Solving $\partial_tu-\Delta u+\vec{V}\cdot\nabla u=0$

Question

How can I solve the equation $\partial_tu-\Delta u+\vec{V}\cdot\nabla u=0,$ where $\vec{V}$ is a constant vector, for a given initial condition $u_0$? The only method I'm really aware of is the method of characteristic curves, which led nowhere.

Answer 1

Hint: let $u(x,t)=v(x-t\,V,t)$ and obtain an equation for $v$.

Answer 2

You can also use Separation of Variables, the idea is, if you assume $u({\bf r}, t) = X({\bf r})T(t)$, then the equation can be rewritten as

$$ \frac{1}{T(t)}\frac{{\rm d}T}{{\rm d}t} = \frac{1}{X({\bf r})}[\Delta X({\bf r}) - V\cdot \nabla X({\bf r})] $$

So you have something on the l.h.s that depends only on $t$, and on the r.h.s something that depends solely on ${\bf r}$, so these two things must be a constant, $\alpha$. Therefore, you're left with two separated differential equations

$$ \frac{1}{T(t)}\frac{{\rm d}T}{{\rm d}t} = \alpha ~~~\mbox{and}~~~ \frac{1}{X({\bf r})}[\Delta X({\bf r}) - V\cdot \nabla X({\bf r})] = \alpha $$

You can apply the same argument to try and separate the last one again, depending on the geometry of the domain you have at hand

Answer 3

Do the Fourier transform of the equation $$u(r, t)=\int_{\mathbb{R}^{n}}\hat{u}(t, q)e^{ir\cdot{q}}d^{n}q$$ To give $$\partial_{t}\hat{u}(t, q)=-(|q|^{2}+iq\cdot{V})\hat{u}(t, q)$$ The solution is $$\hat{u}(t, q)=ce^{-|q|^{2}t}e^{-iq\cdot{V}t}$$ Take the Fourier transform of the initial condition $$\hat{u}_{0}(q)=\int_{\mathbb{R}^{n}}u_{0}(r)e^{-ir\cdot{q}}d^{n}r$$ Then $$c=\hat{u}_{0}(q)$$ And the solution is $$u(r, t)=\int_{\mathbb{R}^{n}}\hat{u}_{0}(q)e^{-|q|^{2}t}e^{iq\cdot(r-Vt)}d^{n}q$$ For example, if $u_{0}(r)=u_{0}\delta^{(n)}(r-r')$ and $n=1$ then $$\hat{u}_{0}(q)=u_{0}e^{-ir'\cdot{q}}$$ and $$u(r, t)=u_{0}\int_{\mathbb{R}}e^{-q^{2}t+iq([r-r']-Vt)}dq=u_{0}\sqrt{\frac{\pi}{t}}\exp\Big(-\frac{([r-r']-Vt)^{2}}{4t}\Big)$$