# The inhomogeneous heat equation on $\mathbb{T}$

Jordan Bell
April 3, 2014

## 1 Introduction

In this note I am working out some material following Steve Shkollerβ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 $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$, and for $1\leq p<\infty$,

 $\left\|f\right\|_{L^{p}}=\left(\frac{1}{2\pi}\int_{\mathbb{T}}|f(t)|^{p}dt% \right)^{1/p}.$

Define

 $\left\|f\right\|_{H^{k}}=\left(\sum_{0\leq j\leq k}\left\|\partial_{x}^{j}f% \right\|_{L^{2}}^{2}\right)^{1/2}.$

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

Fix $T>0$. Let $f\in L^{2}(0,T;L^{2}(\mathbb{T}))$ and $g\in H^{1}(\mathbb{T})$; as $H^{1}(\mathbb{T})\subset C^{0}(\mathbb{T})$, 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

 $f_{n}(x,t)=\sum_{k=-n}^{n}\hat{f}(k,t)e^{ikx},$

and for all $x$ define $g_{n}$ by

 $g_{n}(x)=\sum_{k=-n}^{n}\hat{g}(k)e^{ikx}.$

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

 $f_{n}(x,t)=(D_{n}*f(\cdot,t))(x),\qquad g_{n}(x)=(D_{n}*g)(x),$

where

 $(\phi*\psi)(x)=\frac{1}{2\pi}\int_{\mathbb{T}}\phi(y)\psi(x-y)dy.$

## 2 Truncation

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

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

and for all $x\in\mathbb{T}$,

 $u_{n}(x,0)=g_{n}(x).$

We will thus obtain a formula for $u_{n}$. In fact we will not necessarily have $u_{n}\in C^{\infty}(0,T;C^{\infty}(\mathbb{T}))$, 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 $x$,

 $\sum_{k\in\mathbb{Z}}\widehat{u_{n}}^{\prime}(k,t)e^{ikx}+\sum_{k\in\mathbb{Z}% }k^{2}\widehat{u_{n}}(k,t)e^{ikx}=\sum_{k=-n}^{n}\hat{f}(k,t)e^{ikx}.$

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 $|k|\leq n$, then for almost all $t$, $\widehat{u_{n}}^{\prime}(k,t)+k^{2}\widehat{u_{n}}(k,t)=\hat{f}(k,t)$. 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

 $u_{n}\in C^{0}(0,T;C^{\infty}(\mathbb{T})).$

## 3 π»ΒΉ

For almost all $t$, multiply (1) by $u_{n}(x,t)$ and integrate over $\mathbb{T}$. This is,

 $\int_{\mathbb{T}}u_{nt}(x,t)u_{n}(x,t)dx-\int_{\mathbb{T}}u_{nxx}(x,t)u_{n}(x,% t)dx=\int_{\mathbb{T}}f_{n}(x,t)u_{n}(x,t)dx.$

Integrating by parts this becomes

 $\int_{\mathbb{T}}u_{nt}(x,t)u_{n}(x,t)dx+\int_{\mathbb{T}}u_{nx}(x,t)u_{nx}(x,% t)dx=\int_{\mathbb{T}}f_{n}(x,t)u_{n}(x,t)dx,$

which is

 $\pi\cdot\partial_{t}\frac{1}{2\pi}\int_{\mathbb{T}}u_{n}(x,t)^{2}dx+2\pi\cdot% \frac{1}{2\pi}\int_{\mathbb{T}}u_{nx}(x,t)^{2}dx=\int_{\mathbb{T}}f_{n}(x,t)u_% {n}(x,t)dx.$

Writing this using norms,

 $\pi\cdot\partial_{t}\left\|u_{n}(\cdot,t)\right\|_{L^{2}}^{2}+2\pi\cdot\left\|% u_{nx}(\cdot,t)\right\|_{L^{2}}^{2}=\int_{\mathbb{T}}f_{n}(x,t)u_{n}(x,t)dx.$

