5.3. Applications of Fourier transform to PDEs

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

5.3. Applications of Fourier transform to PDEs


  1. Heat equation
  2. Schrödinger equation
  3. Laplace equation in half-plane
  4. Laplace equation in half-plane. II
  5. Laplace equation in strip
  6. 1D wave equation
  7. Multidimensional equations

In the previous Section 5.1 and Section 5.2 we introduced Fourier transform and Inverse Fourier transform and established some of its properties; we also calculated some Fourier transforms. Now we going to apply to PDEs.

Heat equation

Consider problem \begin{align} & u_t= k u_{xx},&& t>0,\ -\infty< x< \infty, \label{eq-5.3.1}\\[3pt] &u|_{t=0}=g(x).\label{eq-5.3.2} \end{align} Making partial Fourier transform with respect to $x\mapsto\xi$ (so $u(x,t)\mapsto \hat{u}(\xi,t)$) we arrive to \begin{align} & \hat{u}_t= -k\xi^2 \hat{u},\label{eq-5.3.3}\\[3pt] & \hat{u}|_{t=0}=\hat{g}(\xi).\label{eq-5.3.4} \end{align} Indeed, $\partial_x \mapsto i\xi$ and therefore $\partial^2_x \mapsto -\xi^2$.

Note that (\ref{eq-5.3.3}) is an ODE and solving it we arrive to $\hat{u}=A(\xi)e^{-k\xi^2t}$; plugging into (\ref{eq-5.3.4}) we find that $A(\xi)=\hat{g}(\xi)$ and therefore \begin{equation} \hat{u}(\xi,t)= \hat{g}(\xi)e^{-k\xi^2t}. \label{eq-5.3.5} \end{equation} The right-hand expression is a product of two Fourier transforms, one is $\hat{g}(\xi)$ and another is Fourier transform of IFT of $e^{-k\xi^2t}$ (reverse engineering?).

If we had $e^{-\xi^2/2}$ we would have IFT equal to $\sqrt{2\pi}e^{-x^2/2}$; but we can get from $e^{-\xi^2/2}$ to $e^{-k\xi^2t}$ by scaling $\xi \mapsto (2kt)^{\frac{1}{2}}\xi $ and therefore $x \mapsto (2kt)^{-\frac{1}{2}}x$ (and we need to multiply the result by by $(2kt)^{-\frac{1}{2}}$); therefore $e^{-k\xi^2t}$ is a Fourier transform of $\frac{\sqrt{2\pi}}{\sqrt{2kt}}e^{-x^2/4kt}$.

Again: $\hat{u}(\xi,t)$ is a product of FT of $g$ and of $\frac{\sqrt{2\pi}}{\sqrt{2kt}}e^{-x^2/4kt}$ and therefore $u$ is the convolution of these functions (multiplied by $1/(2\pi)$): \begin{equation} u(x,t) = g* \frac{1}{\sqrt{4\pi kt}}e^{-\frac{x^2}{4kt}}= \frac{1}{\sqrt{4\pi kt}}\int _{-\infty}^\infty g(x') e^{-\frac{(x-x')^2}{4kt}}\,dx'. \label{eq-5.3.6} \end{equation} We recovered formula which we had already.

Remark 1. Formula (\ref{eq-5.3.5}) shows that the problem is really ill-posed for $t<0$.

Schrödinger equation

Consider problem \begin{align} & u_t= i ku_{xx},&& t>0,\ -\infty< x<\infty, \label{eq-5.3.7}\\[3pt] & u|_{t=0}=g(x).\label{eq-5.3.8} \end{align} Making partial Fourier transform with respect to $x\mapsto\xi$ (so $u(x,t)\mapsto \hat{u}(\xi,t)$) we arrive to \begin{align} & \hat{u}_t= -ik\xi^2 \hat{u},\label{eq-5.3.9}\\[3pt] & \hat{u}|_{t=0}=\hat{g}(\xi).\label{eq-5.3.10} \end{align}

Note that (\ref{eq-5.3.9}) is an ODE and solving it we arrive to $\hat{u}=A(\xi)e^{-ik\xi^2t}$; plugging into (\ref{eq-5.3.10}) we find that $A(\xi)=\hat{g}(\xi)$ and therefore \begin{equation} \hat{u}(\xi,t)= \hat{g}(\xi)e^{-ik\xi^2t}. \label{eq-5.3.11} \end{equation} The right-hand expression is a product of two Fourier transforms, one is $\hat{g}(\xi)$ and another is Fourier transform of IFT of $e^{-ik\xi^2t}$ (reverse engineering?).

