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[SOLVED] Volterra Equation

dwsmith

Well-known member
Feb 1, 2012
1,673
How do I solve the Volterra integral equation?
\[
f(x) = \sqrt{x} + \lambda\int_0^x\sqrt{xy}f(y)dy
\]
 

zzephod

Well-known member
Feb 3, 2013
134
Re: volterra eq

How do I solve the Volterra integral equation?
\[
f(x) = \sqrt{x} + \lambda\int_0^x\sqrt{xy}f(y)dy
\]
Have you tried differentiating it?

.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Re: volterra eq

Have you tried differentiating it?

.
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

Well-known member
Feb 3, 2013
134
Re: volterra eq

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?
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

Well-known member
Feb 1, 2012
1,673
Re: volterra eq

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)

.
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 isnt trivial.
 
Last edited:

zzephod

Well-known member
Feb 3, 2013
134
Re: volterra eq

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 isnt trivial.
Well for \(x>0\), I get:

\[f'(x)=f(x)\left[\frac{1}{2x}+\lambda x\right]\]

which assuming no (further) mistakes in the manipulations is of variables seperable type with initial condition \(f(0)=0\)

Note the derivative is:

\[\frac{d}{d\,x}\,f\left( x\right) = \frac{\lambda\,\int_{0}^{x}\sqrt{y}\,f\left( y\right) dy}{2\,\sqrt{x}}+\lambda\,x\,f\left( x\right) +\frac{1}{2\,\sqrt{x}}\]

.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
Re: volterra eq

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 \(f(0)=0\)

.
How did you get that expression?
 

zzephod

Well-known member
Feb 3, 2013
134
Re: volterra eq

How did you get that expression?
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

Well-known member
Feb 1, 2012
1,673
Re: volterra eq

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.
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 \(f(0) = 0\), \(f(x) = C\cdot 0 = 0\).
 

zzephod

Well-known member
Feb 3, 2013
134

zzephod

Well-known member
Feb 3, 2013
134
Re: volterra eq

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 \(f(0) = 0\), \(f(x) = C\cdot 0 = 0\).
Which suggests that we go back to the original integral equation and rewrite it in terms of \(h(x)=f(x)/\sqrt{x}\), which if all has gone well is:

\[h(x)=1+\lambda \int_0^x y\; h(y)\ dy\]

which gives initial condition \(h(0)=1\) and:

\[h'(x) = x h(x)\]
which I think results in \(C=1\).

.