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The equation is \begin{align} \partial_{t}\!\operatorname{u}\!\left(x,t\right) & = x^{2}\,\partial_{x}^{2}\operatorname{u}(x,t) + x\,\partial_{x}\operatorname{u}\left(x,t\right),\quad\quad(t,x)\in\left[0,\infty\right)\times\left(0,\infty\right) \\[3mm] \left. u\right\vert_{t\ =\ 0}\ & = f \in \mathcal{S}\left(\mathbb{R}^{+}\right),\quad f\left(0\right) = 0 \end{align}

the question asks to give an expression for $u(x,t)$ with respect to $f$ in terms of coordinate transforms and Fourier transforms.

I have applied the Fourier transform on the equation but it produces the convolution of derivatives of dirac function $\delta^{\prime}(x)$ and $u(x,t)$ and I don't know how to deal with it.

My work show below apply Fourier Transform to $x$ , let $\tilde{u}(\xi,t)=\mathcal{F}[{u}(x,t)],\tilde{f}(\xi)=\mathcal{F}[f(x)]$, then we get: \begin{equation} \begin{cases} \dfrac{\text{d}\tilde{u}}{\text{d}t}=\mathcal{F}[x^2] \ast (-\lambda\tilde{u}^2)+\mathcal{F}[x] \ast (-i\lambda\tilde{u})\\ \tilde{u}|_{t=0}=\tilde{f} \end{cases} \end{equation} for the first equation we can calulate that \begin{equation} \begin{aligned} \dfrac{\text{d}\tilde{u}}{\text{d}t}&=\mathcal{F}[x^2] \ast (-\lambda^2\tilde{u}^2)+\mathcal{F}[x] \ast (-i\lambda\tilde{u})\\ &= \sqrt{2\pi}\left[ \delta^{\prime\prime}\ast (\lambda^2\tilde{u}^2)+\delta^{\prime}\ast (i\lambda\tilde{u}) \right]\\ &=\sqrt{2\pi}\left[ \lambda^2\int_{\mathbb{R}}\tilde{u}(\xi,t)\delta^{\prime\prime}(x-\xi)\text{d}\xi+\lambda\int_{\mathbb{R}}\tilde{u}(\xi,t)\delta^{\prime}(x-\xi)\text{d}\xi \right] \end{aligned} \end{equation}

then I don't how to continue my caculation.

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    $\begingroup$ Why don't you show us your working so we can see where the issue lies? $\endgroup$ Commented Jun 11 at 11:10
  • $\begingroup$ Sorry, that was my mistake, I've added my work briefly $\endgroup$
    – George Lin
    Commented Jun 11 at 13:46
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    $\begingroup$ Given the particular definition domain, it’s convenient to use the Laplace Transform. $\endgroup$ Commented Jun 11 at 15:34
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    $\begingroup$ @GeorgeLin. The equation can be written $(\partial_{t}-x^2\partial_x^2-x\partial_x)u(x,t)=0,$ which makes it easier to see that it is linear (the operator $\partial_{t}-x^2\partial_x^2-x\partial_x$ doesn't contain $u$) and homogeneous. $\endgroup$
    – md2perpe
    Commented Jun 11 at 16:27
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    $\begingroup$ Your current approach can be salvaged via using the distributional derivatives of the delta function, but an easier approach would be to use the coordinate transform $\xi = \log x$ on the original equation, which should remove all of the variable coefficients and then you can proceed by Fourier or Laplace transforms. $\endgroup$
    – whpowell96
    Commented Jun 11 at 17:12

3 Answers 3

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Set $u(x,t)=v(\ln x, t)=:v(s,t).$ Then the equation becomes $\partial_t v = \partial_s^2 v.$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Given the $\ds{\underline{particular}\ Domain\ De\!f\!inition}$, it's convenient to use the $\ds{Laplace\ Trans\!f\!orm:}$ \begin{align} {\LARGE\bullet\,\,} & \partiald{\on{u}\pars{x,t}}{t} = x^{2}\,\partiald[2]{\on{u}\pars{x,t}}{x} + x\,\partiald{\on{u}\pars{x,t}}{x} \\[5mm] {\LARGE\bullet\,\,} & \int_{0}^{\infty}\partiald{\on{u}\pars{x,t}}{t}\expo{-st}\dd t = \int_{0}^{\infty}\bracks{x^{2}\,\partiald[2]{\on{u}\pars{x,t}}{x} + x\,\partiald{\on{u}\pars{x,t}}{x}}\expo{-st}\dd t \\[5mm] {\LARGE\bullet\,\,} & -\ \overbrace{\on{u}\pars{x,0}}^{\ds{\on{f}\pars{x}}}\ +\ s\hat{\on{u}}\pars{x,s} = x^{2}\,\partiald[2]{\hat{\on{u}}\pars{x,s}}{x} + x\,\partiald{\hat{\on{u}}\pars{x,s}}{x} \\ & \,\,\mbox{where}\quad \hat{\on{u}}\pars{x,s} \equiv\int_{0}^{\infty}\on{u}\pars{x,t} \expo{-st}\dd t \\[5mm] {\LARGE\bullet\,\,} & \mbox{So,}\quad x^{2}\,\partiald[2]{\hat{\on{u}}\pars{x,s}}{x} + x\,\partiald{\hat{\on{u}}\pars{x,s}}{x} - s\hat{\on{u}}\pars{x,s} = -\on{f}\pars{x} \\[5mm] {\LARGE\bullet\,\,} & \mbox{So,}\quad x\,\partiald{}{x}\bracks{x\,\partiald{\hat{\on{u}}\pars{x,s}}{x}} - s\hat{\on{u}}\pars{x,s} = -\on{f}\pars{x} \\[5mm] {\LARGE\bullet\,\,} & \mbox{With}\ y \equiv \ln\pars{x} \mbox{, the above equation is reduced to} \\ & \partiald[2]{\on{U}\pars{y,s}}{y} - s\on{U}\pars{y,s} = -\on{F}\pars{y}\quad\mbox{where}\quad \left\{\begin{array}{rcl} \ds{\on{U}\pars{y,s}} & \ds{\equiv} & \ds{\hat{\on{u}}\pars{\expo{y},s}} \\ \ds{\on{F}\pars{y}} & \ds{\equiv} & \ds{\on{f}\pars{\expo{y}}} \end{array}\right. \\ & \mbox{This equation could be solved with the usual}\ Green's\ Function\ \mbox{technique}. \\[5mm] {\LARGE\bullet\,\,} & \mbox{Can you}\,\, \underline{finish}\,\, \mbox{this problem ?.} \end{align}

