Stability region for two step Nyström method












0












$begingroup$


I have the following two step Nyström method: $u_{k+1}=u_{k-1}+2hcdot f(u_k)$. I want to know the stability region, so I wrote this as $w_{k+1} = Acdot w_k$ with $A=left(begin{matrix} 2hlambda & 1 \ 1 & 0end{matrix}right)$ and $w_k = (u_{k},u_{k-1})^t$. I use the sample/test equation $u'=lambdacdot u = f(u)$ with $lambdainmathbb{C}$. To find the stability region I must set some limitations on the matrix A.



Do I want the whole vector $w_{k+1}$ to go to the zero vector as $k$ goes to infinity, or just the first component that contains $u_{k+1}$? I don't think I can let the second component go to zero, since the matrix just has a $1$ in the bottomleft corner. So I think I want $2hlambda$ to be negative in the reals (like Forward Euler) , or at least less than 1? I'm unfortunately just guessing at this point.



For (Forward) Euler we have $u_{k+1}=u_k + hcdot f(u_k) = (1 + hlambda)u_k$ and so we require that $hlambda$ does not have a positive real part.



So I guess I also want some anologue for matrices being less than 1? I do not know how to deal with the matrices since the left hand vector not only contains $u_{k+1}$ but also the old $u_k$.






numerical-methods stability-theory







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I have the following two step Nyström method: $u_{k+1}=u_{k-1}+2hcdot f(u_k)$. I want to know the stability region, so I wrote this as $w_{k+1} = Acdot w_k$ with $A=left(begin{matrix} 2hlambda & 1 \ 1 & 0end{matrix}right)$ and $w_k = (u_{k},u_{k-1})^t$. I use the sample/test equation $u'=lambdacdot u = f(u)$ with $lambdainmathbb{C}$. To find the stability region I must set some limitations on the matrix A.



    Do I want the whole vector $w_{k+1}$ to go to the zero vector as $k$ goes to infinity, or just the first component that contains $u_{k+1}$? I don't think I can let the second component go to zero, since the matrix just has a $1$ in the bottomleft corner. So I think I want $2hlambda$ to be negative in the reals (like Forward Euler) , or at least less than 1? I'm unfortunately just guessing at this point.



    For (Forward) Euler we have $u_{k+1}=u_k + hcdot f(u_k) = (1 + hlambda)u_k$ and so we require that $hlambda$ does not have a positive real part.



    So I guess I also want some anologue for matrices being less than 1? I do not know how to deal with the matrices since the left hand vector not only contains $u_{k+1}$ but also the old $u_k$.






    numerical-methods stability-theory







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have the following two step Nyström method: $u_{k+1}=u_{k-1}+2hcdot f(u_k)$. I want to know the stability region, so I wrote this as $w_{k+1} = Acdot w_k$ with $A=left(begin{matrix} 2hlambda & 1 \ 1 & 0end{matrix}right)$ and $w_k = (u_{k},u_{k-1})^t$. I use the sample/test equation $u'=lambdacdot u = f(u)$ with $lambdainmathbb{C}$. To find the stability region I must set some limitations on the matrix A.



      Do I want the whole vector $w_{k+1}$ to go to the zero vector as $k$ goes to infinity, or just the first component that contains $u_{k+1}$? I don't think I can let the second component go to zero, since the matrix just has a $1$ in the bottomleft corner. So I think I want $2hlambda$ to be negative in the reals (like Forward Euler) , or at least less than 1? I'm unfortunately just guessing at this point.



      For (Forward) Euler we have $u_{k+1}=u_k + hcdot f(u_k) = (1 + hlambda)u_k$ and so we require that $hlambda$ does not have a positive real part.



      So I guess I also want some anologue for matrices being less than 1? I do not know how to deal with the matrices since the left hand vector not only contains $u_{k+1}$ but also the old $u_k$.






      numerical-methods stability-theory







      share|cite|improve this question











      $endgroup$




      I have the following two step Nyström method: $u_{k+1}=u_{k-1}+2hcdot f(u_k)$. I want to know the stability region, so I wrote this as $w_{k+1} = Acdot w_k$ with $A=left(begin{matrix} 2hlambda & 1 \ 1 & 0end{matrix}right)$ and $w_k = (u_{k},u_{k-1})^t$. I use the sample/test equation $u'=lambdacdot u = f(u)$ with $lambdainmathbb{C}$. To find the stability region I must set some limitations on the matrix A.



      Do I want the whole vector $w_{k+1}$ to go to the zero vector as $k$ goes to infinity, or just the first component that contains $u_{k+1}$? I don't think I can let the second component go to zero, since the matrix just has a $1$ in the bottomleft corner. So I think I want $2hlambda$ to be negative in the reals (like Forward Euler) , or at least less than 1? I'm unfortunately just guessing at this point.



