# Changing domain of PDE (should be easy)

Suppose I have a PDE $$u_t(x,t) + f(x,t)\cdot \nabla u(x,t) = 0 \quad \text{on \gamma(t)}$$ where $\gamma(t)$ is a curve for each fixed t in $\mathbb{R}^2$ and $f$ is given.

We have that $\gamma(t)$ is parametrised by the function $X(s,t)$ for each $t$. So $X(s,t) = x \in \gamma(t)$ for some $s$. So $x \in \mathbb{R}^2$, $s \in S^1$ or some interval in the real line.

I want to write the PDE as a function of $(s,t)$ instead of $(x,t).$ So basically I want to solve a PDE for the unknown function $v(p,t) = u(X(p,t), t)$. How can I convert my PDE to this domain? I can't find any expressions for $u_t$ and $\nabla u$ in terms of $v$..

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Start with $$v(s,t) = u(X(s,t),t)$$ By the chain rule you get $$\begin{cases} v_t(s,t) &= X_t(s,t) \cdot \nabla u(X(s,t),t) + u_t(X(s,t),t) \\ v_s(s,t) &= X_s(s,t) \cdot \nabla u(X(s,t),t) \end{cases}$$ To solve for $u_t$ and $\nabla u$, you need to solve for three scalar functions. But you only have two equations here. Clearly the system is underdetermined, so in general one cannot solve for $u_t$ and $\nabla u$ interms of $v$.

Plugging in the above into the equation you get

$$v_t - X_t\cdot \nabla (u\circ X) + f \cdot \nabla (u\circ X) = 0$$

So if $f - X_t \propto X_s$, that is, if $f - X_t = \alpha(t,s) X_s$, we can plug in

$$v_t + \alpha X_s \nabla (u\circ X) = 0 \implies v_t + \alpha v_s = 0$$

The condition that $f - X_t \propto X_s$ above is a integrability condition. Given $\gamma(t)$ it specifies a two-dimensional surface is $\mathbb{R}^3 = \{t\}\times \mathbb{R}^2$. Your equation can be regarded as a transport equation in $\mathbb{R}^3$, that a certain vector field $V = (1,f)$ acts on $u$ to give zero. Generically one may expect the vector field $V$ to be transverse to the surface $\gamma(t)$: in that case your equation is underdetermined: values on the surface $\gamma(t)$ depends on values of $u$ off the surface, and in that case it is impossible to write the equation in terms of $v$, which is essentially just the value of $u$ on the surface $\gamma(t)$.

The condition $f - X_t \propto X_s$ ensures that the vector $(1,f)$ is in fact tangential to the surface $\gamma(t)$. This way your vector field $(1,f)$ restricts to a tangent vector field of $\gamma(t)$ and it makes sense to study the equation restricted to the surface.

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thanks but I think you can't divide by $X_s(s,t)$. Since $X(s,t)$ parametrises a curve in $\mathbb{R}^2$, $X_s$ is a vector since $X$ is a vector. – soup Jul 3 '12 at 10:45
What are $x$ and $s$ taking values in? (You wrote $X(s,t)$ is a function, so I assume you meant it is scalar valued.) – Willie Wong Jul 3 '12 at 10:51
So $x \in \mathbb{R}^2$ is a point on the surface $\gamma(t)$. For fixed $t$, $X(s,t)$ is the parametrisation of $\gamma(t)$, so say $s \in \mathbb{R}$ lies in $S^1$. Then $x = X(s,t)$ for some $s$ and $x \in \gamma(t)$. Sorry for the confusion. – soup Jul 3 '12 at 10:55
Thanks Willie. I will have to read more about vector fields you mentioned but I have a question. If $f$ denotes the normal velocity of the surface, then $f = (X_t\cdot N) N$, where $N$ is the normal. Then $f -X_t = -(X_t\cdot T)T$ where $T$ is the tangent, and $T = \frac{X_s}{|X_s|}$ so in this case I will get the equation for $v$ you wrote down, am i right? – soup Jul 3 '12 at 15:36
If $N$ and $T$ are the normal and tangent vectors to the curve $\gamma(t)$, then you are correct. – Willie Wong Jul 4 '12 at 8:23