I'm having difficulty with multidimensional Fourier Transforms. I have the following problem for $u=u(t,x) \in \mathbb{R}$

$$ \frac{\partial u}{\partial t} = \sum_{m,n=1}^d a_{mn}\frac{\partial^{2}u}{\partial x_m\partial x_n} + \sum_{n=1}^d b_n\frac{\partial u}{\partial x_n} + cu $$

$$ u(0,x) = \phi(x) $$

Where $A=a_{mn}$ is a positive definite, $d\times d$ matrix, $b=b_n \in \mathbb{R}^d$ is a vector of constants, $c$ is a constant and $\phi \in \mathscr{S}^d$ (Schwartz/rapidly decreasing). We are asked to solve this using the Fourier Transform.

There were two approaches I took, but I got stuck on both. The first was to transform $(x, t) \to (k, t)$ space which gives

$$ \frac{\partial \hat{u}}{\partial t} = -(k^TAk)\hat{u} + i\langle b,k\rangle \hat{u} + c\hat{u} $$

Solving for $\hat{u}$

$$ \hat{u}(t,k) = \hat{\phi}(k)\exp([-(k^TAk) + i\langle b,k\rangle + c]t) $$

Now, because A is diagonalisable, we have that

$$ -(k^TAk) = -(k^TQDQ^Tk) $$

Setting $\eta^T = k^TQ$ we find

$$ \hat{u}(t,k) = \hat{\phi}(k)\exp(ct)\exp([-(\eta^TD\eta) + i\langle b,k\rangle]t) $$

Now, $\eta$ is a vector of constants and $D$ is a diagonal matrix of eigenvalues, so we get

$$\exp(-(\eta^TD\eta)t) = \exp (-(\lambda_1\eta_1^2 + \lambda_2\eta_2^2 + .. + \lambda_n\eta_n^2)t) $$

And this is where I get stuck.. I have absolutely no idea how to get this back into $u(t,x)$ in terms of a convolution inside the integral. I don't even know how to write the product of eigenvalues and constants more succinctly, as I don't think it is correct to write $\langle \lambda,\eta^2 \rangle$ nor is it correct to write $\langle \lambda, \lvert \eta\rvert^2 \rangle$. Is the final result just a product of multiple integrals with convolution ie

$$ u(t,x) = \frac{1}{(2\pi)^{\frac{d}{2}}}\int_\mathbb{R^d} \exp(ikx)\exp(-\lambda_1\eta_1^2) dk \int_\mathbb{R^d} \exp(ikx)\exp(-\lambda_2\eta_2^2) dk... $$

with the convolution term somewhere in the integral?

Note: I also tried to ansatz a solution, setting $u(t,x) = v(t,x+bt)$ to remove the

$$ \sum_{n=1}^d b_n\frac{\partial u}{\partial x_n} $$

term but it didn't work.

  • 1
    $\begingroup$ I think I solved this, will post an answer either tonight or tomorrow morning. $\endgroup$ Dec 9, 2014 at 1:54

