Using Newton Raphson's method to solve a non-linear boundary value problem?
$begingroup$
(My specific question is at the end of the problem)
Examine the boundary problem with a nonlinear right hand side $1+u^2(x)$
$$-u''(x) = 1+u^2(x) quad text{on} quad 0 < x < 1 quad text{with} quad u(0) = u(1) = 0$$
use 4 internal points $x_1$, $x_2$, $x_3$, $x_4$. With $u_i = u(x_i)$ we use the finite difference approximation
$$u''(x_i) approx frac{-u(x_{i-1}) + 2u(x_i) - u(x_{i+1})}{h^2} = frac{-u_{i-1} + 2u_i - u_{i+1}}{h^2}$$
where $h = frac{1}{5} = 0.2$ and $u(x_0) = u_0 = u(x_5) = u_5 = 0$.
Thus we obtain a non-linear system of equations.
$$
frac{1}{0.4^2}
begin{bmatrix}
+2 & -1 & 0 & 0 \
-1 & +2 & -1 & 0 \
0 & -1 & +2 & -1 \
0 & 0 & -1 & +2 \
end{bmatrix}
begin{pmatrix}
u_1 \
u_2 \
u_3 \
u_4
end{pmatrix} =
begin{pmatrix}
1 + u_1^2 \
1 + u_2^2 \
1 + u_3^2 \
1 + u_4^2
end{pmatrix}
$$
To solve this non-linear system use methods from non linear problems, e.g. Newton's method: Use the Taylor approximation $(u+phi)^2approx u^2+2uphi$. Start with an initial vector $vec{u}_0$, e.g.$vec{u}_0=0$. Then use the iterationnewline
begin{align}
textbf{A}cdot (vec{u}_k + vec{phi})&=1+vec{u}_k^2+2vec{u}_kcdotvec{phi}\
textbf{A}cdot (vec{u}_k + vec{phi}) - 2vec{u}_kcdotvec{phi}&= 1+vec{u}_k^2\
textbf{A} cdot (vec{u}_k + vec{phi}) - text{diag}(2vec{u}_k)cdot phi &= 1 + vec{u}_k^2\
textbf{A}cdotphi + textbf{A}cdotvec{u}_k- text{diag}(2vec{u}_k)cdotphi &= 1 + vec{u}_k^2\
textbf{A}cdotphi - text{diag}(2vec{u}_k)cdotphi &= 1 + vec{u}_k^2-textbf{A}cdotvec{u}_k\
(textbf{A} - text{diag}(2vec{u}_k))cdot phi &= 1 + vec{u}_k^2-textbf{A}cdotvec{u}_ktag{1}\
vec{u}_{k+1} &=vec{u}_k+vec{phi}tag{2}
end{align}
My question is: how do we get from (1) to (2)?
linear-algebra nonlinear-system newton-raphson
asked Jan 3 at 19:01
ecjbecjb
2858
$endgroup$
add a comment |
$begingroup$
(My specific question is at the end of the problem)
Examine the boundary problem with a nonlinear right hand side $1+u^2(x)$
$$-u''(x) = 1+u^2(x) quad text{on} quad 0 < x < 1 quad text{with} quad u(0) = u(1) = 0$$
use 4 internal points $x_1$, $x_2$, $x_3$, $x_4$. With $u_i = u(x_i)$ we use the finite difference approximation
$$u''(x_i) approx frac{-u(x_{i-1}) + 2u(x_i) - u(x_{i+1})}{h^2} = frac{-u_{i-1} + 2u_i - u_{i+1}}{h^2}$$
where $h = frac{1}{5} = 0.2$ and $u(x_0) = u_0 = u(x_5) = u_5 = 0$.
Thus we obtain a non-linear system of equations.
