- #1
bdforbes
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Using Laplace transforms, find the solution of Abel's integral equation:
[tex]\int^{x}_{0}\frac{f(u)}{\sqrt{x-u}}du = 1 + x + x^2 [/tex]
I recognized that the integral is a Laplace convolution, leading to:
[tex] (f*g)(x) = 1+x+x^2 [/tex]
where [tex] g(x)=x^{-1/2}[/tex]
So:
[tex] L(f*g)=L(1)+L(x)+L(x^2)[/tex]
[tex] L(f)L(g)=\frac{1}{p}+\frac{1}{p^2}+\frac{2}{p^3}[/tex]
I can't figure out the transform of g(x). I tried contour integration in the first quadrant, indenting around the origin and placing the branch cut along the negative real axis, and I got to this:
[tex] L(g)=\int^{\infty}_{0}\frac{e^{-px}}{\sqrt{x}}dx=i\int^{\infty}_{0}\frac{e^{-ipx}}{\sqrt{ix}}dx[/tex]
Can anyone help me solve this last integral, or suggest another way to find the transform?
[tex]\int^{x}_{0}\frac{f(u)}{\sqrt{x-u}}du = 1 + x + x^2 [/tex]
I recognized that the integral is a Laplace convolution, leading to:
[tex] (f*g)(x) = 1+x+x^2 [/tex]
where [tex] g(x)=x^{-1/2}[/tex]
So:
[tex] L(f*g)=L(1)+L(x)+L(x^2)[/tex]
[tex] L(f)L(g)=\frac{1}{p}+\frac{1}{p^2}+\frac{2}{p^3}[/tex]
I can't figure out the transform of g(x). I tried contour integration in the first quadrant, indenting around the origin and placing the branch cut along the negative real axis, and I got to this:
[tex] L(g)=\int^{\infty}_{0}\frac{e^{-px}}{\sqrt{x}}dx=i\int^{\infty}_{0}\frac{e^{-ipx}}{\sqrt{ix}}dx[/tex]
Can anyone help me solve this last integral, or suggest another way to find the transform?