Cauchy problem for diffusion equation and condition $x^2$ if $x\in [0,1]$

Question

Solve the Cauchy problem for diffusion equation $$u_t=u_{xx}, t\ge0,x\in\mathbb R,\\ u(x,0)=\begin{cases}x^2, x\in [0,1]\\0,x\not\in [0,1] \end{cases}$$

Give the solution in terms of $erf(x)=\sqrt{2/\pi}\int_0^xe^{-u^2/2}du$.

We have that the solution is given by $u(x,t) = \frac{1}{\sqrt{4\pi kt}}\int_{0}^1 e^{-(x-y)^2/4kt} y^2 dy$ .

Let $p=\frac{x-y}{\sqrt{4kt}}.$ Then $u(x,t)=\frac{-\sqrt{4kt}}{\sqrt{4\pi kt}}\int_{\frac{x}{\sqrt{4kt}}}^{\frac{x-1}{\sqrt{4kt}}}e^{-p^2}(-\sqrt{4kt}p+x)^2dp$.

From here should I expand the square term $(-\sqrt{4kt}p+x)^2$ and then multiply by $e^{-p^2}$ and then separate the integrals so I would end up having 3 integrals and then solve each of them?

Or what should I do ?

All that I said above sound (and it is in fact) really tedious, maybe I am missing a shortcut to get the solution in term of erf function.

Could someone help please?

Answer 1

One possible way to solve the problem (if it is presented as a show that) is to write $$\eta=\frac1{\sqrt{4\pi kt}},\,\xi=\frac x{\sqrt{4kt}}$$and then look for a separable solution. $$\frac{\partial}{\partial x}=\eta\sqrt\pi\frac{\partial}{\partial\xi}\\\frac{\partial^2}{\partial x^2}=\frac{\partial}{\partial x}\left(\eta\sqrt\pi\frac{\partial}{\partial\xi}\right)=\pi\eta^2\frac{\partial^2}{\partial\xi^2}\\\frac{\partial}{\partial t}=-2\pi k\eta^3\frac{\partial}{\partial\eta}-2k\pi\eta^2\xi\frac{\partial}{\partial \xi}=-2\pi k\eta^2\left(\eta\frac{\partial}{\partial \eta}+\xi\frac{\partial}{\partial\xi}\right)$$ Substituting this in,

$$u_t-u_{xx}=-2\pi k\eta^2(\eta u_\eta+\xi u_\xi)-\pi\eta^2u_{\xi\xi}=0\\\implies2k\eta u_\eta+2k\xi u_\xi+u_{\xi\xi}=0$$Write $u(\xi,\eta)=X(\xi)T(\eta)$.

$$2k\eta XT'+2k\xi X'T+X''T=0\\\implies 2k\xi\frac{X'}X+\frac{X''}X=-2k\eta\frac{T'}T=\text{constant}=c_1$$

Solve for $T$ and $X$, then use the initial condition to get rid of any constants of integration.

E.g., for $T$ you'd get $$\ln T=-\frac {c_1}{2k}\ln\eta+c_2\implies T=c_2\eta^{-c_1/2k}$$

One can also solve $$X''+2k\xi X'-c_1 X=0$$for $X$.

The solution is then $u=X\times T$

Answer 2

We can use Fourier transforms.$$u_t=u_{xx}\\\tilde u_t=-k^2\tilde u\\\tilde u=C(k)e^{-k^2t}$$Here $C(k)=\mathcal F[f(x)]$ where $$f(x)=\begin{cases}x^2&x\in[0,1]\\0&\text{else}\end{cases}$$ So we have that the Fourier transform of $u$ is given by the product of the Fourier transform of some function $f(x)$ described above by the initial conditions, and the function $e^{-k^2 t}$.

Hint: Is this the Fourier transform of another function? If so, then one can use the convolution theorem to write the solution in exactly the form you have given. Using the solution they have given, can you guess what the function you need is, and show that the Fourier transform of it is $e^{-k^2 t}$?