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[SOLVED] Integral equations

dwsmith

Well-known member
Feb 1, 2012
1,673
Checking to see if my solution is correct.

Solve the integral equation
\[
f(x) = 1 + \lambda\int_0^1\big(xy + x^3y^2\big)f(y)dy
\]
using the method of separable kernels

We need to rewrite \(f(x)\).
\[
f(x_i) = 1 + \sum_{i = 1}^2x^{2i - 1}A_i
\]
where \(A_i = \lambda\int_0^1y^if(y)dy\). Let \(f(y_i) = 1 + \sum\limits_{i = 1}^2y^{2i - 1}A_i\). Then
\begin{align*}
A_i &= \lambda\int_0^1y^i\Bigg(1 + \sum_{j = 1}^2y^{2j - 1}A_j\Bigg)dy\\
&= \lambda\int_0^1y^idy +
\lambda\sum_{j = 1}^2A_j\int_0^1y^{2j - 1}y^idy\\
\sum_{j = 1}^2\Bigg(\delta_{ij} - \lambda\int_0^1y^{2j - 1}y^idy\Bigg)A_j
&= \lambda\int_0^1y^idy\\
\begin{pmatrix}
1 - \frac{\lambda}{3} & -\frac{\lambda}{5}\\
-\frac{\lambda}{4} & 1 - \frac{\lambda}{6}
\end{pmatrix}
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix}
&= \lambda
\begin{pmatrix}
\frac{1}{2}\\
\frac{1}{3}
\end{pmatrix}\\
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix} &=
\begin{pmatrix}
-\frac{3\lambda(\lambda - 30)}{180 + \lambda(\lambda - 90)}\\[.3cm]
\frac{5\lambda(\lambda + 24)}{360 + 2\lambda(\lambda - 90)}
\end{pmatrix}
\end{align*}
Therefore, \(f(x_i)\) is
\[
\begin{pmatrix}
f(x_1)\\
f(x_2)
\end{pmatrix} =
\begin{pmatrix}
-x\frac{3\lambda(\lambda - 30)}{180 + \lambda(\lambda - 90)}\\[.3cm]
x^3\frac{5\lambda(\lambda + 24)}{360 + 2\lambda(\lambda - 90)}
\end{pmatrix}.
\]
Additionally, if the determinant of
\[
\begin{pmatrix}
1 - \frac{\lambda}{3} & -\frac{\lambda}{5}\\
-\frac{\lambda}{4} & 1 - \frac{\lambda}{6}
\end{pmatrix}
= 0,
\]
then we only have a solution if \(v_i(y) = \int_0^1y^idy = 0\).
\[
\begin{pmatrix}
1 - \frac{\lambda}{3} & -\frac{\lambda}{5}\\
-\frac{\lambda}{4} & 1 - \frac{\lambda}{6}
\end{pmatrix}
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix} =
\begin{pmatrix}
A_1\Big(1 - \frac{\lambda}{3}\Big) - \frac{\lambda A_2}{5}\\
A_2\Big(1 - \frac{\lambda}{6}\Big) - \frac{\lambda A_1}{4}
\end{pmatrix} =
\mathbf{0}
\]
Therefore, \(A_i\) is
\[
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix} =
\begin{pmatrix}
1\\
0
\end{pmatrix}
\]
and \(f(x_i)\) is
\[
\begin{pmatrix}
f(x_1)\\
f(x_2)
\end{pmatrix} =
\begin{pmatrix}
1 + x\\
1
\end{pmatrix}.
\]
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Checking to see if my solution is correct.

Solve the integral equation
\[
f(x) = 1 + \lambda\int_0^1\big(xy + x^3y^2\big)f(y)dy
\]
using the method of separable kernels

