# Darboux's Theorem Consider the Sturm-Liouville equation: $ -\psi_{xx} + u \psi = \lambda \psi $ We consider a fixed solution $\psi_0$ with eigenvalue $\lambda = \lambda_1$ of this equation, i.e. $\psi_1 = \psi_1(x;\lambda_1)$. We denote the logarithmic derivative by $\sigma_1 = \psi_{1x}\psi_1^{-1}$. Now consider the Darboux Transformation of the equation, $\psi \rightarrow \psi[1]$ defined by: $ \psi[1] = \left(\frac{d}{dx} - \sigma_1\right)\psi = \psi_x - \frac{\psi_{1x}}{\psi_1}\psi = \frac{W(\psi_1, \psi)}{\psi_1} $ Here we use $W(\psi_1, \psi) = \psi_1 \psi_x - \psi_{1x}\psi$ to denote the Wronskian determinant ## Theorem The function $\psi[1]$ satisfies the Sturm-Liouville equation of the form: $ -\psi[1]_{xx} + u[1] \psi[1] = \lambda \psi[1] $ where $ u[1] = u - 2 \sigma_{0x} = u - 2 \frac{d^2}{dx^2} \ln \psi_0 $ i.e. The Darboux Theorem states that the Sturm-Liouville equation is covariant with respect to the transformation shown above $\psi \rightarrow \psi[1]$ and $u \rightarrow u[1]$. ## Proof The proof is straightforward. Taking the second derivative of $\psi[1]$: $ \psi[1]_{xx} = \psi_{xxx} - 2\sigma_{1x} \psi_x - \sigma_{1xx}\psi - \sigma_1 \psi_{xx} $ Recalling that $\psi_{xx} = (u - \lambda) \psi \implies \psi_{xxx} = u_x \psi + (u - \lambda) \psi_x$ We get $ \begin{align} \psi[1]_{xx} &= u_x \psi + (u - \lambda) \psi_x - 2\sigma_{1x} \psi_x - \sigma_{1xx}\psi - \sigma_1 (u - \lambda) \psi \\ &= (u - \lambda - 2\sigma_{1x}) \psi_x - (\sigma_{1xx} - \sigma_1 u - \sigma_1 \lambda + u_x)\psi \end{align} $ Plugging into the Sturm-Liouville Equation we get: $ \begin{align} (-u + \lambda + 2\sigma_{1x}) \psi_x + (\sigma_{1xx} + \sigma_1 u - \sigma_1 \lambda + u_x)\psi + u[1] (\psi_x - \sigma_1 \psi) &= \lambda (\psi_x - \sigma_1 \psi)\\ (-u + 2\sigma_{1x} + u[1] ) \psi_x + (\sigma_{1xx} - \sigma_1 u + u_x - \sigma_1 u[1])\psi &= 0 \end{align} $ Setting the coefficient of the first term to 0, we get the equation we have already seen for $u[1]$ $ u[1] = u - 2\sigma_{1x} $ The second equation is redundant, it is satisfied by virtue of the form of $u_1$. ## Repeated Application It is clear that we can apply the Darboux Transformation to the once transformed equation above: $ -\psi[1]_{xx} + u[1] \psi[1] = \lambda \psi[1] $ to get the twice transformed equation: $ -\psi[2]_{xx} + u[2] \psi[2] = \lambda \psi[2] $ with the solution: $ \psi[2] = \left(\frac{d}{dx} - \frac{\psi_2'[1]}{\psi_2[1]}\right) \left(\frac{d}{dx} - \frac{\psi_1'}{\psi_1}\right)\psi $ Here, $\psi_2[1]$ is a solution of the intermediate once-transformed equation, corresponding to an original equation with a fixed solution $\psi_2 = \psi_2(x;\lambda_2$), i.e. $ \psi_2[1] = \psi_{2x} - \frac{\psi_{1x}}{\psi_1}\psi_2 $ The potential for the twice transformed equation is given by: $ u[2] = u[1] - 2 \frac{d^2}{dx^2} \ln \psi_2[1] = u - 2 \frac{d^2}{dx^2} \ln W(\psi_2, \psi_1) $ # Crum Theorem # Application to More General Linear Equations