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(a) Write down the characteristic equations for the PDE $$u_t+b\cdot Du =f \text{ in } \mathbb{R}^n\times(0,\infty)$$ where $b\in \mathbb{R}^n, f=f(x,t)$.

(b) Use the characteristic ODE to solve the equation above subject to the initial condition $$u=g \text{ on } \mathbb{R}^n\times \{t=0\}$$

What I have done:

Let $B=(1,b)$ then the equation have the following fashion $$F(p,z,x)=B\cdot p-f$$ therefore $$D_pF=B=x(s)$$ $$D_pF\cdot p=B\cdot p=f =z(s)$$

We got the equations

$$\begin{Bmatrix} \dot x(s)=B \\ \dot z(s)=f \\ \dot p= D_x f \end{Bmatrix}$$

In this case $x=(x^1,...,x^n,t)$. This solve (a). Am I right?

For (b) we solve the differential equation of before:

$$x(s)=Bs+C$$ $$z(s)=\int f(s)ds $$

But I do not know how to continue from here. Somebody can explain me?

Thanks!

EQJ
  • 4,369

1 Answers1

2

This is linear equation, you don't really need to introduce the variable $p$. Use the first two equations and separate the $t$ variable from them: $$\begin{Bmatrix} \dot t(s)=1\\ \dot x(s)=b \\ \dot z(s)=f \end{Bmatrix}$$ The $x,t$ are easy to solve: $$t(s)=s+c_1\\ x(s)=bs+c_2$$ The initial values are $$t(0)=0\\x(0)=x_0\\z(0)=g(x_0)$$

Eliminate $s$ and incorporate the initial condition: $$t=s\\x_0=x-bt$$ Now integrate the third equation using the initial value $$z(x,t)-g(x_0)=\int^t_0f(x(s),s)ds$$

where $x(s)=bs+x_0=bs+x-bt=x+b(s-t)$

You should then be able to get the result.

KittyL
  • 16,965