I want to solve $$u_t+\Delta^2 u=0, \quad u(x,0)=u_0(x)\quad \text{in}\ \mathbb{R}^n$$ Doing a Fourier transformation leads me to $$\hat{u}_t(k)+|k|^4\hat{u}(k)=0,\quad \hat{u}(k,0)=\hat{u}_0(k) $$ Now solving the obtained IVP in time leads me to $$\hat{u}(t,k)=e^{-|k|^4t}\hat{u}_0(k) $$ Doing the inverse Fourier transformation gives $$ u(t,x)=\frac{1}{(2\pi)^2}\int_{\mathbb{R}^n}e^{-|k|^4}\hat{u}_0(k)^{ikx} dk $$

Now I want the explicit solution, meaning without the dependence on $\hat{u}_0$ but instead depending on $u_0$.

For the heat equation one obtains a similar integral (but there is a $|k|^2$ in the exponent instead of $|k|^4$), which allows a trick by setting $\omega:=\mathcal{F}^{-1}(e^{-|k|^2t})$ and using the fact that $e^{-|k|^2t}=\hat{\omega}$. Then $\hat{\omega}\hat{u}_0=\widehat{\omega*u_0}$, so $u(t,x)=\omega*u_0$. Then one has just to calculate $\omega$ which is elegant since one can transform to a Gauß-Function.

Now I want to do something similar for my actual problem, but the inverse Fourier transformation of $e^{-|k|^4t}$ seems really ugly. Could you help me?

  • $\begingroup$ Are you working in $\mathbb{R}$ or $\mathbb{R}^n$? The biharmonic operator has mixed partials, so the Fourier transform will not be so nice as you have. $\endgroup$
    – mallan
    Commented Jun 21, 2019 at 18:51
  • $\begingroup$ @AEngineer $\mathbb{R}^n$. Is there some standard method to solve this as there is for the heat-equation? $\endgroup$
    – GEO
    Commented Jun 21, 2019 at 18:57
  • $\begingroup$ Not that I'm aware of. It might be very difficult to solve in $\mathbb{R}^n$ with transforms but it could be interesting to work on for sure. $\endgroup$
    – mallan
    Commented Jun 21, 2019 at 19:02

1 Answer 1


In $\mathbb{R}^n$, the biharmonic operator takes on the form $$ \nabla^4 = \Delta^2 = \left(\frac{\partial^2 }{\partial x_1^2} + \frac{\partial^2 }{\partial x_2^2} + \cdots + \frac{\partial^2 }{\partial x_n^2}\right)\left(\frac{\partial^2 }{\partial x_1^2} + \frac{\partial^2 }{\partial x_2^2} + \cdots + \frac{\partial^2 }{\partial x_n^2}\right). $$ As such, $\Delta^2 u$ is a set of 4th order partial derivatives containing a subset of Laplacian-like derivatives and a subset of mixed terms. These mixed terms are the main drivers of what complicates the process in finding a solution. \begin{align} \Delta^2 u &= \left(\frac{\partial^2 }{\partial x_1^2} + \frac{\partial^2 }{\partial x_2^2} + \cdots + \frac{\partial^2 }{\partial x_n^2}\right)\left(\frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} + \cdots + \frac{\partial^2 u}{\partial x_n^2}\right) \\ &= u_{{xxxx}_1} + u_{{xx}_2{xx}_1} + \cdots + u_{{xx}_n{xx}_1} + u_{{xx}_1{xx}_2} + u_{{xxxx}_2} + \cdots + u_{{xx}_n{xx}_2} + \cdots \\ &+ u_{{xx}_1{xx}_n} + u_{{xx}_2{xx}_n} + \cdots + u_{xxxx_n} \\ &= \sum_{i = 1}^n \sum_{j = 1}^n u_{{xx}_i{xx}_j}. \end{align} We have then $$ u_t + \sum_{i,j = 1}^n u_{{xx}_i{xx}_j} = 0 \ \ \ \text{ with } \ \ \ u(\mathbf{x},0) = f(\mathbf{x}), \ \ \ \mathbf{x} \in \mathbb{R}^n. $$ Since $n$ is finite, interchanging the integral with the summation is fine. Here $\mathrm{i}^2 = -1$ is the complex unit when $i$ is used as an index instead. \begin{align} \mathscr{F}[u_t] + \sum_{i,j = 1}^{n} \mathscr{F}[u_{{xx}_i{xx}_j}] &= \int_{\mathbb{R}^n} u_t \, e^{\mathrm{i} \mathbf{k} \cdot \mathbf{x}} \, \mathrm{d}\mathbf{x} + \sum_{i,j = 1}^{n} \int_{\mathbb{R}^n}u_{{xx}_i{xx}_j} e^{\mathrm{i} \mathbf{k} \cdot \mathbf{x}} \, \mathrm{d}\mathbf{x} \\ &= \frac{\partial}{\partial t}\int_{\mathbb{R}^n} u(\mathbf{x},t) e^{\mathrm{i} \mathbf{k} \cdot \mathbf{x}} \, \mathrm{d}\mathbf{x} + \sum_{i,j = 1}^{n} \int_{\mathbb{R}^n} u(\mathbf{x},t) \frac{\partial^4 e^{\mathrm{i} \mathbf{k} \cdot \mathbf{x}}}{\partial x_i^2 \partial x_j^2} \, \mathrm{d}\mathbf{x} \\ &= \mathscr{F}[u]_t + \sum_{i,j = 1}^{n} k_i^2 k_j^2 \ \mathscr{F}[u] \\ &= 0. \end{align} So we have the first order ordinary differential equation with immediately available solution $$ \mathscr{F}[u](\mathbf{k},t) = \mathscr{F}[f](\mathbf{k},t) \prod_{i,j = 1}^n e^{-k_i^2 k_j^2 t}. $$ This form is actually equivalent to your finding, but provides a means of writing the solution in terms of the given $f$ through the convolution theorem. $$ u(\mathbf{x},t) = \int_{\mathbb{R}^n} f(\mathbf{x} - \mathbf{s}) \, \mathscr{F}^{-1}_{\mathbf{k} \to \mathbf{s}}\left[\prod_{i,j = 1}^n e^{-k_i^2 k_j^2 t}\right] \mathrm{d}\mathbf{s}. $$ By convolution theorem again, we know \begin{align} \mathscr{F}^{-1}\left[\prod_{i,j = 1}^n e^{-k_i^2 k_j^2 t}\right] &= \mathscr{F}^{-1}\left[e^{-k_1^4 t} \prod_{i = 1}^n \prod_{j = 2}^n e^{-k_i^2 k_j^2 t}\right] \\ &= \mathscr{F}^{-1}\left[e^{-k_1^4 t}\right]*\mathscr{F}^{-1}\left[\prod_{i = 1}^n \prod_{j = 2}^n e^{-k_i^2 k_j^2 t}\right] \\ &= \mathscr{F}^{-1}\left[e^{-k_1^4 t}\right] * \mathscr{F}^{-1}\left[e^{-k_1^2 k_2^2 t}\right] * \cdots * \mathscr{F}^{-1}\left[e^{-k_n^4 t}\right] \\ &= \mathscr{F}^{-1}\left[e^{-k_1^4 t}\right] * \cdots * \mathscr{F}^{-1}\left[e^{-k_n^4 t}\right] * \mathscr{F}^{-1}\left[e^{-2k_1^2 k_2^2 t}\right] * \cdots * \mathscr{F}^{-1}\left[e^{-2k_{n-1}^2 k_n^2 t}\right]. \end{align}

