Bounded solution

Markov

Member
Let $u_t=u_xx,\,t>0,\,x\in\mathbb R$ and $u(x,0)=xe^{-|x|}.$ Show that $|u(x,t)|\le \dfrac K{\sqrt t}$ for all $t>0$ and $x\in\mathbb R$ where $K$ is a constant.

So I apply Fourier transform, then $\mathcal F(u_t)=\mathcal F(u_xx)$ then $\dfrac{{\partial \mathcal F(u)(w,t)}}{{\partial t}} = - {w^2}\mathcal F(u)(w,t)$ so $\mathcal F(u)(w,t)=ce^{-w^2t}$ then $\mathcal F(u)(w,0)=c=xe^{-|x|},$ is the initial condition well put? I'm not sure really, I'm confused.

Jose27

New member
Since you have a representation formula for $$u$$, everything's easier. You have

$$u(x,t)= \frac{1}{\sqrt{4\pi t} } \int_{\mathbb{R}} e^{\frac{(x-y)^2}{4t}}g(y)dy$$

This gives, using Hölder's inequality with $$p=1, p'=\infty$$

$$|u(x,t)|\leq \frac{1}{\sqrt{4\pi t} } \int_{\mathbb{R}} |g(y)|dy$$

since $$\sup_x e^{\frac{(x)^2}{4t}} = 1$$ for all $$t>0$$.

Markov

Member
Could you please show me how did you find $u(x,t)$ ? I'm stuck on the initial condition, don't know if I did it right, can you check?

Jose27

New member
I won't write out everything, since apparently my definition of the Fourier transform is different (non-essentially though) from yours. You should arrive at an equation of the form $$\hat{u}_t (x,t)= c(x)\hat{u}$$ ($$c(x)$$ will depend on your definition of the transform) with the initial condition $$u(x,0)=\hat{f}(x)$$ (where $$f$$ is your initial condition). This has a solution given by $$\hat{u}(x,t)=\hat{f}(x) e^{c(x)t}$$. Put the exponential as the Fourier transform of someone and apply the result that says the transform of a convolution is the product of the transforms along with the inversion theorem. This should give you my formula.

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