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$au_x+bu_y+u_t=0$

$u(x,y,0)=g(x,y)$

solve $u(x,y,t)$

Our professor talked about solving this using Method of Characteristics. However, I am confused about this method. Since it's weekend, I think it might be faster to get respond here. In the lecture, he wrote down the followings:

Fix a point$(x,y,t)$ in $\mathbb{R}^3$.

$h(s)=u(x+as,y+bs,t+s)$,line $φ(s)=(x+as,y+bs,t+s)=(x,y,t)+s(a,b,1)$

$h'(s)=u_xa+u_yb+u_t=0$ for all $s$.

$h(-t)=u(x-at,y-bt,0)=g(x-at,y-bt)$ <----- u equal this value for all points on the line $(x+as,y+bs,t+s)$.

$h(0)=u(x,y,t)$

$u(x,y,t)=g(x-at,y-bt)$

The first question I have is that why we want to parametrize $x,y$ and $t$ this way. In addition, what is the characteristic system of this problem. If we have derived the formula already, why do we still need the characteristics system equations? Thank you!

smiley06
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codeedoc
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2 Answers2

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I think it's easiest just to concisely re-explain the method, so that's what I'll do.

The idea: linear, first-order PDEs have preferred lines (generally curved) along which all the action happens. More specifically, because the differential bit takes the form of $\mathbf f \cdot \nabla u$ where in general $\mathbf f$ varies, it is actually always a directional derivative along the vector field $\mathbf f$.

Therefore, along a line given by $\dot{\mathbf x}(s) = \mathbf f$, we expect the PDE to reduce to an ODE involving $\mathrm d u(\mathbf x(s))/ \mathrm d s$. In fact, by the chain rule,

$\mathrm d u(\mathbf x(s))/ \mathrm d s = \dot{\mathbf x}(s) \cdot\nabla u = \mathbf f \cdot\nabla u$

which is exactly the term we said was in the PDE.

So $\mathbf f \cdot\nabla u = h(\mathbf x)$ is equivalent to $$\mathrm d u(\mathbf x(s))/ \mathrm d s = h(\mathbf x(s))$$

Therefore, by finding $\mathbf x(s)$ we can find the ODE $u$ satisfies, and find the initial conditions relevant to each line by saying that at $s=0$ we are on the space where the initial conditions are given.

Does this help? If you have a more specific question, ask away!

not all wrong
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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=a$ , letting $x(0)=x_0$ , we have $x=as+x_0=at+x_0$

$\dfrac{dy}{ds}=b$ , letting $y(0)=y_0$ , we have $y=bs+y_0=bt+y_0$

$\dfrac{du}{dt}=0$ , letting $u(0)=f(x_0,y_0)$ , we have $u(x,y,t)=f(x_0,y_0)=f(x-at,y-bt)$

$u(x,y,0)=g(x,y)$ :

$f(x,y)=g(x,y)$

$\therefore u(x,y,t)=g(x-at,y-bt)$

doraemonpaul
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