### Author Topic: HA3 problem 1  (Read 2215 times)

#### Victor Ivrii

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##### HA3 problem 1
« on: February 05, 2015, 07:23:36 PM »
Using method of continuation obtain formula similar to (1)--(3) of http://www.math.toronto.edu/courses/apm346h1/20151/HA3.html for solution of IBVP for a heat equation on ${x>0,t>0}$ with the initial function $g(x)$ and with

a.  Dirichlet boundary condition $u|_{x=0}=0$;
b.  Neumann boundary condition $u_x|_{x=0}=0$;

#### Yiyun Liu

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##### Re: HA3 problem 1
« Reply #1 on: February 05, 2015, 09:02:10 PM »
Part(a)
Dirichlet boundary
\begin{array}{l}
{u_t} = k{u_{xx}},  x > 0,t > 0\\
u{|_{t = 0}} = g(x)\\
u{|_{x = 0}} = 0\\
u(x,t) = \int\limits_0^\infty  {G(x - y,t)g(y)dy - \int\limits_0^\infty  {G(x + y,t)g(y)dy} } \\
{\rm{           = }}\frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {{e^{\frac{{ - {{(x - y)}^2}}}{{4kt}}}}} g(y)dy - \frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {{e^{\frac{{ - {{(x + y)}^2}}}{{4kt}}}}} g(y)dy\\
{\rm{          = }}\frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {({e^{\frac{{ - {{(x - y)}^2}}}{{4kt}}}} - {e^{\frac{{ - {{(x + y)}^2}}}{{4kt}}}}) }g(y)dy\\

\end{array}

part(b):
Neumann boundary
\begin{array}{l}
{u_t} = k{u_{xx}},  x > 0,t > 0\\
u{|_{t = 0}} = g(x)\\
{u_x}{|_{x = 0}} = 0\\
u(x,t) = \int\limits_0^\infty  {G(x - y,t)g(y)dy + \int\limits_0^\infty  {G(x + y,t)g(y)dy} } \\
{\rm{           = }}\frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {{e^{\frac{{ - {{(x - y)}^2}}}{{4kt}}}}} g(y)dy + \frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {{e^{\frac{{ - {{(x + y)}^2}}}{{4kt}}}}} g(y)dy\\
{\rm{          = }}\frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {({e^{\frac{{ - {{(x - y)}^2}}}{{4kt}}}} + {e^{\frac{{ - {{(x + y)}^2}}}{{4kt}}}}} )g(y)dy\\
\\

\end{array}