Show that the Fredholm integral equation has a unique solution
$begingroup$
Show that the Fredholm integral equation
$$phi(x)=f(x)+ lambda int^pi_0cos(x-s)phi(s),ds$$
If $lambda neq pm frac{2}{pi}$
My solution:
$phi(x) = f(x) + lambda sin x int^pi_0 cos s phi(s) ds + lambda cos x int^pi _0 sin s phi(s), ds tag1$
Let $int^pi_0 cos sphi(s) ds = c_1$ and $int^pi_0 sin s phi(s), ds = c_2$
then
$phi(x) = f(x) + c_1lambda sin x + c_2 lambda cos x tag2$
sub $(2)$ into $(1)$ giving
$c_1sin x+c_2cos x = sin x[f_1 + c_1lambda int^pi_0 cos s sin s space ds + lambda c_2 int^pi_0 cos^2s space ds] $
$qquadqquadqquadqquadquad + cos x [f_2 + c_1lambda int^pi_0 sin^2s space ds + c_2lambda int^pi_0 sin s cos s space ds]$
where $f_1 = int^pi_0 cos sf(x) space ds$ and $f_2 = int^pi_0 sin sf(x) space ds$
since {$sin x, cos x$} are linearly independent we can compare coefficients giving
$c_1 = f_1 + pi lambda c_2$
$c_2 = f_2 + pi lambda c_1$
$begin{pmatrix}1 &lambda pi \ lambda pi &1end{pmatrix}$$begin{pmatrix}c_1 \ c_2 end{pmatrix}=begin{pmatrix}f_1 \ f_2 end{pmatrix}$
Looking at the determinant of the first matrix shown we have
$$1-lambda^2pi^2 = 0$$
Which doesn't correspond with the initial condition given, so I must've gone wrong somewhere.
TIA
integration
edited Jan 2 at 15:33
Bernard
123k741117
asked Jan 2 at 15:14
Ben JonesBen Jones
19511
$endgroup$
add a comment |
$begingroup$
Show that the Fredholm integral equation
$$phi(x)=f(x)+ lambda int^pi_0cos(x-s)phi(s),ds$$
If $lambda neq pm frac{2}{pi}$
My solution:
$phi(x) = f(x) + lambda sin x int^pi_0 cos s phi(s) ds + lambda cos x int^pi _0 sin s phi(s), ds tag1$
Let $int^pi_0 cos sphi(s) ds = c_1$ and $int^pi_0 sin s phi(s), ds = c_2$
then
$phi(x) = f(x) + c_1lambda sin x + c_2 lambda cos x tag2$
sub $(2)$ into $(1)$ giving
$c_1sin x+c_2cos x = sin x[f_1 + c_1lambda int^pi_0 cos s sin s space ds + lambda c_2 int^pi_0 cos^2s space ds] $
$qquadqquadqquadqquadquad + cos x [f_2 + c_1lambda int^pi_0 sin^2s space ds + c_2lambda int^pi_0 sin s cos s space ds]$
where $f_1 = int^pi_0 cos sf(x) space ds$ and $f_2 = int^pi_0 sin sf(x) space ds$
since {$sin x, cos x$} are linearly independent we can compare coefficients giving
$c_1 = f_1 + pi lambda c_2$
$c_2 = f_2 + pi lambda c_1$
$begin{pmatrix}1 &lambda pi \ lambda pi &1end{pmatrix}$$begin{pmatrix}c_1 \ c_2 end{pmatrix}=begin{pmatrix}f_1 \ f_2 end{pmatrix}$
Looking at the determinant of the first matrix shown we have
$$1-lambda^2pi^2 = 0$$
Which doesn't correspond with the initial condition given, so I must've gone wrong somewhere.
TIA
integration
edited Jan 2 at 15:33
Bernard
123k741117
asked Jan 2 at 15:14
Ben JonesBen Jones
19511
$endgroup$
1
$begingroup$
$displaystyle int_0^pi cos^2 s , ds = int_0^pi sin^2 s , ds = fracpi2$, not $pi$.
