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I know that one may find continuous $f$ (say in a ball $\bar B$ centered at $0$) such that there exists no classical ($\mathcal C^2$) solution $u$ to the Poisson equation $\Delta u =f$ in $B$, and obviously this requires to be in dimension 2 or more. (But a classical solution exists as soon as $f$ is assumed Hölder continuous). This counterexample may be found for instance in Qi Han's book "A Basic Course in Partial Differential Equations" (link) pp.136-137.

Now it is easy to see that for such an $f$ there exists no classical solution to the inhomogeneous heat equation $$u_t - \Delta u = - f(x) \;\;\;\mbox{ on } (0,T) \times B.$$

But the case of one space dimension seems to require a new argument.

So my question is this :

Is there a continuous function $f$ on $[0,T] \times [-a,a] $ s.t. the equation $u_t - u_{xx}=f(t,x)$ has no classical solution ?

(I know that if $f$ is Hölder continuous in $x$, uniformly in $t$, then one may prove by explicit formulas that there is a classical solution.)

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@WillieWong : I mean $C^{1,2}((0,T) \times B)$, i.e. the partial derivatives $u_t$, $u_x$, $u_{xx}$ exist and are continuous on $(0,T) \times B$. (Not sure if it coincides with your space). –  pgassiat May 15 '12 at 9:59

1 Answer 1

up vote 3 down vote accepted
+50

An example can be constructed analogically to the elliptic case.

Let $u(t,x)=\left(2 t+x^2\right) \sqrt{\log \left(\frac{1}{t^2+x^4}\right)}\ $ in some neighbourhood of the origin for $(t,x)\ne0$ and $u(0,0)=0\,$. Denote $$ f(t,x)=u_t(t,x)-u_{xx}(t,x)= $$ $$ \frac{\left(-2 t^4+11 t^3 x^2+12 t^2 x^4-5 t x^6+6 x^8\right) \log \left(\frac{1}{t^2+x^4}\right)+4 x^6 \left(2 t+x^2\right)}{\left(t^2+x^4\right)^2 \log ^{\frac{3}{2}}\left(\frac{1}{t^2+x^4}\right)}. $$ Using Young's inequality $$ |ab|\le \frac{|a|^p}p+\frac{|b|^q}q, $$ there $q=p/(p-1)\,$, it is straightforward to estimate monomials in the numerator by the denominator polynomial $(t^2+x^4)^2$. For example, for $t^3 x^2$ we can put $p=4/3$ to obtain $t^4$ from $t^3$. Then $q=4$ and $$ |t^3 x^2|\le \frac{t^4}{4/3}+\frac{(x^2)^4}{4}\le C(t^2+x^4)^2. $$ So, defining $f(0,0)=0$ we have a function, continuous in some neighbourhood of the origin.

From the other hand $$ u_t(t,x)= 2 \sqrt{\log \left(\frac{1}{t^2+x^4}\right)}-\frac{t \left(2 t+x^2\right)}{\left(t^2+x^4\right) \sqrt{\log \left(\frac{1}{t^2+x^4}\right)}} $$ and $$ \lim_{(x,t)\to0}u_t(x,t)=+\infty $$ due to the first summand.

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Perfect, thanks. The bounty will be yours when the obligatory 24 hrs have passed. –  pgassiat May 15 '12 at 13:28

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