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

 $\pi\cdot\left\|u_{n}(\cdot,t)\right\|_{L^{2}}^{2}-\pi\cdot\left\|u_{n}(\cdot,0% )\right\|_{L^{2}}^{2}+2\pi\int_{0}^{t}\left\|u_{nx}(\cdot,s)\right\|_{L^{2}}^{% 2}ds=\int_{0}^{t}\int_{\mathbb{T}}f_{n}(x,s)u_{n}(x,s)dxds.$

For almost all $s$,

 $\displaystyle\int_{\mathbb{T}}|f_{n}(x,s)u_{n}(x,s)|dx$ $\displaystyle=$ $\displaystyle 2\pi\cdot\frac{1}{2\pi}\int_{\mathbb{T}}|f_{n}(x,s)u_{n}(x,s)|dx$ $\displaystyle\leq$ $\displaystyle 2\pi\cdot\left\|f_{n}(\cdot,s)\right\|_{L^{2}}\left\|u_{n}(\cdot% ,s)\right\|_{L^{2}}$ $\displaystyle\leq$ $\displaystyle 2\pi\left(\frac{\left\|f_{n}(\cdot,s)\right\|_{L^{2}}^{2}}{2}+% \frac{\left\|u_{n}(\cdot,s)\right\|_{L^{2}}^{2}}{2}\right)$ $\displaystyle=$ $\displaystyle\pi\cdot\left\|f_{n}(\cdot,s)\right\|_{L^{2}}^{2}+\pi\cdot\left\|% u_{n}(\cdot,s)\right\|_{L^{2}}^{2}.$

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

 $\begin{split}&\displaystyle\pi\cdot\left\|u_{n}(\cdot,t)\right\|_{L^{2}}^{2}-% \pi\cdot\left\|u_{n}(\cdot,0)\right\|_{L^{2}}^{2}+2\pi\int_{0}^{t}\left\|u_{nx% }(\cdot,s)\right\|_{L^{2}}^{2}ds\\ \displaystyle\leq&\displaystyle\int_{0}^{t}\pi\cdot\left\|f_{n}(\cdot,s)\right% \|_{L^{2}}^{2}+\pi\cdot\left\|u_{n}(\cdot,s)\right\|_{L^{2}}^{2}ds,\end{split},$

so, as $u_{n}(x,0)=g_{n}(x)$,

 $\left\|u_{n}(\cdot,t)\right\|_{L^{2}}^{2}+2\int_{0}^{t}\left\|u_{nx}(\cdot,s)% \right\|_{L^{2}}^{2}ds\leq\left\|g_{n}\right\|_{L^{2}}^{2}+\int_{0}^{t}\left\|% f_{n}(\cdot,s)\right\|_{L^{2}}^{2}+\left\|u_{n}(\cdot,s)\right\|_{L^{2}}^{2}ds.$

Let

 $y(t)=\left\|u_{n}(\cdot,t)\right\|_{L^{2}}^{2}+2\int_{0}^{t}\left\|u_{nx}(% \cdot,s)\right\|_{L^{2}}^{2}ds.$

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

 $\displaystyle y(t)$ $\displaystyle\leq$ $\displaystyle\left\|g_{n}\right\|_{L^{2}}^{2}+\int_{0}^{t}\left\|f_{n}(\cdot,s% )\right\|_{L^{2}}^{2}ds+\int_{0}^{t}\left\|u_{n}(\cdot,s)\right\|_{L^{2}}^{2}ds$ $\displaystyle\leq$ $\displaystyle\left\|g_{n}\right\|_{L^{2}}^{2}+\int_{0}^{t}\left\|f_{n}(\cdot,s% )\right\|_{L^{2}}^{2}ds+\int_{0}^{t}y(s)ds.$

By Gronwallβs inequality, we get

 $y(t)\leq\left(\left\|g_{n}\right\|_{L^{2}}^{2}+\int_{0}^{t}\left\|f_{n}(\cdot,% s)\right\|_{L^{2}}^{2}ds\right)e^{t}.$

As $\left\|g_{n}\right\|_{L^{2}}\leq\left\|g\right\|_{L^{2}}$ and $\left\|f_{n}(\cdot,s)\right\|_{L^{2}}\leq\left\|f(\cdot,s)\right\|_{L^{2}}$ (these two facts follow from Parsevalβs identity), it follows that