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  • $\begingroup$ I'm not very familiar with the Laplace technique, I'll try to learn it and work it out, thanks for your help! $\endgroup$
    – George Lin
    Commented Jun 12 at 14:45
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Firstly, let's recall that the multiplicative operator $x$ is mapped to the derivative $\partial_k$ by the Fourier transform and vice versa. More precisely, one has $\mathscr{F}[xu(x)](\lambda) = i\tilde{u}'(\lambda)$ and $\mathscr{F}[u'(x)](\lambda) = i\lambda\tilde{u}(\lambda)$.

These well-known properties can be proven from the definition of the Fourier transform straightforwardly. Actually, you tried to rederive them in a more complicated and awkward manner with the help of the convolution theorem. Indeed, for instance, the term $x\partial_xu$ became $\delta' * \lambda\tilde{u}$, which can be simplified to $\partial_\lambda(\lambda\tilde{u}) = \tilde{u} + \lambda\partial_\lambda\tilde{u}$ (leaving aside the constant prefactor) thanks an integration by parts or, equivalently, by invoking the identity $\delta' \equiv -\delta\,\partial_\lambda$ and the fact the Dirac delta is the identity element with respect to the convolution product.

In consequence, one understands that Fourier-transforming the initial PDE with respect to the spatial coordinate is a kind of useless move, because the differential operator on the right-hand side will be mapped to itself, or more precisely $x^2\partial_x^2 + x\partial_x \mapsto \partial_\lambda^2\lambda^2 - \partial_\lambda\lambda = \lambda^2\partial_\lambda^2 + \lambda\partial_\lambda + 1$, in such a way that it doesn't help us to solve the equation. That is why I would recommend you to consider the Fourier transform with respect to time only.

Since the problem is defined on $\Bbb{R}_+$ only, it is more appropriate to use the Laplace transform actually, as discussed in the comment section $-$ the procedure would be the same with the Fourier transform anyway, with extra $i$ factors. The initial PDE becomes thus $$ x^2\partial_x^2\hat{u}(x,s) + x\partial_x\hat{u}(x,s) - s\hat{u}(x,s) = -f(x), $$ where $\hat{u}(x,s) = \mathscr{L}[u(x,t)](s)$. It has the advantage to make the initial condition appear thanks to the property $\mathscr{L}[f'(t)](s) = s\hat{f}(s) - f(0)$. Now, we face an inhomogeneous second-order linear differential equation with respect to space only. A particular solution can be derived with the help of Green's function technique, as mentioned by Felix Marin in his answer, or simply thanks to the method of variation of the parameters, but I leave this work to you.

As to the homogeneous problem, it turn out to be a Euler equation. It is traditionally solved through a monomial ansatz, i.e. $\hat{u}(x,s) \propto x^n$, hence the auxiliary equation $n(n-1) + n - s = 0$ in the present case, whose roots are given by $\pm\sqrt{s}$, implying the following homogeneous solutions : $\hat{u}_h(x,s) = Ax^{\sqrt{s}} + Bx^{-\sqrt{s}}$ when $s \neq 0$ and $\hat{u}_h(x,s) = A + B\ln x$ when $s = 0$, with $A$ and $B$. It is to be noted that the change of coordinate $z := \ln x$ permits to rewrite the Euler equation as a standard second-order ODE with constant coefficients for the function $\phi(z) = u(x)$.

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  • $\begingroup$ If I do coordinate transform before Laplace transforms, will the equation become easier? $\endgroup$
    – George Lin
    Commented Jun 12 at 14:46
  • $\begingroup$ Not in that case, because the change of coordinates affects $x$ only, when the Laplace transform acts on $t$. $\endgroup$
    – Abezhiko
    Commented Jun 12 at 18:41

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