Proving that a solution to a differential equation is monotonic












3












$begingroup$


The answer that ws given on a previous question of mine, stated that the solution to this DE:



$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dotstag1$$



Must be monotonic.



Is there a way to proof that that is the case, that the function $x(t)$ is monotonic?!




Background:



I've to find (the average of a function over a particular interval, where $t_1>0$, $t_2>0$ and $t_2>t_1$):



$$frac{1}{t_2-t_1}int_{t_1}^{t_2}x(t)dttag2$$



Where $x(t)$ in equation $(2)$ is the solution to the DE in equation $(1)$.






ordinary-differential-equations definite-integrals logarithms mathematical-physics average







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    The answer that ws given on a previous question of mine, stated that the solution to this DE:



    $$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dotstag1$$



    Must be monotonic.



    Is there a way to proof that that is the case, that the function $x(t)$ is monotonic?!




    Background:



    I've to find (the average of a function over a particular interval, where $t_1>0$, $t_2>0$ and $t_2>t_1$):



    $$frac{1}{t_2-t_1}int_{t_1}^{t_2}x(t)dttag2$$



    Where $x(t)$ in equation $(2)$ is the solution to the DE in equation $(1)$.






    ordinary-differential-equations definite-integrals logarithms mathematical-physics average







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      The answer that ws given on a previous question of mine, stated that the solution to this DE:



      $$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dotstag1$$



      Must be monotonic.



      Is there a way to proof that that is the case, that the function $x(t)$ is monotonic?!




      Background:



      I've to find (the average of a function over a particular interval, where $t_1>0$, $t_2>0$ and $t_2>t_1$):



      $$frac{1}{t_2-t_1}int_{t_1}^{t_2}x(t)dttag2$$



      Where $x(t)$ in equation $(2)$ is the solution to the DE in equation $(1)$.






      ordinary-differential-equations definite-integrals logarithms mathematical-physics average







      share|cite|improve this question











      $endgroup$




      The answer that ws given on a previous question of mine, stated that the solution to this DE:



      $$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dotstag1$$



      Must be monotonic.



      Is there a way to proof that that is the case, that the function $x(t)$ is monotonic?!




      Background:



      I've to find (the average of a function over a particular interval, where $t_1>0$, $t_2>0$ and $t_2>t_1$):



      $$frac{1}{t_2-t_1}int_{t_1}^{t_2}x(t)dttag2$$



      Where $x(t)$ in equation $(2)$ is the solution to the DE in equation $(1)$.






      ordinary-differential-equations definite-integrals logarithms mathematical-physics average




      ordinary-differential-equations definite-integrals logarithms mathematical-physics average






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 31 '18 at 17:29









      David G. Stork

      11k41432




      11k41432










      asked Dec 31 '18 at 17:23









      KlopjasKlopjas

      714




      714






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Considering the DE



          $$
          l x'+r x+alnleft(1+frac xaright)=0
          $$



          taking it's derivative (assuming $x$ is twice diferentiable)



          $$
          frac{a x'}{b left(frac{x}{b}+1right)}+l x''+r x' = 0
          $$



          after substituting $x'$ from the first equation we get



          $$
          l x''= frac{(a+b r+r x) left(a ln left(frac{b+x}{b}right)+r xright)}{l (b+x)}
          $$



          and now solving for $x$ the condition $lx'' = 0$ we obtain



          $$
          x^* = {frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b,-left(b+frac arright)}
          $$



          also regarding the first equation we have



          $$
          l x'=-left(r x+alnleft(1+frac xaright)right) = 0Rightarrow x^* = frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b
          $$



          so the only stationary point has null $x''$ so it is an inflexion point indicating that $x$ is monotonic..






