How to prove the existence and uniqueness of the solution of a second order linear ODE?












4












$begingroup$


Considering homogeneous, linear, second order differential equation
$$y'' + p(x)y' + q(x)y = r(x), a<x<b$$
$p, q, r$ are given. We already know $y(a) = A, y(b) = B$ and $q(x) < o$. How to prove that this equation has unique solution in $[a,b]$ if it has a solution?






ordinary-differential-equations analysis







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Considering homogeneous, linear, second order differential equation
    $$y'' + p(x)y' + q(x)y = r(x), a<x<b$$
    $p, q, r$ are given. We already know $y(a) = A, y(b) = B$ and $q(x) < o$. How to prove that this equation has unique solution in $[a,b]$ if it has a solution?






    ordinary-differential-equations analysis







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      3



      $begingroup$


      Considering homogeneous, linear, second order differential equation
      $$y'' + p(x)y' + q(x)y = r(x), a<x<b$$
      $p, q, r$ are given. We already know $y(a) = A, y(b) = B$ and $q(x) < o$. How to prove that this equation has unique solution in $[a,b]$ if it has a solution?






      ordinary-differential-equations analysis







      share|cite|improve this question











      $endgroup$




      Considering homogeneous, linear, second order differential equation
      $$y'' + p(x)y' + q(x)y = r(x), a<x<b$$
      $p, q, r$ are given. We already know $y(a) = A, y(b) = B$ and $q(x) < o$. How to prove that this equation has unique solution in $[a,b]$ if it has a solution?






      ordinary-differential-equations analysis




      ordinary-differential-equations analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 24 '18 at 13:19









      amWhy

      1




      1










      asked Dec 24 '18 at 5:18









      X.T ChenX.T Chen

      517




      517






















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          Suppose $y_1(x)$ and $y_2(x)$ both satisfy



          $y'' + p(x) y' + q(x)y = r(x) tag 1$



          with



          $y_1(a) = y_2(a) = A, ; y_1(b) = y_2(b) = B; tag 2$



          then we have



          $y_1'' + p(x) y_1' + q(x)y_1 = r(x) tag 3$



          and



          $y_2'' + p(x) y_2' + q(x)y_2 = r(x), tag 4$



          and, subtracting,



          $(y_1 - y_2)'' + p(x) (y_1 - y_2)' + q(x)(y_1 - y_2) = 0, tag 5$



          and



          $(y_1 - y_2)(a) = y_1(a) - y_2(a) = 0 = y_1(b) - y_2(b) = (y_1 - y_2)(b); tag 6$



          setting



          $z(x) = y_1(x) - y_2(x), tag 7$



          we have



          $z'' + p(x) z' + q(x) z = 0, ; z(a) = z(b) = 0; tag 8$



          now if $z'(a) = 0$, then by uniqueness of solutions we must have $z(x) = 0$, $x in [a, b]$, since $z(a) = 0$, and thus $y_1(x) = y_2(x)$, and we are done; so we assume $z'(a) ne 0$; in fact, since (8) is linear, we may take $z'(a) > 0$, and thus, by the continuity of $z'(x)$, there is some $delta > 0$ such that



          $z'(x) > 0, ; x in [a, a + delta); tag 9$



          we then have



          $z(x) = z(x) - z(a) = displaystyle int_0^x z'(s) ; ds > 0, x in (a, a + delta); tag{10}$



          since $z(x) > 0$ somewhere on $[a, b]$, it must have a positive absolute maximum $x_M in (a, b)$; at such a point



          $z'(x_M) = 0, tag{11}$



          and thus by (8),



          $z''(x_M) = -q(x_M)z(x_M) > 0 tag{12}$



          by virtue of the facts that $z(x_M) > 0$, $q(x_M) < 0$; but (12) is impossible at a maximum of $z(x)$, where we must have $z''(x_M) le 0$; this contradiction forces $z'(a) = 0$ and thus also



          $z(x) = 0, ; x in [a, b], tag{13}$



          and hence



          $y_1(x) - y_2(x) = z(x) = 0, ; x in [a, b]; tag{14}$



          we conclude the equation (1) with boundary conditions (2) has at most one solution on $[a, b]$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            why does $z'(a)=0$ lead to $z(x)=0$ and what does "by the uniqueness of solutions" mean?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:17












