Fredholm integral - degenerate kernel method
I have started answering a fredholm integral equation of the second kind and do not know where to go from here.
The answer has to be written in the form
$$ sum a_jx^{j-1} $$
The Fredholm integral equation is
$$ x^3+frac16x^2+frac15x = g(x) + int_0^1(x^2y+xy^2) f(y) dy$$.
My method so far:
Let: $$C_1 = int_0^1yf(y)dy$$ and $$C_2 = int_0^1y^2f(y)dy$$
Then
$$ x^3+frac16x^2+frac15x = (C_1x^2 +C_2x) + g(x)$$.
Eliminating f(y) to get
$$C_1 = (frac14C_1 + frac13C_2) + int_0^1yg(y)dy$$
and
$$C_2 = (frac15C_1 + frac14C_2) + int_0^1y^2g(y)dy$$
I don't know where to go from here to get it into the form
$$ sum a_jx^{j-1} $$
Do I put it into matrix form, and solve simultaneously, (not sure how to do this)
If I have gotten anything wrong here please let me know.
Or if you need any more information I may be able to provide. (Like how I got to a specific equation)
Any help will be appreciated
Thank you very much
linear-algebra integration definite-integrals
edited Jan 3 at 23:03
Bernard
118k639112
asked Jan 3 at 22:53
p s
357
add a comment |
I have started answering a fredholm integral equation of the second kind and do not know where to go from here.
The answer has to be written in the form
$$ sum a_jx^{j-1} $$
The Fredholm integral equation is
$$ x^3+frac16x^2+frac15x = g(x) + int_0^1(x^2y+xy^2) f(y) dy$$.
My method so far:
Let: $$C_1 = int_0^1yf(y)dy$$ and $$C_2 = int_0^1y^2f(y)dy$$
Then
$$ x^3+frac16x^2+frac15x = (C_1x^2 +C_2x) + g(x)$$.
Eliminating f(y) to get
$$C_1 = (frac14C_1 + frac13C_2) + int_0^1yg(y)dy$$
and
$$C_2 = (frac15C_1 + frac14C_2) + int_0^1y^2g(y)dy$$
I don't know where to go from here to get it into the form
$$ sum a_jx^{j-1} $$
Do I put it into matrix form, and solve simultaneously, (not sure how to do this)
If I have gotten anything wrong here please let me know.
Or if you need any more information I may be able to provide. (Like how I got to a specific equation)
Any help will be appreciated
Thank you very much
linear-algebra integration definite-integrals
edited Jan 3 at 23:03
Bernard
118k639112
asked Jan 3 at 22:53
p s
357
add a comment |
I have started answering a fredholm integral equation of the second kind and do not know where to go from here.
The answer has to be written in the form
$$ sum a_jx^{j-1} $$
The Fredholm integral equation is
$$ x^3+frac16x^2+frac15x = g(x) + int_0^1(x^2y+xy^2) f(y) dy$$.
My method so far:
Let: $$C_1 = int_0^1yf(y)dy$$ and $$C_2 = int_0^1y^2f(y)dy$$
Then
$$ x^3+frac16x^2+frac15x = (C_1x^2 +C_2x) + g(x)$$.
Eliminating f(y) to get
$$C_1 = (frac14C_1 + frac13C_2) + int_0^1yg(y)dy$$
and
$$C_2 = (frac15C_1 + frac14C_2) + int_0^1y^2g(y)dy$$
I don't know where to go from here to get it into the form
$$ sum a_jx^{j-1} $$
Do I put it into matrix form, and solve simultaneously, (not sure how to do this)
If I have gotten anything wrong here please let me know.
Or if you need any more information I may be able to provide. (Like how I got to a specific equation)
Any help will be appreciated
Thank you very much
linear-algebra integration definite-integrals
edited Jan 3 at 23:03
Bernard
118k639112
asked Jan 3 at 22:53
p s
357
I have started answering a fredholm integral equation of the second kind and do not know where to go from here.
The answer has to be written in the form
$$ sum a_jx^{j-1} $$
The Fredholm integral equation is
$$ x^3+frac16x^2+frac15x = g(x) + int_0^1(x^2y+xy^2) f(y) dy$$.
My method so far:
Let: $$C_1 = int_0^1yf(y)dy$$ and $$C_2 = int_0^1y^2f(y)dy$$
Then
$$ x^3+frac16x^2+frac15x = (C_1x^2 +C_2x) + g(x)$$.
Eliminating f(y) to get
$$C_1 = (frac14C_1 + frac13C_2) + int_0^1yg(y)dy$$
and
$$C_2 = (frac15C_1 + frac14C_2) + int_0^1y^2g(y)dy$$
I don't know where to go from here to get it into the form
$$ sum a_jx^{j-1} $$
Do I put it into matrix form, and solve simultaneously, (not sure how to do this)
If I have gotten anything wrong here please let me know.
