Series extension of the superposition principle for ODEs
$begingroup$
Take the superposition principle for linear ODEs of the form $y'(t)=A(t,y(t)) + g(t)$ ($yin mathbb{R}^n$, $A$ a linear function in y). If $g(t)=sum _{k=1}^N g_k(t)$ then $y(t)=sum _{k=1}^N y_k(t)$ solves the system, where each $y_k$ solves the ODE with $g(t)=g_k(t)$.
Can this be extended from finite sums to series?
sequences-and-series ordinary-differential-equations sequence-of-function
edited Dec 11 '18 at 16:07
Alex Ravsky
42.7k32383
asked Dec 10 '18 at 15:32
LeonardoLeonardo
3339
$endgroup$
add a comment |
$begingroup$
Take the superposition principle for linear ODEs of the form $y'(t)=A(t,y(t)) + g(t)$ ($yin mathbb{R}^n$, $A$ a linear function in y). If $g(t)=sum _{k=1}^N g_k(t)$ then $y(t)=sum _{k=1}^N y_k(t)$ solves the system, where each $y_k$ solves the ODE with $g(t)=g_k(t)$.
Can this be extended from finite sums to series?
sequences-and-series ordinary-differential-equations sequence-of-function
edited Dec 11 '18 at 16:07
Alex Ravsky
42.7k32383
asked Dec 10 '18 at 15:32
LeonardoLeonardo
3339
$endgroup$
add a comment |
$begingroup$
Take the superposition principle for linear ODEs of the form $y'(t)=A(t,y(t)) + g(t)$ ($yin mathbb{R}^n$, $A$ a linear function in y). If $g(t)=sum _{k=1}^N g_k(t)$ then $y(t)=sum _{k=1}^N y_k(t)$ solves the system, where each $y_k$ solves the ODE with $g(t)=g_k(t)$.
Can this be extended from finite sums to series?
sequences-and-series ordinary-differential-equations sequence-of-function
edited Dec 11 '18 at 16:07
Alex Ravsky
42.7k32383
asked Dec 10 '18 at 15:32
LeonardoLeonardo
3339
$endgroup$
Take the superposition principle for linear ODEs of the form $y'(t)=A(t,y(t)) + g(t)$ ($yin mathbb{R}^n$, $A$ a linear function in y). If $g(t)=sum _{k=1}^N g_k(t)$ then $y(t)=sum _{k=1}^N y_k(t)$ solves the system, where each $y_k$ solves the ODE with $g(t)=g_k(t)$.
Can this be extended from finite sums to series?
sequences-and-series ordinary-differential-equations sequence-of-function
sequences-and-series ordinary-differential-equations sequence-of-function
edited Dec 11 '18 at 16:07
Alex Ravsky
42.7k32383
asked Dec 10 '18 at 15:32
LeonardoLeonardo
3339
edited Dec 11 '18 at 16:07
Alex Ravsky
42.7k32383
asked Dec 10 '18 at 15:32
LeonardoLeonardo
3339
edited Dec 11 '18 at 16:07
Alex Ravsky
42.7k32383
edited Dec 11 '18 at 16:07
Alex Ravsky
42.7k32383
edited Dec 11 '18 at 16:07
Alex Ravsky
42.7k32383
42.7k32383
asked Dec 10 '18 at 15:32
LeonardoLeonardo
3339
asked Dec 10 '18 at 15:32
LeonardoLeonardo
3339
asked Dec 10 '18 at 15:32
LeonardoLeonardo
3339
3339
add a comment |
add a comment |
1 Answer
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$begingroup$
I am wondered that no one of ODE guys answered this question, because it is a common method to look for a solution of a differential equation in a form of a series which is a linear combination of single functions (for instance, $x^n$ or $sin nx$). So I recalled a theorem from the big book [Fich, Ch. 12, $S 1$, 435. Theorem 8]. It concerns one-dimensional case, but I expect that for high-dimensional case the situation is similar.
Theorem. Let functions $y_k(t)$ have bounded derivatives in a bounded segment $I$. If a series $y(t)=sum_{k=1}^infty y_k(t)$ converges at some point $t_0in I$ and a series $y^*(t)=sum_{k=1}^infty y’_k(t)$ converges on $I$ uniformly then the series $y(t)$ converges on $I$ uniformly and $y’(t)= y^*(t)$ for each $tin I$.
It easily implies a positive answer to your question for a fixed $tin I$ when $n=1$, the functions $y_k$ satisfy the conditions of the theorem, the function $A(t,y)$ is continuous, and the series $g(t)=sum _{k=1}^N g_k(t)$ converges to $g(t)$.