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You assumed it's double differentiable.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 18:37










          • $begingroup$
            @LucasHenrique Yes. I assumed that. (Included in the last edit).
            $endgroup$
            – Cesareo
            Dec 31 '18 at 18:53










          • $begingroup$
            Why did you take $lx'' = 0$?
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 20:51










          • $begingroup$
            @LucasHenrique I think that now the conclusion is the correct one.
            $endgroup$
            – Cesareo
            Dec 31 '18 at 21:53










          • $begingroup$
            @Cesareo So $x(t)$ is monotonous? As I conclude from your answer?!
            $endgroup$
            – Klopjas
            Dec 31 '18 at 21:54



















          1












          $begingroup$

          A differentiable function is not monotonic if there are two different points whose derivatives have different signs. Since $x' = -frac{rx + aln ( 1 + x/b)}{l}$ is a composition of differentiable functions, it's differentiable and thus continuous; if not monotonic, $x'$ it must have a root; and at least one root is not an inflection point. Simbolically, $$begin{cases} x'(t) = 0 \ x''(t) neq 0end{cases}$$ for some $t$. But $x'' = -x'left(r + frac{a}{b + x}right)$ so the second derivative is zero on all the roots of the first derivative. So every root is an inflection point and thus, in fact, $x$ is monotonic.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            OP: your function is monotonic.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 23:54











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Considering the DE



          $$
          l x'+r x+alnleft(1+frac xaright)=0
          $$



          taking it's derivative (assuming $x$ is twice diferentiable)



          $$
          frac{a x'}{b left(frac{x}{b}+1right)}+l x''+r x' = 0
          $$



          after substituting $x'$ from the first equation we get



          $$
          l x''= frac{(a+b r+r x) left(a ln left(frac{b+x}{b}right)+r xright)}{l (b+x)}
          $$



          and now solving for $x$ the condition $lx'' = 0$ we obtain



          $$
          x^* = {frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b,-left(b+frac arright)}
          $$



          also regarding the first equation we have



          $$
          l x'=-left(r x+alnleft(1+frac xaright)right) = 0Rightarrow x^* = frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b
          $$



          so the only stationary point has null $x''$ so it is an inflexion point indicating that $x$ is monotonic..






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You assumed it's double differentiable.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 18:37










          • $begingroup$
            @LucasHenrique Yes. I assumed that. (Included in the last edit).
            $endgroup$
            – Cesareo
            Dec 31 '18 at 18:53










          • $begingroup$
            Why did you take $lx'' = 0$?
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 20:51










          • $begingroup$
            @LucasHenrique I think that now the conclusion is the correct one.
            $endgroup$
            – Cesareo
            Dec 31 '18 at 21:53










          • $begingroup$
            @Cesareo So $x(t)$ is monotonous? As I conclude from your answer?!
            $endgroup$
            – Klopjas
            Dec 31 '18 at 21:54
















          2












          $begingroup$

          Considering the DE



          $$
          l x'+r x+alnleft(1+frac xaright)=0
          $$



          taking it's derivative (assuming $x$ is twice diferentiable)



          $$
          frac{a x'}{b left(frac{x}{b}+1right)}+l x''+r x' = 0
          $$



          after substituting $x'$ from the first equation we get



          $$
          l x''= frac{(a+b r+r x) left(a ln left(frac{b+x}{b}right)+r xright)}{l (b+x)}
          $$



          and now solving for $x$ the condition $lx'' = 0$ we obtain



          $$
          x^* = {frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b,-left(b+frac arright)}
          $$



          also regarding the first equation we have



          $$
          l x'=-left(r x+alnleft(1+frac xaright)right) = 0Rightarrow x^* = frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b
          $$



          so the only stationary point has null $x''$ so it is an inflexion point indicating that $x$ is monotonic..






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You assumed it's double differentiable.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 18:37










          • $begingroup$
            @LucasHenrique Yes. I assumed that. (Included in the last edit).
            $endgroup$
            – Cesareo
            Dec 31 '18 at 18:53










          • $begingroup$
            Why did you take $lx'' = 0$?
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 20:51










          • $begingroup$
            @LucasHenrique I think that now the conclusion is the correct one.
            $endgroup$
            – Cesareo
            Dec 31 '18 at 21:53










          • $begingroup$
            @Cesareo So $x(t)$ is monotonous? As I conclude from your answer?!
            $endgroup$
            – Klopjas
            Dec 31 '18 at 21:54














