The inhomogeneous heat equation on $𝕋$

1 Introduction

In this note I am working out some material following Steve Shkoller’s MAT218: Lecture Notes on Partial Differential Equations. However, I have written out a number of details that were not in the original notes, and may thus have introduced errors that were not in the notes on which this is based.

Write $𝕋=ℝ/2\pi \mathbb{Z}$, and for ,

$${\parallel f\parallel }_{L^p}={\left(\frac{1}{2\pi }{\int }_{𝕋}{|f(t)|}^p𝑑t\right)}^{1/p}.$$

Define

$${\parallel f\parallel }_{H^k}={\left(\sum _{0\le j\le k}{\parallel {\partial }_x^{j}f\parallel }_{L^2}^2\right)}^{1/2}.$$

If $u$ is a distribution on $𝕋$, ${\partial }_{x}u$ is also a distribution on $𝕋$, and in particular, if $u\in L^2(𝕋)$ then ${\partial }_{x}u$ is a distribution on $𝕋$. But if $u\in H^2(𝕋)$, for example, then ${\partial }_x^{2}u$ is an element of $L^2(𝕋)$, rather than merely being a distribution.

Fix $T>0$. Let $f\in L^2(0,T;L^2(𝕋))$ and $g\in H^1(𝕋)$; as $H^1(𝕋)\subset C^0(𝕋)$, we can speak about the value of $g$ at every point rather than merely almost all points.

For almost all $t$ and for all $x$, define $f_n$ by

$$f_n(x,t)=\sum _{k=-n}^n\widehat{f}(k,t)e^{ikx},$$

and for all $x$ define $g_n$ by

$$g_n(x)=\sum _{k=-n}^n\widehat{g}(k)e^{ikx}.$$

In other words, if $D_n(x)={\sum }_{k=-n}^n{e}^{ikx}$, then

$$f_n(x,t)=(D_n*f(\cdot ,t))(x),g_n(x)=(D_n*g)(x),$$

where

$$(\varphi *\psi )(x)=\frac{1}{2\pi }{\int }_{𝕋}\varphi (y)\psi (x-y)𝑑y.$$

2 Truncation

For each $n$, assume that there is some $u_n\in C^{\mathrm{\infty }}(0,T;C^{\mathrm{\infty }}(𝕋))$ such that for almost all $t$ and for all $x\in 𝕋$,

$$u_{nt}(x,t)-u_{nxx}(x,t)=f_n(x,t),$$ (1)

and for all $x\in 𝕋$,

$$u_n(x,0)=g_n(x).$$

We will thus obtain a formula for $u_n$. In fact we will not necessarily have $u_n\in C^{\mathrm{\infty }}(0,T;C^{\mathrm{\infty }}(𝕋))$, but once we have an expression for $u_n$ we can determine the function space of which it is an element. We will then show that there is some $u$ in a certain function space such that $u_n(x,t)=(D_n*u(\cdot ,t))(x)$ for all $x$ and $t$.

For all $t$ and $x$,

$$u_n(x,t)=\sum _{k\in \mathbb{Z}}\widehat{u_n}(k,t)e^{ikx}.$$

Then (1) becomes the statement that for almost all $t$ and for all $x$,

$$\sum _{k\in \mathbb{Z}}{\widehat{u_n}}^{\prime }(k,t)e^{ikx}+\sum _{k\in \mathbb{Z}}{k}^2\widehat{u_n}(k,t)e^{ikx}=\sum _{k=-n}^n\widehat{f}(k,t)e^{ikx}.$$

If $|k|>n$, then ${\widehat{u_n}}^{\prime }(k,t)+k^2\widehat{u_n}(k,t)=0$, which is a linear ordinary differential equation, whose solution satisfies $\widehat{u_n}(k,t)=e^{-k^{2}t}\widehat{u_n}(k,0)$. Since $u_n(x,0)=g_n(x)$, $\widehat{u_n}(k,0)=0$. Hence if $|k|>n$ then $\widehat{u_n}(k,t)=0$. If $|k|\le n$, then for almost all $t$, ${\widehat{u_n}}^{\prime }(k,t)+k^2\widehat{u_n}(k,t)=\widehat{f}(k,t)$. The solution of this is, for all $t$ and for all $x$,

$$\widehat{u_n}(k,t)=e^{-k^{2}t}\widehat{g_n}(k)+e^{-k^{2}t}{\int }_0^t{e}^{k^{2}s}\widehat{f_n}(k,s)𝑑s.$$