As it was explained in Section 5.2 that we need just to plug $ik$ instead of $k$ (as $t >0$) into the formulae we got before; so instead of $\frac{1}{\sqrt{4\pi kt}}e^{-\frac{x^2}{4kt}}$ we get $\frac{1}{\sqrt{-4\pi ki t}}e^{\frac{x^2}{4kit}}= \frac{1}{\sqrt{4\pi k t}}e^{-\frac{\pi i}{4}+\frac{i x^2}{4kt}}$ because we need to take a correct branch of $\sqrt{i}=e^{\frac{i\pi}{4}}$. As $t <0$ we need to replace $t$ by $-t$ and $i$ by $-i$ resulting in $\frac{1}{\sqrt{4\pi k |t|}}e^{\frac{\pi i}{4}+\frac{i x^2}{4kt}}$.

Therefore \begin{equation} u(x,t) = \frac{1}{\sqrt{4\pi k|t|}}\int _{-\infty}^\infty g(x') e^{\mp \frac{i\pi}{4}+\frac{i(x-x')^2}{4kt}}\,dx ' \label{eq-5.3.12} \end{equation} as $\pm t>0$.

Remark 2. Formula (\ref{eq-5.3.11}) shows that the problem is well-posed for both $t >0$ and $t <0$.

Laplace equation in half-plane

Consider problem \begin{align} & \Delta u:=u_{xx}+u_{yy}=0,&& y>0,\ -\infty< x<\infty, \label{eq-5.3.13}\\[3pt] & u|_{y=0}=g(x).\label{eq-5.3.14} \end{align} This problem definitely is not uniquely solvable (f.e. $u=y$ satisfies homogeneous boundary condition) and to make it uniquely solvable we need to add condition $|u|\le M$.

Making partial Fourier transform with respect to $x\mapsto\xi$ (so $u(x,t)\mapsto \hat{u}(\xi,t)$) we arrive to \begin{align} & \hat{u}_{yy}-\xi^2 \hat{u}=0,\label{eq-5.3.15}\\[3pt] & \hat{u}|_{y=0}=\hat{g}(\xi).\label{eq-5.3.16} \end{align}

Note that (\ref{eq-5.3.15}) is an ODE and solving it we arrive to \begin{equation} \hat{u}(\xi,y)=A(\xi)e^{-|\xi|y}+B(\xi)e^{|\xi|y}. \label{eq-5.3.17}\end{equation} Indeed, characteristic equation $\alpha^2 -\xi^2$ has two roots $\alpha_{1,2}=\pm |\xi|$; we take $\pm |\xi|$ instead of just $\pm \xi$ because we need to control signs.

We discard the second term in the right-hand expression of (\ref{eq-5.3.17}) because it is unbounded. However if we had Cauchy problem (i.e. $u|_{y=0}=g(x)$, $u_y|_{y=0}=h(x)$) we would not be able to do this and this problem will be ill-posed.

So, $\hat{u}=A(\xi)e^{-|\xi|y}$ and (\ref{eq-5.3.16}) yields $A(\xi)=\hat{g}(\xi)$: \begin{equation} \hat{u}(\xi,y)=\hat{g}(\xi) e^{-|\xi|y}. \label{eq-5.3.18} \end{equation} Now we need to find the IFT of $e^{-|\xi|y}$. This calculations are easy (do them!) and IFT is $2y (x^2+y^2)^{-1}$. Then

\begin{equation} u(x,y) =\frac{1}{\pi}\int _{-\infty}^\infty g(x') \frac{y}{(x-x')^2+y^2}\,dx '. \label{eq-5.3.19} \end{equation}

Remark 3. Setting $y=0$ we see that $u|_{y=0}=0$. Contradiction?--No, we cannot just set $y=0$. We need to find a limit as $y\to +0$, and note that $\frac{y}{(x-x')^2+y^2}\to 0$ except as $x'=x$ and $\frac{1}{\pi}\int_{-\infty}^\infty \frac{y}{(x-x')^2+y^2}\,dx'=1$ so the limit will be $g(x)$ as it should be.

Laplace equation in half-plane. II

Replace Dirichlet boundary condition by Robin boundary condition \begin{align} & \Delta u:=u_{xx}+u_{yy}=0,&& y>0,\ -\infty< x<\infty,\tag{\ref{eq-5.3.13}}\\[3pt] & (u_y-\alpha u)|_{y=0}=h(x). \label{eq-5.3.20} \end{align} Then (\ref{eq-5.3.16}) should be replaced by \begin{equation} (\hat{u}_y-\alpha \hat{u})|_{y=0}=\hat{h}(\xi). \label{eq-5.3.21} \end{equation} and then \begin{equation} A(\xi)= -(|\xi|+\alpha)^{-1}\hat{h}(\xi) \label{eq-5.3.22} \end{equation} and \begin{equation} \hat{u}(\xi,y)=-\hat{h}(\xi)(|\xi|+\alpha)^{-1} e^{-|\xi|y}. \label{eq-5.3.23} \end{equation} The right-hand expression is a nice function provided $\alpha>0$ (and this is correct from the physical point of view) and therefore everything is fine (but we just cannot calculate explicitely IFT of $(|\xi|+\alpha)^{-1} e^{-|\xi|y}$).

