Consider the following problem: $$∂g/∂t − ∂/∂x(x^{2−a} ∂g/∂x) = δ(x − ξ)δ(t − τ )$$ with $0 < x, ξ < ∞, 0 < t, τ,$ where the solution remans finite over the entire interval and initially $g(x, 0|ξ, τ ) = 0$. Find the Green's function for this problem.
First we performed a laplace transform for the time dependent term. Afterwards the following hint was given: (Hint: take the Laplace transform and introduce new variables for $a≠0$ as $y=\frac{2\sqrt{s}}{|a|}x^{\frac{a}{2}}$ and $ η = \frac{2\sqrt{s}}{|a|}ξ^{\frac{a}{2}}$ and $ G(x, s|ξ, τ ) = \frac{2(xξ)^(\frac{a−1}{2})}{|a|}F(y,η)e^{-sτ}$ Next transform back.)
We did the derivation in the original problem and put in the expression for $G$. This yielded a long expression and we were not able to express everything in terms of $y$ and $η$.
Could someone help us solving this problem?
Thanks!
This question is more complicated than I thought but the hint is correct and leads to a right direction. Let us Laplace-transform the original PDE first. We define $\tilde{g}(x,s) := \int\limits_0^\infty g(x,t) \exp(-t s) d t$ and we have: \begin{equation} -g(x,0) + s \tilde{g}(x,s) - \frac{\partial}{\partial x} \left[x^{2-a} \frac{\partial \tilde{g}}{\partial x} \right] = \delta(x - \xi) e^{-\tau s} \end{equation} The first term on the left hand side vanishes by assumption. Now we change variables. We have: \begin{eqnarray} \frac{\partial}{\partial x} = \frac{d y}{d x} \cdot \frac{\partial }{\partial y} = \sqrt{s} \cdot \mbox{sign(a)} x^{a/2-1} \frac{\partial }{\partial y} \\ \delta(y - \eta) = \frac{|a|}{2 \sqrt{s}} \delta(x^{a/2} - \xi^{a/2}) = \frac{\delta(x-\xi)}{\sqrt{s} \xi^{a/2-1}} \\ \tilde{g}(x,s) = \frac{2 (x \xi)^{(a-1)/2}}{|a|} F(y,\eta) e^{-s \tau} \quad (i) \end{eqnarray} Inserting all this into the Laplace-transformed PDE we get: \begin{equation} F(y,\eta) - y^{-1/a} \frac{\partial}{\partial y} \left[ y^{2/a-1} \frac{\partial}{\partial y} y^{1-1/a} F(y,\eta) \right] = \frac{1}{y} \cdot \left(\frac{y}{\eta}\right)^{1/a} \cdot \delta(y-\eta) \end{equation} We expand the second term on the left hand side and we get: \begin{equation} F(y,\eta) - \frac{-(-1+a)^2 F(y,\eta) + a^2 y \partial_y F(y,\eta) + a^2 y^2 \partial_y^2 F(y,\eta)}{a^2 y^2} = \frac{1}{y} \cdot \left(\frac{y}{\eta}\right)^{1/a} \cdot \delta(y-\eta) \end{equation} We multiply both sides by $a^2 y^2$ and we get: \begin{equation} y^2 \partial_y^2 F(y,\eta) + y \partial_y F(y,\eta) - \left[y^2 + \left(\frac{-1+a}{a}\right)^2\right] F(y,\eta) = - \left(\frac{y}{\eta}\right)^{1/a} y \delta(y-\eta) \end{equation} We recognise the Bessel equation with a purely imaginary argument and a source term. Therefore the solution reads: \begin{equation} F(y,\eta) = \left\{ \begin{array}{cc} C^{+}_+ I_{(-1+a)/a}(y) + C^{+}_- I_{(1-a)/a}(y) & \mbox{if $ y> \eta$} \\ C^{-}_+ I_{(-1+a)/a}(y) + C^{-}_- I_{(1-a)/a}(y) & \mbox{if $ y< \eta$} \end{array} \right. \quad (ii) \end{equation} where the constants $C^{\pm}_{\pm}$ depend on $\eta$ only. Out of the four constants two are eliminated from the following relations: \begin{eqnarray} F(\eta+ \epsilon, \eta) - F(\eta - \epsilon, \eta) = 0 \\ F^{(1,0)}(\eta+ \epsilon, \eta) - F^{(1,0)}(\eta - \epsilon, \eta) = - \frac{1}{\eta} \end{eqnarray} which give: \begin{equation} \left( \begin{array}{c} C_+^+ - C_+^{-} \\ C_-^+ - C_-^-\end{array} \right) = \left( \begin{array}{cc} I_{+\theta}(\eta) & I_{-\theta}(\eta) \\ I^{'}_{+\theta}(\eta) & I^{'}_{-\theta}(\eta) \end{array} \right)^{-1} \cdot \left( \begin{array}{c} 0 \\ -\frac{1}{\eta}\end{array} \right) \quad (iii) \end{equation} where $\theta := (-1+a)/a$. By combining equations (i), (ii) and (iii) we get the final solution in Laplace domain. Going to the time domain is trivial because the inverse Laplace transform of $e^{-s \tau}$ is just $\delta(t-\tau)$.