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I would like to know why the following equation is ill posed . $u_t=-ku_{xx}$ and $u(0,t)=u(L,t)=0$

Here , $u_t, u_{xx} $ denotes the partial derivatives of u with respect to time and space respectively . Thank you for your help.

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If that is of any help, here is a non-trivial family of solutions to your PDE : $$ u_n(x,t) = \sinh(2\pi N x/L) e^{-k \frac{4N^2\pi^2 }{L^2}t} $$ Don't ask me how I got the solution, I actually just tried things some functions and found this. Does that help understanding why the problem is ill-posed? (I don't know what "ill-posed" stands for.) –  Patrick Da Silva Jun 26 '12 at 9:28
    
@PatrickDaSilva: Not quite right, since the condition at $x=L$ is not satisfied. Take sin instead of sinh, and change the sign in the exponent to plus (this sign is the reason for the illposedness, by the way). –  Hans Lundmark Jun 26 '12 at 9:32
    
@PatrickDaSilva : I am not able to conclude with the example even after changing $sinh$ to $sin$. Can you give a brief discription. –  Theorem Jun 26 '12 at 10:10
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1 Answer

up vote 6 down vote accepted

The notion of wellposedness and illposedness depends strongly on which function space in which you are studying the equation.

For example, let $X_M$ denote the space of functions $$ \left\{ f \in C^\infty([0,L]) \mid f(0) = 0,\ f(L) = 0,\ \text{and} \forall |\xi| > M, \hat{f}(\xi) = 0 \right\} $$ where for $\xi \in \mathbb{Z}$, $$ \hat{f}(\xi) = \int_0^L \exp\left( -\frac{ 2\pi i \xi x}{L}\right) f(x) \mathrm{d}x $$ is the Fourier coefficient of $f$ with frequency $2\pi \xi/ L$.

Then the time-reversed heat equation $u_t = -k u_{xx}$ is in fact well posed for initial data in $X_M$. Very simply, we see that by taking the Fourier transform the equation becomes $$ \hat{u}_t(\xi) = k \xi^2 \hat{u}(\xi) $$ a finite system of decoupled ordinary differential equations. So we can indeed write that, for initial data $u(0,x) = \sum_{|\xi| \leq M} a_\xi \exp \frac{2\pi i \xi x}{L}$ the solution is

$$ u(t,x) = \sum_{|\xi| \leq M} a_\xi \exp \left(\frac{2\pi i \xi x}{L}\right) \exp \left( k\xi^2 t\right)$$

which, while growing exponentially, is nonetheless a $C^\infty$ function on $[0,T]\times [0,L]$ for any $T < \infty$, solves the equations, and fits the boundary conditions.


On the other hand, we often think of the time-reversed heat equation as a prototypical example of an ill-posed PDE for which existence is not guaranteed: for this we will have to use a more "typical" function space. (The function space $X_M$ above for every $M$ is a subspace of the real analytic functions, and so we expect it to behave much nicer than "typical" function spaces.)

Let $X^s_p$ be the Sobolev space $$ \left\{ f \in C^\infty([0,L]) \mid f(0) = f(L) = 0, \|( (1 + \xi^2)^{s/2} \hat{f}(\xi) )\|_p < \infty \right\} $$ where $$ \| (1+\xi^2)^{s/2} \hat{f}(\xi))\|_p = \begin{cases} \left(\sum_{\xi \in \mathbb{Z}} (1+\xi^2)^{sp/2} |\hat{f}(\xi)|^{p}\right)^{\frac1p} & 1 \leq p < \infty, 0 \leq s < \infty \\ \sup_{\xi\in\mathbb{Z}}(1 + \xi^2)^{s/2}|\hat{f}(\xi)| & p = \infty, 0 \leq s < \infty \\ \sup_{\xi \in \mathbb{Z}} \left(\exp |\xi|\right) |\hat{f}(\xi)| & s = \infty \end{cases}$$ The $s = \infty$ case is more-or-less the space of $C^\infty$ functions on $[0,L]$ vanishing on the boundaries. The $p = 1$ case when $s$ is a positive integer roughly corresponds to the $C^s$ functions. And with $p\in (1,\infty)$ and $s \in [0,\infty)$ the $X^s_p$ spaces corresponds roughly to the $W^{s,q}_0([0,L])$ Sobolev space with $q = p / (p-1)$.

Ill-posedness now takes the following form: For any $(s,p) \in [0,\infty]\times[1,\infty]$, we can find some initial data $u(0,x) \in X^s_p$ such that at any time $t > 0$, any possible solution to the equation cannot possibly remain in the space $X^s_p$. This follows immediately from the Fourier transform version of the equation: Let

$$ u(0,x) = \sum \exp ( - |\xi|^{3/2} ) \exp ( 2\pi i \xi x / L ) $$

a solution will formally be obtained as

$$ u(t,x) = \sum \exp ( kt|\xi|^2 - |\xi|^{3/2}) \exp (2\pi i \xi x / L) $$

which for any $t > 0$ means that the Fourier coefficients of $u(t,x)$ must grow superexponentially as $|\xi| \to \infty$, and hence cannot belong to any Sobolev space.

Exercise/food for thought: show that for any natural number $n$, there exists an initial data in $C^n$ such that the formal solution above (derived from the Fourier transform) does not (cannot) converge. Show that this also implies that any "solution" to this initial data cannot be continuous. (Hint: Riemann-Lebesgue Lemma).


What is the intuition? By looking at the Fourier transform version of the equation, we see that each frequency mode evolves independently of other frequencies. Furthermore, the magnitude of each frequency grow exponentially with rate proportional to the square of the frequency. In the first case we constrained our space so that our functions have bounded frequencies. This forces the solution to grow at most exponentially with an upper bound on the rate of growth. If you are in a more typical function space where there can be small but nonvanishing contributions to arbitrarily high frequencies, however, those extremely high frequencies gets immediately (for arbitrarily small but positive $t$) magnified by an extremely large amount, causing a sort of instability in the problem.

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:Thank you Willie. Can you tell me why as Patrick gave a solution does indeed say that the problem is ill posed ? –  Theorem Jun 26 '12 at 10:05
    
Patrick's solution fits into the case one above: for that particular data ($u(0,x) = \sin 2\pi N x / L$) the solution exists; it in fact is a special case of the frequency limited case. (That data belongs to $X_M$ for any $M > N$.) What he (and Hans) is getting at is that the exponential growth rate in time can become extremely large as the frequency $N$ gets large, which after some fiddling (as in the second case above) gives you ill-posedness. –  Willie Wong Jun 26 '12 at 10:46
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