Hence, for all $t$ and for all $x$,

$$u_n(x,t)=\sum _{k=-n}^n\left(e^{-k^{2}t}\widehat{g_n}(k)+e^{-k^{2}t}{\int }_0^t{e}^{k^{2}s}\widehat{f_n}(k,s)𝑑s\right)e^{ikx}.$$

We merely know that $\widehat{f_n}(k,t)$ is defined for almost all $t$, thus we only know for almost all $t_0\in (0,T)$ and for all $x$ that $u_{nt}(x,t_0)$ exists. We do have that

$$u_n\in C^0(0,T;C^{\mathrm{\infty }}(𝕋)).$$

3 𝐻¹

For almost all $t$, multiply (1) by $u_n(x,t)$ and integrate over $𝕋$. This is,

$${\int }_{𝕋}{u}_{nt}(x,t)u_n(x,t)𝑑x-{\int }_{𝕋}{u}_{nxx}(x,t)u_n(x,t)𝑑x={\int }_{𝕋}{f}_n(x,t)u_n(x,t)𝑑x.$$

Integrating by parts this becomes

$${\int }_{𝕋}{u}_{nt}(x,t)u_n(x,t)𝑑x+{\int }_{𝕋}{u}_{nx}(x,t)u_{nx}(x,t)𝑑x={\int }_{𝕋}{f}_n(x,t)u_n(x,t)𝑑x,$$

which is

$$\pi \cdot {\partial }_t\frac{1}{2\pi }{\int }_{𝕋}{u}_n{(x,t)}^2𝑑x+2\pi \cdot \frac{1}{2\pi }{\int }_{𝕋}{u}_{nx}{(x,t)}^2𝑑x={\int }_{𝕋}{f}_n(x,t)u_n(x,t)𝑑x.$$

Writing this using norms,

$$\pi \cdot {\partial }_t{\parallel u_n(\cdot ,t)\parallel }_{L^2}^2+2\pi \cdot {\parallel u_{nx}(\cdot ,t)\parallel }_{L^2}^2={\int }_{𝕋}{f}_n(x,t)u_n(x,t)𝑑x.$$

Integrating from $0$ to $t$, for any $t$,

$$\pi \cdot {\parallel u_n(\cdot ,t)\parallel }_{L^2}^2-\pi \cdot {\parallel u_n(\cdot ,0)\parallel }_{L^2}^2+2\pi {\int }_0^t{\parallel u_{nx}(\cdot ,s)\parallel }_{L^2}^2𝑑s={\int }_0^t{\int }_{𝕋}{f}_n(x,s)u_n(x,s)𝑑x𝑑s.$$

For almost all $s$,

${\displaystyle {\int }_{𝕋}}|f_n(x,s)u_n(x,s)|𝑑x$ $=$ $2\pi \cdot {\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{𝕋}}|f_n(x,s)u_n(x,s)|𝑑x$

$\le$ $2\pi \cdot {\parallel f_n(\cdot ,s)\parallel }_{L^2}{\parallel u_n(\cdot ,s)\parallel }_{L^2}$

$\le$ $2\pi \left({\displaystyle \frac{{\parallel f_n(\cdot ,s)\parallel }_{L^2}^2}{2}}+{\displaystyle \frac{{\parallel u_n(\cdot ,s)\parallel }_{L^2}^2}{2}}\right)$

$=$ $\pi \cdot {\parallel f_n(\cdot ,s)\parallel }_{L^2}^2+\pi \cdot {\parallel u_n(\cdot ,s)\parallel }_{L^2}^2.$

It follows that for all $t$ (not just almost all $t$)

so, as $u_n(x,0)=g_n(x)$,

$${\parallel u_n(\cdot ,t)\parallel }_{L^2}^2+2{\int }_0^t{\parallel u_{nx}(\cdot ,s)\parallel }_{L^2}^2𝑑s\le {\parallel g_n\parallel }_{L^2}^2+{\int }_0^t{\parallel f_n(\cdot ,s)\parallel }_{L^2}^2+{\parallel u_n(\cdot ,s)\parallel }_{L^2}^{2}ds.$$

