# uniqueness of the solution of heat equation in convolution form

I wish show the solution expressed in the form $$u(t,x)=\int_{\mathbb{R}}\Phi_t(x-y)f(y)\,dy$$ is unique for any $f\in\mathcal{S}(\mathbb{R})$ (the space of rapidly decreasing functions in $\mathbb{R}$.)

Here, $u$ is a classical solution satisfies Consider an equation on $\mathbb{R}$: \begin{align*}\tag{1} \begin{cases} \frac{\partial u}{\partial t}(t,x)&=\frac{\partial^2 u}{\partial x^2}(t,x) \qquad t>0, x\in\mathbb{R}^d, \\ u(0,x)&=f(x),\qquad x\in\mathbb{R}^n \end{cases} \end{align*}

Here we start with assuming there exist two solutions, $u_1$ and $u_2$ both satisfies (1). Then I want to find the differential inequality for the function $W(t)=|u_2(t)-u_1(t)|=\int_{\mathbb{R}}|u_2(t,x)-u_1(t,x)|\,dx$ and the Gronwall inequality.

So I have \begin{align*} w(t)&=\int_{\mathbb{R}}|u_2(t)-u_1(t)|\,dx\\ \Longrightarrow [w(t)]^2&=\left(\int_{\mathbb{R}}|u_2(t)-u_1(t)|\,dx\right)^2 \end{align*}

Then $$\frac{1}{2}\frac{d}{dt}F^2(t)=F(t)\frac{d}{dt}F(t),$$ now I am lost...any help appreciated.

• What is that $S$? – user99914 Oct 13 '15 at 8:52
• please see edited – math101 Oct 13 '15 at 8:55

## 2 Answers

Uniqueness does not hold for the heat equation. An example was constructed by Tychonov. You can find it in chapter 7 of Fritz John's classical book Partial Differential Equations.

There is uniqueness in the class of functions bounded by a quadratic exponential. There is also uniqueness in the class of non-negative functions (Widder's theorem.)

• I wish to complete what I was doing in the sense of the 2nd part of your answer...hope you could help – math101 Oct 13 '15 at 11:25
• The standard proof uses the maximum principle. – Julián Aguirre Oct 13 '15 at 11:32
• yes, it is in evans book and in everywhere...but I want something in ODE sense, like the uniqueness of weak solution...something similar to bottom p.187 to top p.188...math.ucdavis.edu/~hunter/pdes/ch6.pdf – math101 Oct 13 '15 at 11:36
• You can apply the $L^2$ estimate on the ball of radius $R$ and let $R\to\infty$. You will need to justify that the boundary terms allow for this. – Julián Aguirre Oct 13 '15 at 16:14
• Does the equality $|u_2(t)-u_1(t)|=\int_{\mathbb{R}}|u_2(t,x)-u_1(t,x)|\,dx$ hold? – math101 Oct 14 '15 at 13:56

\begin{eqnarray*} \partial _{t}u(t,x) &=&\partial _{x}^{2}u(t,x) \\ u(0,x) &=&f(x) \end{eqnarray*} \begin{eqnarray*} v(t,x) &=&u_{1}(t,x)-u_{2}(t,x) \\ \partial _{t}v(t,x) &=&\partial _{x}^{2}v(t,x) \\ v(0,x) &=&0 \end{eqnarray*} For the Fourier transform $\tilde{v}(t,k)$we have

\begin{eqnarray*} \partial _{t}\tilde{v}(t,k) &=&-k^{2}\tilde{v}(t,k) \\ \tilde{v}(0,k) &=&0 \end{eqnarray*} Then \begin{equation*} \tilde{v}(t,k)=\exp [-k^{2}t]\tilde{v}(0,k)=0 \end{equation*} so $v(t,x)$ must vanish.

• You did not mention that. – Urgje Oct 13 '15 at 11:38