 $y(t)\leq\left(\left\|g\right\|_{L^{2}}^{2}+\int_{0}^{t}\left\|f(\cdot,s)\right% \|_{L^{2}}^{2}ds\right)e^{t}.$

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

 $\displaystyle\left\|u_{n}(\cdot,t)\right\|_{L^{2}}^{2}+2\int_{0}^{t}\left\|u_{% nx}(\cdot,s)\right\|_{L^{2}}^{2}ds$ $\displaystyle\leq$ $\displaystyle\left(\left\|g\right\|_{L^{2}}^{2}+\int_{0}^{t}\left\|f(\cdot,s)% \right\|_{L^{2}}^{2}ds\right)e^{T}$ $\displaystyle\leq$ $\displaystyle\left(\left\|g\right\|_{L^{2}}^{2}+\left\|f\right\|_{L^{2}(0,T;L^% {2}(\mathbb{T})}^{2}\right)e^{T}$ $\displaystyle=$ $\displaystyle M.$

By Parsevalβs identity,

 $\sum_{k\in\mathbb{Z}}|\widehat{u_{n}}(k,t)|^{2}+2\int_{0}^{t}\sum_{k\in\mathbb% {Z}}|\widehat{u_{nx}}(k,s)|^{2}ds\leq M,$

hence for all $t$,

 $\sum_{k\in\mathbb{Z}}|\widehat{u_{n}}(k,t)|^{2}+2\int_{0}^{t}\sum_{k\in\mathbb% {Z}}k^{2}|\widehat{u_{n}}(k,s)|^{2}ds\leq M.$

If $k\leq n\leq m$, then $\widehat{u_{n}}(k,t)=\widehat{u_{m}}(k,t)$ for all $t$. Define $\hat{u}(k,t)$ by

 $\hat{u}(k,t)=\lim_{n\to\infty}\widehat{u_{n}}(k,t)=\widehat{u_{k}}(k,t).$

Thus, for all $t$,

 $\sum_{k\in\mathbb{Z}}|\hat{u}(k,t)|^{2}+2\int_{0}^{t}\sum_{k\in\mathbb{Z}}k^{2% }|\hat{u}(k,s)|^{2}ds\leq M.$ (2)

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

 $\int_{0}^{T}\sum_{k\in\mathbb{Z}}|\hat{u}(k,t)|^{2}+\sum_{k\in\mathbb{Z}}k^{2}% |\hat{u}(k,t)|^{2}dt\leq M^{\prime}.$

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

 $\int_{0}^{T}\left\|u(\cdot,t)\right\|_{H^{1}}^{2}dt\leq M^{\prime}.$

We have

 $\lim_{n\to\infty}\int_{0}^{T}\left\|u_{n}(\cdot,t)-u(\cdot,t)\right\|_{H^{1}}^% {2}dt=0,$

i.e.

 $\lim_{n\to\infty}\left\|u_{n}-u\right\|_{L^{2}(0,T;H^{1}(\mathbb{T}))}^{2}=0.$

## 4 π»Β²

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

 $\int_{\mathbb{T}}u_{nt}(x,t)u_{nxx}(x,t)dx-\int_{\mathbb{T}}u_{nxx}(x,t)u_{nxx% }(x,t)dx=\int_{\mathbb{T}}f_{n}(x,t)u_{nxx}(x,t)dx.$

As

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

we have

 $-\pi\frac{d}{dt}\left\|u_{nx}(\cdot,t)\right\|_{L^{2}}^{2}-2\pi\left\|u_{nxx}(% \cdot,t)\right\|_{L^{2}}^{2}=\int_{\mathbb{T}}f_{n}(x,t)u_{nxx}(x,t)dx.$

Integrating from $0$ to $t$,

 $\begin{split}&\displaystyle-\pi\left\|u_{nx}(\cdot,t)\right\|_{L^{2}}^{2}+\pi% \left\|u_{nx}(\cdot,0)\right\|_{L^{2}}^{2}-2\pi\int_{0}^{t}\left\|u_{nxx}(% \cdot,s)\right\|_{L^{2}}^{2}ds\\ \displaystyle=&\displaystyle\int_{0}^{t}\int_{\mathbb{T}}f_{n}(x,s)u_{nxx}(x,s% )dx.\end{split}$

