Finite element method for nonlinear differential equation
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I encounter this problem $$frac{df(u(x))}{dx} = g(x)$$ with $$u(0) = u(1) = 0$$ I first convert it to weak form $f(u(x))v(x)]^1_0 - int^{1}_0 frac{dv(x)}{dx}f(u(x))dx =- int^{1}_0 frac{dv(x)}{dx}f(u(x))dx = int^{1}_0v(x)g(x)dx $ by multiplying a test function $v(x)$. Then how should I write it in matrix form? The right hand side does not depend on $u$ explicit.
Thanks in advance.
ordinary-differential-equations numerical-methods finite-element-method galerkin-methods
asked Jan 6 at 10:59
metricspacemetricspace
408210
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add a comment |
$begingroup$
I encounter this problem $$frac{df(u(x))}{dx} = g(x)$$ with $$u(0) = u(1) = 0$$ I first convert it to weak form $f(u(x))v(x)]^1_0 - int^{1}_0 frac{dv(x)}{dx}f(u(x))dx =- int^{1}_0 frac{dv(x)}{dx}f(u(x))dx = int^{1}_0v(x)g(x)dx $ by multiplying a test function $v(x)$. Then how should I write it in matrix form? The right hand side does not depend on $u$ explicit.
Thanks in advance.
ordinary-differential-equations numerical-methods finite-element-method galerkin-methods
asked Jan 6 at 10:59
metricspacemetricspace
408210
$endgroup$
$begingroup$
Set $G(x)=f(0)+int_0^xg(s),ds$, then you just have to solve the inverse function problem $f(u(x))=G(x)$ using Newton and some version of path-following. There is no reason why a first order ODE should satisfy two boundary conditions.
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– LutzL
Jan 6 at 11:15
$begingroup$
@LutzL variable is x! Sorry about that.
$endgroup$
– metricspace
Jan 6 at 19:43
add a comment |
$begingroup$
I encounter this problem $$frac{df(u(x))}{dx} = g(x)$$ with $$u(0) = u(1) = 0$$ I first convert it to weak form $f(u(x))v(x)]^1_0 - int^{1}_0 frac{dv(x)}{dx}f(u(x))dx =- int^{1}_0 frac{dv(x)}{dx}f(u(x))dx = int^{1}_0v(x)g(x)dx $ by multiplying a test function $v(x)$. Then how should I write it in matrix form? The right hand side does not depend on $u$ explicit.
Thanks in advance.
ordinary-differential-equations numerical-methods finite-element-method galerkin-methods
asked Jan 6 at 10:59
metricspacemetricspace
408210
$endgroup$
I encounter this problem $$frac{df(u(x))}{dx} = g(x)$$ with $$u(0) = u(1) = 0$$ I first convert it to weak form $f(u(x))v(x)]^1_0 - int^{1}_0 frac{dv(x)}{dx}f(u(x))dx =- int^{1}_0 frac{dv(x)}{dx}f(u(x))dx = int^{1}_0v(x)g(x)dx $ by multiplying a test function $v(x)$. Then how should I write it in matrix form? The right hand side does not depend on $u$ explicit.
Thanks in advance.
ordinary-differential-equations numerical-methods finite-element-method galerkin-methods
ordinary-differential-equations numerical-methods finite-element-method galerkin-methods
asked Jan 6 at 10:59
metricspacemetricspace
408210
asked Jan 6 at 10:59
metricspacemetricspace
408210
edited Jan 6 at 19:42
metricspace
asked Jan 6 at 10:59
metricspacemetricspace
408210
asked Jan 6 at 10:59
metricspacemetricspace
408210
asked Jan 6 at 10:59
metricspacemetricspace
408210
408210
$begingroup$
Set $G(x)=f(0)+int_0^xg(s),ds$, then you just have to solve the inverse function problem $f(u(x))=G(x)$ using Newton and some version of path-following. There is no reason why a first order ODE should satisfy two boundary conditions.
$endgroup$
– LutzL
Jan 6 at 11:15
$begingroup$
@LutzL variable is x! Sorry about that.
$endgroup$
– metricspace
Jan 6 at 19:43
add a comment |
$begingroup$
Set $G(x)=f(0)+int_0^xg(s),ds$, then you just have to solve the inverse function problem $f(u(x))=G(x)$ using Newton and some version of path-following. There is no reason why a first order ODE should satisfy two boundary conditions.
$endgroup$
– LutzL
Jan 6 at 11:15
$begingroup$
@LutzL variable is x! Sorry about that.
$endgroup$
– metricspace
Jan 6 at 19:43
$begingroup$
Set $G(x)=f(0)+int_0^xg(s),ds$, then you just have to solve the inverse function problem $f(u(x))=G(x)$ using Newton and some version of path-following. There is no reason why a first order ODE should satisfy two boundary conditions.
$endgroup$
– LutzL
Jan 6 at 11:15
$begingroup$
Set $G(x)=f(0)+int_0^xg(s),ds$, then you just have to solve the inverse function problem $f(u(x))=G(x)$ using Newton and some version of path-following. There is no reason why a first order ODE should satisfy two boundary conditions.
$endgroup$
– LutzL
Jan 6 at 11:15
$begingroup$
@LutzL variable is x! Sorry about that.
$endgroup$
– metricspace
Jan 6 at 19:43
$begingroup$
@LutzL variable is x! Sorry about that.
$endgroup$
– metricspace
Jan 6 at 19:43
add a comment |
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$begingroup$
Set $G(x)=f(0)+int_0^xg(s),ds$, then you just have to solve the inverse function problem $f(u(x))=G(x)$ using Newton and some version of path-following. There is no reason why a first order ODE should satisfy two boundary conditions.
$endgroup$
– LutzL
Jan 6 at 11:15
$begingroup$
@LutzL variable is x! Sorry about that.
$endgroup$
– metricspace
Jan 6 at 19:43