Problem solving initial value problems using Fourier transform

Question

Consider the following homogeneous linear system of equations in vector form, $$ \frac{d}{dt}\mathbf{v}(t) = M\mathbf{v}(t) \hspace{0.6cm} ; \hspace{0.6cm} \mathbf{v}(0)=\mathbf{v}_0 $$ where $M$ is an invertible matrix with constant coefficients. I know the solution is given by $\mathbf{v}(t) = \exp(Mt)\mathbf{v}_0$ but I am trying to obtain it via an alternative route using the Fourier transform $\mathcal{F}[f(t)]=\int_{-\infty}^{\infty} dt e^{-i\omega t}f(t)$. If I apply it to both sides of the equation I obtain $$ (-i\omega\mathbb{1} - M)\mathbf{v}(\omega)=0 $$ with $\mathbb{1}$ the identity matrix. This implies $\mathbf{v}(\omega)=0$ and thus $\mathbf{v}(t)=0$ which is clearly wrong. Can anybody help me find where I'm making a mistake, and how to impose the initial condition in this Fourier method? Thank you!

Answer 1

The answer is two-fold:

1. the fourier transform is defined only for "somewhat integrable" functions. (Usually $L^1$ but can be extended e.g. to $L^2$ or tempered distributions). Therefore applying the fourier transform usually reduces the family of solutions to something integrable. If we look at your onedimensional problem we see that the zero solution is indeed the only solution the fourier tranform gives and you can not apply the intial condition. (Since any $v(0)\neq0$ will lead to exponential growth and therefore non integrability.)

2. As pointed out by geetha290km, inn higher dimensions the equation $(i\omega\mathbf{1}+M)\mathbf{v}=0$ does not necessarily imply $\mathbf{v}=0$ but rather $\mathbf{v}\in \ker (i\omega\mathbf{1}+M)$. However in this case I don't think that matters too much since $-i\omega$ can not be an eigenvalue of $M$ for all $\omega$ if $M$ does not depend on $\omega$.

Remark: If you consider IVPs a better suitable integral tranformation for your problem would be the Laplace transform, which directly includes the initial condition. $$ s\mathbf{v}(s)-\mathbf{v}_0=M\mathbf{v}(s)\\ (\mathbf{1}s-M)\mathbf{v}(s)=\mathbf{v}_0 $$

Answer 2

Actually you cannot deduce $v(\omega) = 0$ from $(M-i\omega)v(\omega) = 0$. Why ? In such a context, without further informations, the space of Fourier transforms has to be understood as a space of distributions, whose supports may be reduced to atoms (i.e. isolated points). Indeed, $v(\omega)$ must vanish, except when $M = i\omega$, hence $v(\omega) \propto \delta(\omega + iM) \not\equiv 0$, where $\delta$ denotes the Dirac delta function. Usually we would consider this solution as equivalent to the zero function, since it is non-zero only at a single point, hence a vanishing (Lebesgue) integral, but here integration doesn't use the standard Lebesgue measure. In consequence, one has : $$ v(t) = \int_\mathbb{R} \frac{\mathrm{d}\omega}{2\pi i} e^{i\omega t} v(\omega) = \int_\mathbb{R} \frac{\mathrm{d}\omega}{2\pi i} e^{i\omega t} A\,\delta(\omega+iM) = \frac{e^{Mt}A}{2\pi i} $$ with $A = 2\pi iv_0$. This reasoning is also valid when $v$ is a vector and $M$ a matrix/operator.

Remark. Here the argument of the Dirac delta is complex, which is not super formal as such, but it could be formalized more properly $-$ this problem can be avoided with the help of the Laplace transform for instance. In the present case, it has to be seen more as a symbol for an atomic distribution than the standard real Dirac delta function.