Solving Partial Differential Equations with Fourier transforms

Question

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Question

$ \text{1.* Use the Fourier transform to solve the initial value problem}\\ \left\{\begin{array}{l} u_{t}-u_{x x}+u_{x}=0, x \in \mathbb{R}, t>0, \\ u(x, 0)=g(x), x \in \mathbb{R}, \text { where } g \in \mathscr{L}^{1}(\mathbb{R}) \end{array}\right. $

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My Solution

$ \text { Note: }\left(\left(\frac{\partial}{\partial x_{j}}\right)^{\alpha_{j}} u\right)^{\wedge}(\xi)=(i \xi)^{\alpha_{j}} \hat{u}(\xi) \\ \Rightarrow\left(u_{t}-u_{xx}+u_{x}\right)^{\wedge} \\ \Rightarrow \quad \partial_{t} \hat{u}=-i \xi \hat{u}-\xi^{2} \hat{u}=\hat{u}(\xi)\left[-i \xi-\xi^{2}\right] \\ \Rightarrow \int_{\hat{g}(\xi)}^{t} \frac{1}{\hat{v}(\xi)} d \hat{v}(\xi)=-\left(i \xi+\xi^{2}\right) \int_{0}^{t} d s $

$ \Rightarrow \hat{u}(\xi)=\hat{g}(\xi) e^{-\left(i \xi+\xi^{2}\right) t} \cdot \text { Hence, we have } \ u(x, t)=\left(\hat{g}(\xi) e^{-\left(i \xi+\xi^{2}\right) t}\right)\check{} $

$ \text{Note, the convolution} \ \ g * f(x)=\int_{\mathbb{R}} g(x-y) f(y) d y, \\ \text { with }(g * f(x))^{\wedge}=\sqrt{2 \pi} \hat{g}(\xi) \hat{f}(\xi) \Rightarrow u(x, t)=\left(\frac{1}{\sqrt{2} \pi}(g * f(x))^{\wedge}\right)\check{}=\frac{1}{\sqrt{2 \pi}} g * f(x) $

$\therefore u(x, t)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} g(x-y) f(y) d y$ $=\frac{1}{2 \pi} \int_{\mathbb{R}} g(x-y) \int_{\mathbb{R}} e^{i y \xi} e^{-\left(i \xi+\xi^{2}\right) t} d \xi d y$

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This is as far as I got and I don't really know what to do beyond this point. I've had ideas of switching the integrals around a little. Another idea I had was rewriting the last integral as $\frac{1}{2 \pi} \int_{\mathbb{R}} g(x-y) \int_{\mathbb{R}} e^{i(y-t) \xi} e^{-t \xi^{2}} d \xi d y$ and using the fact that for $f(x)=e^{-\frac{\alpha}{2}|x|^{2}}$, this gives $\check{\mathcal{f}}(\xi)=\frac{1}{(\alpha)^{\frac{n}{2}}} e^{-\frac{|\xi|^{2}}{2 \alpha}}$. Any help would be greatly appreciated.

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Answer 1

Use the Fourier transform to solve the initial value problem

$ \\ \left\{\begin{array}{l} u_{t}-u_{xx}+u_{x}=0, x \in \mathbb{R}, t>0, \\ u(x, 0)=g(x), x \in \mathbb{R}. \end{array}\right.\\ \\ \text { where } g \in \mathscr{L}^{1}(\mathbb{R})\\ $

Solution:

Note:

$ \left(\left(\frac{\partial}{\partial x_{j}}\right)^{\alpha_{j}} u\right)^{\wedge}(\xi)=(i \xi)^{\alpha_{j}} \hat{u}(\xi) \Rightarrow\left(u_{t}-u_{xx}+u_{x}\right)^{\wedge} \\ $

$ \Rightarrow \partial_{t} \hat{u}=-i \xi \hat{u}-\xi^{2} \hat{u}=\hat{u}(\xi)\left[-i \xi-\xi^{2}\right]\\ $

$ \Rightarrow \int_{\hat{g}(\xi)}^{t} \frac{1}{\hat{v}(\xi)} d \hat{v}(\xi)=-\left(i \xi+\xi^{2}\right) \int_{0}^{t} d s\\ $

$ \Rightarrow \hat{u}(\xi)=\hat{g}(\xi) e^{-\left(i \xi+\xi^{2}\right) t}.\\ $

Hence, we have

$ u(x, t)=\left(\hat{g}(\xi) e^{-\left(i \xi+\xi^{2}\right) t}\right)\check{}.\\ $

