Well-posedness for Heat Equation with Robin Boundary Condition












1












$begingroup$


Can anyone help me prove the well-posedness of the following heat equation with Robin boundary condition?



$u_t(x,t)=u_{xx}(x,t)$



$u(0,t)=0$



$u_x(1,t)=-au(1,t)$



where $a>0$.



The existence of the solution may be simply obtained by separation of variable. Are there any good references on this problem?






pde heat-equation







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$endgroup$

















    1












    $begingroup$


    Can anyone help me prove the well-posedness of the following heat equation with Robin boundary condition?



    $u_t(x,t)=u_{xx}(x,t)$



    $u(0,t)=0$



    $u_x(1,t)=-au(1,t)$



    where $a>0$.



    The existence of the solution may be simply obtained by separation of variable. Are there any good references on this problem?






    pde heat-equation







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Can anyone help me prove the well-posedness of the following heat equation with Robin boundary condition?



      $u_t(x,t)=u_{xx}(x,t)$



      $u(0,t)=0$



      $u_x(1,t)=-au(1,t)$



      where $a>0$.



      The existence of the solution may be simply obtained by separation of variable. Are there any good references on this problem?






      pde heat-equation







      share|cite|improve this question









      $endgroup$




      Can anyone help me prove the well-posedness of the following heat equation with Robin boundary condition?



      $u_t(x,t)=u_{xx}(x,t)$



      $u(0,t)=0$



      $u_x(1,t)=-au(1,t)$



      where $a>0$.



      The existence of the solution may be simply obtained by separation of variable. Are there any good references on this problem?






      pde heat-equation




      pde heat-equation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 8 '18 at 1:08









      D. ZhD. Zh

      61




      61






















          2 Answers
          2






          active

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          0












          $begingroup$

          Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
          $$
          v_t = v_{xx} \
          v(x,0)=0 \
          v(0,t)=0,;;v_x(1,t)=-av(1,t).
          $$
          Then,
          begin{align}
          &frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
          &=2int_0^1v(x,t)v_t(x,t)dx \
          &= 2int_0^1v(x,t)v_{xx}(x,t)dx \
          &= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
          &= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
          &= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
          &= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
          end{align}
          Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
          $$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$



          This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
            $endgroup$
            – DaveNine
            Dec 8 '18 at 20:26










          • $begingroup$
            @DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
            $endgroup$
            – DisintegratingByParts
            Dec 8 '18 at 21:04










          • $begingroup$
            @DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
            $endgroup$
            – D. Zh
            Dec 9 '18 at 18:46



















          0












          $begingroup$

          your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
          where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
          $$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
          It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            active

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            0












            $begingroup$

            Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
            $$
            v_t = v_{xx} \
            v(x,0)=0 \
            v(0,t)=0,;;v_x(1,t)=-av(1,t).
            $$
            Then,
            begin{align}
            &frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
            &=2int_0^1v(x,t)v_t(x,t)dx \
            &= 2int_0^1v(x,t)v_{xx}(x,t)dx \
            &= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
            &= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
            &= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
            &= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
            end{align}
            Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
            $$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$



            This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
              $endgroup$
              – DaveNine
              Dec 8 '18 at 20:26










            • $begingroup$
              @DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
              $endgroup$
              – DisintegratingByParts
              Dec 8 '18 at 21:04










            • $begingroup$
              @DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
              $endgroup$
              – D. Zh
              Dec 9 '18 at 18:46
















            0












            $begingroup$

            Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
            $$
            v_t = v_{xx} \
            v(x,0)=0 \
            v(0,t)=0,;;v_x(1,t)=-av(1,t).
            $$
            Then,
            begin{align}
            &frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
            &=2int_0^1v(x,t)v_t(x,t)dx \
            &= 2int_0^1v(x,t)v_{xx}(x,t)dx \
            &= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
            &= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
            &= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
            &= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
            end{align}
            Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
            $$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$



            This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
              $endgroup$
              – DaveNine
              Dec 8 '18 at 20:26










            • $begingroup$
              @DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
              $endgroup$
              – DisintegratingByParts
              Dec 8 '18 at 21:04










            • $begingroup$
              @DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
              $endgroup$
              – D. Zh
              Dec 9 '18 at 18:46














            0












            0








            0





            $begingroup$

            Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
            $$
            v_t = v_{xx} \
            v(x,0)=0 \
            v(0,t)=0,;;v_x(1,t)=-av(1,t).
            $$
            Then,
            begin{align}
            &frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
            &=2int_0^1v(x,t)v_t(x,t)dx \
            &= 2int_0^1v(x,t)v_{xx}(x,t)dx \
            &= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
            &= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
            &= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
            &= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
            end{align}
            Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
            $$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$



            This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.






            share|cite|improve this answer









            $endgroup$



            Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
            $$
            v_t = v_{xx} \
            v(x,0)=0 \
            v(0,t)=0,;;v_x(1,t)=-av(1,t).
            $$
            Then,
            begin{align}
            &frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
            &=2int_0^1v(x,t)v_t(x,t)dx \
            &= 2int_0^1v(x,t)v_{xx}(x,t)dx \
            &= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
            &= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
            &= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
            &= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
            end{align}
            Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
            $$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$



            This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 8 '18 at 4:12









            DisintegratingByPartsDisintegratingByParts

            58.9k42580




            58.9k42580












            • $begingroup$
              One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
              $endgroup$
              – DaveNine
              Dec 8 '18 at 20:26










            • $begingroup$
              @DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
              $endgroup$
              – DisintegratingByParts
              Dec 8 '18 at 21:04










            • $begingroup$
              @DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
              $endgroup$
              – D. Zh
              Dec 9 '18 at 18:46


















            • $begingroup$
              One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
              $endgroup$
              – DaveNine
              Dec 8 '18 at 20:26










            • $begingroup$
              @DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
              $endgroup$
              – DisintegratingByParts
              Dec 8 '18 at 21:04










            • $begingroup$
              @DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
              $endgroup$
              – D. Zh
              Dec 9 '18 at 18:46
















            $begingroup$
            One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
            $endgroup$
            – DaveNine
            Dec 8 '18 at 20:26




            $begingroup$
            One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
            $endgroup$
            – DaveNine
            Dec 8 '18 at 20:26












            $begingroup$
            @DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
            $endgroup$
            – DisintegratingByParts
            Dec 8 '18 at 21:04




            $begingroup$
            @DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
            $endgroup$
            – DisintegratingByParts
            Dec 8 '18 at 21:04












            $begingroup$
            @DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
            $endgroup$
            – D. Zh
            Dec 9 '18 at 18:46




            $begingroup$
            @DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
            $endgroup$
            – D. Zh
            Dec 9 '18 at 18:46











            0












            $begingroup$

            your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
            where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
            $$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
            It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
              where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
              $$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
              It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
                where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
                $$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
                It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)






                share|cite|improve this answer









                $endgroup$



                your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
                where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
                $$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
                It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 8 '18 at 9:54









                GustaveGustave

                725211




                725211






























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