Or if you need any more information I may be able to provide. (Like how I got to a specific equation)
Any help will be appreciated
Thank you very much
linear-algebra integration definite-integrals
linear-algebra integration definite-integrals
edited Jan 3 at 23:03
Bernard
118k639112
asked Jan 3 at 22:53
p s
357
edited Jan 3 at 23:03
Bernard
118k639112
asked Jan 3 at 22:53
p s
357
edited Jan 3 at 23:03
Bernard
118k639112
edited Jan 3 at 23:03
Bernard
118k639112
edited Jan 3 at 23:03
Bernard
118k639112
118k639112
asked Jan 3 at 22:53
p s
357
asked Jan 3 at 22:53
p s
357
asked Jan 3 at 22:53
p s
357
357
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$$g(x) = x^3 - frac{38}{1077} x^2 + frac{58}{1795} x$$
By rearranging the original equation we can see that g(x) = x^3 + Ax^2 + Bx. This is because both of the integrals on the right evaluate to a constant multiple of x^2 or x respectively. By using this formula for g(x) in each of the integrals we get:
The integral from 0 to 1 of
$$x^2y(y^3+Ay^2+By) dy = x^2[frac{y^5}{5} + frac{Ay^4}{4} + frac{By^3}{3}] = x^2(frac{1}{5} + frac{A}{4} + frac{B}{3})$$
and the integral from 0 to 1 of
$$xy^2(y^3+Ay^2+By) dy = x[y^6/6 + frac{Ay^5}{5} + frac{By^4}{4}] = x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
So the equality becomes:
$$x^3 + frac{1}{6} x^2 + frac{1}{5} x = (x^3 + Ax^2 + Bx) + x^2(frac{1}{5} + frac{A}{4} + frac{B}{3}) + x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
Rearranging gives:
$$x^2(frac{-1}{30} - frac{5A}{4} - frac{B}{3}) + x(frac{1}{30} - frac{A}{5}- frac{5B}{4}) = 0$$
So, by comparing coefficients;
$$frac{-1}{30} - frac{5A}{4} - frac{B}{3} = 0$$ and
$$frac{1}{30} - frac{A}{5} - frac{5B}{4} = 0$$
Solving simultaneously gives
$$A = - 38/1077 $$ and
$$B = frac{58}{1795} $$
$$ g(x)= x^3-frac{38}{1077} x^2 + frac{58}{1795}$$
edited 2 days ago
p s
357
answered Jan 3 at 23:51
Peter Foreman
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If i was to write it in the form of a sum of functions, how would I write it as $$ sum a_jx^{j-1} $$.
– p s
2 days ago
If we let $a_1=frac{58}{1795}$, $a_2=0$, $a_3=-frac{38}{1077}$ and $a_4=1$ we have:$$g(x)=sum_{j=1}^4{a_jx^{j-1}}$$
– Peter Foreman
2 days ago
Thank you for your help
– p s
22 hours ago
Thank you for your help. If I was to continue my method would I still get the same answer? I didn't know what to do next in my method to get an answer.
– p s
22 hours ago
add a comment |
Your Answer
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$$g(x) = x^3 - frac{38}{1077} x^2 + frac{58}{1795} x$$
By rearranging the original equation we can see that g(x) = x^3 + Ax^2 + Bx. This is because both of the integrals on the right evaluate to a constant multiple of x^2 or x respectively. By using this formula for g(x) in each of the integrals we get:
The integral from 0 to 1 of
$$x^2y(y^3+Ay^2+By) dy = x^2[frac{y^5}{5} + frac{Ay^4}{4} + frac{By^3}{3}] = x^2(frac{1}{5} + frac{A}{4} + frac{B}{3})$$
and the integral from 0 to 1 of
$$xy^2(y^3+Ay^2+By) dy = x[y^6/6 + frac{Ay^5}{5} + frac{By^4}{4}] = x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
So the equality becomes:
$$x^3 + frac{1}{6} x^2 + frac{1}{5} x = (x^3 + Ax^2 + Bx) + x^2(frac{1}{5} + frac{A}{4} + frac{B}{3}) + x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
Rearranging gives:
$$x^2(frac{-1}{30} - frac{5A}{4} - frac{B}{3}) + x(frac{1}{30} - frac{A}{5}- frac{5B}{4}) = 0$$
So, by comparing coefficients;
$$frac{-1}{30} - frac{5A}{4} - frac{B}{3} = 0$$ and
$$frac{1}{30} - frac{A}{5} - frac{5B}{4} = 0$$
Solving simultaneously gives
$$A = - 38/1077 $$ and
$$B = frac{58}{1795} $$
$$ g(x)= x^3-frac{38}{1077} x^2 + frac{58}{1795}$$
edited 2 days ago
p s
357
answered Jan 3 at 23:51
Peter Foreman
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If i was to write it in the form of a sum of functions, how would I write it as $$ sum a_jx^{j-1} $$.
