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I have been trying to solve the following equation via Green's functions: $$ \frac{\mathrm{d}^2u}{\mathrm{d}x} = \begin{cases} 1, x < \frac{1}{2} \\ 0, x > \frac{1}{2} \end{cases} $$ with boundary conditions $u(0) = u(1) = 0$. The Green's function for this case is given by $$ G(x, y) = \begin{cases} x(y-1), x \leq y \\ y(x-1), x \geq y \end{cases} $$

I understand that the solution is given by $$ u(x) = \int_0^1 G(x, y)f(y)\,\mathrm{d}y $$ and have tried to subdivide the integral to be able to integrate it: $\begin{align} u(x) &= \int_0^x G(x, y)f(y)\,\mathrm{d}y + \int_x^{1/2} G(x, y)f(y)\,\mathrm{d}y + \int_{1/2}^1 G(x, y)f(y) \,\mathrm{d}y \\ &= \int_0^x x(y-1)\,\mathrm{d}y + \int_x^{1/2} y(x-1)\,\mathrm{d}y + \int_{1/2}^1 0 \,\mathrm{d}y \\ &= -\frac{x^2}{2}+\frac{x}{8} - \frac{1}{8} \end{align} $

which, while close, is still off by a factor of $-1$. Is there something wrong with the method? I am not certain about how to subdivide the interval. Any help would be appreciated. Thank you!

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You have substituted the wrong branch of the Green's function on the integral. Note that when you are integrating $y$ from $0$ to $x$, $y\leq x$ (so $x\geq y)$, so you should use $y(x-1)$ when integrating. So: $$ u(x)=\int_0^xy(x-1)dy+\int_x^\frac{1}{2}x(y-1)dy=\frac{x^2}{2}-\frac{3x}{8} $$ Note that the second derivative of this is 1 and u satisfies only the first boundary condition. This happens because you should consider the cases $y>\frac{1}{2}$ and $y<\frac{1}{2}$ separately because we don't know the exact value of y (don't know how rigorous this claim is).

The $u(x)$ you have calculated works when $y<1/2$, but if we consider the other case: $$ u(x)=\int_0^\frac{1}{2}y(x-1)dy+\int_\frac{1}{2}^x0dy+\int_x^10dy=\frac{1}{8}(1-x) $$ This satisfies the conditions that the other $u$ couldn't. So: $$ u(x)=\begin{cases} \frac{x^2}{2}-\frac{3x}{8} & x<\frac{1}{2} \\ \frac{1}{8}(1-x) & x>\frac{1}{2} \end{cases} $$ Hope this helped.

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