I have the given problem:
A very long pipe is understood simplistically as semi-infinite ( $0<x<\infty$ ). The tube initially contains clean water and is closed. It is lowered into a container with polluted water. At $t=0$ it is opened at the end at $x=0$. Contaminated water then diffuses into the pipe. The concentration of the polluted substance at $x=0$ is then assumed to be $q_0$. a) Set up a mathematical model for the concentration q(x,t) in the tube of the substance as a function of position and time. b) Solve this math problem.
This problem is clearly the diffusion equation. The answer to a) is:
\begin{equation} \alpha u_{xx}=u_t \ \ \ \ \ \ 0\leq x<\infty \\ u(0,t)=q_0\\ u(x,0)=0 \end{equation}
The answer to b), can be solved by using convolution formula:
\begin{equation} u(x,t)=\int_{-\infty}^\infty S(x-y,t)\phi(y)dy=\frac{1}{\sqrt{4\alpha\pi t}}\int_{-\infty}^\infty e^{\frac{-(x-y)^2}{4\alpha t}}\phi(y)dy \end{equation}
Since the convolution formula is for the infinite domain, we define $u(x)$ on an infinite domain using the reflection method:
\begin{equation} \phi(y)=\begin{cases} u(x)\ \ \ \ \ x>0 \\ u(-x) \ \ \ \ x<0 \end{cases} \end{equation}
Then we evaluate $u(x,t)=\phi(x,t)_{-\infty}^\infty$ and modify accordingly to include only the semi-finite domain. However, this gives:
\begin{equation} u(x,t)=\frac{1}{2\sqrt{\pi t}}\int_{-\infty}^\infty e^{-(x-y)^2/4t}q_0dy \end{equation}
Trying to solve this using the substitution $p=\frac{-(x-y)}{2\sqrt{t}}$, $dy=-2\sqrt{t}dp$ we get:
\begin{equation} u(x,t)=-\frac{q_02\sqrt{t}}{2\sqrt{\pi t}}\int_{-\infty}^\infty e^{p^2}dp\\ u(x,t)=-\frac{q_0}{\sqrt{\pi}}\int_{-\infty}^\infty e^{p^2}dp\\ u(x,t)=-\frac{q_0}{\sqrt{\pi}}\bigg[\int_{-\infty}^0 e^{p^2}dp+\int_{0}^\infty e^{p^2}dp\bigg]\\\\ \end{equation}
Since the original function is defined only on the latter domain, we can omit the integral to the left, and get:
\begin{equation} u(x,t)=-\frac{q_0}{\sqrt{\pi}}\bigg[\int_{0}^\infty e^{p^2}dp\bigg]\\ u(x,t)=-\frac{q_0}{2} \end{equation}
But this is clearly not right.
Is there a way to use this approach and get the correct answer?
Thanks
UPDATE:
Following K.defaoite's suggestion we use:
\begin{equation} \int_{0}^\infty \Phi(x-\xi,t)f(\xi)\mathrm d\xi+\int_0^t\int_{0}^\infty\Phi(x-\xi,t-\tau)h(\xi,\tau)\mathrm d\xi\mathrm d\tau \tag{3} \end{equation}
which gives with $f(\xi)=q_0$ and with $p=\frac{x-\xi}{2\sqrt{t}}$:
\begin{equation} \int_{0}^\infty \Phi(x-\xi,t)f(\xi)\mathrm d\xi=\int_0^\infty e^{-p^2}dp=\frac{\sqrt{\pi}}{2} \end{equation}
Then for the second integral, we set $p=\frac{x-\xi}{\sqrt{4(t-\tau)}}$, and still $f(\xi)=q_0$:
\begin{equation} \int_0^t\int_{0}^\infty\Phi(x-\xi,t-\tau)h(\xi,\tau)\mathrm d\xi\mathrm d\tau =-\int_0^t \frac{\sqrt{4(t-\tau)}}{\sqrt{4\pi t}} \int_{0}^\infty e^{-p^2}dpd\tau \end{equation}
since $\int_{0}^\infty e^{-p^2}dp=\frac{\sqrt{\pi}}{2}$, we get:
\begin{equation} -\frac{\sqrt{\pi}}{2}\int_0^t \frac{\sqrt{4(t-\tau)}}{\sqrt{4\pi t}} d\tau=-\frac{t}{3} \end{equation}
Summing up with the first integral above, we get:
\begin{equation} u(x,t)=\frac{\sqrt{\pi}}{2}-\frac{t}{3} \end{equation} However, this is also wrong, as it can be inserted in the PDE and gives an incorrect answer.
Where is the error here?
Thanks
Full solution.
Referring to my other answer, we let $p(t)=q_0$ and let $u(x,0)=\phi(x)=0$. We then let $$U(x,t)=u(x,t)-p(t)=u(x,t)-q_0$$ Which, you can check, satisfies $$\begin{cases}\partial_t U-k\partial_x^2U=0 \\ U(0,t)=0 \\ U(x,0)=-q_0\equiv \varphi(x)\end{cases}$$ We now extend this to the whole real line $x\in (-\infty,\infty)$ by constructing the odd extension of $\varphi(x)$: $$\varphi_{\text{ext}}(x)=\begin{cases}\varphi(x)=-q_0 & x\geq 0 \\ -\varphi(-x)=q_0 & x<0\end{cases}$$ We then solve $$\partial_tU-k\partial_x^2U=0 ~~~~~x\in\Bbb R,t\in[0,\infty) \\ U(x,0)=\varphi_{\text{ext}}(x)$$ We convolve with the heat kernel: $$U(x,t)=\int_{\Bbb R}\frac{\exp\left(\frac{-(x-\xi)^2}{4kt}\right)}{\sqrt{4\pi kt}}\varphi_{\text{ext}}(\xi)\mathrm d\xi$$ Finally, $$u(x,t)=U(x,t)+q_0=q_0+\int_{\Bbb R}\frac{\exp\left(\frac{-(x-\xi)^2}{4kt}\right)}{\sqrt{4\pi kt}}\varphi_{\text{ext}}(\xi)\mathrm d\xi$$
Shown below is a few snapshots of the solution at $t=10^{-1},10^0,10^1,10^2$, taking $k=q_0=1$, plotted in $x\in[0,10]$.
As we expect the heat traverses down the pipe.
In terms of constructing the solution analytically, this boils down finding a closed form for the integral $$\int_{\Bbb R}\Phi(x-\xi,t)\varphi_{\text{ext}}(\xi)\mathrm d\xi =\int_{\Bbb R}\Phi(\xi,t)\varphi_{\text{ext}}(x-\xi)\mathrm d\xi \\ $$ We can write this as $$\int_{-\infty}^x \Phi(\xi,t)\varphi_{\text{ext}}(x-\xi)\mathrm d\xi + \int_{x}^\infty \Phi(\xi,t)\varphi_{\text{ext}}(x-\xi)\mathrm d\xi \\ =-q_0\int_{-\infty}^x \Phi(\xi,t)\mathrm d\xi +q_0\int_{x}^\infty \Phi(\xi,t)\mathrm d\xi \\ =q_0\left(\int_x^\infty \Phi(\xi,t)\mathrm d\xi -\int_{-\infty}^x \Phi(\xi,t)\mathrm d\xi\right) \\ =-q_0\operatorname{erf}\left(\frac{x}{\sqrt{4kt}}\right)$$ Hence $$\boxed{u(x,t)=q_0\left(1-\operatorname{erf}\left(\frac{x}{\sqrt{4kt}}\right)\right)}$$
If you want to play with this solution you can do so here.