For almost all $s$,

 $\displaystyle\int_{\mathbb{T}}|f_{n}(x,s)u_{nxx}(x,s)|dx$ $\displaystyle\leq$ $\displaystyle 2\pi\left\|f_{n}(\cdot,s)\right\|_{L^{2}}\left\|u_{nxx}(\cdot,s)% \right\|_{L^{2}}$ $\displaystyle\leq$ $\displaystyle 2\pi\left(\frac{\left\|f_{n}(\cdot,s)\right\|_{L^{2}}^{2}}{2}+% \frac{\left\|u_{nxx}(\cdot,s)\right\|_{L^{2}}^{2}}{2}\right)$ $\displaystyle=$ $\displaystyle\pi\left\|f_{n}(\cdot,s)\right\|_{L^{2}}^{2}+\pi\left\|u_{nxx}(% \cdot,s)\right\|_{L^{2}}^{2}.$

It follows that, for all $t$,

 $\begin{split}&\displaystyle\pi\left\|u_{nx}(\cdot,t)\right\|_{L^{2}}^{2}x+2\pi% \int_{0}^{t}\left\|u_{nxx}(\cdot,s)\right\|_{L^{2}}^{2}ds\\ \displaystyle\leq&\displaystyle\pi\left\|g_{n}^{\prime}\right\|_{L^{2}}^{2}+% \pi\int_{0}^{t}\left\|f_{n}(\cdot,s)\right\|_{L^{2}}^{2}+\left\|u_{nxx}(\cdot,% s)\right\|_{L^{2}}^{2}ds.\end{split}$

Hence

 $\displaystyle\left\|u_{nx}(\cdot,t)\right\|_{L^{2}}^{2}+\int_{0}^{t}\left\|u_{% nxx}(\cdot,s)\right\|_{L^{2}}^{2}ds$ $\displaystyle\leq$ $\displaystyle\left\|g_{n}^{\prime}\right\|_{L^{2}}^{2}+\int_{0}^{t}\left\|f_{n% }(\cdot,s)\right\|_{L^{2}}^{2}ds$ $\displaystyle\leq$ $\displaystyle\left\|g\right\|_{H^{1}}^{2}+\int_{0}^{t}\left\|f(\cdot,s)\right% \|_{L^{2}}^{2}ds$ $\displaystyle\leq$ $\displaystyle\left\|g\right\|_{H^{1}}^{2}+\int_{0}^{T}\left\|f(\cdot,s)\right% \|_{L^{2}}^{2}ds.$

Using Parsevalβs identity we have, for all $t$,

 $\sum_{k\in\mathbb{Z}}|\widehat{u_{nx}}(k,t)|^{2}+\int_{0}^{t}\sum_{k\in\mathbb% {Z}}|\widehat{u_{nxx}}(k,s)|^{2}ds\leq\left\|g\right\|_{H^{1}}^{2}+\int_{0}^{T% }\left\|f(\cdot,s)\right\|_{L^{2}}^{2}ds,$

hence

 $\sum_{k\in\mathbb{Z}}k^{2}|\widehat{u_{n}}(k,t)|^{2}+\int_{0}^{t}\sum_{k\in% \mathbb{Z}}k^{4}|\widehat{u_{n}}(k,s)|^{2}ds\leq\left\|g\right\|_{H^{1}}^{2}+% \int_{0}^{T}\left\|f(\cdot,s)\right\|_{L^{2}}^{2}ds,$

so

 $\sum_{k\in\mathbb{Z}}k^{2}|\widehat{u}(k,t)|^{2}+\int_{0}^{t}\sum_{k\in\mathbb% {Z}}k^{4}|\widehat{u}(k,s)|^{2}ds\leq\left\|g\right\|_{H^{1}}^{2}+\int_{0}^{T}% \left\|f(\cdot,s)\right\|_{L^{2}}^{2}ds.$

It follows that, for almost all $t$,11 1 The reason I see that this follows involves the fact that the intersection of two sets of full measure is itself a set of full measure.