          • $begingroup$
            @X.TChen: it's the standard result on uniqueness of solutions to ODEs with initial conditions. $z(x) =0$ is the only solution with $z(a) = z'(a) = 0$. See en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:20












          • $begingroup$
            @X.TChen: "uniqueness of solutions" means there is at most one solution for given initial conditions. So since solutions are unique, we have $z(x) = 0$ if $z(a) = z'(a) = 0$.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:23












          • $begingroup$
            picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:39










          • $begingroup$
            @X.TChen: Picard-Lindeloef extends to all orders. $z(b) = 0$ is what forces a maximum of $z(x)$ to exist; it means that $z(x)$, once positive, must return to $0$ and not just keep growing.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:44











          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%2f3050962%2fhow-to-prove-the-existence-and-uniqueness-of-the-solution-of-a-second-order-line%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Suppose $y_1(x)$ and $y_2(x)$ both satisfy



          $y'' + p(x) y' + q(x)y = r(x) tag 1$



          with



          $y_1(a) = y_2(a) = A, ; y_1(b) = y_2(b) = B; tag 2$



          then we have



          $y_1'' + p(x) y_1' + q(x)y_1 = r(x) tag 3$



          and



          $y_2'' + p(x) y_2' + q(x)y_2 = r(x), tag 4$



          and, subtracting,



          $(y_1 - y_2)'' + p(x) (y_1 - y_2)' + q(x)(y_1 - y_2) = 0, tag 5$



          and



          $(y_1 - y_2)(a) = y_1(a) - y_2(a) = 0 = y_1(b) - y_2(b) = (y_1 - y_2)(b); tag 6$



          setting



          $z(x) = y_1(x) - y_2(x), tag 7$



          we have



          $z'' + p(x) z' + q(x) z = 0, ; z(a) = z(b) = 0; tag 8$



          now if $z'(a) = 0$, then by uniqueness of solutions we must have $z(x) = 0$, $x in [a, b]$, since $z(a) = 0$, and thus $y_1(x) = y_2(x)$, and we are done; so we assume $z'(a) ne 0$; in fact, since (8) is linear, we may take $z'(a) > 0$, and thus, by the continuity of $z'(x)$, there is some $delta > 0$ such that



          $z'(x) > 0, ; x in [a, a + delta); tag 9$



          we then have



          $z(x) = z(x) - z(a) = displaystyle int_0^x z'(s) ; ds > 0, x in (a, a + delta); tag{10}$



          since $z(x) > 0$ somewhere on $[a, b]$, it must have a positive absolute maximum $x_M in (a, b)$; at such a point



          $z'(x_M) = 0, tag{11}$



          and thus by (8),



          $z''(x_M) = -q(x_M)z(x_M) > 0 tag{12}$



          by virtue of the facts that $z(x_M) > 0$, $q(x_M) < 0$; but (12) is impossible at a maximum of $z(x)$, where we must have $z''(x_M) le 0$; this contradiction forces $z'(a) = 0$ and thus also



          $z(x) = 0, ; x in [a, b], tag{13}$



          and hence



          $y_1(x) - y_2(x) = z(x) = 0, ; x in [a, b]; tag{14}$



          we conclude the equation (1) with boundary conditions (2) has at most one solution on $[a, b]$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            why does $z'(a)=0$ lead to $z(x)=0$ and what does "by the uniqueness of solutions" mean?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:17












          • $begingroup$
            @X.TChen: it's the standard result on uniqueness of solutions to ODEs with initial conditions. $z(x) =0$ is the only solution with $z(a) = z'(a) = 0$. See en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:20












          • $begingroup$
            @X.TChen: "uniqueness of solutions" means there is at most one solution for given initial conditions. So since solutions are unique, we have $z(x) = 0$ if $z(a) = z'(a) = 0$.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:23












          • $begingroup$
            picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:39










          • $begingroup$
            @X.TChen: Picard-Lindeloef extends to all orders. $z(b) = 0$ is what forces a maximum of $z(x)$ to exist; it means that $z(x)$, once positive, must return to $0$ and not just keep growing.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:44
















