04 - Variation of Parameters

Variation of Parameters

Let $(N)$ denote the Inhomogeneous Equation ( $L y = f$ )

Using Variation of Parameters then
$y (x) = - \int^{x} \frac{f ( ξ ) y _{2} ( ξ ) y _{1} ( x )}{P _{2} ( ξ ) W ( ξ )} d ξ + \int^{x} \frac{f ( ξ ) y _{1} ( ξ ) y _{2} ( x )}{P _{2} ( ξ ) W ( ξ )} d ξ$

Proof

General way to find the general solution for the inhomogeneous version $(N)$

Suppose $(H)$ is solved by
$y (x) = c_{1} y_{1} (x) + c_{2} y_{2} (x)$
where $y_{1}, y_{2}$ are linearly independent

Seeking a solution to $(N)$ in the form
$y (x) = c_{1} (x) y_{1} (x) + c_{2} (x) y_{2} (x)$
(that is we vary the parameter $c_{1}$ and $c_{2}$ )

Differentiating
$y^{'} = c_{1} y_{1}^{'} + c_{2} y_{2}^{'} + c_{1}^{'} y_{1} + c_{2}^{'} y_{2}$
Eliminating the highest derivative of $c_{i}$ , impose condition
$c_{1}^{'} y_{1} + c_{2}^{'} y_{2} = 0$
on $c_{1}$ and $c_{2}$

Since two functions $c_{1}$ and $c_{2}$ define $y$ , should be enough freedom to satisfy constraint

Under assumption then $y^{'}$ simplifies to
$y^{'} = c_{1} y_{1}^{'} + c_{2} y_{2}^{'}$
Differentiating and substituting into $(H)$ then
$L [y] = P_{2} (c_{1} y_{1}^{''} + c_{2} y_{2}^{''} + c_{1}^{'} y_{1}^{'} + c_{2}^{'} y_{2}^{'}) + P_{1} (c_{1} y_{1}^{'} + c_{2} y_{2}^{'}) + P_{0} (c_{1} y_{1} + c_{2} y_{2}) = c_{1} L [y_{1}] + c_{2} L [y_{2}] + P_{2} (c_{1}^{'} y_{1}^{'} + c_{2}^{'} y_{2}^{'})$
Since $y_{i}$ satisfy $(H)$ then inhomogeneous ODE $(N)$ becomes
$L [y] = P_{2} (c_{1}^{'} y_{1}^{'} + c_{2}^{'} y_{2}^{'}) = f$
So then
$(y_{1} y_{1}^{'} y_{2} y_{2}^{'}) (c_{1}^{'} c_{2}^{'}) = (0 \frac{f}{P _{2}})$
Determinant of matrix on left is Wronskian $W$ so it is non-zero (by linear independence)
By inverting then
$(c_{1}^{'} c_{2}^{'}) = \frac{1}{W} (y_{2}^{'} - y_{1}^{'} - y_{2} y_{1}) (0 \frac{f}{P _{2}}) = \frac{f}{P _{2} W} (- y_{2} y_{1})$
Hence by integrating
$c_{1} (x) = - \int^{x} \frac{f ( ξ ) y _{2} ( ξ )}{P _{2} ( ξ ) W ( ξ )} d ξ, c_{2} (x) = \int^{x} \frac{f ( ξ ) y _{1} ( ξ )}{P _{2} ( ξ ) W ( ξ )} d ξ$
Hence by substitution
$y (x) = - \int^{x} \frac{f ( ξ ) y _{2} ( ξ ) y _{1} ( x )}{P _{2} ( ξ ) W ( ξ )} d ξ + \int^{x} \frac{f ( ξ ) y _{1} ( ξ ) y _{2} ( x )}{P _{2} ( ξ ) W ( ξ )} d ξ$

Variation of Parameters - $c_{i}$ definition

$c_{1} (x) = - \int^{x} \frac{f ( ξ ) y _{2} ( ξ )}{P _{2} ( ξ ) W ( ξ )} d ξ, c_{2} (x) = \int^{x} \frac{f ( ξ ) y _{1} ( ξ )}{P _{2} ( ξ ) W ( ξ )} d ξ$

Boundary Conditions

Attempting Variation of Parameters to solve $(N)$ with homogenous boundary conditions

Consider BVP
$P_{2} (x) y^{''} (x) + P_{1} (x) y^{'} (x) + P_{0} (x) y (x) = f (x) a < x < b$
with boundary data
$y (a) = 0, y (b) = 0$
Then
$y (x) = \int_{a}^{x} \frac{f ( ξ ) y _{1} ( ξ ) y _{2} ( x )}{P _{2} ( ξ ) W ( ξ )} d ξ + \int_{x}^{b} \frac{f ( ξ ) y _{2} ( ξ ) y _{1} ( x )}{P _{2} ( ξ ) W ( ξ )} d ξ$
which can be concisely written as
$y (x) = \int_{a}^{b} g (x, ξ) f (ξ) d ξ$
where $g (x, ξ)$ is Green’s Function

Proof

Use Variation of Parameters but with condition
$y_{1} (a) = 0 and y_{2} (b) = 0$
where $y_{1}$ and $y_{2}$ are the two basis solutions

Assuming they are linearly independent so that
$W [y_{1}, y_{2}] \neq = 0$
Hence
$y_{1} (b) \neq = 0 and y_{2} (a) \neq = 0$
Thus solution takes form
$y (x) = c_{1} (x) y_{1} (x) + c_{2} (x) y_{2} (x)$
where $c_{1} (x)$ and $c_{2} (x)$ come from above

Hence
$y (a) y (b) = c_{1} (a) y_{1} (a) + c_{2} (a) y_{2} (a) = c_{2} (a) y_{2} (a) = 0 = c_{1} (b) y_{1} (b) + c_{2} (b) y_{2} (b) = c_{1} (b) y_{1} (b) = 0$
Thus
$c_{2} (a) = 0 = c_{1} (b)$
Imposing conditions on the formulae for $c_{i}$ to get explicit unique forms
$c_{1} (x) = \int_{x}^{b} \frac{f ( ξ ) y _{2} ( ξ )}{P _{2} ( ξ ) W ( ξ )} d ξ, c_{2} (x) = \int_{a}^{x} \frac{f ( ξ ) y _{1} ( ξ )}{P _{2} ( ξ ) W ( ξ )} d ξ$
Note that the limits in the integral of $c_{1}$ are switched

Hence the solution to the BVP can be written as
$y (x) = \int_{a}^{x} \frac{f ( ξ ) y _{1} ( ξ ) y _{2} ( x )}{P _{2} ( ξ ) W ( ξ )} d ξ + \int_{x}^{b} \frac{f ( ξ ) y _{2} ( ξ ) y _{1} ( x )}{P _{2} ( ξ ) W ( ξ )} d ξ$
which can be concisely written as
$y (x) = \int_{a}^{b} g (x, ξ) f (ξ) d ξ$
where $g (x, ξ)$ is Green’s Function

Oxford Maths Notes

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04 - Variation of Parameters

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