$$
frac{1}{0.4^2}
begin{bmatrix}
+2 & -1 & 0 & 0 \
-1 & +2 & -1 & 0 \
0 & -1 & +2 & -1 \
0 & 0 & -1 & +2 \
end{bmatrix}
begin{pmatrix}
u_1 \
u_2 \
u_3 \
u_4
end{pmatrix} =
begin{pmatrix}
1 + u_1^2 \
1 + u_2^2 \
1 + u_3^2 \
1 + u_4^2
end{pmatrix}
$$
To solve this non-linear system use methods from non linear problems, e.g. Newton's method: Use the Taylor approximation $(u+phi)^2approx u^2+2uphi$. Start with an initial vector $vec{u}_0$, e.g.$vec{u}_0=0$. Then use the iterationnewline
begin{align}
textbf{A}cdot (vec{u}_k + vec{phi})&=1+vec{u}_k^2+2vec{u}_kcdotvec{phi}\
textbf{A}cdot (vec{u}_k + vec{phi}) - 2vec{u}_kcdotvec{phi}&= 1+vec{u}_k^2\
textbf{A} cdot (vec{u}_k + vec{phi}) - text{diag}(2vec{u}_k)cdot phi &= 1 + vec{u}_k^2\
textbf{A}cdotphi + textbf{A}cdotvec{u}_k- text{diag}(2vec{u}_k)cdotphi &= 1 + vec{u}_k^2\
textbf{A}cdotphi - text{diag}(2vec{u}_k)cdotphi &= 1 + vec{u}_k^2-textbf{A}cdotvec{u}_k\
(textbf{A} - text{diag}(2vec{u}_k))cdot phi &= 1 + vec{u}_k^2-textbf{A}cdotvec{u}_ktag{1}\
vec{u}_{k+1} &=vec{u}_k+vec{phi}tag{2}
end{align}
My question is: how do we get from (1) to (2)?
linear-algebra nonlinear-system newton-raphson
asked Jan 3 at 19:01
ecjbecjb
2858
$endgroup$
$begingroup$
(1) is a system of linear equations. What exactly do you want to know? How to solve systems of linear equations in general? Or which method is the best one for this special type of matrix (sparse band matrix)? Or if there is an explicit solution?
$endgroup$
– Reinhard Meier
Jan 5 at 20:29
$begingroup$
thank you for your answer @ReinhardMeier. Sorry I think I put too much for a rather focused question. I think solving a linear problem is clear for me as well as why you use Newton's method. My only question is how do you get at the end of the first of newton to the second one from (1) to (2) (the 2 last line of the question)
$endgroup$
– ecjb
Jan 5 at 21:20
$begingroup$
Solve (1) for $phi$ and plug $phi$ into (2)
$endgroup$
– Reinhard Meier
Jan 5 at 21:35
$begingroup$
indeed because $phi$ is the unknown and is isolated in the last equation. Many thanks @ReinhardMeier
$endgroup$
– ecjb
Jan 5 at 21:47
add a comment |
$begingroup$
(My specific question is at the end of the problem)
Examine the boundary problem with a nonlinear right hand side $1+u^2(x)$
$$-u''(x) = 1+u^2(x) quad text{on} quad 0 < x < 1 quad text{with} quad u(0) = u(1) = 0$$
use 4 internal points $x_1$, $x_2$, $x_3$, $x_4$. With $u_i = u(x_i)$ we use the finite difference approximation
$$u''(x_i) approx frac{-u(x_{i-1}) + 2u(x_i) - u(x_{i+1})}{h^2} = frac{-u_{i-1} + 2u_i - u_{i+1}}{h^2}$$
where $h = frac{1}{5} = 0.2$ and $u(x_0) = u_0 = u(x_5) = u_5 = 0$.
Thus we obtain a non-linear system of equations.
$$
frac{1}{0.4^2}
begin{bmatrix}
+2 & -1 & 0 & 0 \
-1 & +2 & -1 & 0 \
0 & -1 & +2 & -1 \
0 & 0 & -1 & +2 \
end{bmatrix}
begin{pmatrix}
u_1 \
u_2 \
u_3 \
u_4
end{pmatrix} =
begin{pmatrix}
1 + u_1^2 \
1 + u_2^2 \
1 + u_3^2 \
1 + u_4^2
end{pmatrix}
$$
To solve this non-linear system use methods from non linear problems, e.g. Newton's method: Use the Taylor approximation $(u+phi)^2approx u^2+2uphi$. Start with an initial vector $vec{u}_0$, e.g.$vec{u}_0=0$. Then use the iterationnewline
begin{align}
textbf{A}cdot (vec{u}_k + vec{phi})&=1+vec{u}_k^2+2vec{u}_kcdotvec{phi}\
textbf{A}cdot (vec{u}_k + vec{phi}) - 2vec{u}_kcdotvec{phi}&= 1+vec{u}_k^2\
textbf{A} cdot (vec{u}_k + vec{phi}) - text{diag}(2vec{u}_k)cdot phi &= 1 + vec{u}_k^2\
textbf{A}cdotphi + textbf{A}cdotvec{u}_k- text{diag}(2vec{u}_k)cdotphi &= 1 + vec{u}_k^2\
textbf{A}cdotphi - text{diag}(2vec{u}_k)cdotphi &= 1 + vec{u}_k^2-textbf{A}cdotvec{u}_k\
(textbf{A} - text{diag}(2vec{u}_k))cdot phi &= 1 + vec{u}_k^2-textbf{A}cdotvec{u}_ktag{1}\
vec{u}_{k+1} &=vec{u}_k+vec{phi}tag{2}
end{align}
My question is: how do we get from (1) to (2)?