Let

$$y(t)={\parallel u_n(\cdot ,t)\parallel }_{L^2}^2+2{\int }_0^t{\parallel u_{nx}(\cdot ,s)\parallel }_{L^2}^2𝑑s.$$

By the inequality we just established we have, for all $t$,

$y(t)$ $\le$ ${\parallel g_n\parallel }_{L^2}^2+{\displaystyle {\int }_0^t}{\parallel f_n(\cdot ,s)\parallel }_{L^2}^2𝑑s+{\displaystyle {\int }_0^t}{\parallel u_n(\cdot ,s)\parallel }_{L^2}^2𝑑s$

$\le$ ${\parallel g_n\parallel }_{L^2}^2+{\displaystyle {\int }_0^t}{\parallel f_n(\cdot ,s)\parallel }_{L^2}^2𝑑s+{\displaystyle {\int }_0^t}y(s)𝑑s.$

By Gronwall’s inequality, we get

$$y(t)\le \left({\parallel g_n\parallel }_{L^2}^2+{\int }_0^t{\parallel f_n(\cdot ,s)\parallel }_{L^2}^2𝑑s\right)e^t.$$

As ${\parallel g_n\parallel }_{L^2}\le {\parallel g\parallel }_{L^2}$ and ${\parallel f_n(\cdot ,s)\parallel }_{L^2}\le {\parallel f(\cdot ,s)\parallel }_{L^2}$ (these two facts follow from Parseval’s identity), it follows that

$$y(t)\le \left({\parallel g\parallel }_{L^2}^2+{\int }_0^t{\parallel f(\cdot ,s)\parallel }_{L^2}^2𝑑s\right)e^t.$$

Therefore, if $0\le t\le T$ then

${\parallel u_n(\cdot ,t)\parallel }_{L^2}^2+2{\displaystyle {\int }_0^t}{\parallel u_{nx}(\cdot ,s)\parallel }_{L^2}^2𝑑s$ $\le$ $\left({\parallel g\parallel }_{L^2}^2+{\displaystyle {\int }_0^t}{\parallel f(\cdot ,s)\parallel }_{L^2}^2𝑑s\right)e^T$

$\le$ $\left({\parallel g\parallel }_{L^2}^2+{\parallel f\parallel }_{L^2(0,T;L^2(𝕋)}^2\right)e^T$

$=$ $M.$

By Parseval’s identity,

$$\sum _{k\in \mathbb{Z}}{|\widehat{u_n}(k,t)|}^2+2{\int }_0^t\sum _{k\in \mathbb{Z}}{|\widehat{u_{nx}}(k,s)|}^{2}ds\le M,$$

hence for all $t$,

$$\sum _{k\in \mathbb{Z}}{|\widehat{u_n}(k,t)|}^2+2{\int }_0^t\sum _{k\in \mathbb{Z}}{k}^2{|\widehat{u_n}(k,s)|}^{2}ds\le M.$$

If $k\le n\le m$, then $\widehat{u_n}(k,t)=\widehat{u_m}(k,t)$ for all $t$. Define $\widehat{u}(k,t)$ by

$$\widehat{u}(k,t)=\underset{n\to \mathrm{\infty }}{lim}\widehat{u_n}(k,t)=\widehat{u_k}(k,t).$$

Thus, for all $t$,

$$\sum _{k\in \mathbb{Z}}{|\widehat{u}(k,t)|}^2+2{\int }_0^t\sum _{k\in \mathbb{Z}}{k}^2{|\widehat{u}(k,s)|}^{2}ds\le M.$$ (2)

Then, for some $M^{\prime }=M^{\prime }(f,g,T)$,

$${\int }_0^T\sum _{k\in \mathbb{Z}}{|\widehat{u}(k,t)|}^2+\sum _{k\in \mathbb{Z}}{k}^2{|\widehat{u}(k,t)|}^{2}dt\le M^{\prime }.$$

It follows that for almost all $t$, there is some $u\in H^1(𝕋)$ whose Fourier coefficients are $\widehat{u}(k,t)$, and that we have

$${\int }_0^T{\parallel u(\cdot ,t)\parallel }_{H^1}^2𝑑t\le M^{\prime }.$$

We have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^T{\parallel u_n(\cdot ,t)-u(\cdot ,t)\parallel }_{H^1}^2𝑑t=0,$$

i.e.

