Question

The following integral equation for f : [-a, a] -----> R  arises in a model of the motion of gas particles on a line:

Answer 1

Let us take an integral equation for f:[-a,a]--->R

Equation for the motion of gas particles on a line

H(x)=1+(1/π) |_-a---->a(1/(1+(x-t)²))f(t)dt. Here,I am taking integration symbol as |.

Now,let us assume x-t=y and t=x-y.

Apply differentiation with respect to t,then above equation becomes

-dt=dy.

and limits will be changed as

If t=-a,then y=x+a.and if t=a,then y=x-a.

Now substitute above conditions in the H(x) equation

H(x)=1+|_{(x-a)---->(x+a)}(1/(1+y²))f(x-y)-dt.

Apply integral uv formula for the second part in above equation.

Let us take u=f(x-y) and v=(1/(1+y²)).

We know that | uv=u|v-| u^'|v.

du/dy=-f^'(x-y). And we know formula | (1/(a²+y²))dy=1/a arctan(x/a)+c.

Here a=1,so,| (1/(1+y²))dy=arctany+c.

Now, apply integration with proper substitutions

We will get

H(x)=1+(f(x-y)arctany-| (-f'(x-y)arctany dy)_{(x-a)--->(x+a).}

Again we have to apply |uv formula and above equation becomes

H(x)=1+(f(x-y).arctany+f^'(x-y)(-ln arccosy)-|(-f^'(x-y)(-ln arccosy)dy)_{(x-a)---->(x+a).}

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In this way integration is continuous depending on the given function f.Given boundary condition is a is varies between 0 and infinity(&).

And now limits become x and &. But solution is continuously changing with function f_.

So,From above explanation and proof,the given equation is unique bounded and has Continuous solution for 0<a<&.