# Backwards heat equation (stability analysis)

## 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?

• What conditions are there on $f(x)$? – DaveNine May 16 '15 at 8:28
• 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 $A_m=\frac{2}{L}\exp((\frac{\pi m}{L})\lambda^2T)\int_0^L\sin(\frac{\pi mx}{L})f(x)\,dx$. – Ellya May 16 '15 at 21:58
• @ellya thanks for the correction – Vlad May 18 '15 at 1:40

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.
• 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. – Vlad May 18 '15 at 1:56