mx/ramon/docs/RenRog.pdf



      I am looking to use semigroup methods. In particular I am trying to verify the conditions of Lumer-Phillips Theorem, Theorem 12.22 in the book.



      Let my operator $A=Delta$. Then if I can show that $A$ satisfies the conditions of Lumer-Phillips I will have that $A$ is the infinitesimal generator of a $C_0$-semigroup and so a solution will exist.



      My Hilbert Space is $L^2(Omega)$, and I have that $D(A)=H^2(Omega)cap H^1_0(Omega)$, which is dense in $L^2(Omega)$. So the first condition of Lumer-Phillips is satisfied.



      I still need to show the following:
      $$
      (2) text Re(x,Ax)leq w (x,x), text for some w, text for every xin D(A),
      $$
      $$
      (3) text There exists a lambda_0>w text such that A-lambda_0 I text is onto.
      $$
      Here $(bullet,bullet)$ denotes the $L^2(Omega)$ inner product, i.e. $(u,v)=int_Omega u(x)overlinev(x) dx$.



      I'm not sure how to show (2) or (3) and could use some help.



      Thanks!









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 17 '15 at 1:27

























      asked Nov 16 '15 at 23:44









      User112358

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          For $uin D(A)$, an integration by parts yields



          beginalign(u,Au)&=int_Omega uoverlineAu;dx
          =int_Omega uoverlineDelta u;dx=int_Omega uDeltaoverlineu;dx\
          &=sum_i int_Omega uoverlineu_x_ix_i;dx
          =-sum_i int_Omega u_x_ioverlineu_x_i;dx
          =-sum_i int_Omega |u_x_i|^2;dx\
          &=-int_Omega |Du|^2; dxleq 0
          endalign



          and thus we have $(2)$ with $w=0$.



          Let $fin L^2(Omega)$. The sesquilinear form $B:H^1_0(Omega)times H^1_0(Omega)to mathbbC$ given by
          $$B[u,v]=int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx$$
          is continuous and coercive. Furthermore, the functional linear $Lambda:H_0^1(Omega)to mathbbC$ given by
          $$Lambda(u)=-int_Omega foverlineu;dx$$
          is continuous. Thus, by the Lax-Milgram Theorem (see Theorem 1, p. 376, in Dautray's book), there exists an unique $uin H_0^1(Omega)$ such that
          $$int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx=-int_Omega foverlineu;dx,qquadforall vin H_0^1(Omega).$$
          It follows from elliptic regularity (see Theorem 9.25 in Brezis book) that $uin H^2(Omega)$ and thus
          $$u-Delta u=-f.$$



          This argumment shows that, for any $fin L^2(Omega)$, there exists (an unique) $uin D(A)$ such that $Au-Iu=f$. Therefore, $A-I$ is onto and we have $(3)$ with $lambda_0=1$.






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            up vote
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            down vote













            For $uin D(A)$, an integration by parts yields



            beginalign(u,Au)&=int_Omega uoverlineAu;dx
            =int_Omega uoverlineDelta u;dx=int_Omega uDeltaoverlineu;dx\
            &=sum_i int_Omega uoverlineu_x_ix_i;dx
            =-sum_i int_Omega u_x_ioverlineu_x_i;dx
            =-sum_i int_Omega |u_x_i|^2;dx\
            &=-int_Omega |Du|^2; dxleq 0
            endalign



            and thus we have $(2)$ with $w=0$.



            Let $fin L^2(Omega)$. The sesquilinear form $B:H^1_0(Omega)times H^1_0(Omega)to mathbbC$ given by
            $$B[u,v]=int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx$$
            is continuous and coercive. Furthermore, the functional linear $Lambda:H_0^1(Omega)to mathbbC$ given by
            $$Lambda(u)=-int_Omega foverlineu;dx$$
            is continuous. Thus, by the Lax-Milgram Theorem (see Theorem 1, p. 376, in Dautray's book), there exists an unique $uin H_0^1(Omega)$ such that
            $$int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx=-int_Omega foverlineu;dx,qquadforall vin H_0^1(Omega).$$
            It follows from elliptic regularity (see Theorem 9.25 in Brezis book) that $uin H^2(Omega)$ and thus
            $$u-Delta u=-f.$$



            This argumment shows that, for any $fin L^2(Omega)$, there exists (an unique) $uin D(A)$ such that $Au-Iu=f$. Therefore, $A-I$ is onto and we have $(3)$ with $lambda_0=1$.






            share|cite|improve this answer


























              up vote
              1
              down vote













              For $uin D(A)$, an integration by parts yields



              beginalign(u,Au)&=int_Omega uoverlineAu;dx
              =int_Omega uoverlineDelta u;dx=int_Omega uDeltaoverlineu;dx\
              &=sum_i int_Omega uoverlineu_x_ix_i;dx
              =-sum_i int_Omega u_x_ioverlineu_x_i;dx
              =-sum_i int_Omega |u_x_i|^2;dx\
              &=-int_Omega |Du|^2; dxleq 0
              endalign



              and thus we have $(2)$ with $w=0$.