      For (Forward) Euler we have $u_{k+1}=u_k + hcdot f(u_k) = (1 + hlambda)u_k$ and so we require that $hlambda$ does not have a positive real part.



      So I guess I also want some anologue for matrices being less than 1? I do not know how to deal with the matrices since the left hand vector not only contains $u_{k+1}$ but also the old $u_k$.






      numerical-methods stability-theory




      numerical-methods stability-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 21:17







      The Coding Wombat

















      asked Jan 15 at 20:54









      The Coding WombatThe Coding Wombat

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          $begingroup$

          The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.



          In conclusion, the method is nowhere stable.



          Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.



          Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:16










          • $begingroup$
            If I impose conditions on the real part of $λ$, it obviously can be complex.
            $endgroup$
            – LutzL
            Jan 15 at 21:21










          • $begingroup$
            Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:40






          • 1




            $begingroup$
            For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
            $endgroup$
            – LutzL
            Jan 15 at 22:39






          • 1




            $begingroup$
            Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
            $endgroup$
            – LutzL
            Jan 16 at 21:51












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          $begingroup$

          The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.



          In conclusion, the method is nowhere stable.



          Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.



          Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:16










          • $begingroup$
            If I impose conditions on the real part of $λ$, it obviously can be complex.
            $endgroup$
            – LutzL
            Jan 15 at 21:21










          • $begingroup$
            Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:40






          • 1




            $begingroup$
            For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
            $endgroup$
            – LutzL
            Jan 15 at 22:39






          • 1




            $begingroup$
            Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
            $endgroup$
            – LutzL
            Jan 16 at 21:51
















          1












          $begingroup$

          The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.



          In conclusion, the method is nowhere stable.



          Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.



          Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:16










          • $begingroup$
            If I impose conditions on the real part of $λ$, it obviously can be complex.
            $endgroup$
            – LutzL
            Jan 15 at 21:21










          • $begingroup$
            Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:40






          • 1




            $begingroup$
            For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
            $endgroup$
            – LutzL
            Jan 15 at 22:39






          • 1




            $begingroup$
            Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
            $endgroup$
            – LutzL
            Jan 16 at 21:51














          1












          1








          1





          $begingroup$

          The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.



          In conclusion, the method is nowhere stable.



          Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.



          Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.






          share|cite|improve this answer











          $endgroup$



          The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.



          In conclusion, the method is nowhere stable.



          Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.



          Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 15 at 21:33

























          answered Jan 15 at 21:02









          LutzLLutzL

          60.1k42057




          60.1k42057












          • $begingroup$
            Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:16










          • $begingroup$
            If I impose conditions on the real part of $λ$, it obviously can be complex.
            $endgroup$
            – LutzL
            Jan 15 at 21:21










          • $begingroup$
            Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:40






          • 1




            $begingroup$
            For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
            $endgroup$
            – LutzL
            Jan 15 at 22:39






          • 1




            $begingroup$
            Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
            $endgroup$
            – LutzL
            Jan 16 at 21:51


















          • $begingroup$
            Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:16










          • $begingroup$
            If I impose conditions on the real part of $λ$, it obviously can be complex.
            $endgroup$
            – LutzL
            Jan 15 at 21:21










          • $begingroup$
            Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
            $endgroup$
            – The Coding Wombat
            Jan 15 at 21:40






          • 1




            $begingroup$
            For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
            $endgroup$
            – LutzL
            Jan 15 at 22:39






          • 1




            $begingroup$
            Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
            $endgroup$
            – LutzL
            Jan 16 at 21:51
















          $begingroup$
          Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
          $endgroup$
          – The Coding Wombat
          Jan 15 at 21:16




          $begingroup$
          Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
          $endgroup$
          – The Coding Wombat
          Jan 15 at 21:16












          $begingroup$
          If I impose conditions on the real part of $λ$, it obviously can be complex.
          $endgroup$
          – LutzL
          Jan 15 at 21:21




          $begingroup$
          If I impose conditions on the real part of $λ$, it obviously can be complex.
          $endgroup$
          – LutzL
          Jan 15 at 21:21












          $begingroup$
          Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
          $endgroup$
          – The Coding Wombat
          Jan 15 at 21:40




          $begingroup$
          Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
          $endgroup$
          – The Coding Wombat
          Jan 15 at 21:40




          1




          1




          $begingroup$
          For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
          $endgroup$
          – LutzL
          Jan 15 at 22:39




          $begingroup$
          For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
          $endgroup$
          – LutzL
          Jan 15 at 22:39




          1




          1




          $begingroup$
          Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
          $endgroup$
          – LutzL
          Jan 16 at 21:51




          $begingroup$
          Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
          $endgroup$
          – LutzL
          Jan 16 at 21:51


















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          Human spaceflight

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          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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          張江高科駅

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          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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