Representation formula for the $\Delta^2 u =f$

Question

I already know the representation formula of $\Delta u =f$ is the that $u(x)$ equal to the $\phi * f(x)$, where $\phi$ is fandamental solution.

Now I have wonder konw the representation formula of $\Delta^2 u =f$ in $R^2$, is there some relative to above? And how I can find it?

I'm find that convolution the property that $D(f*g)=Df*g=f*Dg$, so I can do the same thing here, right? That is $\Delta \phi * f(x).$

Answer 1

In general, solving $Lu = f$ by solving $L\Phi = \delta$ and exploiting $\delta * f = f$ to say $u = \Phi * f$ is known as finding the Green's function of the operator $L$.

In your case, $L = \Delta^2$ is called the biharmonic operator. So you are looking to find the Green's function of the Biharmonic operator.

An answer to this question along with a reference is provided here, along with the generalised problem for $L = \Delta^k$

Answer 2

You could solve the system $$ \begin{cases} \Delta u&=g \\ \Delta g&=f \end{cases} $$ by applying the fundamental solution twice, i.e. $$ u=\phi * \phi *f, $$ provided this integral converges. The other way is done via the Fourier transformation of the equation $$ \Delta^2u=\delta(x) $$ and it depends on the underlying dimension of $\mathbb{R}^n$. You can find the formula $$ \phi(x)=\begin{cases}Cr^{4-n}\ln r \; \text{if $n$ is even and $4>n$} \\ Cr^{4-n} \; \text{otherwise}\end{cases} $$ by using the Fourier transform and a bit of complex analysis in the book "Generalized Functions Vol 1 Properties And Operations" in section 4 by Gelfand and Shilov.