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Find $v(x)$ if $v(1)=0$ and $-\frac{dv}{dx}=f(x)$:

(a) $f=x$; (b) $f=U\left(x-\frac12\right)$; (c) $f=\delta\left(x-\frac12\right)$

Find $u(x)$ for the three $v$'s if $k\frac{du}{dx}=v(x)$ with the end condition $u(0)=0$.

The problem was slightly modified for the purpose of my question. First, the three $v$'s:

(a) $v(x)=-\frac12x^2+\frac12$

(b) $v(x)=\begin{cases}\frac12&\text{if}\;x<\frac12\\1-x&\text{if}\;x\ge\frac12\end{cases}$

(c) $v(x)=\begin{cases}1&\text{if}\;x<\frac12\\0&\text{if}\;x\ge\frac12\end{cases}$

I hope I am right at this point. My questions are for the $u's$. The general solution is $u(x)=\frac1k\int_0^xv(x)dx$.

(a) is easy. $u(x)=\frac1k\left(-\frac{x^3}{6}+\frac{x}{2}\right)$.

For (b), I think it should be $u(x)=\begin{cases}\frac{x}{2k}&\text{if}\;x<\frac12\\\frac1k\left(x-\frac{x^2}{2}\right)&\text{if}\;x\ge\frac12\end{cases}$, but the answer key says "$\frac{x}{k}$ for $x\le\frac12$, $\frac1k\left(2x-x^2-\frac14\right)$ for $x\ge\frac12$".

For (c), my answer is $u(x)=\begin{cases}\frac{x}{k}&\text{if}\;x<\frac12\\0&\text{if}\;x\ge\frac12\end{cases}$, but the answer key says "$\frac{x}{k}$ for $x\le\frac12$, $\frac{1}{2k}$ for $x\ge\frac12$".

Where was I wrong?

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    The solution needs to be continuous, which yours aren't. – 2017-01-03
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    @Chappers Thank you. Now I see. But my answer for (b) is still different from the answer key. My answer becomes $u(x)=\begin{cases}\frac{x}{2k}&\text{if}\;x<\frac12\\\frac1k\left(x-\frac{x^2}{2}-\frac18\right)&\text{if}\;x\ge\frac12\end{cases}$. My function is twice as flat. – 2017-01-04
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    Yes, if we differentiate the given answer, it looks like it is for $ f(x) = 2U(x-1/2)$. I think you've actually got the right answer now. – 2017-01-04

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