How to prove the existence and uniqueness of the solution of a second order linear ODE?
$begingroup$
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
edited Dec 24 '18 at 13:19
amWhy
1
asked Dec 24 '18 at 5:18
X.T ChenX.T Chen
517
$endgroup$
add a comment |
$begingroup$
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
edited Dec 24 '18 at 13:19
amWhy
1
asked Dec 24 '18 at 5:18
X.T ChenX.T Chen
517
$endgroup$
add a comment |
$begingroup$
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
edited Dec 24 '18 at 13:19
amWhy
1
asked Dec 24 '18 at 5:18
X.T ChenX.T Chen
517
$endgroup$
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
edited Dec 24 '18 at 13:19
amWhy
1
asked Dec 24 '18 at 5:18
X.T ChenX.T Chen
517
edited Dec 24 '18 at 13:19
amWhy
1
asked Dec 24 '18 at 5:18
X.T ChenX.T Chen
517
edited Dec 24 '18 at 13:19
amWhy
1
edited Dec 24 '18 at 13:19
amWhy
1
edited Dec 24 '18 at 13:19
amWhy
1
1
asked Dec 24 '18 at 5:18
X.T ChenX.T Chen
517
asked Dec 24 '18 at 5:18
X.T ChenX.T Chen
517
asked Dec 24 '18 at 5:18
X.T ChenX.T Chen
517
517
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $[a, b]$.
answered Dec 24 '18 at 6:00
Robert LewisRobert Lewis
44.9k22964
$endgroup$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 $[a, b]$.
answered Dec 24 '18 at 6:00
Robert LewisRobert Lewis
44.9k22964
$endgroup$
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
add a comment |
$begingroup$
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 $[a, b]$.
answered Dec 24 '18 at 6:00
Robert LewisRobert Lewis
44.9k22964
$endgroup$
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
add a comment |
$begingroup$
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 $[a, b]$.
answered Dec 24 '18 at 6:00
Robert LewisRobert Lewis
44.9k22964
$endgroup$
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 $[a, b]$.
answered Dec 24 '18 at 6:00
Robert LewisRobert Lewis
44.9k22964
edited Dec 24 '18 at 6:15
answered Dec 24 '18 at 6:00
Robert LewisRobert Lewis
44.9k22964
answered Dec 24 '18 at 6:00
Robert LewisRobert Lewis
44.9k22964
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
add a comment |
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
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– Robert Lewis
Dec 24 '18 at 6:20
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@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
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– Robert Lewis
Dec 24 '18 at 6:20
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@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 $z(a) = z'(a) = 0$.
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– Robert Lewis
Dec 24 '18 at 6:23
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@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 $z(a) = z'(a) = 0$.
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– Robert Lewis
Dec 24 '18 at 6:23
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picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
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– X.T Chen
Dec 24 '18 at 6:39
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picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for?
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– X.T Chen
Dec 24 '18 at 6:39
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@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.
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– Robert Lewis
Dec 24 '18 at 6:44
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@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.
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– Robert Lewis
Dec 24 '18 at 6:44
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