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Consider the following expression where $x(s)$ and $y(s)$ are continuous as is necessary on the closed interval [a,b]. (This is a functional analysis question -- see below for details.)

$$x(s) = \int\limits_a^s\left(\int\limits_a^uy(t)dt\right)du-\frac{s-a}{b-a}\int\limits_a^b\left(\int\limits_a^uy(t)dt\right)du$$

Could someone please show the mechanical steps to transform the above expression to the following expression:

$$x(s) = \int\limits_a^bK\left(s,t\right)y\left(t\right)dt\text{,}$$

where

$$K(s,t) = \frac{(s-b)(t-a)}{b-a}\;\;\;\;\;\text{if}\;\;a \le t \le s \le b\text{.}$$

Here is some additional information that might, or might not, be useful:

$$\frac{d^2x}{ds^2} = y(s)\text{,}\;\;x(a)=x(b)=0$$

The above is an example that appears in Angus E. Taylor's book "Introduction to Functional Analysis, Second Edition" (page 17) however, Taylor does NOT give the details of the transformation. Taylor gives a second expression for $K(s,t)$, for $a \le s \le t \le b$, but I would like to derive that expression once I see how the above case is handled. Thanks.

Martini, I appreciate your hint, however the crux of the problem for me is transforming the expression into one where the limits of integration is from $a$ to $b$ (i.e., not from $a$ to $s$). With the help of your hint I was able to go forward -- let me show you what I was able to do. I am deadlocked on the last equation.

$$\int\limits_a^S\int\limits_a^uy(t)dt\;du-\frac{S-a}{b-a}\int\limits_a^b\int\limits_a^uy(t)dt\;du$$ $$\int\limits_a^S\int\limits_a^tdu\;y(t)dt-\frac{S-a}{b-a}\int\limits_a^b\int\limits_a^tdu\;y(t)dt$$ $$\int\limits_a^S(t-a)\;y(t)dt-\frac{S-a}{b-a}\int\limits_a^b(t-a)\;y(t)dt$$ $$\frac{1}{b-a}\int\limits_a^S(b-a)(t-a)y(t)dt-\frac{S-a}{b-a}\int\limits_a^b(t-a)y(t)dt$$ $$\frac{1}{b-a}\int\limits_a^S(b-a)(t-a)y(t)dt-\frac{S-a}{b-a}\int\limits_a^S(t-a)y(t)dt-\frac{S-a}{b-a}\int\limits_S^b(t-a)y(t)dt$$ $$\frac{1}{b-a}\int\limits_a^S(b-a-S+a)(t-a)y(t)dt-\frac{S-a}{b-a}\int\limits_S^b(t-a)y(t)dt$$ $$\frac{b-S}{b-a}\int\limits_a^S(t-a)y(t)dt-\frac{S-a}{b-a}\int\limits_S^b(t-a)y(t)dt$$

And I'm deadlocked here. In addition, I see lots of things that do not line up like they should at this point. I should see a term like $S-b$ not $b-S$ and there should not be any $S-a$ term. Obviously, I've strayed down the wrong path. Can you or anyone else point out where I went wrong? Please remember the goal is to derive an integral that is evaluated from $a$ to $b$ and where the function $K(s,t)$ is as defined above. Thanks again.

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  • $\begingroup$ Draw the triangular region of integration in $t$, $u$. $\endgroup$ – DisintegratingByParts Jul 19 '14 at 21:04
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The region of integration for $\int_{a}^{s}\int_{a}^{u}\cdots\,dt\,du$ is a triangle; if you draw the region then you can see how to write the integral as one where the orders of integration of swapped. However, you can approach this purely algebraically so that you don't have to rely on visualizing the region. Let $H(t)$ be the Heaviside step function where $H(t)=1$ for $t > 0$ and is $0$ otherwise. Then $$ \begin{align} \int_{a}^{s}\int_{a}^{u}y(t)\,dt\,du & = \int_{a}^{s}\int_{a}^{s}y(t)H(u-t)\,dt\,du \\ & = \int_{a}^{s}\int_{a}^{s}y(t)H(u-t)\,du\,dt \\ & = \int_{a}^{s}y(t)\int_{a}^{s}H(u-t)\,du\,dt \\ & = \int_{a}^{s}y(t)\int_{t}^{s}\,du\,dt \\ & = \int_{a}^{s}y(t)(s-t)\,dt \\ & = \int_{a}^{b}y(t)(s-t)H(s-t)\,dt \end{align} $$ The other integral works the same way except that $s=b$: $$ \int_{a}^{b}\int_{a}^{u}y(t)\,dt\,du=\int_{a}^{b}y(t)(b-t)\,dt. $$ So you get $$ x(s)=\int_{a}^{b}\left[(s-t)H(s-t)-\frac{s-a}{b-a}(b-t)\right]y(t)\,dt. $$

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Hint. We have by Fubini $$\int_a^s \int_a^u y(t) \,dt\,du = \int_a^s\int_t^a y(t) \,du \, dt = \int_a^s (t-a)y(t)\,dt.$$

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