5.3. Applications of Fourier transform to PDEs


## 5.3. Applications of Fourier transform to PDEs

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 $$\hat{u}(\xi,t)= \hat{g}(\xi)e^{-k\xi^2t}. \label{eq-5.3.5}$$ 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 $1/(2\pi)$): $$u(x,t) = g* \frac{1}{\sqrt{4\pi kt}}e^{-\frac{x^2}{4kt}}= \frac{1}{\sqrt{4\pi kt}}\int _{-\infty}^\infty g(x') e^{-\frac{(x-x')^2}{4kt}}\,dx'. \label{eq-5.3.6}$$ 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 $$\hat{u}(\xi,t)= \hat{g}(\xi)e^{-ik\xi^2t}. \label{eq-5.3.11}$$ 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 $$u(x,t) = \frac{1}{\sqrt{4\pi k|t|}}\int _{-\infty}^\infty g(x') e^{\mp \frac{i\pi}{4}+\frac{i(x-x')^2}{4kt}}\,dx ' \label{eq-5.3.12}$$ 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 $$\hat{u}(\xi,y)=A(\xi)e^{-|\xi|y}+B(\xi)e^{|\xi|y}. \label{eq-5.3.17}$$ 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 $A(\xi)=\hat{g}(\xi)$: $$\hat{u}(\xi,y)=\hat{g}(\xi) e^{-|\xi|y}. \label{eq-5.3.18}$$ 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

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

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 $$(\hat{u}_y-\alpha \hat{u})|_{y=0}=\hat{h}(\xi). \label{eq-5.3.21}$$ and then $$A(\xi)= -(|\xi|+\alpha)^{-1}\hat{h}(\xi) \label{eq-5.3.22}$$ and $$\hat{u}(\xi,y)=-\hat{h}(\xi)(|\xi|+\alpha)^{-1} e^{-|\xi|y}. \label{eq-5.3.23}$$ 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 $$\int_{-\infty}^\infty h(x)\,dx=0 \label{eq-5.3.24}$$ 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 $$u(x,y) = \frac{1}{2\pi}\int_{-\infty}^\infty h(x') \log\bigl((x-x')^2+y^2\bigr)\,dx '+C. \label{eq-5.3.25}$$

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 $$\hat{u}(\xi,y)=A(\xi)e^{-|\xi|y}+B(\xi)e^{|\xi|y} \tag{\ref{eq-5.3.17}}$$ 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 $$\hat{u}(\xi,y)=\frac{\sinh (|\xi|(b-y))}{\sinh (|\xi|b)}\hat{g}(\xi)+ \frac{\sinh (|\xi|y)}{\sinh (|\xi|b)}\hat{h}(\xi). \label{eq-5.3.31}$$ 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 $$\hat{u}(\xi,t)= \hat{g}(\xi)\cos (c\xi t) + \hat{h}(\xi)\cdot \frac{1}{c\xi}\sin (c\xi t). \label{eq-5.3.38}$$ 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 $$u(x,t)= \frac{1}{2}\bigl(g(x+ct)+g(x-ct)\bigr) + \frac{1}{2c}\int_{x-ct}^{x+ct}h(x')\,dx' \label{eq-5.3.39}$$ 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 $y$.