Heat equation in infinite space

[mathmari]

Hey!

I have to solve the following problem:
$$u_t=u_{xx}, x \in \mathbb{R}, t>0$$
$$u(x,0)=f(x)=H(x)=\left\{\begin{matrix}
1, x>0\\
0, x<0
\end{matrix}\right.$$

I have done the following:

We use the method separation of variables, $u(x,t)=X(x)T(t)$.

I have found that the eigenfunctions are $X_k(x)=e^{ikx}, \lambda=k^2, k \in \mathbb{R}$

$T_k(t)=e^{-k^2t}$

$$u(x,t)=\frac{1}{2\pi} \int_{-\infty}^{+\infty}{\widetilde{f}(k) e^{ikx} e^{-k^2t}}dk$$

$$u(x,0)=H(x) \Rightarrow H(x)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{ikx}}dk$$

$$\widetilde{f}(k)=\int_0^{\infty}{e^{-ikx}}dx=\frac{1}{ik}$$

$$u(x,t)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\frac{1}{ik} e^{ikx} e^{-k^2t}}dk$$

To find the last integral,I thought the following:

$$u_x= \frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^2}{4t}}$$

$$u(x,t)=\int_{-\infty}^{+\infty} \frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^2}{4t}} dx=\frac{1}{\sqrt{4 \pi t}} \int_{-\infty}^{\infty}{ e^{-\frac{x^2}{4t}}}=1$$
since:

$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-\frac{x^2}{4t}}e^{-\frac{y^2}{4t}}dxdy= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-\frac{x^2+y^2}{4t}}=\int_{0}^{+\infty} \int_{0}^{2 \pi} e^{\frac{-r^2}{4t}}r d \theta dr=\int_{0}^{+\infty} 2 \pi e^{\frac{-r^2}{4t}}r dr=- \pi [4t e^{\frac{-r^2}{4t}}]_0^{+\infty}=4t \pi \Rightarrow \int_{\infty}^{+\infty} e^{\frac{-x^2}{4t}}=\sqrt{4t \pi}$$

But the solution $u(x,t)=1$ does not satisfy the conditions. Where is my mistake? (Wondering)

 

[I like Serena]

Hi! :D

$$\widetilde{f}(k)=\int_0^{\infty}{e^{-ikx}}dx=\frac{1}{ik}$$

This integral is only convergent if $k>0$.
If for instance $k=0$ you'll need to find a different approach.

$$u(x,t)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\frac{1}{ik} e^{ikx} e^{-k^2t}}dk$$

To find the last integral,I thought the following:

$$u_x= \frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^2}{4t}}$$

Where did you get that formula? (Confused)

$$u(x,t)=\int_{-\infty}^{+\infty} \frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^2}{4t}} dx=\frac{1}{\sqrt{4 \pi t}} \int_{-\infty}^{\infty}{ e^{-\frac{x^2}{4t}}}=1$$
since:

$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-\frac{x^2}{4t}}e^{-\frac{y^2}{4t}}dxdy= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-\frac{x^2+y^2}{4t}}=\int_{0}^{+\infty} \int_{0}^{2 \pi} e^{\frac{-r^2}{4t}}r d \theta dr=\int_{0}^{+\infty} 2 \pi e^{\frac{-r^2}{4t}}r dr=- \pi [4t e^{\frac{-r^2}{4t}}]_0^{+\infty}=4t \pi \Rightarrow \int_{\infty}^{+\infty} e^{\frac{-x^2}{4t}}=\sqrt{4t \pi}$$

But the solution $u(x,t)=1$ does not satisfy the conditions. Where is my mistake? (Wondering)

The solution $u(x,t)=1$ follows from $k=0$, which is a special case.
Take another look at your derivation to see what happens if $k=0$.

 

[mathmari]

Hi! :D
This integral is only convergent if $k>0$.
If for instance $k=0$ you'll need to find a different approach.

Where did you get that formula? (Confused)

The solution $u(x,t)=1$ follows from $k=0$, which is a special case.
Take another look at your derivation to see what happens if $k=0$.

I found that the solution of the problem has the following form:
$$u(x,t)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\frac{1}{ik} e^{ikx}e^{-k^2t}}dk$$

Then, I derivated this in respect to $x$:
$$u_x=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\frac{1}{ik} e^{ikx}e^{-k^2t}}dk$$

Then, from the formula:
$$\int_{-\infty}^{+\infty}{e^{ikb}e^{-k^2a}}dk=\sqrt{\frac{\pi}{a}}e^{-\frac{b^2}{4a}}$$

we get the following:
$$u_x=\frac{1}{2 \pi} \sqrt{\frac{\pi}{t}}e^{-\frac{x^2}{4t}}=\frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^2}{4t}}$$

So to find $u(x,t)$, I have to calculate the integral $\displaystyle{\int u_x dx}$ but with what limits?? (Wondering)