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integral equation Fredholm series

dwsmith

Well-known member
Feb 1, 2012
1,673
Given
\[
f(x) = 1 + \lambda\int_0^1(xy + x^3y^2)f(y)dy.
\]
To use the Fredholm series, we need to find \(D(\lambda)\) wich is \(1 - \text{order }\lambda + \cdots\). I have already calculated the orders and of order 3 and 4 the terms are 0 so the terms greater than 3 are all zero since we have repeated zeros.
\[
\text{order }\lambda:\quad\int_0^1K(y,y)dy = \frac{1}{2}
\]
and
\[
\text{order }\lambda^2:\quad\iint_0^1
\begin{vmatrix}
K(y,y) & K(y,y')\\
K(y',y) & K(y',y')
\end{vmatrix}
dydy' = \frac{1}{90}
\]
Therefore,
\[
D(\lambda) = 1 - \frac{\lambda}{2} + \frac{\lambda^2}{90}.
\]
Next we need to find \(\mathcal{D}(x,y;\lambda)\). In the Taylor series, we obtain repeated zeros after order lambda squared and lambda cubed. Again, repeated zeros so
\[
\mathcal{D}(x,y;\lambda) = K(x,y) - \lambda\int_0^1
\begin{vmatrix}
K(x,y) & K(x,z)\\
K(z,y) & K(z,z)
\end{vmatrix}dz =
-\lambda \left(\frac{x^3 y^2}{3}-\frac{x^3 y}{4}-\frac{x y^2}{5}+\frac{x y}{6}\right)+x^3 y^2+x y
\]
I get
\[
f(x) = 1 + \frac{\lambda}{D(\lambda)}\int_0^1\mathcal{D}(x,y;\lambda)dy
= \text{see second post}
\]
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
The solution I get for the Fredholm method is
\[
f(x) = 1 - 3x\lambda\frac{\lambda - 30}{180 + \lambda(2\lambda - 45)} +
5x^3\lambda\frac{\lambda + 24}{360 + 2\lambda(2\lambda - 45)},
\]
and when I use the sum of separable kernels, I get
\[
f(x) = 1 - 3x\lambda\frac{\lambda - 30}{180 + \lambda(\lambda - 90)} +
5x^3\lambda\frac{\lambda + 24}{360 + 2\lambda(\lambda - 90)},
\]
The math for the Fredholm method is above and the math for the separable kernels is in this thread second post. They should yield the same result but aren't. They are just off a little. Which one is wrong? Where is there a mistake?