# [SOLVED]Volterra Equation

#### dwsmith

##### Well-known member
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
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?

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#### dwsmith

##### Well-known member
Re: volterra eq

Have you tried differentiating it?

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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
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)

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#### dwsmith

##### Well-known member
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)

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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.

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#### zzephod

##### Well-known member
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:

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

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#### dwsmith

##### Well-known member
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 $$f(0)=0$$

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How did you get that expression?

#### zzephod

##### Well-known member
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
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
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 $$C=1$$.

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