We need to rewrite \(f(x)\).
\[
f(x_i) = 1 + \sum_{i = 1}^2x^{2i - 1}A_i
\]
where \(A_i = \lambda\int_0^1y^if(y)dy\). Let \(f(y_i) = 1 + \sum\limits_{i = 1}^2y^{2i - 1}A_i\). Then
\begin{align*}
A_i &= \lambda\int_0^1y^i\Bigg(1 + \sum_{j = 1}^2y^{2j - 1}A_j\Bigg)dy\\
&= \lambda\int_0^1y^idy +
\lambda\sum_{j = 1}^2A_j\int_0^1y^{2j - 1}y^idy\\
\sum_{j = 1}^2\Bigg(\delta_{ij} - \lambda\int_0^1y^{2j - 1}y^idy\Bigg)A_j
&= \lambda\int_0^1y^idy\\
\begin{pmatrix}
1 - \frac{\lambda}{3} & -\frac{\lambda}{5}\\
-\frac{\lambda}{4} & 1 - \frac{\lambda}{6}
\end{pmatrix}
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix}
&= \lambda
\begin{pmatrix}
\frac{1}{2}\\
\frac{1}{3}
\end{pmatrix}\\
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix} &=
\begin{pmatrix}
-\frac{3\lambda(\lambda - 30)}{180 + \lambda(\lambda - 90)}\\[.3cm]
\frac{5\lambda(\lambda + 24)}{360 + 2\lambda(\lambda - 90)}
\end{pmatrix}
\end{align*}
Therefore, \(f(x_i)\) is
\[
\begin{pmatrix}
f(x_1)\\
f(x_2)
\end{pmatrix} =
\begin{pmatrix}
-x\frac{3\lambda(\lambda - 30)}{180 + \lambda(\lambda - 90)}\\[.3cm]
x^3\frac{5\lambda(\lambda + 24)}{360 + 2\lambda(\lambda - 90)}
\end{pmatrix}.
\]
Additionally, if the determinant of
\[
\begin{pmatrix}
1 - \frac{\lambda}{3} & -\frac{\lambda}{5}\\
-\frac{\lambda}{4} & 1 - \frac{\lambda}{6}
\end{pmatrix}
= 0,
\]
then we only have a solution if \(v_i(y) = \int_0^1y^idy = 0\).
\[
\begin{pmatrix}
1 - \frac{\lambda}{3} & -\frac{\lambda}{5}\\
-\frac{\lambda}{4} & 1 - \frac{\lambda}{6}
\end{pmatrix}
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix} =
\begin{pmatrix}
A_1\Big(1 - \frac{\lambda}{3}\Big) - \frac{\lambda A_2}{5}\\
A_2\Big(1 - \frac{\lambda}{6}\Big) - \frac{\lambda A_1}{4}
\end{pmatrix} =
\mathbf{0}
\]
Therefore, \(A_i\) is
\[
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix} =
\begin{pmatrix}
1\\
0
\end{pmatrix}
\]
and \(f(x_i)\) is
\[
\begin{pmatrix}
f(x_1)\\
f(x_2)
\end{pmatrix} =
\begin{pmatrix}
1 + x\\
1
\end{pmatrix}.
\]
olve the integral equation
\[
f(x) = 1 + \lambda\int_0^1\big(xy + x^3y^2\big)f(y)dy
\]
using the method of separable kernels

We need to rewrite \(f(x)\).
\[
f(x) = 1 + \sum_{i = 1}^2x^{2i - 1}A_i
\]
where \(A_i = \lambda\int_0^1y^if(y)dy\). Let \(f(y_i) = 1 + \sum\limits_{i = 1}^2y^{2i - 1}A_i\). Then
\begin{align*}
A_i &= \lambda\int_0^1y^i\Bigg(1 + \sum_{j = 1}^2y^{2j - 1}A_j\Bigg)dy\\
&= \lambda\int_0^1y^idy +
\lambda\sum_{j = 1}^2A_j\int_0^1y^{2j - 1}y^idy\\
\sum_{j = 1}^2\Bigg(\delta_{ij} - \lambda\int_0^1y^{2j - 1}y^idy\Bigg)A_j
&= \lambda\int_0^1y^idy\\
\begin{pmatrix}
1 - \frac{\lambda}{3} & -\frac{\lambda}{5}\\
-\frac{\lambda}{4} & 1 - \frac{\lambda}{6}
\end{pmatrix}
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix}
&= \lambda
\begin{pmatrix}
\frac{1}{2}\\
\frac{1}{3}
\end{pmatrix}\\
\begin{pmatrix}
A_1\\
A_2
\end{pmatrix} &=
\begin{pmatrix}
-\frac{3\lambda(\lambda - 30)}{180 + \lambda(\lambda - 90)}\\[.3cm]
\frac{5\lambda(\lambda + 24)}{360 + 2\lambda(\lambda - 90)}
\end{pmatrix}
\end{align*}
Then the solution is
\[
f(x) = 1 -x\frac{3\lambda(\lambda - 30)}{180 + \lambda(\lambda - 90)} + x^3 \frac{5\lambda(\lambda + 24)}{360 + 2\lambda(\lambda - 90)}.
\]