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Suppose $f(x)\in C^1([0,1])$ and $f(0)=0$. Let

$$\phi(x)= \begin{cases} \int_0^x\frac{f(t)}{\sqrt{x-t}}dt &\quad\text{if}\quad x\in(0,1]\\ 0&\quad\text{if}\quad x=0 \end{cases} $$

(a) Prove that $\phi(x)\in C^1([0,1])$ and $$ \phi'(x)=\int_0^x\frac{f'(t)}{\sqrt{x-t}}dt $$

(b) Prove that $$ f(x)=\frac1{\pi}\int_0^x\frac{\phi'(t)}{\sqrt{x-t}}dt $$

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1 Answer 1

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This is Abel integral equation. To find $\phi'(x)$ using the Leibniz rule directly will create a singularity. Instead, integrate by parts to get a better form,

$$\phi(x)=\int_{0}^{x}\frac{f(t)}{\sqrt{x-t}}dt = 2\int_{0}^{x}\sqrt{x-t} \,f'(t)\,dt \,.$$

To solve the integral equation, you can use the Laplace transform technique or just verify directly.

Notice that, the integral

$$ \phi'(x)=\int_0^x\frac{f'(t)}{\sqrt{x-t}}dt $$

is of convolution type. Then we can apply Laplace transform to both sides of the above equation, and use the fact that the Laplace of the convolution equals the product of the Laplace.

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Could you provide the details on applying Laplace transform? Thanks. –  Christmas Bunny Oct 14 '12 at 5:23
    
@YifengXu: Do you have any back ground on Laplace transform? –  Mhenni Benghorbal Oct 14 '12 at 6:04
    
I do see how to use convolution now. That's really cool. But how to verify this directly without Laplace Transform? Also I don't see how (b) follows from (a). Thank you! –  Christmas Bunny Oct 20 '12 at 3:04

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