# [SOLVED]Integral equations

#### dwsmith

##### Well-known member
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}.
$\begin{pmatrix} 1 - \frac{\lambda}{3} & -\frac{\lambda}{5}\\ -\frac{\lambda}{4} & 1 - \frac{\lambda}{6} \end{pmatrix} = 0,$
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
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}.
$\begin{pmatrix} 1 - \frac{\lambda}{3} & -\frac{\lambda}{5}\\ -\frac{\lambda}{4} & 1 - \frac{\lambda}{6} \end{pmatrix} = 0,$
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)}.