Wronskian method

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Toby Wood

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Toby Wood 2 months ago

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Toby Wood 2 months ago

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Question statement

Consider the homogeneous, 2^nd-order ODE

$\begin{equation}x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - \var{2r-1}x\dfrac{\mathrm{d}y}{\mathrm{d}x} + \var{r^2}y = 0\,. \qquad\qquad(1)\end{equation}$

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This equation has a solution $y=x^r$ , where $r$ is a constant. Enter the correct value for this constant below.

$r =$ Unnamed gap

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a) If we substitute $y = x^r$ into the ODE, we find that $r(r-1)x^r - \var{2r-1}x^r + \var{r^2}x^r = 0$ . This is satisfied (for all values of $x$ ) if and only if the constant $r$ satisfies the quadratic equation $(r-\var{r})^2 = 0$ , i.e. if $r = \var{r}$ .

b) For any two particular solutions of this ODE, say $y = y_1(x)$ and $y = y_2(x)$ , the "Wronskian" $W(x) \equiv y_1y_2^{\prime} - y_1^{\prime}y_2$ must satisfy

$\dfrac{\mathrm{d}W}{\mathrm{d}x} = a(x)\,W\,,$ where here

$a(x) = \dfrac{\var{2r-1}}{x}$ . Solving this equation for

$W$ , we find that

$W(x) = Ax^{\var{2r-1}}$ , where

$A$ is an arbitrary constant. We know that one solution of our ODE is

$y_1 = x^\var{r}$ , and so if

$y_2$ is any other solution then it must satisfy

$y_1y_2^{\prime} - y_1^{\prime}y_2 = W$ and so

$x^\var{r}y_2 - \var{r}x^\var{r-1}y_2 = Ax^\var{2r-1}\,.$ We can solve this first-order equation for

$y_2$ using an integrating factor, which leads to

$\begin{align}\dfrac{\mathrm{d}}{\mathrm{d}x}[x^\var{-r}y_2] &= \dfrac{A}{x} \\ \Rightarrow x^\var{-r}y_2 &= A\ln(x) + B \\ \Rightarrow y_2 &= Ax^\var{r}\ln(x) + Bx^\var{r}\,,\end{align}$ where

$B$ is another arbitrary constant. This is, in fact, the general solution of the original ODE.

The only solution that satisfies the given boundary conditions has $A=B=1$ , and so $y = x^\var{r}\ln(x)$ .

c) The adjoint of equation (1) is

$\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}(x^2z) + \var{2r-1}\dfrac{\mathrm{d}}{\mathrm{d}x}(xz) + \var{r^2}z = 0\,.$ Expanding out the derivative terms (using the product rule repeatedly) and simplifying, we end up with

$x^2\dfrac{\mathrm{d}^2z}{\mathrm{d}x^2} + \var{2r+3}x\dfrac{\mathrm{d}z}{\mathrm{d}x} + \var{(r+1)^2}z = 0\,.$ So

$c = \var{2r+3}$ and

$d = \var{(r+1)^2}$ .

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This question is used in the following exam:

Assignment 1 by Toby Wood in NCL MSP3801.

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Toby Wood 2 months ago

Toby Wood 2 months ago

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Wronskian method

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Toby Wood 2 months ago

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