          5












          $begingroup$

          Suppose $y_1(x)$ and $y_2(x)$ both satisfy



          $y'' + p(x) y' + q(x)y = r(x) tag 1$



          with



          $y_1(a) = y_2(a) = A, ; y_1(b) = y_2(b) = B; tag 2$



          then we have



          $y_1'' + p(x) y_1' + q(x)y_1 = r(x) tag 3$



          and



          $y_2'' + p(x) y_2' + q(x)y_2 = r(x), tag 4$



          and, subtracting,



          $(y_1 - y_2)'' + p(x) (y_1 - y_2)' + q(x)(y_1 - y_2) = 0, tag 5$



          and



          $(y_1 - y_2)(a) = y_1(a) - y_2(a) = 0 = y_1(b) - y_2(b) = (y_1 - y_2)(b); tag 6$



          setting



          $z(x) = y_1(x) - y_2(x), tag 7$



          we have



          $z'' + p(x) z' + q(x) z = 0, ; z(a) = z(b) = 0; tag 8$



          now if $z'(a) = 0$, then by uniqueness of solutions we must have $z(x) = 0$, $x in [a, b]$, since $z(a) = 0$, and thus $y_1(x) = y_2(x)$, and we are done; so we assume $z'(a) ne 0$; in fact, since (8) is linear, we may take $z'(a) > 0$, and thus, by the continuity of $z'(x)$, there is some $delta > 0$ such that



          $z'(x) > 0, ; x in [a, a + delta); tag 9$



          we then have



          $z(x) = z(x) - z(a) = displaystyle int_0^x z'(s) ; ds > 0, x in (a, a + delta); tag{10}$



          since $z(x) > 0$ somewhere on $[a, b]$, it must have a positive absolute maximum $x_M in (a, b)$; at such a point



          $z'(x_M) = 0, tag{11}$



          and thus by (8),



          $z''(x_M) = -q(x_M)z(x_M) > 0 tag{12}$



          by virtue of the facts that $z(x_M) > 0$, $q(x_M) < 0$; but (12) is impossible at a maximum of $z(x)$, where we must have $z''(x_M) le 0$; this contradiction forces $z'(a) = 0$ and thus also



          $z(x) = 0, ; x in [a, b], tag{13}$



          and hence



          $y_1(x) - y_2(x) = z(x) = 0, ; x in [a, b]; tag{14}$



          we conclude the equation (1) with boundary conditions (2) has at most one solution on $[a, b]$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            why does $z'(a)=0$ lead to $z(x)=0$ and what does "by the uniqueness of solutions" mean?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:17












          • $begingroup$
            @X.TChen: it's the standard result on uniqueness of solutions to ODEs with initial conditions. $z(x) =0$ is the only solution with $z(a) = z'(a) = 0$. See en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:20












          • $begingroup$
            @X.TChen: "uniqueness of solutions" means there is at most one solution for given initial conditions. So since solutions are unique, we have $z(x) = 0$ if $z(a) = z'(a) = 0$.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:23












          • $begingroup$
            picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:39










          • $begingroup$
            @X.TChen: Picard-Lindeloef extends to all orders. $z(b) = 0$ is what forces a maximum of $z(x)$ to exist; it means that $z(x)$, once positive, must return to $0$ and not just keep growing.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:44














          5












          5








          5





          $begingroup$

          Suppose $y_1(x)$ and $y_2(x)$ both satisfy



          $y'' + p(x) y' + q(x)y = r(x) tag 1$



          with



          $y_1(a) = y_2(a) = A, ; y_1(b) = y_2(b) = B; tag 2$



          then we have



          $y_1'' + p(x) y_1' + q(x)y_1 = r(x) tag 3$



          and



          $y_2'' + p(x) y_2' + q(x)y_2 = r(x), tag 4$



          and, subtracting,



          $(y_1 - y_2)'' + p(x) (y_1 - y_2)' + q(x)(y_1 - y_2) = 0, tag 5$



          and



          $(y_1 - y_2)(a) = y_1(a) - y_2(a) = 0 = y_1(b) - y_2(b) = (y_1 - y_2)(b); tag 6$



          setting



          $z(x) = y_1(x) - y_2(x), tag 7$



          we have



          $z'' + p(x) z' + q(x) z = 0, ; z(a) = z(b) = 0; tag 8$



          now if $z'(a) = 0$, then by uniqueness of solutions we must have $z(x) = 0$, $x in [a, b]$, since $z(a) = 0$, and thus $y_1(x) = y_2(x)$, and we are done; so we assume $z'(a) ne 0$; in fact, since (8) is linear, we may take $z'(a) > 0$, and thus, by the continuity of $z'(x)$, there is some $delta > 0$ such that



