How can I proof that this aplication is coercive?
0
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I have the next problem. I have to proof that the aplication $a$ is coercive, defined as $$a: H times H longrightarrow R$$
$$ a(x,y) = 2 int_0^1 frac{x'y'}{1+t}dt + x(1) $$
Where $H = {x in H^1(0,1),x(0) = 0} $. I also have to proof that is continous, and I did it, but I not sure that I made it in the right way. Pls HELP!!!
functional-analysis
asked Dec 7 '18 at 22:16
Fernando cañizaresFernando cañizares
11
$endgroup$
add a comment |
0
$begingroup$
I have the next problem. I have to proof that the aplication $a$ is coercive, defined as $$a: H times H longrightarrow R$$
$$ a(x,y) = 2 int_0^1 frac{x'y'}{1+t}dt + x(1) $$
Where $H = {x in H^1(0,1),x(0) = 0} $. I also have to proof that is continous, and I did it, but I not sure that I made it in the right way. Pls HELP!!!
functional-analysis
asked Dec 7 '18 at 22:16
Fernando cañizaresFernando cañizares
11
$endgroup$
$begingroup$
Pelas, can you show what have you tried? Maybe it would be useful to give an answer
$endgroup$
– Tito Eliatron
Dec 7 '18 at 22:49
$begingroup$
I'm sorry, I made a mistake when i was capying the problem. I'll send the whole problem. I have to proof that exists a unique solution for the next problem: $min {int_0^1( frac{|x'|^2}{1+t} -2x)dt + x(1), x in H^1(0,1),x(0) = 0}$. My mistake was that the bilinear form is $a(x,y) = 2int_0^1( frac{x'y'}{1+t})$. $x(1)$ is a part of the linear form.
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:28
$begingroup$
For find the solution i have to use the Lax-Milgram Theorem, and i need to check the condición. That the space is a Hilbert space, but that is obvious, $a(x,y)$ is bilinear, continuous, coercive, and simetric, and that $f(y) = int_0^1 2y(t)dt - y(1)$ is continuous
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:36
add a comment |
0
0
0
$begingroup$
I have the next problem. I have to proof that the aplication $a$ is coercive, defined as $$a: H times H longrightarrow R$$
$$ a(x,y) = 2 int_0^1 frac{x'y'}{1+t}dt + x(1) $$
Where $H = {x in H^1(0,1),x(0) = 0} $. I also have to proof that is continous, and I did it, but I not sure that I made it in the right way. Pls HELP!!!
functional-analysis
asked Dec 7 '18 at 22:16
Fernando cañizaresFernando cañizares
11
$endgroup$
I have the next problem. I have to proof that the aplication $a$ is coercive, defined as $$a: H times H longrightarrow R$$
$$ a(x,y) = 2 int_0^1 frac{x'y'}{1+t}dt + x(1) $$
Where $H = {x in H^1(0,1),x(0) = 0} $. I also have to proof that is continous, and I did it, but I not sure that I made it in the right way. Pls HELP!!!
functional-analysis
functional-analysis
asked Dec 7 '18 at 22:16
Fernando cañizaresFernando cañizares
11
asked Dec 7 '18 at 22:16
Fernando cañizaresFernando cañizares
11
asked Dec 7 '18 at 22:16
Fernando cañizaresFernando cañizares
11
asked Dec 7 '18 at 22:16
Fernando cañizaresFernando cañizares
11
asked Dec 7 '18 at 22:16
Fernando cañizaresFernando cañizares
11
11
$begingroup$
Pelas, can you show what have you tried? Maybe it would be useful to give an answer
$endgroup$
– Tito Eliatron
Dec 7 '18 at 22:49
$begingroup$
I'm sorry, I made a mistake when i was capying the problem. I'll send the whole problem. I have to proof that exists a unique solution for the next problem: $min {int_0^1( frac{|x'|^2}{1+t} -2x)dt + x(1), x in H^1(0,1),x(0) = 0}$. My mistake was that the bilinear form is $a(x,y) = 2int_0^1( frac{x'y'}{1+t})$. $x(1)$ is a part of the linear form.
