boundary problem: use main theorem of monotone operators
I am trying to investigate for which $alpha in mathbb R$ the boundary problem
$$-u''(x)+alpha sin(u(x))u'(x)=f(x)$$
$$u(a) = u(b) = 0$$
is weakly solvable using the main theorem of monotone operators which states that if
$A:Vrightarrow V^*$ be monotone and radially continuous
$B:Vrightarrow V^*$ be strongly continuous, such that $A+B :Vrightarrow V^*$ is coercive, then for any $fin V^*$ there is a solution $uin V$ to $(A+B)u=f in V^*$
So far I don't have problems showing A is monoton/rad. cont. and for B with $<Bu,v> = alpha sin(u(x))u'(x)v(x)$ with $v,u in H_0^1(a,b)$. I've showed that B is strongly cont. and coercive but I didn't had to consider specific values of $alpha$, did I made a mistake somewhere?
My attempt for B is $textbf{strongly continuous}$: Let $u_nrightharpoonup u$ in $H_0^1(a,b)$ and since $H_0^1(a,b)$ is a compact embedding in $C([a,b])$ one can show that $u_n rightarrow u$ in $C([a,b])$ by using the subsequence principle.
$<Bu_n-Bu,v> = alpha (int_a^b sin(u_n(x))u_n'(x)v(x)-sin(u(x))u'(x)v(x) stackrel{text{partial integration}}{=} alpha (int_a^bcos(u_n(x))v'(x)-cos(u(x))v'(x)dx) leq alpha (||cos(u_n(x)) - cos(u(x))||_{C([a.b])}sqrt{b-a}|v|_{1,2} rightarrow0$
and for the $textbf{coercivity}$: $<Bu,u>=int_a^balpha sin(u(x))u'(x)u(x)dx = alpha[-cos(u(x))u(x)]_a^b +int_a^bcos(u(x))u'(x)dx = 0 $
functional-analysis differential-equations boundary-value-problem nonlinear-analysis
asked Nov 29 at 20:05
newbie
15
add a comment |
I am trying to investigate for which $alpha in mathbb R$ the boundary problem
$$-u''(x)+alpha sin(u(x))u'(x)=f(x)$$
$$u(a) = u(b) = 0$$
is weakly solvable using the main theorem of monotone operators which states that if
$A:Vrightarrow V^*$ be monotone and radially continuous
$B:Vrightarrow V^*$ be strongly continuous, such that $A+B :Vrightarrow V^*$ is coercive, then for any $fin V^*$ there is a solution $uin V$ to $(A+B)u=f in V^*$
So far I don't have problems showing A is monoton/rad. cont. and for B with $<Bu,v> = alpha sin(u(x))u'(x)v(x)$ with $v,u in H_0^1(a,b)$. I've showed that B is strongly cont. and coercive but I didn't had to consider specific values of $alpha$, did I made a mistake somewhere?
My attempt for B is $textbf{strongly continuous}$: Let $u_nrightharpoonup u$ in $H_0^1(a,b)$ and since $H_0^1(a,b)$ is a compact embedding in $C([a,b])$ one can show that $u_n rightarrow u$ in $C([a,b])$ by using the subsequence principle.
$<Bu_n-Bu,v> = alpha (int_a^b sin(u_n(x))u_n'(x)v(x)-sin(u(x))u'(x)v(x) stackrel{text{partial integration}}{=} alpha (int_a^bcos(u_n(x))v'(x)-cos(u(x))v'(x)dx) leq alpha (||cos(u_n(x)) - cos(u(x))||_{C([a.b])}sqrt{b-a}|v|_{1,2} rightarrow0$
and for the $textbf{coercivity}$: $<Bu,u>=int_a^balpha sin(u(x))u'(x)u(x)dx = alpha[-cos(u(x))u(x)]_a^b +int_a^bcos(u(x))u'(x)dx = 0 $
functional-analysis differential-equations boundary-value-problem nonlinear-analysis
asked Nov 29 at 20:05
newbie
15
add a comment |
I am trying to investigate for which $alpha in mathbb R$ the boundary problem
$$-u''(x)+alpha sin(u(x))u'(x)=f(x)$$
$$u(a) = u(b) = 0$$
is weakly solvable using the main theorem of monotone operators which states that if
$A:Vrightarrow V^*$ be monotone and radially continuous
$B:Vrightarrow V^*$ be strongly continuous, such that $A+B :Vrightarrow V^*$ is coercive, then for any $fin V^*$ there is a solution $uin V$ to $(A+B)u=f in V^*$
So far I don't have problems showing A is monoton/rad. cont. and for B with $<Bu,v> = alpha sin(u(x))u'(x)v(x)$ with $v,u in H_0^1(a,b)$. I've showed that B is strongly cont. and coercive but I didn't had to consider specific values of $alpha$, did I made a mistake somewhere?