          $z'(x) > 0, ; x in [a, a + delta); tag 9$



          we then have



          $z(x) = z(x) - z(a) = displaystyle int_0^x z'(s) ; ds > 0, x in (a, a + delta); tag{10}$



          since $z(x) > 0$ somewhere on $[a, b]$, it must have a positive absolute maximum $x_M in (a, b)$; at such a point



          $z'(x_M) = 0, tag{11}$



          and thus by (8),



          $z''(x_M) = -q(x_M)z(x_M) > 0 tag{12}$



          by virtue of the facts that $z(x_M) > 0$, $q(x_M) < 0$; but (12) is impossible at a maximum of $z(x)$, where we must have $z''(x_M) le 0$; this contradiction forces $z'(a) = 0$ and thus also



          $z(x) = 0, ; x in [a, b], tag{13}$



          and hence



          $y_1(x) - y_2(x) = z(x) = 0, ; x in [a, b]; tag{14}$



          we conclude the equation (1) with boundary conditions (2) has at most one solution on $[a, b]$.






          share|cite|improve this answer











          $endgroup$



          Suppose $y_1(x)$ and $y_2(x)$ both satisfy



          $y'' + p(x) y' + q(x)y = r(x) tag 1$



          with



          $y_1(a) = y_2(a) = A, ; y_1(b) = y_2(b) = B; tag 2$



          then we have



          $y_1'' + p(x) y_1' + q(x)y_1 = r(x) tag 3$



          and



          $y_2'' + p(x) y_2' + q(x)y_2 = r(x), tag 4$



          and, subtracting,



          $(y_1 - y_2)'' + p(x) (y_1 - y_2)' + q(x)(y_1 - y_2) = 0, tag 5$



          and



          $(y_1 - y_2)(a) = y_1(a) - y_2(a) = 0 = y_1(b) - y_2(b) = (y_1 - y_2)(b); tag 6$



          setting



          $z(x) = y_1(x) - y_2(x), tag 7$



          we have



          $z'' + p(x) z' + q(x) z = 0, ; z(a) = z(b) = 0; tag 8$



          now if $z'(a) = 0$, then by uniqueness of solutions we must have $z(x) = 0$, $x in [a, b]$, since $z(a) = 0$, and thus $y_1(x) = y_2(x)$, and we are done; so we assume $z'(a) ne 0$; in fact, since (8) is linear, we may take $z'(a) > 0$, and thus, by the continuity of $z'(x)$, there is some $delta > 0$ such that



          $z'(x) > 0, ; x in [a, a + delta); tag 9$



          we then have



          $z(x) = z(x) - z(a) = displaystyle int_0^x z'(s) ; ds > 0, x in (a, a + delta); tag{10}$



          since $z(x) > 0$ somewhere on $[a, b]$, it must have a positive absolute maximum $x_M in (a, b)$; at such a point



          $z'(x_M) = 0, tag{11}$



          and thus by (8),



          $z''(x_M) = -q(x_M)z(x_M) > 0 tag{12}$



          by virtue of the facts that $z(x_M) > 0$, $q(x_M) < 0$; but (12) is impossible at a maximum of $z(x)$, where we must have $z''(x_M) le 0$; this contradiction forces $z'(a) = 0$ and thus also



          $z(x) = 0, ; x in [a, b], tag{13}$



          and hence



          $y_1(x) - y_2(x) = z(x) = 0, ; x in [a, b]; tag{14}$



          we conclude the equation (1) with boundary conditions (2) has at most one solution on $[a, b]$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 24 '18 at 6:15

























          answered Dec 24 '18 at 6:00









          Robert LewisRobert Lewis

          44.9k22964




          44.9k22964








          • 1




            $begingroup$
            why does $z'(a)=0$ lead to $z(x)=0$ and what does "by the uniqueness of solutions" mean?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:17












          • $begingroup$
            @X.TChen: it's the standard result on uniqueness of solutions to ODEs with initial conditions. $z(x) =0$ is the only solution with $z(a) = z'(a) = 0$. See en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:20












          • $begingroup$
            @X.TChen: "uniqueness of solutions" means there is at most one solution for given initial conditions. So since solutions are unique, we have $z(x) = 0$ if $z(a) = z'(a) = 0$.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:23












          • $begingroup$
            picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:39