          2












          2








          2





          $begingroup$

          Considering the DE



          $$
          l x'+r x+alnleft(1+frac xaright)=0
          $$



          taking it's derivative (assuming $x$ is twice diferentiable)



          $$
          frac{a x'}{b left(frac{x}{b}+1right)}+l x''+r x' = 0
          $$



          after substituting $x'$ from the first equation we get



          $$
          l x''= frac{(a+b r+r x) left(a ln left(frac{b+x}{b}right)+r xright)}{l (b+x)}
          $$



          and now solving for $x$ the condition $lx'' = 0$ we obtain



          $$
          x^* = {frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b,-left(b+frac arright)}
          $$



          also regarding the first equation we have



          $$
          l x'=-left(r x+alnleft(1+frac xaright)right) = 0Rightarrow x^* = frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b
          $$



          so the only stationary point has null $x''$ so it is an inflexion point indicating that $x$ is monotonic..






          share|cite|improve this answer











          $endgroup$



          Considering the DE



          $$
          l x'+r x+alnleft(1+frac xaright)=0
          $$



          taking it's derivative (assuming $x$ is twice diferentiable)



          $$
          frac{a x'}{b left(frac{x}{b}+1right)}+l x''+r x' = 0
          $$



          after substituting $x'$ from the first equation we get



          $$
          l x''= frac{(a+b r+r x) left(a ln left(frac{b+x}{b}right)+r xright)}{l (b+x)}
          $$



          and now solving for $x$ the condition $lx'' = 0$ we obtain



          $$
          x^* = {frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b,-left(b+frac arright)}
          $$



          also regarding the first equation we have



          $$
          l x'=-left(r x+alnleft(1+frac xaright)right) = 0Rightarrow x^* = frac ar Wleft(frac{b e^{frac{b r}{a}} r}{a}right)-b
          $$



          so the only stationary point has null $x''$ so it is an inflexion point indicating that $x$ is monotonic..







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 31 '18 at 21:48

























          answered Dec 31 '18 at 18:19









          CesareoCesareo

          9,1063517




          9,1063517












          • $begingroup$
            You assumed it's double differentiable.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 18:37










          • $begingroup$
            @LucasHenrique Yes. I assumed that. (Included in the last edit).
            $endgroup$
            – Cesareo
            Dec 31 '18 at 18:53










          • $begingroup$
            Why did you take $lx'' = 0$?
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 20:51










          • $begingroup$
            @LucasHenrique I think that now the conclusion is the correct one.
            $endgroup$
            – Cesareo
            Dec 31 '18 at 21:53










          • $begingroup$
            @Cesareo So $x(t)$ is monotonous? As I conclude from your answer?!
            $endgroup$
            – Klopjas
            Dec 31 '18 at 21:54


















          • $begingroup$
            You assumed it's double differentiable.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 18:37










          • $begingroup$
            @LucasHenrique Yes. I assumed that. (Included in the last edit).
            $endgroup$
            – Cesareo
            Dec 31 '18 at 18:53










          • $begingroup$
            Why did you take $lx'' = 0$?
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 20:51










          • $begingroup$
            @LucasHenrique I think that now the conclusion is the correct one.
            $endgroup$
            – Cesareo
            Dec 31 '18 at 21:53










          • $begingroup$
            @Cesareo So $x(t)$ is monotonous? As I conclude from your answer?!
            $endgroup$
            – Klopjas
            Dec 31 '18 at 21:54
















          $begingroup$
          You assumed it's double differentiable.
          $endgroup$
          – Lucas Henrique
          Dec 31 '18 at 18:37




          $begingroup$
          You assumed it's double differentiable.
          $endgroup$
          – Lucas Henrique
          Dec 31 '18 at 18:37












          $begingroup$
          @LucasHenrique Yes. I assumed that. (Included in the last edit).
          $endgroup$
          – Cesareo
          Dec 31 '18 at 18:53




          $begingroup$
          @LucasHenrique Yes. I assumed that. (Included in the last edit).
          $endgroup$
          – Cesareo
          Dec 31 '18 at 18:53












          $begingroup$
          Why did you take $lx'' = 0$?
          $endgroup$
          – Lucas Henrique
          Dec 31 '18 at 20:51




          $begingroup$
          Why did you take $lx'' = 0$?
          $endgroup$
          – Lucas Henrique
          Dec 31 '18 at 20:51