My attempt for B is $textbf{strongly continuous}$: Let $u_nrightharpoonup u$ in $H_0^1(a,b)$ and since $H_0^1(a,b)$ is a compact embedding in $C([a,b])$ one can show that $u_n rightarrow u$ in $C([a,b])$ by using the subsequence principle.
$<Bu_n-Bu,v> = alpha (int_a^b sin(u_n(x))u_n'(x)v(x)-sin(u(x))u'(x)v(x) stackrel{text{partial integration}}{=} alpha (int_a^bcos(u_n(x))v'(x)-cos(u(x))v'(x)dx) leq alpha (||cos(u_n(x)) - cos(u(x))||_{C([a.b])}sqrt{b-a}|v|_{1,2} rightarrow0$
and for the $textbf{coercivity}$: $<Bu,u>=int_a^balpha sin(u(x))u'(x)u(x)dx = alpha[-cos(u(x))u(x)]_a^b +int_a^bcos(u(x))u'(x)dx = 0 $
functional-analysis differential-equations boundary-value-problem nonlinear-analysis
asked Nov 29 at 20:05
newbie
15
I am trying to investigate for which $alpha in mathbb R$ the boundary problem
$$-u''(x)+alpha sin(u(x))u'(x)=f(x)$$
$$u(a) = u(b) = 0$$
is weakly solvable using the main theorem of monotone operators which states that if
$A:Vrightarrow V^*$ be monotone and radially continuous
$B:Vrightarrow V^*$ be strongly continuous, such that $A+B :Vrightarrow V^*$ is coercive, then for any $fin V^*$ there is a solution $uin V$ to $(A+B)u=f in V^*$
So far I don't have problems showing A is monoton/rad. cont. and for B with $<Bu,v> = alpha sin(u(x))u'(x)v(x)$ with $v,u in H_0^1(a,b)$. I've showed that B is strongly cont. and coercive but I didn't had to consider specific values of $alpha$, did I made a mistake somewhere?
My attempt for B is $textbf{strongly continuous}$: Let $u_nrightharpoonup u$ in $H_0^1(a,b)$ and since $H_0^1(a,b)$ is a compact embedding in $C([a,b])$ one can show that $u_n rightarrow u$ in $C([a,b])$ by using the subsequence principle.
$<Bu_n-Bu,v> = alpha (int_a^b sin(u_n(x))u_n'(x)v(x)-sin(u(x))u'(x)v(x) stackrel{text{partial integration}}{=} alpha (int_a^bcos(u_n(x))v'(x)-cos(u(x))v'(x)dx) leq alpha (||cos(u_n(x)) - cos(u(x))||_{C([a.b])}sqrt{b-a}|v|_{1,2} rightarrow0$
and for the $textbf{coercivity}$: $<Bu,u>=int_a^balpha sin(u(x))u'(x)u(x)dx = alpha[-cos(u(x))u(x)]_a^b +int_a^bcos(u(x))u'(x)dx = 0 $
functional-analysis differential-equations boundary-value-problem nonlinear-analysis
functional-analysis differential-equations boundary-value-problem nonlinear-analysis
asked Nov 29 at 20:05
newbie
15
asked Nov 29 at 20:05
newbie
15
asked Nov 29 at 20:05
newbie
15
asked Nov 29 at 20:05
newbie
15
asked Nov 29 at 20:05
newbie
15
15
add a comment |
add a comment |
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