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












4














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?






differential-equations analysis







share|cite|improve this question





























    4














    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?






    differential-equations analysis







    share|cite|improve this question



























      4












      4








      4


      3





      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?






      differential-equations analysis







      share|cite|improve this question















      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?






      differential-equations analysis




      differential-equations analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      amWhy

      191k28224439




      191k28224439










      asked Dec 24 at 5:18









      X.T Chen

      576




      576






















          1 Answer
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          5














          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



















          • 1




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












          • @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
            – Robert Lewis
            Dec 24 at 6:20












          • @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 $z(a) = z'(a) = 0$.
            – Robert Lewis
            Dec 24 at 6:23












          • picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            – X.T Chen
            2 days ago










          • @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.
            – Robert Lewis
            2 days ago











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          1 Answer
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          5














          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



















          • 1




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












          • @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
            – Robert Lewis
            Dec 24 at 6:20












          • @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 $z(a) = z'(a) = 0$.
            – Robert Lewis
            Dec 24 at 6:23












          • picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            – X.T Chen
            2 days ago










          • @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.
            – Robert Lewis
            2 days ago
















          5














          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



















          • 1




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












          • @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
            – Robert Lewis
            Dec 24 at 6:20












          • @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 $z(a) = z'(a) = 0$.
            – Robert Lewis
            Dec 24 at 6:23












          • picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            – X.T Chen
            2 days ago










          • @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.
            – Robert Lewis
            2 days ago














          5












          5








          5






          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














          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 at 6:15

























          answered Dec 24 at 6:00









          Robert Lewis

          43.5k22863




          43.5k22863








          • 1




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












          • @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
            – Robert Lewis
            Dec 24 at 6:20












          • @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 $z(a) = z'(a) = 0$.
            – Robert Lewis
            Dec 24 at 6:23












          • picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            – X.T Chen
            2 days ago










          • @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.
            – Robert Lewis
            2 days ago














          • 1




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












          • @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
            – Robert Lewis
            Dec 24 at 6:20












          • @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 $z(a) = z'(a) = 0$.
            – Robert Lewis
            Dec 24 at 6:23












          • picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
            – X.T Chen
            2 days ago










          • @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.
            – Robert Lewis
            2 days ago








          1




          1




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






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














          @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
          – Robert Lewis
          Dec 24 at 6:20






          @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
          – Robert Lewis
          Dec 24 at 6:20














          @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 $z(a) = z'(a) = 0$.
          – Robert Lewis
          Dec 24 at 6:23






          @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 $z(a) = z'(a) = 0$.
          – Robert Lewis
          Dec 24 at 6:23














          picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
          – X.T Chen
          2 days ago




          picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
          – X.T Chen
          2 days ago












          @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.
          – Robert Lewis
          2 days ago




          @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.
          – Robert Lewis
          2 days ago


















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