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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.

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

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}

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  • $\begingroup$ 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

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