1 Answer 1


As you say, taking the Fourier transform yields $$\dot{\hat{u}}=\left(-\mathbf{k}^\mathsf{T}A\mathbf{k}+i\mathbf{k}\cdot\mathbf{b}+c\right)\hat{u}$$ and solving for $\hat{u}$ gives $$\hat{u}=\hat{\phi}(\mathbf{k})\exp\left[\left(-\mathbf{k}^\mathsf{T}A\mathbf{k}+i\mathbf{k}\cdot\mathbf{b}+c\right)t\right].$$ Letting $A=QDQ^\mathsf{T}$, we can rewrite the expression as $$\hat{u}=\hat{\phi}(\mathbf{k})e^{ct}\exp\left[\left(-\mathbf{k}^\mathsf{T}QDQ^\mathsf{T}\mathbf{k}+i\mathbf{b}^\mathsf{T}QQ^\mathsf{T}\mathbf{k}\right)t\right]$$ and we can inverse Fourier transform to get $$u=\phi(\mathbf{r})e^{ct}*\mathcal{F}^{-1}\left(\exp\left[\left(-\mathbf{k}^\mathsf{T}QDQ^\mathsf{T}\mathbf{k}+i\mathbf{b}^\mathsf{T}QQ^\mathsf{T}\mathbf{k}\right)t\right],\mathbf{k},\mathbf{r}\right).$$ Using the scaling property $$\mathcal{F}\left(f(A\mathbf{k}),\mathbf{k},\mathbf{r}\right)=\frac{1}{\text{det}(A)}\mathcal{F}\left(f(\mathbf{k}),\mathbf{k},\left(A^{-1}\right)^\mathsf{T}\mathbf{r}\right)$$ where $\mathcal{F}$ is any Fourier operator and $A$ is any matrix, we obtain (letting $Q^\mathsf{T}\mathbf{b}=\mathbf{b}^*)$ $$u=\phi(\mathbf{r})e^{ct}*\frac{1}{\left|Q^\mathsf{T}\right|}\mathcal{F}^{-1}\left(\exp\left[\left(-\mathbf{k}^\mathsf{T}D\mathbf{k}+i\mathbf{b}^*\cdot\mathbf{k}\right)t\right],\mathbf{k},Q^\mathsf{T}\mathbf{r}\right).$$ Using the shift property of $\mathcal{F}^{-1}$, we get $$u=\phi(\mathbf{r})e^{ct}*\frac{1}{\left|Q^\mathsf{T}\right|}\mathcal{F}^{-1}\left(\exp\left[\left(-\mathbf{k}^\mathsf{T}D\mathbf{k}\right)t\right],\mathbf{k},Q^\mathsf{T}\mathbf{r}+Q^\mathsf{T}\mathbf{b}t\right).$$ Using the scaling property once more, we get $$u=\phi(\mathbf{r})e^{ct}*\frac{1}{\left|Q^\mathsf{T}\right|}\frac{1}{\left|\sqrt{tD}\right|}\mathcal{F}^{-1}\left(\exp\left[\left(-\mathbf{k}\cdot\mathbf{k}\right)\right],\mathbf{k},\left(tD\right)^{-1/2}Q^\mathsf{T}(\mathbf{r}+\mathbf{b}t)\right).$$ Since $$\mathcal{F}^{-1}\left(\exp\left[\left(-\mathbf{k}\cdot\mathbf{k}\right)\right],\mathbf{k},\mathbf{r}\right)=\frac{1}{2^{d/2}}\exp\left(-\frac{1}{4}\mathbf{r}\cdot\mathbf{r}\right)$$ and since $Q$ is unitary, we finally get \begin{align} u&=\phi(\mathbf{r})e^{ct}*\frac{1}{2^{d/2}\left|\sqrt{tD}\right|}\exp\left(-\frac{1}{4}\left(\left(tD\right)^{-1/2}Q^\mathsf{T}(\mathbf{r}+\mathbf{b}t)\right)\cdot\left(\left(tD\right)^{-1/2}Q^\mathsf{T}(\mathbf{r}+\mathbf{b}t)\right)\right)\\ &=\phi(\mathbf{r})e^{ct}*\frac{1}{2^{d/2}\left|\sqrt{tD}\right|}\exp\left(-\frac{1}{4t}(\mathbf{r}+\mathbf{b}t)^\mathsf{T}A^{-1}(\mathbf{r}+\mathbf{b}t)\right). \end{align} There might be a faster way to do this, although I'm not sure (and by all means check my work for errors).

Edit 1: Faster method

As before, we have $$u=\phi(\mathbf{r})e^{ct}*\mathcal{F}^{-1}\left(\exp\left[\left(-\mathbf{k}^\mathsf{T}A\mathbf{k}+i\mathbf{b}^\mathsf{T}\mathbf{k}\right)t\right],\mathbf{k},\mathbf{r}\right).$$ By the shift property, this is $$u=\phi(\mathbf{r})e^{ct}*\mathcal{F}^{-1}\left(\exp\left[-\mathbf{k}^\mathsf{T}At\mathbf{k}\right],\mathbf{k},\mathbf{r}+\mathbf{b}t\right).$$

Since $A$ is positive definite, it admits Cholesky factorization $LL^\mathsf{T}$. Using the scaling property with $M=\sqrt{t}L^\mathsf{T}$ gives \begin{align} u&=\phi(\mathbf{r})e^{ct}*\frac{1}{\left|\sqrt{t}L^\mathsf{T}\right|}\mathcal{F}^{-1}\left(\exp\left[-\mathbf{k}^\mathsf{T}\mathbf{k}\right],\mathbf{k},t^{-1/2}L^{-1}\left(\mathbf{r}+\mathbf{b}t\right)\right)\\ &=\frac{\phi(\mathbf{r})e^{ct}}{\left|L\right|\sqrt{2^dt}}*\exp\left(-\left(\mathbf{r}+\mathbf{b}t\right)^\mathsf{T}\frac{A^{-1}}{4t}\left(\mathbf{r}+\mathbf{b}t\right)\right). \end{align}

  • $\begingroup$ Thankyou very much, that was a great answer although the faster method didn't quite make sense to me. But your first answer was perfect. $\endgroup$ Dec 11, 2014 at 5:28
  • $\begingroup$ @Mattos: The faster method using Cholesky factorizations does exactly the same thing as the slower method using eigenfactorizations, except that it combines two of the scaling properties into one. Note that the determinant of $L$ is the same as the determinant of the square root of $D$, since the $D$-factor in an $LDL^\dagger$ decomposition gets absorbed in both directions when forming the $LL^\dagger$ decomposition. Thus the final answer in both cases is the same, but the Cholesky factorization approach makes the derivation easier. $\endgroup$ Dec 11, 2014 at 15:14
  • $\begingroup$ @Mattos: By the way, I'm using $*$ to denote convolution in my answer; I think this is common practice in the signal processing community, but I just wanted to make sure this was clear. $\endgroup$ Dec 11, 2014 at 15:17
  • $\begingroup$ Yeah the convolution symbol is fine, I just haven't used Cholesky factorisation before. But now that you explained that $detL$ is the same as $det\sqrt D$ it makes much more sense. Thankyou very much for the solution. $\endgroup$ Dec 13, 2014 at 1:53

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