Consider Neumann boundary condition i.e. set $\alpha=0$. Then we have a trouble: $-\hat{h}(\xi)(|\xi|+\alpha)^{-1} e^{-|\xi|y}$ could be singular at $\xi=0$ and to avoid it we assume that $\hat{h}(0)=0$. This means exactly that \begin{equation} \int_{-\infty}^\infty h(x)\,dx=0 \label{eq-5.3.24} \end{equation} and this condition we really need and it is justified from the physical point of view: f.e. if we are looking for stationary heat distribution and we have heat flow defined, we need to assume that the total flow is $0$ (otherwise the will be no stationary distribution!).

So we need to calculate IFT of $-|xi|^{-1}e^{-|\xi|y}$. Note that derivative of this with respect to $y$ is $e^{-|\xi|y}$ which has an IFT $\frac{1}{\pi}\frac{y}{x^2+y^2}$; integrating with respect to $y$ we get $\frac{1}{2\pi}\log (x^2+y^2)+c$ and therefore \begin{equation} u(x,y) = \frac{1}{2\pi}\int_{-\infty}^\infty h(x') \log\bigl((x-x')^2+y^2\bigr)\,dx '+C. \label{eq-5.3.25} \end{equation}

Remark 4.

  1. Here $C$ is an arbitrary constant. Again, the same physical interpretation: knowing heat flow we define solution up to a constant as the total heat energy is arbitrary.
  2. Formula (\ref{eq-5.3.25}) gives us a solution which can grow as $|x|\to \infty$ even if $h$ is fast decaying there (or even if $h(x)=0$ as $|x|\ge c$). However as $|x|\gg 1$ and $h$ is fast decaying $\bigl((x-x')^2+y^2\bigr)\approx \bigl(x^2+y^2\bigr)$ (with a small error) and growing part of $u$ is $\frac{1}{2\pi}\log \bigl(x^2+y^2\bigr)\int _{-\infty}^\infty h(x') \,dx '$ which is $0$ precisely because of condition (\ref{eq-5.3.24}).

Laplace equation in strip

Consider problem \begin{align} & \Delta u:=u_{xx}+u_{yy}=0,&& 0< y< b,\ -\infty< x<\infty, \label{eq-5.3.26}\\[3pt] &u|_{y=0}=g(x),\label{eq-5.3.27}\\[3pt] &u|_{y=b}=h(x)\label{eq-5.3.28}. \end{align} Then we get (\ref{eq-5.3.17}) again \begin{equation} \hat{u}(\xi,y)=A(\xi)e^{-|\xi|y}+B(\xi)e^{|\xi|y} \tag{\ref{eq-5.3.17}} \end{equation} but with two boundary condition we cannot diacard anything; we get instead \begin{align} & A(\xi) && + B (\xi)&& =\hat{g}(\xi),\label{eq-5.3.29}\\[3pt] & A (\xi)e^{-|\xi|b} && + B (\xi)e^{|\xi|b}&& =\hat{h}(\xi)\label{eq-5.3.30} \end{align} which implies \begin{align*} & A(\xi)= \frac{e^{|\xi|b}}{2\sinh (|\xi|b)}\hat{g}(\xi)- \frac{1}{2\sinh (|\xi|b)}\hat{h}(\xi) ,\\[3pt] & B(\xi)= -\frac{e^{-|\xi|b}}{2\sinh (|\xi|b)}\hat{g}(\xi)+ \frac{1}{2\sinh (|\xi|b)}\hat{h}(\xi) \end{align*} and therefore \begin{equation} \hat{u}(\xi,y)=\frac{\sinh (|\xi|(b-y))}{\sinh (|\xi|b)}\hat{g}(\xi)+ \frac{\sinh (|\xi|y)}{\sinh (|\xi|b)}\hat{h}(\xi). \label{eq-5.3.31} \end{equation} One can see easily that $\frac{\sinh (|\xi|(b-y))}{\sinh (|\xi|b)}$ and $\frac{\sinh (|\xi|y)}{\sinh (|\xi|b)}$ are bounded as $0\le y\le b$ and fast decaying as $|\xi|\to \infty$ as $y\ge \epsilon$ ($y\le b-\epsilon$ respectively) with arbitrarily small $\epsilon>0$.