$endgroup$
– tilper
Jan 2 at 15:39
$begingroup$
I'm not sure I understand the first step of your solution also, or what the determinant of that matrix at the end is supposed to show. The standard way to show a solution is unique is to assume there are two solutions and then show that they're equal, usually by showing their difference is zero. Although I think for this particular thing the standard is to use contraction mappings. Is that allowed in your case? See here - math.stackexchange.com/questions/2045232/…
$endgroup$
– tilper
Jan 2 at 15:41
add a comment |
$begingroup$
Show that the Fredholm integral equation
$$phi(x)=f(x)+ lambda int^pi_0cos(x-s)phi(s),ds$$
If $lambda neq pm frac{2}{pi}$
My solution:
$phi(x) = f(x) + lambda sin x int^pi_0 cos s phi(s) ds + lambda cos x int^pi _0 sin s phi(s), ds tag1$
Let $int^pi_0 cos sphi(s) ds = c_1$ and $int^pi_0 sin s phi(s), ds = c_2$
then
$phi(x) = f(x) + c_1lambda sin x + c_2 lambda cos x tag2$
sub $(2)$ into $(1)$ giving
$c_1sin x+c_2cos x = sin x[f_1 + c_1lambda int^pi_0 cos s sin s space ds + lambda c_2 int^pi_0 cos^2s space ds] $
$qquadqquadqquadqquadquad + cos x [f_2 + c_1lambda int^pi_0 sin^2s space ds + c_2lambda int^pi_0 sin s cos s space ds]$
where $f_1 = int^pi_0 cos sf(x) space ds$ and $f_2 = int^pi_0 sin sf(x) space ds$
since {$sin x, cos x$} are linearly independent we can compare coefficients giving
$c_1 = f_1 + pi lambda c_2$
$c_2 = f_2 + pi lambda c_1$
$begin{pmatrix}1 &lambda pi \ lambda pi &1end{pmatrix}$$begin{pmatrix}c_1 \ c_2 end{pmatrix}=begin{pmatrix}f_1 \ f_2 end{pmatrix}$
Looking at the determinant of the first matrix shown we have
$$1-lambda^2pi^2 = 0$$
Which doesn't correspond with the initial condition given, so I must've gone wrong somewhere.
TIA
integration
edited Jan 2 at 15:33
Bernard
123k741117
asked Jan 2 at 15:14
Ben JonesBen Jones
19511
$endgroup$
Show that the Fredholm integral equation
$$phi(x)=f(x)+ lambda int^pi_0cos(x-s)phi(s),ds$$
If $lambda neq pm frac{2}{pi}$
My solution:
$phi(x) = f(x) + lambda sin x int^pi_0 cos s phi(s) ds + lambda cos x int^pi _0 sin s phi(s), ds tag1$
Let $int^pi_0 cos sphi(s) ds = c_1$ and $int^pi_0 sin s phi(s), ds = c_2$
then
$phi(x) = f(x) + c_1lambda sin x + c_2 lambda cos x tag2$
sub $(2)$ into $(1)$ giving
$c_1sin x+c_2cos x = sin x[f_1 + c_1lambda int^pi_0 cos s sin s space ds + lambda c_2 int^pi_0 cos^2s space ds] $
$qquadqquadqquadqquadquad + cos x [f_2 + c_1lambda int^pi_0 sin^2s space ds + c_2lambda int^pi_0 sin s cos s space ds]$
where $f_1 = int^pi_0 cos sf(x) space ds$ and $f_2 = int^pi_0 sin sf(x) space ds$
since {$sin x, cos x$} are linearly independent we can compare coefficients giving
$c_1 = f_1 + pi lambda c_2$
$c_2 = f_2 + pi lambda c_1$
$begin{pmatrix}1 &lambda pi \ lambda pi &1end{pmatrix}$$begin{pmatrix}c_1 \ c_2 end{pmatrix}=begin{pmatrix}f_1 \ f_2 end{pmatrix}$
Looking at the determinant of the first matrix shown we have
$$1-lambda^2pi^2 = 0$$
Which doesn't correspond with the initial condition given, so I must've gone wrong somewhere.