linear-algebra nonlinear-system newton-raphson
asked Jan 3 at 19:01
ecjbecjb
2858
$endgroup$
(My specific question is at the end of the problem)
Examine the boundary problem with a nonlinear right hand side $1+u^2(x)$
$$-u''(x) = 1+u^2(x) quad text{on} quad 0 < x < 1 quad text{with} quad u(0) = u(1) = 0$$
use 4 internal points $x_1$, $x_2$, $x_3$, $x_4$. With $u_i = u(x_i)$ we use the finite difference approximation
$$u''(x_i) approx frac{-u(x_{i-1}) + 2u(x_i) - u(x_{i+1})}{h^2} = frac{-u_{i-1} + 2u_i - u_{i+1}}{h^2}$$
where $h = frac{1}{5} = 0.2$ and $u(x_0) = u_0 = u(x_5) = u_5 = 0$.
Thus we obtain a non-linear system of equations.
$$
frac{1}{0.4^2}
begin{bmatrix}
+2 & -1 & 0 & 0 \
-1 & +2 & -1 & 0 \
0 & -1 & +2 & -1 \
0 & 0 & -1 & +2 \
end{bmatrix}
begin{pmatrix}
u_1 \
u_2 \
u_3 \
u_4
end{pmatrix} =
begin{pmatrix}
1 + u_1^2 \
1 + u_2^2 \
1 + u_3^2 \
1 + u_4^2
end{pmatrix}
$$
To solve this non-linear system use methods from non linear problems, e.g. Newton's method: Use the Taylor approximation $(u+phi)^2approx u^2+2uphi$. Start with an initial vector $vec{u}_0$, e.g.$vec{u}_0=0$. Then use the iterationnewline
begin{align}
textbf{A}cdot (vec{u}_k + vec{phi})&=1+vec{u}_k^2+2vec{u}_kcdotvec{phi}\
textbf{A}cdot (vec{u}_k + vec{phi}) - 2vec{u}_kcdotvec{phi}&= 1+vec{u}_k^2\
textbf{A} cdot (vec{u}_k + vec{phi}) - text{diag}(2vec{u}_k)cdot phi &= 1 + vec{u}_k^2\
textbf{A}cdotphi + textbf{A}cdotvec{u}_k- text{diag}(2vec{u}_k)cdotphi &= 1 + vec{u}_k^2\
textbf{A}cdotphi - text{diag}(2vec{u}_k)cdotphi &= 1 + vec{u}_k^2-textbf{A}cdotvec{u}_k\
(textbf{A} - text{diag}(2vec{u}_k))cdot phi &= 1 + vec{u}_k^2-textbf{A}cdotvec{u}_ktag{1}\
vec{u}_{k+1} &=vec{u}_k+vec{phi}tag{2}
end{align}
My question is: how do we get from (1) to (2)?
linear-algebra nonlinear-system newton-raphson
linear-algebra nonlinear-system newton-raphson
asked Jan 3 at 19:01
ecjbecjb
2858
asked Jan 3 at 19:01
ecjbecjb
2858
asked Jan 3 at 19:01
ecjbecjb
2858
asked Jan 3 at 19:01
ecjbecjb
2858
asked Jan 3 at 19:01
ecjbecjb
2858
2858
$begingroup$
(1) is a system of linear equations. What exactly do you want to know? How to solve systems of linear equations in general? Or which method is the best one for this special type of matrix (sparse band matrix)? Or if there is an explicit solution?
