A Boundary-Value Problem Transformed |

Show that the boundary-value problem

$\begin{cases} v''(x)+p(x)v'(x)+q(x)v(x) = f(x) \\ v(0) = \alpha, v(1)=\beta\end{cases}$

can be converted to a problem with zero boundary conditions:

$\begin{cases} u''(x)+p(x)u'(x)+q(x)u(x)=f(x)-\alpha q(x) - (\beta-\alpha)(p(x) + x q(x)) \\ u(0)=0, u(1)=0\end{cases}$

Define a new function $u(x)$ in terms of $v(x)$ as follows:

$u(x) = v(x) - \underbrace{(kx + m)}_{\text{for simplicity}}$

where $k$ and $m$ are constants to be determined such that $u(x)$ satisfies zero boundary conditions.

To dertermine $k$ and $m$ , consider the boundary conditions:

$u(0) = v(0) - (k\cdot 0 + m) = \alpha-m =0 \implies m = \alpha$ ,

$u(1) = v(1) - (k\cdot 1 +m) = \beta -(k+m) = \beta -(k+\alpha) =0 \implies k = \beta-\alpha$ .

Hence,

$u(x) = v(x) -( (\beta-\alpha)x + \alpha)$

or equivalently,

$v(x) = u(x) +(\beta - \alpha)x +\alpha$ .

Submitting $v(x) = u(x) + (\beta-\alpha)x +\alpha$ into the original differential equation, we get

$u''(x)+p(x)(u'(x)+(\beta-\alpha)) + q(x)(u(x)+(\beta-\alpha)x+\alpha)=f(x)$

which simplifies to

$u''(x)+p(x)u'(x)+q(x)u(x) = f(x)-\alpha q(x) - (\beta-\alpha)(p(x)+x q(x))$ .

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