Answer to Question #103468 in Electricity and Magnetism for Abhishek kumar

Question

Consider a 2nd order differential equation
d
2
dx2
u(x) = f(x), 0 ≤ x ≤ a . (1)
Find the solutions of the above equation with the boundary conditions u(0) = 0 and u(a) = 0 using the
Green’s function technique.

Answer 1

As per the given question,

$\dfrac{d^2u(x)}{dx^2}+u(x)=f(x)$ $,0\leq x\leq a$

let $u(x)=A (x)\cos kx+B (x)\sin kx$ ----(i)

now, taking the differenciation twice with respect to x equation (i) with

$-kA'\sin kx+kB' \cos kx=f(x)$

$-k^2(A'\cos kx+B' \sin kx)=u''(x)$

So, $A'(x)=\dfrac{f(x)\sin kx}{k}$ and $B'(x)=\dfrac{-f(x)\cos kx}{k}$

Now, We can write this equation as per the below

$u(x)=\dfrac{\sin kx}{k}\int^x_af(y)\sin ky dy-\dfrac{\cos kx}{k}\int^x_af(y)\cos ky dy$

Now u(0)=0

$\int^0_af(y)\sin ky dy=0$

So a=0

similarly,

$\int^x_af(y)\cos ky dy=0$

so, u(a)=0

Hence, it is satisfying the condition.