I'm trying to solve the following Cauchy problem in ${\rm \Bbb R}$ without using the fundamental solution. $$ \begin{cases} u_t = u_{xx} &\text{ for }\;\,(x,t)\in\Bbb R\times\{ 0<t<\infty\}\\ u|_{t=0}=g, \end{cases} $$ where $g(x)$ is defined by $$ g(x)= \begin{cases} 0, & x < 0 \\ 1, & x > 0 \\ \frac{1}{2} & x=0 \end{cases} $$ I have a hint to look for a solution in the form $u(x,t)=\phi\big(\frac{x}{\sqrt{t}}\big)$, but I don't know how to apply this hint and get started! Any help, or skeleton of a solution would be appreciated!

Edit: I think we can use the hint to write $\phi$ as an ODE, but then it would be a function of both x and t, so I don't know if this would help.

  • $\begingroup$ g(x) is defined for all x in the question $\endgroup$ – chilin Feb 19 at 21:10
  • $\begingroup$ Did you manage to derive the ODE in $\phi$ and solve it? $\endgroup$ – mattos Feb 20 at 8:51
  • $\begingroup$ $\phi '' + \frac{x}{2\sqrt{t}} \phi ' = 0$ would be the ODE, but then $\phi$ depends on both x and t (i.e we have $\phi(\frac{x}{\sqrt{t}})$), can you just let $v=\frac{x}{\sqrt{t}}$ and then solve as a regular ODE? $\endgroup$ – chilin Feb 20 at 16:41
  • $\begingroup$ Yes, setting $v = x/\sqrt{t}$ and solving the problem as an ODE in $v$ is exactly what you should do. $\endgroup$ – mattos Feb 21 at 0:13

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\partiald{\mrm{u}\pars{x,t}}{t} = \partiald[2]{\mrm{u}\pars{x,t}}{x}.\qquad\mrm{u}\pars{x,0} = \mrm{g}\pars{x} \equiv \left\{\begin{array}{lrcl} \ds{0\,,} & \ds{x} & \ds{<} & \ds{0} \\ \ds{{1 \over 2}\,,} & \ds{x} & \ds{=} & \ds{0} \\ \ds{1\,,} & \ds{x} & \ds{>} & \ds{0} \end{array}\right.}$.

Hint: \begin{align} \int_{0}^{\infty}\partiald{\mrm{u}\pars{x,t}}{t}\expo{-st}\dd t & = \int_{0}^{\infty}\partiald[2]{\mrm{u}\pars{x,t}}{x}\expo{-st}\dd t \\[5mm] -\mrm{g}\pars{x} +s\hat{\mrm{u}}\pars{x,s}& = \partiald[2]{\hat{\mrm{u}}\pars{x,t}}{x}\,,\quad \left\{\begin{array}{rcl} \ds{\hat{\mrm{u}}\pars{x,s}} & \ds{\equiv} & \ds{\int_{0}^{\infty}\mrm{u}\pars{x,t}\expo{-st}\dd t} \\[2mm] \ds{\mrm{u}\pars{x,t}} & \ds{=} & \ds{\int_{c - \infty\ic}^{c + \infty\ic}}\hat{\mrm{u}}\pars{x,s}\expo{ts}{\dd s \over 2\pi\ic} \end{array}\right. \end{align}

  • $\begingroup$ what is the $i$ that appears in your bounds and with the $2\pi i $? $\endgroup$ – chilin Feb 20 at 16:53
  • $\begingroup$ @chilin $\displaystyle \mathrm{i} = \,\sqrt{\,{-1}\,}\,$. $\endgroup$ – Felix Marin Feb 20 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.