 $\sum_{k\in\mathbb{Z}}k^{2}|\widehat{u}(k,t)|^{2}+\sum_{k\in\mathbb{Z}}k^{4}|% \widehat{u}(k,t)|^{2}<\infty,$

thus $u(\cdot,t)\in H^{2}(\mathbb{T})$.

We have

 $\lim_{n\to\infty}\int_{0}^{T}\left\|u_{n}(\cdot,t)-u(\cdot,t)\right\|_{H^{2}}^% {2}dt=0,$

i.e.

 $\lim_{n\to\infty}\left\|u_{n}-u\right\|_{L^{2}(0,T;H^{2}(\mathbb{T}))}^{2}=0.$

## 5 Solution

For all $t$ we have $u(\cdot,t)\in H^{1}(\mathbb{T})$, and $H^{1}(\mathbb{T})\subset C^{0}(\mathbb{T})$, 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 $H^{k}(\mathbb{T})\subset C^{\alpha}(\mathbb{T})$. So, being specific, we have $H^{1}(\mathbb{T})\subset C^{1/4}(\mathbb{T})$. It is a fact that if $h\in C^{\alpha}(\mathbb{T})$, $\alpha>0$, then the partial sums of the Fourier series of $h$ converge to $h$ in the supremum norm.

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

 $\hat{u}(k,t)=e^{-k^{2}t}\widehat{g}(k)+e^{-k^{2}t}\int_{0}^{t}e^{k^{2}s}% \widehat{f}(k,s)ds.$

It follows that, for all $x$,

 $u(x,0)=\lim_{N\to\infty}\sum_{|k|\leq N}\widehat{g}(k)e^{ikx}.$

On the other hand,

 $g(x)=\lim_{N\to\infty}\sum_{|k|\leq N}\widehat{g}(k)e^{ikx}.$

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

We have

 $\displaystyle\left\|u_{t}-u_{xx}-f\right\|_{L^{2}(0,T;L^{2}(\mathbb{T}))}$ $\displaystyle\leq$ $\displaystyle\left\|u_{t}-u_{nt}\right\|_{L^{2}(0,T;L^{2}(\mathbb{T}))}$ $\displaystyle+\left\|u_{xx}-u_{nxx}\right\|_{L^{2}(0,T;L^{2}(\mathbb{T}))}$ $\displaystyle+\left\|f-f_{n}\right\|_{L^{2}(0,T;L^{2}(\mathbb{T}))}$ $\displaystyle+\left\|u_{nt}-u_{nxx}-f_{n}\right\|_{L^{2}(0,T;L^{2}(\mathbb{T})% )}.$

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

 $\displaystyle\widehat{u_{t}}(k,t)-\widehat{u_{nt}}(k,t)$ $\displaystyle=$ $\displaystyle\sum_{|k|>n}-k^{2}e^{-k^{2}t}\hat{g}(k)-k^{2}e^{-k^{2}t}\int_{0}^% {t}e^{k^{2}s}\hat{f}(k,s)ds$ $\displaystyle+e^{-k^{2}t}e^{k^{2}t}\hat{f}(k,t)$ $\displaystyle=$ $\displaystyle\sum_{|k|>n}-k^{2}\hat{u}(k,t)+\hat{f}(k,t).$

Then using Parsevalβs identity,

 $\displaystyle\left\|u_{t}-u_{nt}\right\|_{L^{2}(0,T;L^{2}(\mathbb{T}))}^{2}$ $\displaystyle=$ $\displaystyle\int_{0}^{T}\sum_{|k|>n}|-k^{2}\hat{u}(k,t)+\hat{f}(k,t)|^{2}dt$ $\displaystyle\leq$ $\displaystyle 2\int_{0}^{T}\sum_{|k|>n}k^{2}|\hat{u}(k,t)|^{2}+|\hat{f}(k,t)|^% {2}dt.$

Then,

 $\left\|u_{t}-u_{xx}-f\right\|_{L^{2}(0,T;L^{2}(\mathbb{T}))}^{2}=0.$

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

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