Conclude $u$ cannot obtain its maximum if it fulfills a linear differential inequality $Lu geq 0$
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I have a function $u in C^2((a,b)) cap C^0([a,b])$ and a bounded function $g : (a,b) rightarrow mathbb{R}$, whereas $(a,b) subset mathbb{R}$
Now consider the linear operator $L := frac{d^2}{dx^2} + gfrac{d}{dx}$ . I would like to show that if $Lu geq 0$ then either $u$ is constant or it cannot obtain its maximum in $(a,b)$.
I can show that if $Lu >0$ then $u$ cannot obtain its maximum,
since if it did obtain a maximum in $(a,b)$ at, let's say $x_0$, then $Lu(x_0) = u''(x_0) + g(x_0)u'(x_0) leq 0$ (using the fact that the second derivative at a maximum must be negative and that the first derivative must be zero)
I am not quite sure how to prove the case $Lu geq 0$ and I am also quite uncertain where the boundedness of $g$ would come into play. Any hints are greatly apreciated.
real-analysis differential-equations
asked Nov 14 at 19:24
AxiomaticApproach
661512
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up vote
2
down vote
favorite
I have a function $u in C^2((a,b)) cap C^0([a,b])$ and a bounded function $g : (a,b) rightarrow mathbb{R}$, whereas $(a,b) subset mathbb{R}$
Now consider the linear operator $L := frac{d^2}{dx^2} + gfrac{d}{dx}$ . I would like to show that if $Lu geq 0$ then either $u$ is constant or it cannot obtain its maximum in $(a,b)$.
I can show that if $Lu >0$ then $u$ cannot obtain its maximum,
since if it did obtain a maximum in $(a,b)$ at, let's say $x_0$, then $Lu(x_0) = u''(x_0) + g(x_0)u'(x_0) leq 0$ (using the fact that the second derivative at a maximum must be negative and that the first derivative must be zero)
I am not quite sure how to prove the case $Lu geq 0$ and I am also quite uncertain where the boundedness of $g$ would come into play. Any hints are greatly apreciated.
real-analysis differential-equations
asked Nov 14 at 19:24
AxiomaticApproach
661512
1
This is called the (one-dimensional) strong maximum principle, a not so trivial result. For a proof, see, e.g., Theorem 1 on pp. 2-3 of Protter and Weinberger Maximum Principles in Differential Equations.
– user539887
Nov 14 at 22:28
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a function $u in C^2((a,b)) cap C^0([a,b])$ and a bounded function $g : (a,b) rightarrow mathbb{R}$, whereas $(a,b) subset mathbb{R}$
Now consider the linear operator $L := frac{d^2}{dx^2} + gfrac{d}{dx}$ . I would like to show that if $Lu geq 0$ then either $u$ is constant or it cannot obtain its maximum in $(a,b)$.
I can show that if $Lu >0$ then $u$ cannot obtain its maximum,
since if it did obtain a maximum in $(a,b)$ at, let's say $x_0$, then $Lu(x_0) = u''(x_0) + g(x_0)u'(x_0) leq 0$ (using the fact that the second derivative at a maximum must be negative and that the first derivative must be zero)
I am not quite sure how to prove the case $Lu geq 0$ and I am also quite uncertain where the boundedness of $g$ would come into play. Any hints are greatly apreciated.
real-analysis differential-equations
asked Nov 14 at 19:24
AxiomaticApproach
661512
I have a function $u in C^2((a,b)) cap C^0([a,b])$ and a bounded function $g : (a,b) rightarrow mathbb{R}$, whereas $(a,b) subset mathbb{R}$
Now consider the linear operator $L := frac{d^2}{dx^2} + gfrac{d}{dx}$ . I would like to show that if $Lu geq 0$ then either $u$ is constant or it cannot obtain its maximum in $(a,b)$.
I can show that if $Lu >0$ then $u$ cannot obtain its maximum,
since if it did obtain a maximum in $(a,b)$ at, let's say $x_0$, then $Lu(x_0) = u''(x_0) + g(x_0)u'(x_0) leq 0$ (using the fact that the second derivative at a maximum must be negative and that the first derivative must be zero)
I am not quite sure how to prove the case $Lu geq 0$ and I am also quite uncertain where the boundedness of $g$ would come into play. Any hints are greatly apreciated.
real-analysis differential-equations
real-analysis differential-equations
asked Nov 14 at 19:24
AxiomaticApproach
661512
asked Nov 14 at 19:24
AxiomaticApproach
661512
asked Nov 14 at 19:24
AxiomaticApproach
661512
asked Nov 14 at 19:24
AxiomaticApproach
661512
asked Nov 14 at 19:24
AxiomaticApproach
661512
661512
1
This is called the (one-dimensional) strong maximum principle, a not so trivial result. For a proof, see, e.g., Theorem 1 on pp. 2-3 of Protter and Weinberger Maximum Principles in Differential Equations.
– user539887
Nov 14 at 22:28
add a comment |
1
This is called the (one-dimensional) strong maximum principle, a not so trivial result. For a proof, see, e.g., Theorem 1 on pp. 2-3 of Protter and Weinberger Maximum Principles in Differential Equations.
– user539887
Nov 14 at 22:28
1
1
This is called the (one-dimensional) strong maximum principle, a not so trivial result. For a proof, see, e.g., Theorem 1 on pp. 2-3 of Protter and Weinberger Maximum Principles in Differential Equations.
– user539887
Nov 14 at 22:28
This is called the (one-dimensional) strong maximum principle, a not so trivial result. For a proof, see, e.g., Theorem 1 on pp. 2-3 of Protter and Weinberger Maximum Principles in Differential Equations.
– user539887
Nov 14 at 22:28
add a comment |
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This is called the (one-dimensional) strong maximum principle, a not so trivial result. For a proof, see, e.g., Theorem 1 on pp. 2-3 of Protter and Weinberger Maximum Principles in Differential Equations.
– user539887
Nov 14 at 22:28