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Is there a way to get an analytical solution for $\psi(y)$ for this second order ODE that contains the integral of $\psi$ over the domain:

$$\frac{d^2\psi}{dy^2}=a+b\int^{L}_{0}{\psi dy}$$

where $a$ and $b$ are constants, and with boundary conditions:

$\psi(0)=\psi_1$

$\dfrac{d\psi}{dy}(L)=0$

If not, how would you do it numerically? Do Matlab or Maple handle this?

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  • $\begingroup$ Are you sure the equation is right? Do you mean for the right hand side to be constant? $\endgroup$ Oct 1, 2022 at 9:11
  • $\begingroup$ The right hand side is constant, so....? $\endgroup$
    – K.defaoite
    Oct 6, 2022 at 13:31

2 Answers 2

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Well, I think that $$ \frac{d^3 \psi}{dy^3}=0 $$ since it is a definite integral. Now you only have to write $\psi(y)$ like a degree 2 polynomial and compute coefficients.

Edit:

Elaborating a bit more, $\psi(y)=Ay^2+By+C$. But $C=\psi_1$, because of the initial condition. And $$ \frac{d\psi}{dy}(L)=2AL+B=0, $$ and then $B=-2AL$.

Therefore $\psi(y)=Ay^2-2ALy+\psi_1$.

Substituting in the original equation, $$ \frac{d^2\psi}{dy^2}=2A=a+b\int_0^L Ay^2-2ALy+\psi_1 dy $$ so $$ 2A=a+b(AL^3/3-2AL^3/2+\psi_1 L) $$ and you solve for $A$.

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Let us solve $$\frac{d^2y}{dx}=a+b\int_{0}^{L} y(x) dx, dy/dx, y(0)=y_0, \frac{dy}{dx}(x=L)=0.....(1)$$ Let $$\int_{0}^{L} y dx=c,......(2)$$ then $$\frac{d^2y}{dx^2}=a+bc \implies \frac{dy}{dx}=(a+bc)x+u \implies y(x)=(a+bc)x^2/2+ux+v......(3)$$ We get $v=y_0$ and $(a+bc)L=-u \implies c=\frac{\frac{-u}{L}-a}{b}.$ Next from (2) and (3), $$c=\int_{0}^{L} [(a+bc)x^2/2+ux+v] dx,$$ you solve for $c$ and get it, then get $u$ and $v$ is already $y_0$.. Here, $y_0,a,b, L$ are fixed,

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