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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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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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    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!2012-10-20