Solve Bounded $u_t=u_xx$ Let $u_t=u_xx,\,t>0,\,x\in\mathbb R$

[Markov2]

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]

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 \(t>0\).

 

[Markov2]

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]

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.

 

[blue_raver22]

Yes, your approach is correct. Applying Fourier transform to the PDE gives $\frac{\partial \mathcal{F}(u)}{\partial t} = -w^2\mathcal{F}(u)$, which has the solution $\mathcal{F}(u)(w,t) = ce^{-w^2t}$. Plugging in the initial condition $\mathcal{F}(u)(w,0) = c = xe^{-|x|}$, we get $\mathcal{F}(u)(w,t) = xe^{-|x|}e^{-w^2t}$.

To find the constant $K$, we can use the fact that the Fourier transform is a unitary operator, meaning it preserves the $L^2$ norm. So we have $\|u(x,t)\|_{L^2} = \|\mathcal{F}(u)(w,t)\|_{L^2}$. Using the Parseval's identity, we have
$$\|u(x,t)\|_{L^2} = \|\mathcal{F}(u)(w,t)\|_{L^2} = \left(\int_{-\infty}^{\infty}|\mathcal{F}(u)(w,t)|^2dw\right)^{1/2} = \left(\int_{-\infty}^{\infty}|xe^{-|x|}e^{-w^2t}|^2dw\right)^{1/2} = \|xe^{-|x|}\|_{L^2}.$$

Using the fact that $\|xe^{-|x|}\|_{L^2} = \sqrt{\frac{2}{\pi}}$, we have $\|u(x,t)\|_{L^2} = \sqrt{\frac{2}{\pi}}$. Therefore, we can choose $K = \sqrt{\frac{2}{\pi}}$ and we have $|u(x,t)| \leq \frac{K}{\sqrt{t}}$ for all $t>0$ and $x\in\mathbb{R}$, as required.