The reverse implication in the stability theory











up vote
0
down vote

favorite












Let us consider the system of differential equation $dot x=Ax+h(t),$ where A is a constant matrix of dimension $ngeq2$ and $h$ is continuous on $[0,infty).$ All eigenvalues of $A$ have a negative real part. It is known, that $h(t)to 0$ for $ttoinfty$ implies that all solutions $x(t)$ of $dot x=Ax+h(t)$ converge to $0$ for $ttoinfty.$
Is the converse implication true too? (All solutions $x(t)to 0$ implies $h(t)to 0$).



Thank you.










share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    Let us consider the system of differential equation $dot x=Ax+h(t),$ where A is a constant matrix of dimension $ngeq2$ and $h$ is continuous on $[0,infty).$ All eigenvalues of $A$ have a negative real part. It is known, that $h(t)to 0$ for $ttoinfty$ implies that all solutions $x(t)$ of $dot x=Ax+h(t)$ converge to $0$ for $ttoinfty.$
    Is the converse implication true too? (All solutions $x(t)to 0$ implies $h(t)to 0$).



    Thank you.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let us consider the system of differential equation $dot x=Ax+h(t),$ where A is a constant matrix of dimension $ngeq2$ and $h$ is continuous on $[0,infty).$ All eigenvalues of $A$ have a negative real part. It is known, that $h(t)to 0$ for $ttoinfty$ implies that all solutions $x(t)$ of $dot x=Ax+h(t)$ converge to $0$ for $ttoinfty.$
      Is the converse implication true too? (All solutions $x(t)to 0$ implies $h(t)to 0$).



      Thank you.










      share|cite|improve this question













      Let us consider the system of differential equation $dot x=Ax+h(t),$ where A is a constant matrix of dimension $ngeq2$ and $h$ is continuous on $[0,infty).$ All eigenvalues of $A$ have a negative real part. It is known, that $h(t)to 0$ for $ttoinfty$ implies that all solutions $x(t)$ of $dot x=Ax+h(t)$ converge to $0$ for $ttoinfty.$
      Is the converse implication true too? (All solutions $x(t)to 0$ implies $h(t)to 0$).



      Thank you.







      differential-equations stability-in-odes






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 23 at 18:27









      Robert

      112




      112






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote













          More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
          begin{equation}
          suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
          end{equation}

          as $ttoinfty.$



          For the details and (long) proof see Theorem B in:



          Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.






          share|cite|improve this answer




























            up vote
            0
            down vote













            No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
            $$
            h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
            $$

            Consider now for instance $x'(t)=-x(t)+h(t)$.
            All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
            $$
            x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
            $$

            From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.






            share|cite|improve this answer





















              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%2f3010682%2fthe-reverse-implication-in-the-stability-theory%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote













              More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
              begin{equation}
              suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
              end{equation}

              as $ttoinfty.$



              For the details and (long) proof see Theorem B in:



              Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.






              share|cite|improve this answer

























                up vote
                1
                down vote













                More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
                begin{equation}
                suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
                end{equation}

                as $ttoinfty.$



                For the details and (long) proof see Theorem B in:



                Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
                  begin{equation}
                  suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
                  end{equation}

                  as $ttoinfty.$



                  For the details and (long) proof see Theorem B in:



                  Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.






                  share|cite|improve this answer












                  More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
                  begin{equation}
                  suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
                  end{equation}

                  as $ttoinfty.$



                  For the details and (long) proof see Theorem B in:



                  Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 at 9:54









                  Robert

                  112




                  112






















                      up vote
                      0
                      down vote













                      No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
                      $$
                      h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
                      $$

                      Consider now for instance $x'(t)=-x(t)+h(t)$.
                      All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
                      $$
                      x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
                      $$

                      From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
                        $$
                        h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
                        $$

                        Consider now for instance $x'(t)=-x(t)+h(t)$.
                        All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
                        $$
                        x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
                        $$

                        From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
                          $$
                          h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
                          $$

                          Consider now for instance $x'(t)=-x(t)+h(t)$.
                          All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
                          $$
                          x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
                          $$

                          From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.






                          share|cite|improve this answer












                          No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
                          $$
                          h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
                          $$

                          Consider now for instance $x'(t)=-x(t)+h(t)$.
                          All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
                          $$
                          x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
                          $$

                          From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 23 at 18:51









                          Federico

                          4,184512




                          4,184512






























                              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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              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%2f3010682%2fthe-reverse-implication-in-the-stability-theory%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

                              Ellipse (mathématiques)

                              Quarter-circle Tiles

                              Mont Emei