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel u_n-u\parallel }_{L^2(0,T;H^1(𝕋))}^2=0.$$

4 𝐻²

Multiply (1) by $u_{nxx}(x,t)$ and integrate over $𝕋$. We get, for almost all $t$,

$${\int }_{𝕋}{u}_{nt}(x,t)u_{nxx}(x,t)𝑑x-{\int }_{𝕋}{u}_{nxx}(x,t)u_{nxx}(x,t)𝑑x={\int }_{𝕋}{f}_n(x,t)u_{nxx}(x,t)𝑑x.$$

As

$${\int }_{𝕋}{u}_{nt}(x,t)u_{nxx}(x,t)𝑑x=-{\int }_{𝕋}{u}_{ntx}(x,t)u_{nx}(x,t)𝑑x=-\frac{1}{2}\frac{d}{dt}{\int }_{𝕋}{u}_{nx}{(x,t)}^2𝑑x,$$

we have

$$-\pi \frac{d}{dt}{\parallel u_{nx}(\cdot ,t)\parallel }_{L^2}^2-2\pi {\parallel u_{nxx}(\cdot ,t)\parallel }_{L^2}^2={\int }_{𝕋}{f}_n(x,t)u_{nxx}(x,t)𝑑x.$$

Integrating from $0$ to $t$,

For almost all $s$,

${\displaystyle {\int }_{𝕋}}|f_n(x,s)u_{nxx}(x,s)|𝑑x$ $\le$ $2\pi {\parallel f_n(\cdot ,s)\parallel }_{L^2}{\parallel u_{nxx}(\cdot ,s)\parallel }_{L^2}$

$\le$ $2\pi \left({\displaystyle \frac{{\parallel f_n(\cdot ,s)\parallel }_{L^2}^2}{2}}+{\displaystyle \frac{{\parallel u_{nxx}(\cdot ,s)\parallel }_{L^2}^2}{2}}\right)$

$=$ $\pi {\parallel f_n(\cdot ,s)\parallel }_{L^2}^2+\pi {\parallel u_{nxx}(\cdot ,s)\parallel }_{L^2}^2.$

It follows that, for all $t$,

Hence

${\parallel u_{nx}(\cdot ,t)\parallel }_{L^2}^2+{\displaystyle {\int }_0^t}{\parallel u_{nxx}(\cdot ,s)\parallel }_{L^2}^2𝑑s$ $\le$ ${\parallel g_n^{\prime }\parallel }_{L^2}^2+{\displaystyle {\int }_0^t}{\parallel f_n(\cdot ,s)\parallel }_{L^2}^2𝑑s$

$\le$ ${\parallel g\parallel }_{H^1}^2+{\displaystyle {\int }_0^t}{\parallel f(\cdot ,s)\parallel }_{L^2}^2𝑑s$

$\le$ ${\parallel g\parallel }_{H^1}^2+{\displaystyle {\int }_0^T}{\parallel f(\cdot ,s)\parallel }_{L^2}^2𝑑s.$

Using Parseval’s identity we have, for all $t$,

$$\sum _{k\in \mathbb{Z}}{|\widehat{u_{nx}}(k,t)|}^2+{\int }_0^t\sum _{k\in \mathbb{Z}}{|\widehat{u_{nxx}}(k,s)|}^{2}ds\le {\parallel g\parallel }_{H^1}^2+{\int }_0^T{\parallel f(\cdot ,s)\parallel }_{L^2}^2𝑑s,$$

hence

$$\sum _{k\in \mathbb{Z}}{k}^2{|\widehat{u_n}(k,t)|}^2+{\int }_0^t\sum _{k\in \mathbb{Z}}{k}^4{|\widehat{u_n}(k,s)|}^{2}ds\le {\parallel g\parallel }_{H^1}^2+{\int }_0^T{\parallel f(\cdot ,s)\parallel }_{L^2}^2𝑑s,$$

so

$$\sum _{k\in \mathbb{Z}}{k}^2{|\widehat{u}(k,t)|}^2+{\int }_0^t\sum _{k\in \mathbb{Z}}{k}^4{|\widehat{u}(k,s)|}^{2}ds\le {\parallel g\parallel }_{H^1}^2+{\int }_0^T{\parallel f(\cdot ,s)\parallel }_{L^2}^2𝑑s.$$

It follows that, for almost all $t$,

thus $u(\cdot ,t)\in H^2(𝕋)$.

