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How can I write poisson's equation

$\partial_{xx} u = f$

solution in 1d explicitly?

I have seen somewhere I can write $u(x) = \int^{x}_{0}\int^{y}_{0} f(z) dz dy - \int^{1}_{0}\int^{x}_{0}\int^{y}_{0} f(z) dz dy dx$

in $[0,1]$ ?

and say u is comparable?

How do you get this explicit form?some integration?what regularity does $f$ need?

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  • $\begingroup$ Certainly for Riemann integrable $f$, the first term of the purported solution will have second derivative equal to $f$ (to get $-f$, you'll need to negate it) by the fundamental theorem of calculus. The second term of what you've written makes no sense, because $x$ is both the argument of $u$ and used as a variable of integration. BTW, if $u$ is any solution, so is $v(x) = u(x) + ax + b$, and the freedom of choosing $a$ and $b$ lets you fit boundary conditions. $\endgroup$ Jul 21, 2016 at 14:45
  • $\begingroup$ so what's the solution? $\endgroup$
    – Geo
    Jul 24, 2016 at 18:30
  • $\begingroup$ $u(x) = \int_0^x \int_0^y -f(z) ~dz ~dy + ax + b$, where $a$ and $b$ are chosen to satisfy the boundary conditions. E.g., if you know that $u(0) = 0$ and $u'(0) = 1$, then $a = 1$ and $b = 0$. $\endgroup$ Jul 24, 2016 at 19:20
  • $\begingroup$ Please specify the boundary conditions. $\endgroup$
    – user65203
    Jul 24, 2016 at 19:44

1 Answer 1

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A complete solution is $$ u(x) = \int_0^x \int_0^y -f(z) ~dz ~dy + ax + b, $$ where $a$ and $b$ are chosen to satisfy the boundary conditions. E.g., if you know that $u(0) = 0$ and $u'(0) = 1$, then $a = 1$ and $b = 0$. On the other hand, if you know that $u(0) = 3$ and $u(1) = 7$, then you have $$ b = 3\\ a = -\left(\int_0^1 \int_0^y -f(z) ~dz ~dy + 3 \right) + 7. $$

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