Problem 1. Investigate other boundary conditions (Robin, Neumann, mixed--Dirichlet at $y=0$ and Neumann at $y=b$ and so on.).

1D wave equation

Consider problem \begin{align} & u_{tt}= c^2u_{xx},&& -\infty< x<\infty,\label{eq-5.3.32}\\[3pt] & u|_{t=0}=g(x),\label{eq-5.3.33}\\[3pt] &u_t|_{t=0}=h(x). \label{eq-5.3.34} \end{align} Making partial Fourier transform with respect to $x\mapsto\xi$ we arrive to \begin{align} & \hat{u}_{tt}= -c^2\xi^2\hat{u}_{xx}, \label{eq-5.3.35}\\[3pt] & \hat{u}|_{t=0}=\hat{g}(\xi),\label{eq-5.3.36}\\[3pt] & \hat{u}_t|_{t=0}=\hat{h}(\xi).\label{eq-5.3.37} \end{align} Then characteristic equation for ODE (\ref{eq-5.3.35}) is $\alpha^2=-c^2\xi^2$ and $\alpha_{1,2}=\pm ic\xi$, \begin{equation*} \hat{u}(\xi,t)= A(\xi)\cos (c\xi t) + B(\xi)\sin (c\xi t) \end{equation*} with initial conditions implying $A(\xi)=\hat{g}(\xi)$, $B(\xi)=1/(ci\xi)\cdot \hat{h}(\xi)$ and \begin{equation} \hat{u}(\xi,t)= \hat{g}(\xi)\cos (c\xi t) + \hat{h}(\xi)\cdot \frac{1}{c\xi}\sin (c\xi t). \label{eq-5.3.38} \end{equation} Rewriting $\cos (c\xi t)=\frac{1}{2}\bigl( e^{ic\xi t}+ e^{-ic\xi t}\bigr)$ and recalling that multiplication of FT by $e^{i\xi b}$ is equivalen to to shifting original to the left by $b$ we conclude that $\hat{g}(\xi)\cos (c\xi t)$ is a Fourier transform of $\frac{1}{2}\bigl(g(x+ct)+g(x-ct)\bigr)$.

If we denote $H$ as a primitive of $h$ then $\hat{h}(\xi)\cdot \frac{1}{c \xi} \sin (c\xi t)= \hat{H}(\xi)\cdot \frac{1}{c} i\sin (c\xi t)$ which in virtue of the same arguments is FT of $\frac{1}{2c}\bigl(H(x+ct)-H(x-ct)\bigr)= \frac{1}{2c}\int_{x-ct}^{x+ct}h(x')\,dx'$.

Therefore \begin{equation} u(x,t)= \frac{1}{2}\bigl(g(x+ct)+g(x-ct)\bigr) + \frac{1}{2c}\int_{x-ct}^{x+ct}h(x')\,dx' \label{eq-5.3.39} \end{equation} and we arrive again to d'Alembert formula.

Multidimensional equations

Multidimensional equations are treated in the same way:

Heat and Schrödinger equations

We make partial FT (with respect to spatial variables) and we get \begin{align*} &\hat u(\xi,t)= \hat g(\xi) e^{-k|\xi|^2t},\\ &\hat u(\xi,t)= \hat g(\xi) e^{-ik|\xi|^2t} \end{align*} respectively where $\xi=(\xi_1,\ldots,\xi_n)$, $|\xi|=(\xi_1^2+\ldots\xi_n^2)^{\frac{1}{2}}$; in this case $e^{-k|\xi|^2t}=\prod_{j=1}^n e^{-k|\xi_j|^2t}$ is a product of functions depending on different variables, so IFT will be again such product and we have IFT equal \begin{equation*} \prod_{j=1}^n\frac{1}{\sqrt{4\pi k t}} e^{-|x_j|^2/4kt}= (4\pi kt)^{-\frac{n}{2}}e^{-|x|^2 /4kt} \end{equation*} for heat equation and similarly for Schrödinger equation and we get a solution as a multidimensional convolution. Here $x=(x_1,\ldots,x_n)$ and $|x|=(x_1^2+\ldots+x_n^2)^{\frac{1}{2}}$.

Wave equation

We do the same but now \begin{equation*} \hat{u}(\xi,t)= \hat{g}(\xi)\cos (c|\xi| t) + \hat{h}(\xi)\cdot \frac{1}{c|\xi|}\sin (c|\xi| t). \end{equation*} Finding IFT is not easy.

Laplace equation

We consider it in $\mathbb{R}^{n}\times I\ni (x;y)$ with either $I=\{y:\, y>0\}$ or $I=\{y:\, 0< y< b\}$ and again make partial FT with respect to $x$ but not $y$.


$\Leftarrow$  $\Uparrow$  $\downarrow$  $\Rightarrow$