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:28
$begingroup$
For find the solution i have to use the Lax-Milgram Theorem, and i need to check the condición. That the space is a Hilbert space, but that is obvious, $a(x,y)$ is bilinear, continuous, coercive, and simetric, and that $f(y) = int_0^1 2y(t)dt - y(1)$ is continuous
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:36
add a comment |
$begingroup$
Pelas, can you show what have you tried? Maybe it would be useful to give an answer
$endgroup$
– Tito Eliatron
Dec 7 '18 at 22:49
$begingroup$
I'm sorry, I made a mistake when i was capying the problem. I'll send the whole problem. I have to proof that exists a unique solution for the next problem: $min {int_0^1( frac{|x'|^2}{1+t} -2x)dt + x(1), x in H^1(0,1),x(0) = 0}$. My mistake was that the bilinear form is $a(x,y) = 2int_0^1( frac{x'y'}{1+t})$. $x(1)$ is a part of the linear form.
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:28
$begingroup$
For find the solution i have to use the Lax-Milgram Theorem, and i need to check the condición. That the space is a Hilbert space, but that is obvious, $a(x,y)$ is bilinear, continuous, coercive, and simetric, and that $f(y) = int_0^1 2y(t)dt - y(1)$ is continuous
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:36
$begingroup$
Pelas, can you show what have you tried? Maybe it would be useful to give an answer
$endgroup$
– Tito Eliatron
Dec 7 '18 at 22:49
$begingroup$
Pelas, can you show what have you tried? Maybe it would be useful to give an answer
$endgroup$
– Tito Eliatron
Dec 7 '18 at 22:49
$begingroup$
I'm sorry, I made a mistake when i was capying the problem. I'll send the whole problem. I have to proof that exists a unique solution for the next problem: $min {int_0^1( frac{|x'|^2}{1+t} -2x)dt + x(1), x in H^1(0,1),x(0) = 0}$. My mistake was that the bilinear form is $a(x,y) = 2int_0^1( frac{x'y'}{1+t})$. $x(1)$ is a part of the linear form.
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:28
$begingroup$
I'm sorry, I made a mistake when i was capying the problem. I'll send the whole problem. I have to proof that exists a unique solution for the next problem: $min {int_0^1( frac{|x'|^2}{1+t} -2x)dt + x(1), x in H^1(0,1),x(0) = 0}$. My mistake was that the bilinear form is $a(x,y) = 2int_0^1( frac{x'y'}{1+t})$. $x(1)$ is a part of the linear form.
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:28
$begingroup$
For find the solution i have to use the Lax-Milgram Theorem, and i need to check the condición. That the space is a Hilbert space, but that is obvious, $a(x,y)$ is bilinear, continuous, coercive, and simetric, and that $f(y) = int_0^1 2y(t)dt - y(1)$ is continuous
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:36
$begingroup$
For find the solution i have to use the Lax-Milgram Theorem, and i need to check the condición. That the space is a Hilbert space, but that is obvious, $a(x,y)$ is bilinear, continuous, coercive, and simetric, and that $f(y) = int_0^1 2y(t)dt - y(1)$ is continuous
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:36
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$begingroup$
Pelas, can you show what have you tried? Maybe it would be useful to give an answer
$endgroup$
– Tito Eliatron
Dec 7 '18 at 22:49
$begingroup$
I'm sorry, I made a mistake when i was capying the problem. I'll send the whole problem. I have to proof that exists a unique solution for the next problem: $min {int_0^1( frac{|x'|^2}{1+t} -2x)dt + x(1), x in H^1(0,1),x(0) = 0}$. My mistake was that the bilinear form is $a(x,y) = 2int_0^1( frac{x'y'}{1+t})$. $x(1)$ is a part of the linear form.
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:28
$begingroup$
For find the solution i have to use the Lax-Milgram Theorem, and i need to check the condición. That the space is a Hilbert space, but that is obvious, $a(x,y)$ is bilinear, continuous, coercive, and simetric, and that $f(y) = int_0^1 2y(t)dt - y(1)$ is continuous
$endgroup$
– Fernando cañizares
Dec 8 '18 at 11:36