Note, the convolution gives the following

$ u(x, t)=\left(\hat{g}(\xi) e^{-i \xi t} \cdot e^{-\xi^{2} t}\right)\check{}.\\ $

Let $ \hat{\phi}(\xi)=\hat{g}(\xi) e^{-i \xi t} \textbf{ and }\hat{\psi}(\xi)=e^{-\xi^{2} t}\\ $

$ \Rightarrow(\phi * \psi)^{\wedge}(\xi)=\sqrt{2 \pi} \hat{\phi}(\xi) \hat{\psi}(\xi)\\ $

$ \Rightarrow u(x, t)=\left(\frac{(\phi * \psi)^{\wedge}(\xi)}{\sqrt{2 \pi}}\right)\check{} =\frac{1}{\sqrt{2 \pi}}(\phi * \psi)(x)\\ \\ $

$ \textbf{We now need to find } \phi \textbf{ and } \psi...\\ $

First, lets find $\phi$ :

$ \phi(x)=\int_{\mathbb{R}} \hat{g}(\xi) e^{-i \xi t} e^{i \xi x} d \xi =\int_{\mathbb{R}} \hat{g}(\xi) e^{i(x-t) \xi} d \xi=g(x-t).\\ $

Now for $\psi$ :

We know for

$f=e^{-\frac{\alpha}{2}(x)^{2}} \Rightarrow f(\xi)=\frac{1}{(\alpha)^{\frac{n}{2}}} e^{-\frac{\xi^{2}}{2 \alpha}}.\\ $

$ \operatorname{Let} \alpha=2 t \\ $

$ \Rightarrow \psi(x)=\frac{1}{\sqrt{2 t}} e^{-\frac{x^{2}}{4 t}}\\ $

$ \therefore u(x, t)=\frac{1}{\sqrt{2 \pi}}(\phi * \psi)(x) \\ $

$ =\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} \phi(x-\tilde{y}) \psi(\tilde{y}) d \tilde{y} \\ $

$ =\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} g(x-\tilde{y}-t) \cdot \frac{1}{\sqrt{2 t}} e^{-\frac{\tilde{y}^2}{4 t}} d \tilde{y}\\ $

Let $\tilde{y}=y+t ; \quad d \tilde{y}=d y$. So we get

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

$ =\int_{\mathbb{R}} g(x-y) \cdot \frac{1}{\sqrt{4 \pi t}} e^{-\frac{\left(y-t\right)^{2}}{4 t}} d y \\ $

$ \equiv\left[g(x) * \frac{1}{\sqrt{4 \pi t}} e^{-\frac{(x-t)^{2}}{4 t}}\right](x, t) $

Answer 2

Taking the Fourier transform in the $x$ coordinate of the PDE gives $\hat{u}_{t}+\xi^2\hat{u}+i\xi\hat{u}=0$ i.e. $\hat{u}_{t} + (\xi^2+i\xi)\hat{u}=0,$ so $$ \hat{u}(\xi,t) = C(\xi) e^{-(\xi^2+i\xi)t}. $$ At $t=0$ we have $\hat{u}=\hat{g}$ so $C=\hat{g}.$ Thus, $$ \hat{u}(\xi,t) = \hat{g}(\xi)e^{-(\xi^2+i\xi)t} = \hat{g}(\xi) e^{-\xi^2} e^{-i\xi t}. $$

This gives $$ u(x,t) = \mathcal{F}^{-1}\{ \hat{g}(\xi) e^{-\xi^2} e^{-i\xi t} \} = \tau_{t}(g(x)*_x\frac{1}{\sqrt{4\pi}}e^{-x^2/4}) \\ = g(x)*_x\frac{1}{\sqrt{4\pi}}\tau_{t}(e^{-x^2/4}) = g(x)*_x\frac{1}{\sqrt{4\pi}}e^{-(x-t)^2/4}. $$ Here I used that ...

• $\mathcal{F}^{-1}\{\hat{f}\hat{g}\}=f*g$,
• $\mathcal{F}^{-1}\{ e^{-\xi^2} \} = \frac{1}{\sqrt{4\pi}}e^{-x^2/4}$,
• $\mathcal{F}^{-1}\{ \hat{f}(\xi) e^{-ia\xi} \} = \tau_{a}f(x) = f(x-a)$

See tables section on Wikipedia Fourier transform page.