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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

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  • $\begingroup$ Odd functions automatically vanish at the origin. The Dirichlet boundary condition is automatically satisfied. Even extensions can be used for Neumann boundary conditions. $\endgroup$ – suresh May 9 '14 at 2:36
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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*}

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  • $\begingroup$ @ Jeb : Can we tak even extension of $\phi$ $\endgroup$ – user120386 May 9 '14 at 2:40
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    $\begingroup$ No, otherwise the Boundary Condition won't be satisfied. $\endgroup$ – Jeb May 9 '14 at 2:43
  • $\begingroup$ see my comment above -- even extensions will be incompatible with your boundary condition. $\endgroup$ – suresh May 9 '14 at 2:43
  • $\begingroup$ Thank you so much jeb and Suresh for help $\endgroup$ – user120386 May 9 '14 at 3:01

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