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Solving Fredholm Equation of the second kind
How to solve a non-linear integral equation?Solving homogeneous Fredholm Equation of the second kindNon-linear integral equationFredholm integral equation of the second kind with kernel containing Bessel and Struve functionsSolving Fredholm Equation of the first kindNumerical solution of singular non-linear integral equationRecursive Function Numerical IntegrationSolving an equation with no analytical form (NIntegrate)Fredholm Integral Equation of the 2nd Kind with a Singular Difference KernelNumerical Integration with Inequality Condition
$begingroup$
Consider the Fredholm Equation of the second kind,
$$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$
Where the analytical solution is found as,
$$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$
How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?
numerical-integration integral-equations numerical-value
asked 1 hour ago
user57401user57401
534
$endgroup$
add a comment |
$begingroup$
Consider the Fredholm Equation of the second kind,
$$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$
Where the analytical solution is found as,
$$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$
How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?
numerical-integration integral-equations numerical-value
asked 1 hour ago
user57401user57401
534
$endgroup$
add a comment |
$begingroup$
Consider the Fredholm Equation of the second kind,
$$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$
Where the analytical solution is found as,
$$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$
How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?
numerical-integration integral-equations numerical-value
asked 1 hour ago
user57401user57401
534
$endgroup$
Consider the Fredholm Equation of the second kind,
$$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$
Where the analytical solution is found as,
$$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$
How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?
numerical-integration integral-equations numerical-value
numerical-integration integral-equations numerical-value
asked 1 hour ago
user57401user57401
534
asked 1 hour ago
user57401user57401
534
asked 1 hour ago
user57401user57401
534
asked 1 hour ago
user57401user57401
534
asked 1 hour ago
user57401user57401
534
534
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Use DSolve
PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
(*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)
The solution can be further used in the form PHI[x].
answered 1 hour ago
Ulrich NeumannUlrich Neumann
9,261516
$endgroup$
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
1 hour ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
1 hour ago
add a comment |
$begingroup$
Following
Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
[Phi][x_, 0] = 3;
Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]
The last term in the series [Phi][x,n] is the approximation to [Phi][x].
Here is what Mathematica returns for [Phi][x,10].
To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].
answered 1 hour ago
mjwmjw
3116
$endgroup$
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– user57401
53 mins ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
28 mins ago
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
24 mins ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use DSolve
PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
(*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)
The solution can be further used in the form PHI[x].
answered 1 hour ago
Ulrich NeumannUlrich Neumann
9,261516
$endgroup$
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
1 hour ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
1 hour ago
add a comment |
$begingroup$
Use DSolve
PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
(*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)
The solution can be further used in the form PHI[x].
answered 1 hour ago
Ulrich NeumannUlrich Neumann
9,261516
$endgroup$
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
1 hour ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
1 hour ago
add a comment |
$begingroup$
Use DSolve
PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
(*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)
The solution can be further used in the form PHI[x].
answered 1 hour ago
Ulrich NeumannUlrich Neumann
9,261516
$endgroup$
Use DSolve
PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
(*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)
The solution can be further used in the form PHI[x].
answered 1 hour ago
Ulrich NeumannUlrich Neumann
9,261516
edited 1 hour ago
answered 1 hour ago
Ulrich NeumannUlrich Neumann
9,261516
answered 1 hour ago
Ulrich NeumannUlrich Neumann
9,261516
answered 1 hour ago
Ulrich NeumannUlrich Neumann
9,261516
9,261516
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
1 hour ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
1 hour ago
add a comment |
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
1 hour ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
1 hour ago
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
1 hour ago
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
1 hour ago
1
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
1 hour ago
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
1 hour ago
add a comment |
$begingroup$
Following
Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
[Phi][x_, 0] = 3;
Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]
The last term in the series [Phi][x,n] is the approximation to [Phi][x].
Here is what Mathematica returns for [Phi][x,10].
To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].
answered 1 hour ago
mjwmjw
3116
$endgroup$
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– user57401
53 mins ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
28 mins ago
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
24 mins ago
add a comment |
$begingroup$
Following
Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
[Phi][x_, 0] = 3;
Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]
The last term in the series [Phi][x,n] is the approximation to [Phi][x].
Here is what Mathematica returns for [Phi][x,10].
To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].
answered 1 hour ago
mjwmjw
3116
$endgroup$
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– user57401
53 mins ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
28 mins ago
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
24 mins ago
add a comment |
$begingroup$
Following
Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
[Phi][x_, 0] = 3;
Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]
The last term in the series [Phi][x,n] is the approximation to [Phi][x].
Here is what Mathematica returns for [Phi][x,10].
To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].
answered 1 hour ago
mjwmjw
3116
$endgroup$
Following
Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
[Phi][x_, 0] = 3;
Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]
The last term in the series [Phi][x,n] is the approximation to [Phi][x].
Here is what Mathematica returns for [Phi][x,10].
To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].
answered 1 hour ago
mjwmjw
3116
edited 26 mins ago
answered 1 hour ago
mjwmjw
3116
answered 1 hour ago
mjwmjw
3116
answered 1 hour ago
mjwmjw
3116
3116
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– user57401
53 mins ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
28 mins ago
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
24 mins ago
add a comment |
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– user57401
53 mins ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
28 mins ago
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
24 mins ago
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– user57401
53 mins ago
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– user57401
53 mins ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:
[Phi][x, n].$endgroup$
– mjw
28 mins ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:
[Phi][x, n].$endgroup$
– mjw
28 mins ago
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it is
x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.$endgroup$
– mjw
24 mins ago
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it is
x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.$endgroup$
– mjw
24 mins ago
add a comment |
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