          $begingroup$
          @LucasHenrique I think that now the conclusion is the correct one.
          $endgroup$
          – Cesareo
          Dec 31 '18 at 21:53




          $begingroup$
          @LucasHenrique I think that now the conclusion is the correct one.
          $endgroup$
          – Cesareo
          Dec 31 '18 at 21:53












          $begingroup$
          @Cesareo So $x(t)$ is monotonous? As I conclude from your answer?!
          $endgroup$
          – Klopjas
          Dec 31 '18 at 21:54




          $begingroup$
          @Cesareo So $x(t)$ is monotonous? As I conclude from your answer?!
          $endgroup$
          – Klopjas
          Dec 31 '18 at 21:54











          1












          $begingroup$

          A differentiable function is not monotonic if there are two different points whose derivatives have different signs. Since $x' = -frac{rx + aln ( 1 + x/b)}{l}$ is a composition of differentiable functions, it's differentiable and thus continuous; if not monotonic, $x'$ it must have a root; and at least one root is not an inflection point. Simbolically, $$begin{cases} x'(t) = 0 \ x''(t) neq 0end{cases}$$ for some $t$. But $x'' = -x'left(r + frac{a}{b + x}right)$ so the second derivative is zero on all the roots of the first derivative. So every root is an inflection point and thus, in fact, $x$ is monotonic.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            OP: your function is monotonic.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 23:54
















          1












          $begingroup$

          A differentiable function is not monotonic if there are two different points whose derivatives have different signs. Since $x' = -frac{rx + aln ( 1 + x/b)}{l}$ is a composition of differentiable functions, it's differentiable and thus continuous; if not monotonic, $x'$ it must have a root; and at least one root is not an inflection point. Simbolically, $$begin{cases} x'(t) = 0 \ x''(t) neq 0end{cases}$$ for some $t$. But $x'' = -x'left(r + frac{a}{b + x}right)$ so the second derivative is zero on all the roots of the first derivative. So every root is an inflection point and thus, in fact, $x$ is monotonic.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            OP: your function is monotonic.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 23:54














          1












          1








          1





          $begingroup$

          A differentiable function is not monotonic if there are two different points whose derivatives have different signs. Since $x' = -frac{rx + aln ( 1 + x/b)}{l}$ is a composition of differentiable functions, it's differentiable and thus continuous; if not monotonic, $x'$ it must have a root; and at least one root is not an inflection point. Simbolically, $$begin{cases} x'(t) = 0 \ x''(t) neq 0end{cases}$$ for some $t$. But $x'' = -x'left(r + frac{a}{b + x}right)$ so the second derivative is zero on all the roots of the first derivative. So every root is an inflection point and thus, in fact, $x$ is monotonic.






          share|cite|improve this answer











          $endgroup$



          A differentiable function is not monotonic if there are two different points whose derivatives have different signs. Since $x' = -frac{rx + aln ( 1 + x/b)}{l}$ is a composition of differentiable functions, it's differentiable and thus continuous; if not monotonic, $x'$ it must have a root; and at least one root is not an inflection point. Simbolically, $$begin{cases} x'(t) = 0 \ x''(t) neq 0end{cases}$$ for some $t$. But $x'' = -x'left(r + frac{a}{b + x}right)$ so the second derivative is zero on all the roots of the first derivative. So every root is an inflection point and thus, in fact, $x$ is monotonic.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 31 '18 at 23:50

























          answered Dec 31 '18 at 22:16









          Lucas HenriqueLucas Henrique

          1,026414




          1,026414












          • $begingroup$
            OP: your function is monotonic.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 23:54


















          • $begingroup$
            OP: your function is monotonic.
            $endgroup$
            – Lucas Henrique
            Dec 31 '18 at 23:54
















          $begingroup$
          OP: your function is monotonic.
          $endgroup$
          – Lucas Henrique
          Dec 31 '18 at 23:54




          $begingroup$
          OP: your function is monotonic.
          $endgroup$
          – Lucas Henrique
          Dec 31 '18 at 23:54


















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          Storängen

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          För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
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          VLT Carioca

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          VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
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