### Author Topic: Halloween Challenge 3  (Read 418 times)

#### Victor Ivrii

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##### Halloween Challenge 3
« on: October 21, 2016, 12:01:48 PM »

#### Roro Sihui Yap

• Full Member
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##### Re: Halloween Challenge 3
« Reply #1 on: October 21, 2016, 04:40:13 PM »
Problem 3j)

u(x,t)=\int_0^\infty G_D(x,y,t)g(y)\,dy+k
\int_0^{t} G_{D\,y}(x,0,t-\tau)h(\tau) \,d\tau;

Since $g(y) = 0$ and h(t)=\left\{\begin{aligned} &1 &&t<1, \\&0 &&t\ge 1.\end{aligned}\right.

u(x,t) =\left\{\begin{aligned} &k\int_0^{t} G_{D\,y}(x,0,t-\tau) \,d\tau &&t<1, \\&k\int_0^{1} G_{D\,y}(x,0,t-\tau) \,d\tau &&t\ge 1.\end{aligned}\right.
$$G_{D\,y}(x,y,t) = \partial_y\big[ \frac{1}{2\sqrt{k\pi t}}\big(e^{-\frac{(x-y)^2}{4kt}} - e^{-\frac{(x+y)^2}{4kt}}\big)\big]$$
$$G_{D\,y}(x,y,t) = \big[ \frac{1}{2\sqrt{k\pi t}}\big(e^{-\frac{(x-y)^2}{4kt}}(\frac{x-y}{2kt}) + e^{-\frac{(x+y)^2}{4kt}}(\frac{x+y}{2kt})\big)\big]$$
$$G_{D\,y}(x,0,t-\tau) = \big[ \frac{1}{2\sqrt{k\pi (t-\tau)}}\big(e^{-\frac{x^2}{4k(t-\tau)}}(\frac{x}{k(t-\tau)})\big) \big] = \big[ \frac{x}{2k^{3/2}\sqrt{\pi }}\big(e^{-\frac{x^2}{4k(t-\tau)}}(\frac{1}{(t-\tau)^{3/2}})\big) \big]$$

Let $z = \frac{x}{2\sqrt {k(t-\tau)}}$
$\frac{dz}{d\tau} = \frac{x}{4\sqrt {k}(t-\tau)^{3/2} }$

u(x,t) =\left\{\begin{aligned} &\frac{2}{\sqrt\pi}\int_\frac{x}{2\sqrt {tk}}^{\infty} e^{-z^2}\,dz &&t<1, \\&\frac{2}{\sqrt\pi}\int_\frac{x}{2\sqrt {tk}}^{\frac{x}{2\sqrt {(t-1)k}}} e^{-z^2}\,dz &&t\ge 1.\end{aligned}\right.
u(x,t) =\left\{\begin{aligned} &1 - erf (\frac{x}{2\sqrt {tk}}) &&t<1, \\& erf (\frac{x}{2\sqrt {(t-1)k}}) - erf (\frac{x}{2\sqrt {tk}}) &&t\ge 1.\end{aligned}\right.

Problem 4j)

u(x,t)=\int_0^\infty G_N(x,y,t)g(y)\,dy-
k \int_0^{t} G_{N}(x,0,t-\tau)h(\tau) \,d\tau

Since $g(y) = 0$ and h(t)=\left\{\begin{aligned} &1 &&t<1, \\&0 &&t\ge 1.\end{aligned}\right.

u(x,t) =\left\{\begin{aligned} &- k \int_0^{t} G_{N}(x,0,t-\tau) \,d\tau &&t<1, \\&- k \int_0^{1} G_{N}(x,0,t-\tau)\,d\tau &&t\ge 1.\end{aligned}\right.

$$G_{N}(x,0,t-\tau) = \big[ \frac{1}{\sqrt{k\pi (t-\tau)}} \ e^{-\frac{x^2}{4k(t-\tau)}} \big]$$

Let $z = \frac{x}{2\sqrt {k(t-\tau)}}$
$\frac{dz}{d\tau} = \frac{x}{4\sqrt {k}(t-\tau)^{3/2} }$

u(x,t) =\left\{\begin{aligned} &\int_\frac{x}{2\sqrt {tk}}^{\infty} \frac{-x}{z^2\sqrt \pi}e^{-z^2}\,dz &&t<1, \\&\int_\frac{x}{2\sqrt {tk}}^{\frac{x}{2\sqrt {(t-1)k}}} \frac{-x}{z^2\sqrt \pi}e^{-z^2}\,dz &&t\ge 1.\end{aligned}\right.

Integrate by parts, let $w =\frac{x}{\sqrt \pi}e^{-z^2}$, $\frac{dw}{dz} =\frac{-2xz}{\sqrt \pi}e^{-z^2}$
let $\frac{dv}{dz} = \frac{-1}{z^2}$, $v = \frac{1}{z}$

u(x,t) =\left\{\begin{aligned} &\frac{x}{z\sqrt \pi}e^{-z^2}\big|_{\frac{x}{2\sqrt {tk}}}^{\infty} + \int_\frac{x}{2\sqrt {tk}}^{\infty} \frac{2x}{\sqrt \pi}e^{-z^2}\,dz &&t<1, \\&\frac{x}{z\sqrt \pi}e^{-z^2}\big|_{\frac{x}{2\sqrt {tk}}}^{\frac{x}{2\sqrt {(t-1)k}}} + \int_\frac{x}{2\sqrt {tk}}^{\frac{x}{2\sqrt {(t-1)k}}} \frac{2x}{\sqrt \pi}e^{-z^2}\,dz &&t\ge 1.\end{aligned}\right.

u(x,t) =\left\{\begin{aligned} &-\frac{2\sqrt{tk}}{\sqrt\pi}e^\frac{-x^2}{4tk}+ x (1- erf(\frac{x}{2\sqrt{tk}})) &&t<1, \\&\frac{2\sqrt{(t-1)k}}{\sqrt\pi}e^\frac{-x^2}{4(t-1)k}-\frac{2\sqrt{tk}}{\sqrt\pi}e^\frac{-x^2}{4tk}+ x (erf(\frac{x}{2\sqrt{(t-1)k}}- erf(\frac{x}{2\sqrt{tk}})) &&t\ge 1.\end{aligned}\right.

« Last Edit: October 21, 2016, 06:30:18 PM by Roro Sihui Yap »