Multidimensional Fourier Transform

Question

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.

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}