              Let $fin L^2(Omega)$. The sesquilinear form $B:H^1_0(Omega)times H^1_0(Omega)to mathbbC$ given by
              $$B[u,v]=int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx$$
              is continuous and coercive. Furthermore, the functional linear $Lambda:H_0^1(Omega)to mathbbC$ given by
              $$Lambda(u)=-int_Omega foverlineu;dx$$
              is continuous. Thus, by the Lax-Milgram Theorem (see Theorem 1, p. 376, in Dautray's book), there exists an unique $uin H_0^1(Omega)$ such that
              $$int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx=-int_Omega foverlineu;dx,qquadforall vin H_0^1(Omega).$$
              It follows from elliptic regularity (see Theorem 9.25 in Brezis book) that $uin H^2(Omega)$ and thus
              $$u-Delta u=-f.$$



              This argumment shows that, for any $fin L^2(Omega)$, there exists (an unique) $uin D(A)$ such that $Au-Iu=f$. Therefore, $A-I$ is onto and we have $(3)$ with $lambda_0=1$.






              share|cite|improve this answer
























                up vote
                1
                down vote










                up vote
                1
                down vote









                For $uin D(A)$, an integration by parts yields



                beginalign(u,Au)&=int_Omega uoverlineAu;dx
                =int_Omega uoverlineDelta u;dx=int_Omega uDeltaoverlineu;dx\
                &=sum_i int_Omega uoverlineu_x_ix_i;dx
                =-sum_i int_Omega u_x_ioverlineu_x_i;dx
                =-sum_i int_Omega |u_x_i|^2;dx\
                &=-int_Omega |Du|^2; dxleq 0
                endalign



                and thus we have $(2)$ with $w=0$.



                Let $fin L^2(Omega)$. The sesquilinear form $B:H^1_0(Omega)times H^1_0(Omega)to mathbbC$ given by
                $$B[u,v]=int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx$$
                is continuous and coercive. Furthermore, the functional linear $Lambda:H_0^1(Omega)to mathbbC$ given by
                $$Lambda(u)=-int_Omega foverlineu;dx$$
                is continuous. Thus, by the Lax-Milgram Theorem (see Theorem 1, p. 376, in Dautray's book), there exists an unique $uin H_0^1(Omega)$ such that
                $$int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx=-int_Omega foverlineu;dx,qquadforall vin H_0^1(Omega).$$
                It follows from elliptic regularity (see Theorem 9.25 in Brezis book) that $uin H^2(Omega)$ and thus
                $$u-Delta u=-f.$$



                This argumment shows that, for any $fin L^2(Omega)$, there exists (an unique) $uin D(A)$ such that $Au-Iu=f$. Therefore, $A-I$ is onto and we have $(3)$ with $lambda_0=1$.






                share|cite|improve this answer














                For $uin D(A)$, an integration by parts yields



                beginalign(u,Au)&=int_Omega uoverlineAu;dx
                =int_Omega uoverlineDelta u;dx=int_Omega uDeltaoverlineu;dx\
                &=sum_i int_Omega uoverlineu_x_ix_i;dx
                =-sum_i int_Omega u_x_ioverlineu_x_i;dx
                =-sum_i int_Omega |u_x_i|^2;dx\
                &=-int_Omega |Du|^2; dxleq 0
                endalign



                and thus we have $(2)$ with $w=0$.



                Let $fin L^2(Omega)$. The sesquilinear form $B:H^1_0(Omega)times H^1_0(Omega)to mathbbC$ given by
                $$B[u,v]=int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx$$
                is continuous and coercive. Furthermore, the functional linear $Lambda:H_0^1(Omega)to mathbbC$ given by
                $$Lambda(u)=-int_Omega foverlineu;dx$$
                is continuous. Thus, by the Lax-Milgram Theorem (see Theorem 1, p. 376, in Dautray's book), there exists an unique $uin H_0^1(Omega)$ such that
                $$int_Omega uoverlinev;dx+int_Omega Ducdot Doverlinev;dx=-int_Omega foverlineu;dx,qquadforall vin H_0^1(Omega).$$
                It follows from elliptic regularity (see Theorem 9.25 in Brezis book) that $uin H^2(Omega)$ and thus
                $$u-Delta u=-f.$$



                This argumment shows that, for any $fin L^2(Omega)$, there exists (an unique) $uin D(A)$ such that $Au-Iu=f$. Therefore, $A-I$ is onto and we have $(3)$ with $lambda_0=1$.







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                edited Aug 10 at 15:07

























                answered Nov 17 '15 at 12:13









                Pedro

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