– p s
2 days ago
If we let $a_1=frac{58}{1795}$, $a_2=0$, $a_3=-frac{38}{1077}$ and $a_4=1$ we have:$$g(x)=sum_{j=1}^4{a_jx^{j-1}}$$
– Peter Foreman
2 days ago
Thank you for your help
– p s
22 hours ago
Thank you for your help. If I was to continue my method would I still get the same answer? I didn't know what to do next in my method to get an answer.
– p s
22 hours ago
add a comment |
$$g(x) = x^3 - frac{38}{1077} x^2 + frac{58}{1795} x$$
By rearranging the original equation we can see that g(x) = x^3 + Ax^2 + Bx. This is because both of the integrals on the right evaluate to a constant multiple of x^2 or x respectively. By using this formula for g(x) in each of the integrals we get:
The integral from 0 to 1 of
$$x^2y(y^3+Ay^2+By) dy = x^2[frac{y^5}{5} + frac{Ay^4}{4} + frac{By^3}{3}] = x^2(frac{1}{5} + frac{A}{4} + frac{B}{3})$$
and the integral from 0 to 1 of
$$xy^2(y^3+Ay^2+By) dy = x[y^6/6 + frac{Ay^5}{5} + frac{By^4}{4}] = x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
So the equality becomes:
$$x^3 + frac{1}{6} x^2 + frac{1}{5} x = (x^3 + Ax^2 + Bx) + x^2(frac{1}{5} + frac{A}{4} + frac{B}{3}) + x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
Rearranging gives:
$$x^2(frac{-1}{30} - frac{5A}{4} - frac{B}{3}) + x(frac{1}{30} - frac{A}{5}- frac{5B}{4}) = 0$$
So, by comparing coefficients;
$$frac{-1}{30} - frac{5A}{4} - frac{B}{3} = 0$$ and
$$frac{1}{30} - frac{A}{5} - frac{5B}{4} = 0$$
Solving simultaneously gives
$$A = - 38/1077 $$ and
$$B = frac{58}{1795} $$
$$ g(x)= x^3-frac{38}{1077} x^2 + frac{58}{1795}$$
edited 2 days ago
p s
357
answered Jan 3 at 23:51
Peter Foreman
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If i was to write it in the form of a sum of functions, how would I write it as $$ sum a_jx^{j-1} $$.
– p s
2 days ago
If we let $a_1=frac{58}{1795}$, $a_2=0$, $a_3=-frac{38}{1077}$ and $a_4=1$ we have:$$g(x)=sum_{j=1}^4{a_jx^{j-1}}$$
– Peter Foreman
2 days ago
Thank you for your help
– p s
22 hours ago
Thank you for your help. If I was to continue my method would I still get the same answer? I didn't know what to do next in my method to get an answer.
– p s
22 hours ago
add a comment |
$$g(x) = x^3 - frac{38}{1077} x^2 + frac{58}{1795} x$$
By rearranging the original equation we can see that g(x) = x^3 + Ax^2 + Bx. This is because both of the integrals on the right evaluate to a constant multiple of x^2 or x respectively. By using this formula for g(x) in each of the integrals we get:
The integral from 0 to 1 of
$$x^2y(y^3+Ay^2+By) dy = x^2[frac{y^5}{5} + frac{Ay^4}{4} + frac{By^3}{3}] = x^2(frac{1}{5} + frac{A}{4} + frac{B}{3})$$
and the integral from 0 to 1 of
$$xy^2(y^3+Ay^2+By) dy = x[y^6/6 + frac{Ay^5}{5} + frac{By^4}{4}] = x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
So the equality becomes:
$$x^3 + frac{1}{6} x^2 + frac{1}{5} x = (x^3 + Ax^2 + Bx) + x^2(frac{1}{5} + frac{A}{4} + frac{B}{3}) + x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
Rearranging gives:
$$x^2(frac{-1}{30} - frac{5A}{4} - frac{B}{3}) + x(frac{1}{30} - frac{A}{5}- frac{5B}{4}) = 0$$
So, by comparing coefficients;
$$frac{-1}{30} - frac{5A}{4} - frac{B}{3} = 0$$ and
$$frac{1}{30} - frac{A}{5} - frac{5B}{4} = 0$$
Solving simultaneously gives
$$A = - 38/1077 $$ and
$$B = frac{58}{1795} $$
$$ g(x)= x^3-frac{38}{1077} x^2 + frac{58}{1795}$$
edited 2 days ago
p s
357
answered Jan 3 at 23:51
Peter Foreman
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$$g(x) = x^3 - frac{38}{1077} x^2 + frac{58}{1795} x$$