Alternatively, we can write out the inverse Fourier transform explicity and make use of the commutativity of convolution and say \begin{align} u(\mathbf{x},t) &= \frac{1}{(2\pi)^{n}} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} f(\mathbf{x} - \mathbf{s}) \prod_{i,j = 1}^n e^{-\left(k_i^2 k_j^2 t + \frac{\mathrm{i} s_i k_i}{n} \right)} \mathrm{d}\mathbf{k} \, \mathrm{d}\mathbf{s} \\ &= \frac{1}{(2\pi)^{n}} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \prod_{i,j = 1}^n e^{-\left(k_i^2 k_j^2 t + \frac{\mathrm{i} (x_i - s_i) k_i}{n} \right)} f(\mathbf{s}) \, \mathrm{d}\mathbf{k} \, \mathrm{d}\mathbf{s}. \end{align} This result seems to be about the best one can do. Regardless of which form we use, finding an explicit solution from this integral may be very difficult in general as opposed to resorting to other solving methods. We can still check the solution though by inserting it into the differential equation and verifying the initial condition. \begin{align} u_t + \Delta^2 u &= \left(\frac{\partial}{\partial t} + \Delta^2\right) \frac{1}{(2\pi)^{n}} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} f(\mathbf{x} - \mathbf{s}) \prod_{i,j = 1}^n e^{-\left(k_i^2 k_j^2 t + \frac{\mathrm{i} s_i k_i }{n} \right)} \, \mathrm{d}\mathbf{k} \, \mathrm{d}\mathbf{s} \\ &= \frac{1}{(2\pi)^{n}} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \left(\frac{\partial}{\partial t} + \Delta^2\right) \prod_{i,j = 1}^n e^{-\left(k_i^2 k_j^2 t + \frac{\mathrm{i} (x_i - s_i) k_i}{n} \right)} f(\mathbf{s}) \, \mathrm{d}\mathbf{k} \, \mathrm{d}\mathbf{s} \\ &= \frac{1}{(2\pi)^{n}} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \left(-\sum_{i,j = 1}^n k_i^2 k_j^2 + \sum_{i,j = 1}^n k_i^2 k_j^2\right) \prod_{i,j = 1}^n e^{-\left(k_i^2 k_j^2 t + \frac{\mathrm{i} (x_i - s_i) k_i}{n} \right)} f(\mathbf{s}) \, \mathrm{d}\mathbf{k} \, \mathrm{d}\mathbf{s} \\ &\equiv 0. \\ u(\mathbf{x},0) &= \int_{\mathbb{R}^n} f(\mathbf{x} - \mathbf{s}) \mathscr{F}^{-1}_{\mathbf{k} \to \mathbf{s}}[1] \, \mathrm{d}\mathbf{s} \\ &= \int_{\mathbb{R}^n} f(\mathbf{x} - \mathbf{s}) \delta(\mathbf{s}) \, \mathrm{d}\mathbf{s} \\ &= f(\mathbf{x}). \end{align}

If preferred, expressed in full vector form with no indices \begin{align} u(\mathbf{x},t) &= \frac{1}{(2\pi)^{n}} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} f(\mathbf{x} - \mathbf{s}) e^{-\left(|\mathbf{k}|^4 t + i \mathbf{s} \cdot \mathbf{k} \right)} \mathrm{d}\mathbf{k} \, \mathrm{d}\mathbf{s} \\ &= \frac{1}{(2\pi)^{n}} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} e^{-\left(|\mathbf{k}|^4t + i \left(\mathbf{x} - \mathbf{s}\right) \cdot \mathbf{k} \right)} f(\mathbf{s}) \, \mathrm{d}\mathbf{k} \, \mathrm{d}\mathbf{s}. \end{align}


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