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

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?

• Expand the square. One term will involve erf, on term will be elementary, and the last term can be converted to an erf term by integrating by parts first. – Ian Jun 24 '18 at 0:09
• @Ian did you do the explicit calculations:)? – Isa Jun 25 '18 at 20:20
• @Ian I expanded the square, and I don't see the elementary term. What do you mean by elementary? – Isa Jun 25 '18 at 20:25
• $\int e^{-p^2} p dp$ is an elementary integral (i.e. you can solve it in elementary functions). $\int e^{-p^2} p^2 dp$ and $\int e^{-p^2} dp$ need to be written using erf. – Ian Jun 25 '18 at 20:26
• @Ian ok, it's solved using $u=p^2,du=2pdp,$ right and $\int e^{-p^2} p^2 dp$ is the one solved by parts? How this integral $\int_{\frac{x}{\sqrt{4kt}}}^{\frac{x-1}{\sqrt{4kt}}}e^{-p^2}dp$ can be converted to an erf function if the limits are not symmetric? – Isa Jun 25 '18 at 20:29

## 2 Answers

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 $u=X\times T$

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 $e^{-k^2 t}$?

• :( I do not know Fourier transforms, my instructor did not taught us – Isa Jun 24 '18 at 0:23
• only Fourier series but that's different isn't? – Isa Jun 24 '18 at 0:26
• Yes, Fourier series are different. It seems to me that this question is obviously a question about Fourier Transforms. Regardless an alternative is to do separation of variables using some newly defined variables. But this relies heavily on you knowing what the answer should be (to define these new variables). I'll write a different answer using that method – John Doe Jun 24 '18 at 0:29