          • $begingroup$
            @X.TChen: Picard-Lindeloef extends to all orders. $z(b) = 0$ is what forces a maximum of $z(x)$ to exist; it means that $z(x)$, once positive, must return to $0$ and not just keep growing.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:44














          • 1




            $begingroup$
            why does $z'(a)=0$ lead to $z(x)=0$ and what does "by the uniqueness of solutions" mean?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:17












          • $begingroup$
            @X.TChen: it's the standard result on uniqueness of solutions to ODEs with initial conditions. $z(x) =0$ is the only solution with $z(a) = z'(a) = 0$. See en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:20












          • $begingroup$
            @X.TChen: "uniqueness of solutions" means there is at most one solution for given initial conditions. So since solutions are unique, we have $z(x) = 0$ if $z(a) = z'(a) = 0$.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:23












          • $begingroup$
            picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            $endgroup$
            – X.T Chen
            Dec 24 '18 at 6:39










          • $begingroup$
            @X.TChen: Picard-Lindeloef extends to all orders. $z(b) = 0$ is what forces a maximum of $z(x)$ to exist; it means that $z(x)$, once positive, must return to $0$ and not just keep growing.
            $endgroup$
            – Robert Lewis
            Dec 24 '18 at 6:44








          1




          1




          $begingroup$
          why does $z'(a)=0$ lead to $z(x)=0$ and what does "by the uniqueness of solutions" mean?
          $endgroup$
          – X.T Chen
          Dec 24 '18 at 6:17






          $begingroup$
          why does $z'(a)=0$ lead to $z(x)=0$ and what does "by the uniqueness of solutions" mean?
          $endgroup$
          – X.T Chen
          Dec 24 '18 at 6:17














          $begingroup$
          @X.TChen: it's the standard result on uniqueness of solutions to ODEs with initial conditions. $z(x) =0$ is the only solution with $z(a) = z'(a) = 0$. See en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
          $endgroup$
          – Robert Lewis
          Dec 24 '18 at 6:20






          $begingroup$
          @X.TChen: it's the standard result on uniqueness of solutions to ODEs with initial conditions. $z(x) =0$ is the only solution with $z(a) = z'(a) = 0$. See en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
          $endgroup$
          – Robert Lewis
          Dec 24 '18 at 6:20














          $begingroup$
          @X.TChen: "uniqueness of solutions" means there is at most one solution for given initial conditions. So since solutions are unique, we have $z(x) = 0$ if $z(a) = z'(a) = 0$.
          $endgroup$
          – Robert Lewis
          Dec 24 '18 at 6:23






          $begingroup$
          @X.TChen: "uniqueness of solutions" means there is at most one solution for given initial conditions. So since solutions are unique, we have $z(x) = 0$ if $z(a) = z'(a) = 0$.
          $endgroup$
          – Robert Lewis
          Dec 24 '18 at 6:23














          $begingroup$
          picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
          $endgroup$
          – X.T Chen
          Dec 24 '18 at 6:39




          $begingroup$
          picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
          $endgroup$
          – X.T Chen
          Dec 24 '18 at 6:39












          $begingroup$
          @X.TChen: Picard-Lindeloef extends to all orders. $z(b) = 0$ is what forces a maximum of $z(x)$ to exist; it means that $z(x)$, once positive, must return to $0$ and not just keep growing.
          $endgroup$
          – Robert Lewis
          Dec 24 '18 at 6:44




          $begingroup$
          @X.TChen: Picard-Lindeloef extends to all orders. $z(b) = 0$ is what forces a maximum of $z(x)$ to exist; it means that $z(x)$, once positive, must return to $0$ and not just keep growing.
          $endgroup$
          – Robert Lewis
          Dec 24 '18 at 6:44


















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050962%2fhow-to-prove-the-existence-and-uniqueness-of-the-solution-of-a-second-order-line%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

          "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness. ...
          Read more

          Alcedinidae

          Image
          Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout ...
          Read more

          Origin of the phrase “under your belt”?

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 17 1 Today, I found myself discussing what students should have "under their belts" during a lecture, and I wondered to myself if there was some inappropriate undertone here that I might be unaware of. Question : What is the origin of the phrase "under your belt"? A google search revealed some webpages (e.g. [1]) that assert its origin is related to consumption of food -- once a meal has been eaten, it's under your belt (which is a relief). However, with such websites, I have no way to determine fact from "random guy on the internet making stuff up". etymology idioms ...
          Read more