$endgroup$
– Reinhard Meier
Jan 5 at 20:29
$begingroup$
thank you for your answer @ReinhardMeier. Sorry I think I put too much for a rather focused question. I think solving a linear problem is clear for me as well as why you use Newton's method. My only question is how do you get at the end of the first of newton to the second one from (1) to (2) (the 2 last line of the question)
$endgroup$
– ecjb
Jan 5 at 21:20
$begingroup$
Solve (1) for $phi$ and plug $phi$ into (2)
$endgroup$
– Reinhard Meier
Jan 5 at 21:35
$begingroup$
indeed because $phi$ is the unknown and is isolated in the last equation. Many thanks @ReinhardMeier
$endgroup$
– ecjb
Jan 5 at 21:47
add a comment |
$begingroup$
(1) is a system of linear equations. What exactly do you want to know? How to solve systems of linear equations in general? Or which method is the best one for this special type of matrix (sparse band matrix)? Or if there is an explicit solution?
$endgroup$
– Reinhard Meier
Jan 5 at 20:29
$begingroup$
thank you for your answer @ReinhardMeier. Sorry I think I put too much for a rather focused question. I think solving a linear problem is clear for me as well as why you use Newton's method. My only question is how do you get at the end of the first of newton to the second one from (1) to (2) (the 2 last line of the question)
$endgroup$
– ecjb
Jan 5 at 21:20
$begingroup$
Solve (1) for $phi$ and plug $phi$ into (2)
$endgroup$
– Reinhard Meier
Jan 5 at 21:35
$begingroup$
indeed because $phi$ is the unknown and is isolated in the last equation. Many thanks @ReinhardMeier
$endgroup$
– ecjb
Jan 5 at 21:47
$begingroup$
(1) is a system of linear equations. What exactly do you want to know? How to solve systems of linear equations in general? Or which method is the best one for this special type of matrix (sparse band matrix)? Or if there is an explicit solution?
$endgroup$
– Reinhard Meier
Jan 5 at 20:29
$begingroup$
(1) is a system of linear equations. What exactly do you want to know? How to solve systems of linear equations in general? Or which method is the best one for this special type of matrix (sparse band matrix)? Or if there is an explicit solution?
$endgroup$
– Reinhard Meier
Jan 5 at 20:29
$begingroup$
thank you for your answer @ReinhardMeier. Sorry I think I put too much for a rather focused question. I think solving a linear problem is clear for me as well as why you use Newton's method. My only question is how do you get at the end of the first of newton to the second one from (1) to (2) (the 2 last line of the question)
$endgroup$
– ecjb
Jan 5 at 21:20
$begingroup$
thank you for your answer @ReinhardMeier. Sorry I think I put too much for a rather focused question. I think solving a linear problem is clear for me as well as why you use Newton's method. My only question is how do you get at the end of the first of newton to the second one from (1) to (2) (the 2 last line of the question)
$endgroup$
– ecjb
Jan 5 at 21:20
$begingroup$
Solve (1) for $phi$ and plug $phi$ into (2)
$endgroup$
– Reinhard Meier
Jan 5 at 21:35
$begingroup$
Solve (1) for $phi$ and plug $phi$ into (2)
$endgroup$
– Reinhard Meier
Jan 5 at 21:35
$begingroup$
indeed because $phi$ is the unknown and is isolated in the last equation. Many thanks @ReinhardMeier
$endgroup$
– ecjb
Jan 5 at 21:47
$begingroup$
indeed because $phi$ is the unknown and is isolated in the last equation. Many thanks @ReinhardMeier
$endgroup$
– ecjb
Jan 5 at 21:47
add a comment |
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$begingroup$
(1) is a system of linear equations. What exactly do you want to know? How to solve systems of linear equations in general? Or which method is the best one for this special type of matrix (sparse band matrix)? Or if there is an explicit solution?
$endgroup$
– Reinhard Meier
Jan 5 at 20:29
$begingroup$
thank you for your answer @ReinhardMeier. Sorry I think I put too much for a rather focused question. I think solving a linear problem is clear for me as well as why you use Newton's method. My only question is how do you get at the end of the first of newton to the second one from (1) to (2) (the 2 last line of the question)
$endgroup$
– ecjb
Jan 5 at 21:20
$begingroup$
Solve (1) for $phi$ and plug $phi$ into (2)
$endgroup$
– Reinhard Meier
Jan 5 at 21:35
$begingroup$
indeed because $phi$ is the unknown and is isolated in the last equation. Many thanks @ReinhardMeier
$endgroup$
– ecjb
Jan 5 at 21:47