Modified transport equation (PDE)

[docnet]

Homework Statement

solve the modified transport equation using the method of characteristics.

Relevant Equations

$\partial_t u + <b, Du> + cu=0$
$u(0,x)=f$

Hi all, I

Fix $$(t,x) ∈ (0,\infty) \times R^n$$and consider auxillary function
$$w(s)=u(t+s,x+sb)$$
Then, $$\partial_s w(s)=(\partial_tu)(t+s,x+sb)\frac{d}{ds}(t+s)+<Du(t+s,x+sb)\frac{d}{ds}(x+sb)>$$
$$=(\partial_tu)(t+s,x+sb)+<b,Du(t+s,x+sb)>$$
$$=-cu(t+s,x+sb)$$
$$\partial_sw(s)=-cu(t+s,x+sb)$$
by Fundamental theorem of calculus,
$$u(t,x)-f(x-tb)=u(t,x)-u(0,x-tb)$$
$$=w(0)-w(-t)$$
$$=\int^0_{-t}\partial_sw(s)ds$$
$$=\int^0_{-t}-cu(t+s,x+sb)ds$$
$$s_o=s+t$$
$$=\int^t_{0}cu(s_o,x+(s_o-t)b)ds_o$$

$$u(t,x)=f(x-tb)+\int^t_{0}cu(s_o,x+(s_o-t)b)ds_o$$

 

[pasmith]

You've gone astray somewhere; this one has a closed-form analytic solution in terms of $f$.

Start by setting $u(t,x) = e^{\alpha t}v(t,x)$ and choose $\alpha$ such that $v$ satisfies $$\partial_t v + (b, Dv) = 0$$ subject to $v(0,\cdot) = f$. Hopefully this is the "transport equation" as defined in lectures.

 

[docnet]

You've gone astray somewhere; this one has a closed-form analytic solution in terms of $f$.

Start by setting $u(t,x) = e^{\alpha t}v(t,x)$ and choose $\alpha$ such that $v$ satisfies $$\partial_t v + (b, Dv) = 0$$ subject to $v(0,\cdot) = f$. Hopefully this is the "transport equation" as defined in lectures.

That PDE in $v$ is the homogeneous transport equation we learned in lecture. I think our original problem is a transport equation with an inhomogeneous term. To be honest, I don't know whether that closed-form analytic solution is ultimately different from the answer from my professor's notes.

I looked in his notes and found this:

Given a PDE problem
$\partial_t u+<b,Du>=f$ for $u$ in $t\times R^n$
$u(0,x)=g$ for $g$ on $\left\{t=0\right\}\times R^n$
(The term in <> is the dot product of $b$ in $R^n$ and $Du$ the gradient of $u$.)
The solution is given by $u(t,x)=g(x-tb)+\int^t_0 f(s, x+(s-t)b)ds$

Our problem has $-cu$ as the inhomogeneous term, and I adapted the solution to fit the problem. I am not understanding the proof from his notes yet, that I copied almost directly into my solution. That is the method of characteristics, and it is not coming so easily yet.

 

[docnet]

The method of characteristics turns a PDE problem into an ODE problem (if the PDE is inhomogeneous) along lines parallel to $(1, b)$. If the PDE is homogeneous, then the solutions become constant on lines parallel to $(1, b)$. I can visualize this for the homogeneous case for $u(t,x)$ in $R^2$, but not for higher dimensions.

I made a mistake with my first answer because it should have said $u(t,x)=f(x-tb)-\int^t_{0}cu(s_o,x+(s_o-t)b)ds_o$ instead of $u(t,x)=f(x-tb)+\int^t_{0}cu(s_o,x+(s_o-t)b)ds_o$

edited for grammar

 

[wrobel]

see also https://www.physicsforums.com/threads/the-method-of-characteristics.999332/