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Let $b\in \mathbb{R^n}$ and $c>0$. Assume $g \in C(R^n)$ has compact support and $f = f(x,t)$, $f \in C_1^2(R^n \times [0,\infty))$ has compact support. I'm trying to solve the following IBP via fourier transform: $$ \left\{ \begin{array}{rc} u_t+b \cdot \nabla u-c^2 \Delta u &=& f & \text{ in } \mathbb{R^n} \times(0,\infty)\\ u &=& g& \text{on } \mathbb{R^n}\times \{t=0\} \end{array} \right. $$

Thing's are getting a little iffy for me right around when I'm solving for $\mathcal{F}(u(x,t))=\hat{u}(y,t)$ so that I can apply an inverse fourier transform to get the end solution.

I get to an ODE, namely

$$\left\{ \begin{array} \hat{u_t} + \hat{u}(b \cdot iy - c^2 \left| y \right|^2)=\hat{f}\\ \hat{u}(y,0) = \hat{g} \end{array} \right.$$

However it's getting a little awkward solving for $u$ from here, that is, if what I've done so far is correct. How do I proceed in the calculation here?

EDIT: I'm certain that the solution to the ODE IBP is

$$\hat{u}(y,t) = e^{-t(b \cdot iy - c^2 |y|^2)}\int_0^t \hat{f}(y,s)e^{s(b \cdot iy - c^2 \left| y \right|^2)}ds + \hat{g}(y)e^{-t(b \cdot iy - c^2 |y|^2)}$$

So that I guess I am a little iffy on how to compute the inverse fourier transform of that expression.

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I have since then solved the problem, I thought I'd put it here just in case someone else has the same question. Computing the inverse fourier transform, we have,

$$u(x,t) = \int_0^t \mathcal{F}^{-1}{\left[\hat{f}(y,s)e^{(s-t)(b \cdot iy + c^2 |y|^2} \right]}ds +\mathcal{F}^{-1} \left[ \hat{g}(y)e^{-t(b \cdot iy+c^2|y|^2}\right]$$

To which the inverse fourier transforms can be computed as convolutions, and we must make a change of variable of the form $x-(t-s)b$ in the first and $x-tb$ in the second.This calculation can be seen by directly applying the inverse fourier transform. So we get

$$u(x,t) = \int_0^t \frac{1}{(4 \pi (t-s)c^2)^{\frac{n}{2}}}\left(f(x-b(t-s),s)*\exp{\left[\frac{-|x-b(t-s)|^2}{4(t-s)c^2} \right]}\right)ds+ \frac{1}{(4 \pi tc^2)^{\frac{n}{2}}}\left(g(x-tb) *\exp{\left[\frac{-|x-bt|^2}{4tc^2} \right]} \right)$$

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