We have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^T{\parallel u_n(\cdot ,t)-u(\cdot ,t)\parallel }_{H^2}^2𝑑t=0,$$

i.e.

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel u_n-u\parallel }_{L^2(0,T;H^2(𝕋))}^2=0.$$

5 Solution

For all $t$ we have $u(\cdot ,t)\in H^1(𝕋)$, and $H^1(𝕋)\subset C^0(𝕋)$, so for all $t$ and all $x$, $u(x,t)$ is defined. The Sobolev embedding tells us that if $k>\alpha +\frac{1}{2}$ then $H^k(𝕋)\subset C^{\alpha }(𝕋)$. So, being specific, we have $H^1(𝕋)\subset C^{1/4}(𝕋)$. It is a fact that if $h\in C^{\alpha }(𝕋)$, $\alpha >0$, then the partial sums of the Fourier series of $h$ converge to $h$ in the supremum norm.

For all $t$ and for each $k$,

$$\widehat{u}(k,t)=e^{-k^{2}t}\widehat{g}(k)+e^{-k^{2}t}{\int }_0^t{e}^{k^{2}s}\widehat{f}(k,s)𝑑s.$$

It follows that, for all $x$,

$$u(x,0)=\underset{N\to \mathrm{\infty }}{lim}\sum _{|k|\le N}\widehat{g}(k)e^{ikx}.$$

On the other hand,

$$g(x)=\underset{N\to \mathrm{\infty }}{lim}\sum _{|k|\le N}\widehat{g}(k)e^{ikx}.$$

Thus for all $x$, $u(x,0)=g(x)$.

We have

${\parallel u_t-u_{xx}-f\parallel }_{L^2(0,T;L^2(𝕋))}$ $\le$ ${\parallel u_t-u_{nt}\parallel }_{L^2(0,T;L^2(𝕋))}$

$+{\parallel u_{xx}-u_{nxx}\parallel }_{L^2(0,T;L^2(𝕋))}$

$+{\parallel f-f_n\parallel }_{L^2(0,T;L^2(𝕋))}$

$+{\parallel u_{nt}-u_{nxx}-f_n\parallel }_{L^2(0,T;L^2(𝕋))}.$

Each of the four norms has limit $0$ as $n\to \mathrm{\infty }$. Let me work out the first one. For almost all $t$,

$\widehat{u_t}(k,t)-\widehat{u_{nt}}(k,t)$ $=$ ${\displaystyle \sum _{|k|>n}}-k^2{e}^{-k^{2}t}\widehat{g}(k)-k^2{e}^{-k^{2}t}{\displaystyle {\int }_0^t}{e}^{k^{2}s}\widehat{f}(k,s)𝑑s$

$+e^{-k^{2}t}{e}^{k^{2}t}\widehat{f}(k,t)$

$=$ ${\displaystyle \sum _{|k|>n}}-k^2\widehat{u}(k,t)+\widehat{f}(k,t).$

Then using Parseval’s identity,

${\parallel u_t-u_{nt}\parallel }_{L^2(0,T;L^2(𝕋))}^2$ $=$ ${\displaystyle {\int }_0^T}{\displaystyle \sum _{|k|>n}}{|-k^2\widehat{u}(k,t)+\widehat{f}(k,t)|}^{2}dt$

$\le$ $2{\displaystyle {\int }_0^T}{\displaystyle \sum _{|k|>n}}{k}^2{|\widehat{u}(k,t)|}^2+{|\widehat{f}(k,t)|}^{2}dt.$

Then,

$${\parallel u_t-u_{xx}-f\parallel }_{L^2(0,T;L^2(𝕋))}^2=0.$$

So, for almost all $t$ and for almost all $x$,

$$u_t(x,t)-u_{xx}(x,t)=f(x,t).$$