Dustinsfl
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How do I solve the Volterra integral equation?
\[
f(x) = \sqrt{x} + \lambda\int_0^x\sqrt{xy}f(y)dy
\]
\[
f(x) = \sqrt{x} + \lambda\int_0^x\sqrt{xy}f(y)dy
\]
dwsmith said:How do I solve the Volterra integral equation?
\[
f(x) = \sqrt{x} + \lambda\int_0^x\sqrt{xy}f(y)dy
\]
zzephod said:Have you tried differentiating it?
.
dwsmith said:The derivative is
\[
f'(x) = \frac{1}{2\sqrt{x}} + \lambda\left(xf(x) + \int_0^1\frac{y}{2\sqrt{xy}}f(y)dy\right)
\]
How does this help?
zzephod said:Now you can replace the integral in the expression for the derivative by the expression for it you get from the original equation leaving you with a first order ordinary differential equation.
(and you should still have \(x\) as the upper limit in the integral, you will also find things easier if you take the \(x\) outside the integral)
.
dwsmith said:So
\[
\int_0^x\sqrt{y}f(y)dy = \frac{f(x) - \sqrt{x}}{\lambda\sqrt{x}}
\]
Plugging this in we get
\[
f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2\lambda\sqrt{x}} + f(x)\left(x\lambda + \frac{1}{2\lambda\sqrt{x}}\right).
\]
IWhat am I suppose to do with the \(f'(x)\) equation? The solution to this ODE isn't trivial.
zzephod said:Well for \(x>0\), I get:
\[f'(x)=f(x)\left[-\frac{1}{2x}+\lambda x\right]\]
which assuming no mistakes in the manipulations is of variables seperable type with initial condition \(f(0)=0\)
.
dwsmith said:How did you get that expression?
zzephod said:By substituting \(\frac{f(x) - \sqrt{x}}{\sqrt{x}}\) for \(\lambda \int_0^x\sqrt{y}f(y)dy \) in the expression for the derivative.
dwsmith said:Shouldn't it be + not minus?
dwsmith said:Shouldn't it be + not minus? Also, for +, the ODE evaluates to
\[
f(x) = C \sqrt{x} e^{\frac{\lambda x^2}{2}}.
\]
With the IC of \(f(0) = 0\), \(f(x) = C\cdot 0 = 0\).