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For the heat equation $(\partial_t-\partial_x^2)f(t,x)=0$ defined on $[0,T)\times(-\infty,\infty)$, to obtain uniqueness of the initial value problem, usually it is required to limit the growth of the potential solution at infinity, i.e. $|f(t,x)|<\exp(c\cdot x^2)$. My question is, if we do not impose any such conditions, is uniqueness no longer valid? In particular, is there a well known example of a function $f(t,x)$ that satisfies the heat equation on $[0,T)\times(-\infty,\infty)$, $f(0,x)=0$, but $f$ is not identically zero?

What if we relaxed the conditions a little bit, and only required that $f$ satisfies the heat equation in $(0,T)\times(-\infty,\infty)$ and is continuous on $[0,T)\times(-\infty,\infty)$, is there an example in this case?

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up vote 5 down vote accepted

The type of solutions you are looking for were constructed by Tychonov. You can find them in chapter 7 of Fritz John's classical book Partial Differential Equations.

Uniqueness follows also from one sided inequalities like in Widder's theorem: if $f(x,t)$ is a non-negative solution of the heat equation with $f(x,0)=0$, then $f\equiv0$. You can find a proof again in chapter 7 of Fritz John's book.

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Of course use separation of variables:

Let $f(t,x)=T(t)X(x)$ ,

Then $T'(t)X(x)-T(t)X''(x)=0$

$T'(t)X(x)=T(t)X''(x)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore f(t,x)=C_1x+C_2+\int_sC_3(s)e^{-ts^2}\sin xs~ds+\int_sC_4(s)e^{-ts^2}\cos xs~ds$ or $f(t,x)=C_1x+C_2+\sum_sC_3(s)e^{-ts^2}\sin xs+\sum_sC_4(s)e^{-ts^2}\cos xs$

This is already the general solution on $[0,\infty)\times(-\infty,\infty)$ , not only on $[0,T)\times(-\infty,\infty)$ .

If further $f(0,x)=0$ exist, it is the best to choose $f(t,x)=C_1x+C_2+\int_0^\infty C_3(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds$

Then substitute $f(0,x)=0$ :

$C_1x+C_2+\int_0^\infty C_3(s)\sin xs~ds+\int_0^\infty C_4(s)\cos xs~ds=0$

$C_1x+C_2+\mathcal{F}_{s,s\to x}\{C_3(s)\}+\mathcal{F}_{c,s\to x}\{C_4(s)\}=0$

$\mathcal{F}_{s,s\to x}\{C_3(s)\}=-C_1x-C_2-\mathcal{F}_{c,s\to x}\{C_4(s)\}$

$C_3(s)=C_1\delta'(s)-\dfrac{2C_2}{\pi s}-\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}$

$\therefore f(t,x)=C_1x+C_2+C_1\int_0^\infty\delta'(s)e^{-ts^2}\sin xs~ds-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds=C_1x+C_2-C_1\Biggl.\dfrac{d}{ds}(e^{-ts^2}\sin xs)\Biggr|_{s=0}-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds=C_1x+C_2-C_1x-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds=C_2-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds$

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This answer does not answer the question which asks for the uniqueness. This procedure is not even completed. I do not know where the claim "of course" comes from. – Hans Jan 13 at 18:54

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