I have been tasked with solving $\Delta(\Delta(u)) = f$ in $\mathbb{R}^d$ for $f \in C_c^4(\mathbb{R}^d)$.
Frankly, I don't know where to begin. The "naive" thing to do would be to say we can solve $\Delta v=f$ and then $\Delta u =v$, but the solution to $\Delta v=f$ is $$\int_{\mathbb{R}^d} \Phi(x-y)f(y) \ dy$$ (where $\Phi$ is the fundamental solution to Laplace's equation) which isn't $C_c^2$, which is all we know how to solve for Poisson's equation for (in Evans; I know that there are more general sufficient conditions).
So then a hands-on approach seems like the only way to go. Indeed, my professor said to find a $\Phi$ whose convolution with $f$ gives the solution. Yet, even in the case of $$\Delta(\Delta(u)) = 0$$ in $\mathbb{R}^2$, the problem seems unapproachable. I do not know how to solve $u_{xxxx} + 2u_{xxyy} +u_{yyyy} =0$. I have tried looking up results about this equation, but nothing seems helpful at all for the inhomogeneous $\mathbb{R}^d$ case. Can anyone point me in the right direction? Thanks
The biharmonic operator is a member of the wider class of polyharmonic operators, defined recursively as $$ \Delta^m=\Delta(\Delta^{m-1})\quad\forall m\in\mathbb{N}, m\geq 2 $$ where $\Delta=\frac{\partial^2}{\partial x_1^2}+\cdots+\frac{\partial^2}{\partial x_d^2}$, $d\geq 2$. The inhomogeneous equation involving the polyharmonic operator $$ \Delta^m u=f $$ has been extensively studied for $f$ belonging to different (generalized) function spaces. In particular, being a constant coefficient partial differential operator, by the Hörmander-Malgrange theorem the polyharmonic operator possess a fundamental solution, i.e. a solution of the equation $$ \Delta^m \mathscr{E}(\mathbf{x})=\delta(\mathbf{x})\quad \mathbf{x}\in\mathbb{R}^d $$ A construction of $\mathscr{E}(\mathbf{x})$ by using the theory of Fourier transform is given by Gel'fand & Shilov in their treatise on distributions [1, §4.2, pp. 201-202]. The result is precisely the following one: $$ \mathscr{E}(\mathbf{x})= \begin{cases} C_{2m}r^{2m-d}\ln r & \text{ if $2m>d$ and $d$ is even,}\\ C_{2m}r^{2m-d} & \text{otherwise,} \end{cases} $$ where ([1, §3.3, p. 193, formula (1)]) $$ C_{2m}=2^{2m+d}\pi^{\frac{1}{2}d}\frac{\Gamma\left(\frac{2m+d}{2}\right)}{\Gamma(-m)}, $$ and $r=\sqrt{x_1^2+\dots+x_d^2}$.
Putting $m=2$ gives you the solution you are searching for, since for every $f\in C^4_c(\mathbb{R}^d)$ (and also for more general classes of functions) $$ \Delta^m u(\mathbf{x})=f(\mathbf{x}) \iff u(\mathbf{x})=\int_{\mathbb{R}^d}\mathscr{E}(\mathbf{x}-\mathbf{y})f(\mathbf{y})\mathrm{d}\mathbf{y} $$
[1] Gel'fand, I. M.; Shilov, G. E. (1964), Generalized functions. Vol. I: Properties and operations, Translated by Eugene Saletan, Boston, MA: Academic Press, pp. xviii+423, ISBN 978-0-12-279501-5, MR 0166596, Zbl 0115.33101.