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Let $u(x)$ be a $C^2$ solution to $$-\Delta u(x)=|x|^2 \hspace{0.25cm} \text{ on } \mathbb{R}^n$$

  • Show that $\Phi * |y|^2$ does not make sense.
  • Find a solution nevertheless. Look for a polynomial.
  • Set $m(r)= \mathrel{\int\!\!\!\!\!\!-}_{\partial B(0,r)} u(y)\,dS(y)$. Show that $$m(r)=u(0)+\dfrac{r^4}{4(n+2)}.$$

We "prove" previously,

Let $f\in C^2(\mathbb{R}^n)$. Assume that \begin{equation*} \begin{aligned} \int_{\mathbb{R}^n} |f(y)|(1+|y|)^{2-n}\,dy<\infty & \hspace{1cm} \text{ for } n>3 \\ \int_{\mathbb{R}^n} |f(y)|\ln(1+|y|)\,dy<\infty & \hspace{1cm} \text{ for } n=2 \\ \end{aligned} \end{equation*} Prove that $u(x)=\Phi * f(x)\in C^2(\mathbb{R}^n)$ and $-\Delta u(x)=f(x)$.

The first bullet, I think that the reason why it doesn't make sense is because when you use $f(y)=|y|^2$ both of these integrals diverge. Is this correct?

Also, I can't think of a polynomial that is a solution to this. Can I get some help?

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1 Answers1

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For the Laplacian of $u$ to be a 2nd degree polynomial, we need $u$ to be a 4th degree polynomial. $u(x) = x_1^4+\dots+x_n^4$ does the job, up to a constant ($\Delta u = 12|x|^2$).

To prove the statement about $m(r)$, compute $m'(r)$ using the divergence theorem. Up to some constants, $$ m'(r) = \frac{1}{r^{n-1}} \int_{|x|<r}\Delta u = - \frac{1}{r^{n-1}} \int_{|x|<r} |x|^2 = -\frac{c_n}{r^{n-1}}\int_0^r t^2 t^{n-1}\,dt = -\frac{c_n}{r^{n-1}} \frac{r^{n+2}}{n+2} $$ So the derivative is a multiple of $r^3$, hence the conclusion.

The function $u$ can't be obtained by convolving $|x|^2$ with the fundamental solution $|x|^{2-n}$ because the integral diverges: the contribution from scale $|x-y|\approx 2^k$ is about $2^{(n-1)k}$ (from volume) times $2^{2k}$ (from the function) times $2^{(2-n)k}$ (from the kernel), totaling $ 2^{3k}$. Without computing this, one can observe that $|x|^{2-n}$ is already non-integrable at infinity, so it needs help to converge and $|x|^2$ is no help.