Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to solve $u_t + b \cdot \nabla u + cu =0$ with initial condition $u(x,0) = g(x)$. I started by using the method of characteristics. Define: $$ z(s) := u(x + bs, t+s)$$ Then: $$ \frac{dz}{ds} = b\cdot\nabla u + \frac{du}{dt} = -cu, $$ where I used the PDE in the last equality. Integrating this yields $ z(s) = -cus + K$, where K is a constant. Furthermore: $$ z(0) = u = K $$ and $$ z(-t) = u(x-bt,0) = g(x-bt) = K + cut $$ Substracting these last two and rearranging yields: $$ \boxed{u(x,t) = \frac{g(x-bt)}{1+ct}.} $$ However, plugging this into the PDE yields: $$ (1+ct)^{-2}ct(b\cdot\nabla g + cg)\neq0$$ Does anyone see my mistake?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Your mistake is where you integrated the expression $\frac{\mathrm{d}z}{\mathrm{d}s}$ and treated $u$ as a constant in $s$. That is false, since $z$ is defined in terms of $u$, and so you can invert the condition to write $u$ in terms of $z$. In fact, what you need to do is to keep track of the arguments:

$$ \frac{\mathrm{d}z}{\mathrm{d}s}(s) = (b\cdot \nabla u)(x + bs, t + s) + (\partial_t u)(x+bs, t+s) = - cu(x+bs, t+s) = - c z(s) $$

So integrating the above ODE should give you instead

$$ z = z_0 \exp (-c s) $$

where $z_0$ is fixed by the boundary condition at $s = 0$.

share|improve this answer

$u_t+b\cdot\nabla u+cu=0$ with initial condition $u(x,0)=g(x)$ i.e. $u_t+bu_x+cu=0$ with initial condition $u(x,0)=g(x)$ .

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

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

$\dfrac{dx}{ds}=b$ , letting $x(0)=x_0$ , we have $x=bs+x_0=bt+x_0$

$\dfrac{du}{ds}=-cu$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)e^{-cs}=f(x-bt)e^{-ct}$

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

$f(x)=g(x)$

$\therefore u(x,t)=g(x-bt)e^{-ct}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.