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  1. MHB Master
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    #1
    Hello!!!

    We consider the following Cauchy problem

    $u_t=u_{xx} \text{ in } (0,T) \times \mathbb{R} \\ u(0,x)=\phi(x) \text{ where } \phi(x)=-\phi(-x), x \in \mathbb{R} $

    I want to show that $ u(t,0)=0, \forall t \geq 0 $.

    We have the following theorem:

    Let $\phi \in C^0(\mathbb{R}^n)$ and bounded. Then

    $u(t,x)=\int_{\mathbb{R}^n} \Gamma (t,x-\xi) \phi(\xi)d{\xi} $

    is the solution of the problem

    $ u_t-\Delta u=0 \text{ in } (0,T) \times \mathbb{R}^n, T>0 \\ u(0,x)=\phi(x), x \in \mathbb{R}^n $.

    From this we have that the solution of the given problem is

    $ u(t,x)=\int_{\mathbb{R}} \Gamma(t,x-\xi) \phi(\xi) d{\xi}=-\int_{\mathbb{R}} \Gamma(t,x-\xi) \phi(-\xi) d{\xi}=\int_{\mathbb{R}}\Gamma(t,x+u) \phi(u) du$

    So we have that $u(t,0)=\int_{\mathbb{R}} \Gamma(t,u) \phi(u)du$.

    How can we show that the latter is equal to 0?

  2. MHB Apprentice

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    #2
    Hi evinda,

    If I'm not mistaken, $\Gamma(t,u)$ is an even function of $u$. Combining this with the oddness of $\phi(u)$ should do the trick.

  3. MHB Master
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    #3 Thread Author
    Yes, we have that $\Gamma (t, x-\xi)=\frac{1}{2^n [\pi t]^{\frac{n}{2}}} e^{\frac{|x-\xi|^2}{4t}}$.

    So we have that

    $$u(t,x)=\int_{\mathbb{R}} \Gamma (t,x-\xi) \phi(\xi) d{\xi}$$

    For $x=0$: $u(t,0)=\int_{\mathbb{R}} \Gamma (t, -\xi) \phi(\xi) d{\xi}=\int_{\mathbb{R}^n} \Gamma(t,\xi) (-\phi(-\xi)) d{\xi}=-\int_{\mathbb{R}} \Gamma(t,\xi) \phi(\xi) d{\xi}$

    So we have that $\int_{\mathbb{R}} \Gamma(t,-\xi) \phi(\xi) d{\xi}=\int_{\mathbb{R}} \Gamma(t,\xi) \phi(\xi) d{\xi}=-\int_{\mathbb{R}} \Gamma(t,\xi) \phi(\xi) d{\xi} \Rightarrow \int_{\mathbb{R}} \Gamma(t,\xi) \phi(\xi) d{\xi}=0$.

    Right?

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