- #1

evinda

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Let $\phi \in C^1(\mathbb{R})$ and periodic.

We consider the problem

$u_t=u_{xx}, x \in \mathbb{R}, \ 0<t<\infty$,

with initial data $\phi$.

I want to compute a formula for a solution $u$ and I want to prove strictly that this formula solves the initial value problem. I also want to show that there is no other solution ($C^1(\mathbb{R})$ and periodic).

I have thought the following.

$u(x,t)=X(x) T(t)$

$u(0,x)=\phi(x)$

$u_t=u_{xx} \Rightarrow X(x) T'(t)=X''(x) T(t) \Rightarrow \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda$

So we have the system

$\left\{\begin{matrix}

X''(x)+\lambda X(x)=0\\

T'(t)+\lambda T(t)=0

\end{matrix}\right.$.

So $T(t)$ is of the form $T(t)=c_1 e^{-\lambda t}$.

The characteristic equation of $X''(x)+\lambda X(x)=0$ is $\mu^2=-\lambda$.

For $\lambda<0$ we get that $X(x)=c_1 e^{\sqrt{-\lambda}x}+c_2 e^{-\sqrt{-\lambda}x}$.

We have that $X(x+T)=c_1 e^{\sqrt{-\lambda}x} e^{\sqrt{-\lambda}T}+c_2 e^{-\sqrt{-\lambda}x}e^{-\sqrt{-\lambda} T}$

This holds only if $ e^{\sqrt{-\lambda}T}=1 \Rightarrow \sqrt{-\lambda}T=0 \Rightarrow T=0$, contradiction.

Am I right so far? (Thinking)