Solving Fredholm Equation of the second kindHow to solve a non-linear integral equation?Solving homogeneous...
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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 12 hours ago
user57401user57401
635
$endgroup$
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 12 hours ago
user57401user57401
635
$endgroup$
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 12 hours ago
user57401user57401
635
$endgroup$
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 12 hours ago
user57401user57401
635
asked 12 hours ago
user57401user57401
635
asked 12 hours ago
user57401user57401
635
asked 12 hours ago
user57401user57401
635
asked 12 hours ago
user57401user57401
635
635
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 12 hours ago
Ulrich NeumannUlrich Neumann
9,271516
$endgroup$
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
12 hours ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
11 hours 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 11 hours 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
11 hours ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
10 hours 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
10 hours 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 12 hours ago
Ulrich NeumannUlrich Neumann
9,271516
$endgroup$
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
12 hours ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
11 hours 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 12 hours ago
Ulrich NeumannUlrich Neumann
9,271516
$endgroup$
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
12 hours ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
11 hours 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 12 hours ago
Ulrich NeumannUlrich Neumann
9,271516
$endgroup$
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 12 hours ago
Ulrich NeumannUlrich Neumann
9,271516
edited 12 hours ago
answered 12 hours ago
Ulrich NeumannUlrich Neumann
9,271516
answered 12 hours ago
Ulrich NeumannUlrich Neumann
9,271516
answered 12 hours ago
Ulrich NeumannUlrich Neumann
9,271516
9,271516
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– user57401
12 hours ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
11 hours 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
12 hours ago
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
11 hours 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
12 hours 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
12 hours ago
1
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
11 hours ago
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
11 hours 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 11 hours 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
11 hours ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
10 hours 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
10 hours 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 11 hours 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
11 hours ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
10 hours 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
10 hours 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 11 hours 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 11 hours ago
mjwmjw
3116
edited 10 hours ago
answered 11 hours ago
mjwmjw
3116
answered 11 hours ago
mjwmjw
3116
answered 11 hours 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
11 hours ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
10 hours 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
10 hours 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
11 hours ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
10 hours 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
10 hours 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
11 hours 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
11 hours ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:
[Phi][x, n].$endgroup$
– mjw
10 hours ago
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:
[Phi][x, n].$endgroup$
– mjw
10 hours 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
10 hours 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
10 hours ago
add a comment |
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