The equation$$ u_t = \frac{1}{4} u_{xx}, \; \; \; (x,t) \in \Omega = (- \infty,
\infty) \times (0, \infty) $$can be solved using Fourier transform and the solution is
$$ u(x,t) = \frac{1}{\sqrt{ \pi t } } \int\limits_{-\infty}^{\infty}
g(y) \mathrm{e}^{ – (x-y)^2/ t } \mathrm{d} y $$What is the solution for initial condition $u(x,0) = \delta(x+1)$?
Where $\delta(x)$ is the Dirac delta function
thought:
We know $g(x) = u(x,0)$ is the initial condition. I know $\delta(x+1) = \infty$ when $x=-1$ and zero otherwise. So, if I plug this into my solution I get something of the form
$$ u(x,t) = \frac{1}{\sqrt{\pi t} }* (\infty ) * \mathrm{e}^{ – (x+1)^2/ t }$$
How to simplify such expression?
Best Answer
In your statement, I think you mean that the function $g(y) = u(y,0)$ contains the initial conditions. Then to answer your question, you need to perform the integral with $g(y) = \delta(y+1)$. This can be done quite easily since integration against a $\delta$ simply evaluates the integrand at the support of the $\delta$. So in your case, we would have $u(x,t) = \frac{1}{\sqrt{\pi t}} e^{-(x+1)^{2}/t}$, and there is no need for the "infinity" that you write above. (Edit: I completely agree with the comment by Winther, you should probably not think of the delta function as being "equal to infinity" at the origin, just use the integral property that Winther stated as the defining property of the delta function.)
Edit to answer your comment: If we replace the $\delta(x-1)$ by a Heaviside $H(x-1)$ then since $$H(x-1) = \cases{0,\,\text{for } x<1,\\ 1,\,\text{for } x>1,}$$ we need to evaluate the integral $$u(x,t)=\frac{1}{\sqrt{\pi t}}\int_{1}^{\infty}e^{-(x-y)^{2}/t} \,\mathrm{d}y.$$ I think you should be able to do this in terms of the error function.