Consider the Heat equation and take the Dirichlet boundary condition :
$$v_t – kv_{xx} = =0 \ \ \ \ \ \ ( 0 < x < \infty, \ \ 0 < t < \infty) ,$$
$$ v(x,0) = \phi (x) \ \ \ \ \ \ for \ \ t = 0 $$
$$ v(0,t) = 0 \ \ \ \ \ \ for \ \ x = 0 $$
we convert this problem on Whole Line $\mathbb R$ by taking odd extension of $\phi$ and Solve by using this formulae
$$u_t = ku_{xx} \ \ \ \ \ \ (-\infty < x < \infty , 0 < t < \infty) \ \ \ \ \ \ \ \ (1)$$
$$u(x,0) = \phi_{odd}(x) $$
$u(x,t) = \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty}e^{-(x-y)^2/4kt} \phi_{odd}(y) dy $
Why have we taken odd extension of $\phi$ on dirichlet boundary condition
Please help me i am confusing about the extension of $\phi$ which depend on the boundary condition
Thank You
Best Answer
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 extend 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_{odd}(y) dy \\ =& \frac{1}{\sqrt{4 \pi t}} \left ( \int^{\infty}_0 e^{\frac{-(x-y)^2}{4Dt} } g(y)dy -\int_{-\infty}^0 e^{\frac{-(x-y)^2}{4Dt} } g(-y) dy \right) \\ = & \frac{1}{\sqrt{4 \pi t}} \left ( \int^{\infty}_0 e^{\frac{-(x-y)^2}{4Dt} } g(y)dy -\int^{\infty}_0 e^{\frac{-(x+y)^2}{4Dt} } g(y) 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)dy \\ \end{align*}