I was searching around here for a while but I didn't find any discussion on this topic. We know that a problem is well-posed when

  1. There is a solution (existence),
  2. There is only one solution (uniqueness),
  3. The solution depends continuously on the data for the problem (stability).

I suppose that as soon as either one of these conditions is not satisfied, the problem becomes ill-posed. Now as I was reading about this topic I came across a statement that said

To show ill-posedness of a given problem you must find a family of solutions with unity $L_{2}$ norm of the initial data whose growth rate is unbounded, e.g. backward heat equation.

But what does this statement say exactly? In particular, what does "family of solutions with unity $L_{2}$ norm of the initial data" mean? How do we see this for the backward heat equation for instance?

For example, as an exercise they want you to show that the problem

$$ \begin{align} \frac{\partial u}{\partial t}(x,t) &= \alpha \frac{\partial^{2}u}{\partial x^{2}}(x,t) + \beta \frac{\partial^{4}u}{\partial x^{4}}(x,t), &x\in\mathbb{R},\; t>0,\\[3mm] u(x,0)&=\sin x, &x\in\mathbb{R},\\[3mm] u(x,t)&=u(x+2\pi,t), &x\in\mathbb{R},\;t>0 \end{align} $$

($2\pi$-periodic Cauchy problem) where $\beta>0$ and $\alpha\in\mathbb{R}$ is ill-posed. I made an attempt on a solution, but I am not sure what my conclusions become (or if they are right):

Make the Fourier ansatz $u(x,t)=\sum_{k=0}^{\infty}\hat{u}_{k}(t)\cos(kx)\implies \hat{u}_{k}(t)=\frac{1}{\pi}\int_{-\pi}^{\pi}u(x,t)\cos(kx)dx$.

Thus we have $\hat{u}_{k}(0)=\frac{1}{\pi}\int_{-\pi}^{\pi}u(x,0)\cos(kx)dx=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin x\cos(kx)dx \stackrel{?}{=}0$ for all $k$.

Space derivatives of ansatz become

$$\frac{\partial^{2}u}{\partial x^{2}}(x,t) = -\sum_{k=0}^{\infty}k^{2}\hat{u}_{k}(t)\cos(kx)$$


$$\frac{\partial^{4}u}{\partial x^{4}}(x,t) = \sum_{k=0}^{\infty}k^{4}\hat{u}_{k}(t)\cos(kx)$$

Insertion into PDE yields

$$ \begin{align} \sum_{k=0}^{\infty}\frac{\partial \hat{u}_{k}}{\partial t}(t)\cos(kx) &= -\alpha \sum_{k=0}^{\infty}k^{2}\hat{u}_{k}(t)\cos(kx) + \beta \sum_{k=0}^{\infty}k^{4}\hat{u}_{k}(t)\cos(kx). \end{align} $$

Comparison of coefficients and with initial conditions given by $\hat{u}_{k}(0)$ we obtain the solutions

$$ \begin{align} \hat{u}_{k}(t) = \hat{u}_{k}(0) e^{(-\alpha k^{2} + \beta k^{4})t}. \end{align} $$

We see that this inserted into the ansatz implies that the solution $u(x,t)$ grows unbounded for any $\alpha$ and $\beta>0$, but the ansatz or initial conditions $\hat{u}_{k}(0)$ seem off (everything just becomes zero?). And have I implicitly showed that the problem is ill-posed? Is the $L_{2}$ norm of the initial data at unity?


1 Answer 1


Your calculation is fine. You have discovered data $u(x,0)=\cos(kt)$ whose solution becomes unbounded. Now consider initial $v(x)$ having unit initial norm and combine them into initial data $$U(x,0)=v(x)+\epsilon \cos(kt)$$ You can make $\epsilon$ small enough that the initial data remains as close to unity as you wish. But, however small you make $\epsilon$, you can choose $k$ to make the solution at time $t$ as large as you like. This is ill-posedness.

  • $\begingroup$ I think I understand, but you mean $u(x,0)=\sin(kx)$ right? And why is the $k$ parameter introduced, to construct a family of initial data? And also what is the importance of $v(x)$? Why can we not just choose $v(x)=0$? Thanks for the answer! $\endgroup$ Commented Apr 19, 2020 at 7:14
  • 1
    $\begingroup$ I can choose $v(x)$ so that it satisfies the required initial condition, and then I can perturb it so that it still satisfies the initial condition (to any accuracy that you wish), but grows without limit. To do that I must choose $k$ large enough. $\sin$ or $\cos$ doesnt matter $\endgroup$
    – Philip Roe
    Commented Apr 19, 2020 at 18:49

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