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970 (in Russian).
answered Dec 11 '18 at 17:08
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I am wondered that no one of ODE guys answered this question, because it is a common method to look for a solution of a differential equation in a form of a series which is a linear combination of single functions (for instance, $x^n$ or $sin nx$). So I recalled a theorem from the big book [Fich, Ch. 12, $S 1$, 435. Theorem 8]. It concerns one-dimensional case, but I expect that for high-dimensional case the situation is similar.
Theorem. Let functions $y_k(t)$ have bounded derivatives in a bounded segment $I$. If a series $y(t)=sum_{k=1}^infty y_k(t)$ converges at some point $t_0in I$ and a series $y^*(t)=sum_{k=1}^infty y’_k(t)$ converges on $I$ uniformly then the series $y(t)$ converges on $I$ uniformly and $y’(t)= y^*(t)$ for each $tin I$.
It easily implies a positive answer to your question for a fixed $tin I$ when $n=1$, the functions $y_k$ satisfy the conditions of the theorem, the function $A(t,y)$ is continuous, and the series $g(t)=sum _{k=1}^N g_k(t)$ converges to $g(t)$.
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970 (in Russian).
answered Dec 11 '18 at 17:08
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
I am wondered that no one of ODE guys answered this question, because it is a common method to look for a solution of a differential equation in a form of a series which is a linear combination of single functions (for instance, $x^n$ or $sin nx$). So I recalled a theorem from the big book [Fich, Ch. 12, $S 1$, 435. Theorem 8]. It concerns one-dimensional case, but I expect that for high-dimensional case the situation is similar.
Theorem. Let functions $y_k(t)$ have bounded derivatives in a bounded segment $I$. If a series $y(t)=sum_{k=1}^infty y_k(t)$ converges at some point $t_0in I$ and a series $y^*(t)=sum_{k=1}^infty y’_k(t)$ converges on $I$ uniformly then the series $y(t)$ converges on $I$ uniformly and $y’(t)= y^*(t)$ for each $tin I$.
It easily implies a positive answer to your question for a fixed $tin I$ when $n=1$, the functions $y_k$ satisfy the conditions of the theorem, the function $A(t,y)$ is continuous, and the series $g(t)=sum _{k=1}^N g_k(t)$ converges to $g(t)$.
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970 (in Russian).
answered Dec 11 '18 at 17:08
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
I am wondered that no one of ODE guys answered this question, because it is a common method to look for a solution of a differential equation in a form of a series which is a linear combination of single functions (for instance, $x^n$ or $sin nx$). So I recalled a theorem from the big book [Fich, Ch. 12, $S 1$, 435. Theorem 8]. It concerns one-dimensional case, but I expect that for high-dimensional case the situation is similar.
Theorem. Let functions $y_k(t)$ have bounded derivatives in a bounded segment $I$. If a series $y(t)=sum_{k=1}^infty y_k(t)$ converges at some point $t_0in I$ and a series $y^*(t)=sum_{k=1}^infty y’_k(t)$ converges on $I$ uniformly then the series $y(t)$ converges on $I$ uniformly and $y’(t)= y^*(t)$ for each $tin I$.
It easily implies a positive answer to your question for a fixed $tin I$ when $n=1$, the functions $y_k$ satisfy the conditions of the theorem, the function $A(t,y)$ is continuous, and the series $g(t)=sum _{k=1}^N g_k(t)$ converges to $g(t)$.
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970 (in Russian).
answered Dec 11 '18 at 17:08
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
I am wondered that no one of ODE guys answered this question, because it is a common method to look for a solution of a differential equation in a form of a series which is a linear combination of single functions (for instance, $x^n$ or $sin nx$). So I recalled a theorem from the big book [Fich, Ch. 12, $S 1$, 435. Theorem 8]. It concerns one-dimensional case, but I expect that for high-dimensional case the situation is similar.
Theorem. Let functions $y_k(t)$ have bounded derivatives in a bounded segment $I$. If a series $y(t)=sum_{k=1}^infty y_k(t)$ converges at some point $t_0in I$ and a series $y^*(t)=sum_{k=1}^infty y’_k(t)$ converges on $I$ uniformly then the series $y(t)$ converges on $I$ uniformly and $y’(t)= y^*(t)$ for each $tin I$.
It easily implies a positive answer to your question for a fixed $tin I$ when $n=1$, the functions $y_k$ satisfy the conditions of the theorem, the function $A(t,y)$ is continuous, and the series $g(t)=sum _{k=1}^N g_k(t)$ converges to $g(t)$.
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970 (in Russian).
answered Dec 11 '18 at 17:08
Alex RavskyAlex Ravsky
42.7k32383
edited Dec 11 '18 at 23:11
answered Dec 11 '18 at 17:08
Alex RavskyAlex Ravsky
42.7k32383
answered Dec 11 '18 at 17:08
Alex RavskyAlex Ravsky
42.7k32383
answered Dec 11 '18 at 17:08
Alex RavskyAlex Ravsky
42.7k32383
42.7k32383
add a comment |
add a comment |
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