By rearranging the original equation we can see that g(x) = x^3 + Ax^2 + Bx. This is because both of the integrals on the right evaluate to a constant multiple of x^2 or x respectively. By using this formula for g(x) in each of the integrals we get:
The integral from 0 to 1 of
$$x^2y(y^3+Ay^2+By) dy = x^2[frac{y^5}{5} + frac{Ay^4}{4} + frac{By^3}{3}] = x^2(frac{1}{5} + frac{A}{4} + frac{B}{3})$$
and the integral from 0 to 1 of
$$xy^2(y^3+Ay^2+By) dy = x[y^6/6 + frac{Ay^5}{5} + frac{By^4}{4}] = x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
So the equality becomes:
$$x^3 + frac{1}{6} x^2 + frac{1}{5} x = (x^3 + Ax^2 + Bx) + x^2(frac{1}{5} + frac{A}{4} + frac{B}{3}) + x(frac{1}{6} + frac{A}{5} + frac{B}{4})$$
Rearranging gives:
$$x^2(frac{-1}{30} - frac{5A}{4} - frac{B}{3}) + x(frac{1}{30} - frac{A}{5}- frac{5B}{4}) = 0$$
So, by comparing coefficients;
$$frac{-1}{30} - frac{5A}{4} - frac{B}{3} = 0$$ and
$$frac{1}{30} - frac{A}{5} - frac{5B}{4} = 0$$
Solving simultaneously gives
$$A = - 38/1077 $$ and
$$B = frac{58}{1795} $$
$$ g(x)= x^3-frac{38}{1077} x^2 + frac{58}{1795}$$
edited 2 days ago
p s
357
answered Jan 3 at 23:51
Peter Foreman
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
p s
357
edited 2 days ago
p s
357
edited 2 days ago
p s
357
357
answered Jan 3 at 23:51
Peter Foreman
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 3 at 23:51
Peter Foreman
2056
answered Jan 3 at 23:51
Peter Foreman
2056
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If i was to write it in the form of a sum of functions, how would I write it as $$ sum a_jx^{j-1} $$.
– p s
2 days ago
If we let $a_1=frac{58}{1795}$, $a_2=0$, $a_3=-frac{38}{1077}$ and $a_4=1$ we have:$$g(x)=sum_{j=1}^4{a_jx^{j-1}}$$
– Peter Foreman
2 days ago
Thank you for your help
– p s
22 hours ago
Thank you for your help. If I was to continue my method would I still get the same answer? I didn't know what to do next in my method to get an answer.
– p s
22 hours ago
add a comment |
If i was to write it in the form of a sum of functions, how would I write it as $$ sum a_jx^{j-1} $$.
– p s
2 days ago
If we let $a_1=frac{58}{1795}$, $a_2=0$, $a_3=-frac{38}{1077}$ and $a_4=1$ we have:$$g(x)=sum_{j=1}^4{a_jx^{j-1}}$$
– Peter Foreman
2 days ago
Thank you for your help
– p s
22 hours ago
Thank you for your help. If I was to continue my method would I still get the same answer? I didn't know what to do next in my method to get an answer.
– p s
22 hours ago
If i was to write it in the form of a sum of functions, how would I write it as $$ sum a_jx^{j-1} $$.
– p s
2 days ago
If i was to write it in the form of a sum of functions, how would I write it as $$ sum a_jx^{j-1} $$.
– p s
2 days ago
If we let $a_1=frac{58}{1795}$, $a_2=0$, $a_3=-frac{38}{1077}$ and $a_4=1$ we have:$$g(x)=sum_{j=1}^4{a_jx^{j-1}}$$
– Peter Foreman
2 days ago
If we let $a_1=frac{58}{1795}$, $a_2=0$, $a_3=-frac{38}{1077}$ and $a_4=1$ we have:$$g(x)=sum_{j=1}^4{a_jx^{j-1}}$$
– Peter Foreman
2 days ago
Thank you for your help
– p s
22 hours ago
Thank you for your help
– p s
22 hours ago
Thank you for your help. If I was to continue my method would I still get the same answer? I didn't know what to do next in my method to get an answer.
– p s
22 hours ago
Thank you for your help. If I was to continue my method would I still get the same answer? I didn't know what to do next in my method to get an answer.
– p s
22 hours ago
add a comment |
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