Consider the Heat equation and take the Dirichlet boundary condition:
$$v_t - kv_{xx} = 0 \qquad ( 0 < x < \infty, \ \ 0 < t < \infty) ,$$
$$ v(x,0) = \phi (x) \qquad \text{for } t = 0 $$
$$ v(0,t) = 0 \qquad \text{for } x = 0 $$
We convert this problem over the Whole Line $\mathbb R$ by taking the odd extension of $\phi$ and solve by using this formulae
$$u_t = ku_{xx} \qquad (-\infty < x < \infty , 0 < t < \infty) \tag1$$
$$u(x,0) = \phi_\text{odd}(x) $$ $$u(x,t) = \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty}e^{-(x-y)^2/4kt} \phi_\text{odd}(y) \,\Bbb dy $$
Why have we taken the odd extension of $\phi$ on the Dirichlet boundary condition?
Please help me, I am confused about the extension of $\phi$ which depends on the boundary condition.
Thank you
You’ve taken the odd extension so you can satisfy the boundary condition that $v(0,t)= 0$ for all $x$. You want to use some symmetry of the boundary condition, and the odd system is the obvious choice. Then after you’ve extended to the real line, you can use the Green’s Solution: (I’ve stolen this code from an old assignment, so hopefully the notation difference isn’t confusing)
\begin{align} u(x,t) &= \int_{\mathbb{R}} \Phi (x-y,t) \, g_\text{odd}(y) \,\Bbb dy \\ &= \frac{1}{\sqrt{4 \pi t}} \left ( \int^{\infty}_0 e^{\frac{-(x-y)^2}{4Dt} } g(y) \,\Bbb dy -\int_{-\infty}^0 e^{\frac{-(x-y)^2}{4Dt} } g(-y) \,\Bbb dy \right) \\ &= \frac{1}{\sqrt{4 \pi t}} \left ( \int^{\infty}_0 e^{\frac{-(x-y)^2}{4Dt} } g(y) \,\Bbb dy -\int^{\infty}_0 e^{\frac{-(x+y)^2}{4Dt} } g(y) \,\Bbb dy \right) \quad \text{letting } y \to -y \\ &= \frac{1}{\sqrt{4 \pi t}} \int^{\infty}_0 \left (e^{\frac{-(x-y)^2}{4Dt} } - e^{\frac{-(x+y)^2}{4Dt} } \right) g(y) \,\Bbb dy \end{align}