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Find the solution $u(x, t)$ to the 1-D heat equation

$$ u_t = c^2u_{xx}$$

with initial conditions

$$ u(x,0) = \begin{cases} \hfill 0 \hfill & \text{ if $x < 0$ } \\ \hfill u_0 \hfill & \text{ if $x > 0$} \end{cases} $$

The final answer should be expressed in terms of the error function, which is defined as:

$$ \frac 2 {\sqrt{\pi}} \int _{-\infty}^\infty e^{-w^2} \, dx$$

So far, I have taken the Fourier transform of the PDE and obtained

$$\frac {du}{dt} = -c^2w^2u$$

Then, I took the Fourier transform of the initial conditions and obtained

$$ \frac {u_0}{iw \sqrt{2\pi}}$$

After this part, I am not sure what do. Can someone clarify if what I have done is correct so far and tell me what I should do next?

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  • $\begingroup$ Which transform are you using? $\endgroup$ – Mattos Jan 29 '17 at 22:24
  • $\begingroup$ Fourier Transform! $\endgroup$ – skancharla Jan 30 '17 at 0:00
  • $\begingroup$ I know that. Which version are you using? $\endgroup$ – Mattos Jan 30 '17 at 3:45
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Solving the differential equation $U_t(w,t) = -c^2 w^2 U(w,t)$ gives \begin{equation} U(w,t) = F(w) e^{-c^2w^2t},\ w\in \mathbb{R},\ t>0 \end{equation} where $F(w)$ is the Fourier Transformation of the initial condition function $f(x)$.

Now we shall use the fact that $\mathcal{F}^{-1}(F\cdot G) = f * g$, where $*$ denotes the convolution of $f$ and $g$. \begin{align} u(x,t) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(z)g(x-z,t)\ dz \\ &= \frac{1}{2c\sqrt{\pi t}}\int_{-\infty}^{\infty}f(z)\exp\left(-\frac{(x-z)^2}{4c^2t}\right)\ dz \\ &=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}f(x-2c\tau\sqrt{t})\ e^{-\tau^2}\ d\tau\\ &= \frac{u_0}{\sqrt{\pi}} \int_{-\infty}^{x/(2c\sqrt{t})}e^{-\tau^2}d\tau\\ &= \frac{u_0}{\sqrt{\pi}} \left(\int_{-\infty}^{0}e^{-\tau^2}d\tau \ +\ \int_{0}^{x/(2c\sqrt{t})}e^{-\tau^2}d\tau \right)\\ &= \frac{u_0}{2} \left(1+erf\left(\frac{x}{2c\sqrt{t}}\right)\right) \end{align} where we used the inverse Fourier transform of $G(w,t) = e^{-c^2w^2t}$ and the transformation $\tau = \frac{x-z}{2c\sqrt{t}}$.

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