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Problem

Consider the backwards heat equation of the form $$ \left\{ \begin{aligned} u_{t} &= \lambda^2 u_{xx}, & x\in[0,L], \quad t\in[0,T]\\ u(0,t) &= u(L,t) = 0 \\ u(x,T) &= f(x), \end{aligned} \right.\tag{*}\label{*}$$ Establish whether solution is unique and analyze its stability.

Attempt of proving uniqueness

My attempt to prove uniqueness is provided in this post.

Attempt of (dis)proving stability

The general solution of $\eqref{*}$ is of the form $$ u(x,t) = \sum_{m=1}^{\infty} A_m \sin\bigg( \frac{\pi m }{L}\,x\bigg) \exp \Bigg(\!\!-\!\bigg(\frac{\pi m }{L}\bigg)^2 \lambda^2 \big(T -t \big) \Bigg)\\ A_m = \frac{2}{L} \int_0^L \sin \!\bigg( \frac{\pi m }{L}\,x\bigg)\, f(x)\,dx $$ I think the solution is not stable in $L^p$ sense, so I need to come up with a good counterexample of the sequence of initial data $f_n(x)\to f(x)$ converging in $L_p$, so that it would not be difficult to show that corresponding solutions do not converge in $L_p$.

Could anyone propose such an example?

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  • $\begingroup$ What conditions are there on $f(x)$? $\endgroup$ – DaveNine May 16 '15 at 8:28
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    $\begingroup$ The general solution is of the form $u(x,t)=\sum_{m=1}^\infty A_m\sin(\frac{\pi mx}{L})\exp(-(\frac{\pi m}{L})\lambda^2t)$, where $A_m=\frac{2}{L}\exp((\frac{\pi m}{L})\lambda^2T)\int_0^L\sin(\frac{\pi mx}{L})f(x)\,dx$. $\endgroup$ – Ellya May 16 '15 at 21:58
  • $\begingroup$ @ellya thanks for the correction $\endgroup$ – Vlad May 18 '15 at 1:40
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Use a sequence of harmonics with increasing frequency and decreasing amplitude: $$f_n(x)= A_n \sin \left(\frac{\pi n}{L} x\right)$$ As long as $A_n\to 0$, the values at time $T$ tend to zero in the $L^p$ sense. On the other hand, the solution at time $t<T$ is $$f_n(x)= A_n e^{\lambda (\pi n/L)^2 (T-t)}\sin \left(\frac{\pi n}{L} x\right)$$ which, for most natural choices of $A_n$, blows up in $L^p$ norm.

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  • $\begingroup$ Thank you! I was looking at something like $$f_n(x) = \sin\left(\frac{\pi x}{L}\right) + \frac{1}{n}\ln x,$$ but your example is even easier. $\endgroup$ – Vlad May 18 '15 at 1:56

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