---
title: General solution
tags: Applied_math_method
GA: G-77TT93X4N1
---

# General solution

Consider inhomogeneous problem
$$
\qquad -T u'' = f(x), \tag{1}
$$
with boundary conditions
$$
u(x=0)=a, \quad u(x=\ell)=b.
$$
We should also recall that the corresponding Green's function satisfying
$$
\qquad -T g'' = \delta(x-\xi), \tag{2}
$$
with vanishing boundary conditions
$$
g(x=0)=g(x=\ell)=0,
$$
that has solution
$$
g(x|\xi)=
\begin{cases}
\frac{\ell - \xi}{\ell T}\,x, \quad 0\le x<\xi,\\[.1cm]
\frac{\ell-x}{\ell T}\,\xi, \quad \xi<x\le \ell.
\end{cases}
$$

## Construction of the general function
$$
\int^{\ell}_0 (1)\times g(x|\xi) - (2)\times u(x)\,dx
$$

### With the left hand side of (1) and (2)
$$
-T\int^{\ell}_0 u'' g(x) - g'' u(x)\,dx = -T\left(u'(x)g(x|\xi) - \left.g'(x|\xi)u(x)\right|^{\ell}_0\right)=-\left(\frac{\ell -\xi}{\ell}a+\frac{\xi}{\ell}b\right)
$$

### With the right hand side of (1) and (2)
$$
\int^{\ell}_0 f(x) g(x|\xi) - \delta(x-\xi) u(x)\,dx = \int^{\ell}_0 f(x) g(x|\xi)\,dx - u(\xi)
$$

### Combining the two we obtain
$$
u(\xi) = \int^{\ell}_0 f(x) g(x|\xi)\,dx + \frac{\ell -\xi}{\ell}a+\frac{\xi}{\ell}b.
$$

Finally, we re-name the variables to obtain
$$
u(x) = \int^{\ell}_0 f(\xi) g(\xi|x)\,d\xi + \frac{\ell -x}{\ell}a+\frac{x}{\ell}b = \int^{\ell}_0 f(\xi) g(x|\xi)\,d\xi + \frac{\ell -x}{\ell}a+\frac{x}{\ell}b.
$$

### Remark
The last equality comes from the symmetry of $g(x|\xi)$.