TIA
integration
integration
edited Jan 2 at 15:33
Bernard
123k741117
asked Jan 2 at 15:14
Ben JonesBen Jones
19511
edited Jan 2 at 15:33
Bernard
123k741117
asked Jan 2 at 15:14
Ben JonesBen Jones
19511
edited Jan 2 at 15:33
Bernard
123k741117
edited Jan 2 at 15:33
Bernard
123k741117
edited Jan 2 at 15:33
Bernard
123k741117
123k741117
asked Jan 2 at 15:14
Ben JonesBen Jones
19511
asked Jan 2 at 15:14
Ben JonesBen Jones
19511
asked Jan 2 at 15:14
Ben JonesBen Jones
19511
19511
1
$begingroup$
$displaystyle int_0^pi cos^2 s , ds = int_0^pi sin^2 s , ds = fracpi2$, not $pi$.
$endgroup$
– tilper
Jan 2 at 15:39
$begingroup$
I'm not sure I understand the first step of your solution also, or what the determinant of that matrix at the end is supposed to show. The standard way to show a solution is unique is to assume there are two solutions and then show that they're equal, usually by showing their difference is zero. Although I think for this particular thing the standard is to use contraction mappings. Is that allowed in your case? See here - math.stackexchange.com/questions/2045232/…
$endgroup$
– tilper
Jan 2 at 15:41
add a comment |
1
$begingroup$
$displaystyle int_0^pi cos^2 s , ds = int_0^pi sin^2 s , ds = fracpi2$, not $pi$.
$endgroup$
– tilper
Jan 2 at 15:39
$begingroup$
I'm not sure I understand the first step of your solution also, or what the determinant of that matrix at the end is supposed to show. The standard way to show a solution is unique is to assume there are two solutions and then show that they're equal, usually by showing their difference is zero. Although I think for this particular thing the standard is to use contraction mappings. Is that allowed in your case? See here - math.stackexchange.com/questions/2045232/…
$endgroup$
– tilper
Jan 2 at 15:41
1
1
$begingroup$
$displaystyle int_0^pi cos^2 s , ds = int_0^pi sin^2 s , ds = fracpi2$, not $pi$.
$endgroup$
– tilper
Jan 2 at 15:39
$begingroup$
$displaystyle int_0^pi cos^2 s , ds = int_0^pi sin^2 s , ds = fracpi2$, not $pi$.
$endgroup$
– tilper
Jan 2 at 15:39
$begingroup$
I'm not sure I understand the first step of your solution also, or what the determinant of that matrix at the end is supposed to show. The standard way to show a solution is unique is to assume there are two solutions and then show that they're equal, usually by showing their difference is zero. Although I think for this particular thing the standard is to use contraction mappings. Is that allowed in your case? See here - math.stackexchange.com/questions/2045232/…
$endgroup$
– tilper
Jan 2 at 15:41
$begingroup$
I'm not sure I understand the first step of your solution also, or what the determinant of that matrix at the end is supposed to show. The standard way to show a solution is unique is to assume there are two solutions and then show that they're equal, usually by showing their difference is zero. Although I think for this particular thing the standard is to use contraction mappings. Is that allowed in your case? See here - math.stackexchange.com/questions/2045232/…
$endgroup$
– tilper
Jan 2 at 15:41
add a comment |
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$begingroup$
$displaystyle int_0^pi cos^2 s , ds = int_0^pi sin^2 s , ds = fracpi2$, not $pi$.
$endgroup$
– tilper
Jan 2 at 15:39
$begingroup$
I'm not sure I understand the first step of your solution also, or what the determinant of that matrix at the end is supposed to show. The standard way to show a solution is unique is to assume there are two solutions and then show that they're equal, usually by showing their difference is zero. Although I think for this particular thing the standard is to use contraction mappings. Is that allowed in your case? See here - math.stackexchange.com/questions/2045232/…
$endgroup$
– tilper
Jan 2 at 15:41