# Thread: Partial Differential eqn - method of characteristics

1. ## Partial Differential eqn - method of characteristics

Am going through problems from book and have come across a section of my lecture notes that i don't understand, i'm sure it's something small i'm missing but can't get my head around it.

In lecture notes we solve the wave eqn;

it gets to the General solution which is;

Q(x,t) = F(x-t) +G(x-t)

and then as one of my initial conditions is;

dQ/dx(0,t) = 2 (where this is the partial derivative of Q wrt. x)

i need to find an expression for the dQ/dx, which was written as;

dQ/dx = F'(x+t) + G'(x-t)
i dont understand how this works?
can anyone help?

2. Originally Posted by monster
Am going through problems from book and have come across a section of my lecture notes that i don't understand, i'm sure it's something small i'm missing but can't get my head around it.

In lecture notes we solve the wave eqn;

it gets to the General solution which is;

Q(x,t) = F(x-t) +G(x-t)

and then as one of my initial conditions is;

dQ/dx(0,t) = 2 (where this is the partial derivative of Q wrt. x)

i need to find an expression for the dQ/dx, which was written as;

dQ/dx = F'(x+t) + G'(x-t)
i dont understand how this works?
can anyone help?
It's the chain rule. Let $\displaystyle r = x + t$ and $\displaystyle s = x- t$ so $\displaystyle Q = F(r) + G(s)$ so

$\displaystyle \frac{\partial Q}{\partial x} = \frac{\partial Q}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial Q}{\partial s}\frac{\partial s}{\partial x}$

3. Consider the IVP
$\displaystyle a(x,t,u) u_t + b(x,t,u) u_x = c(x,t,u),\; u(x,0) = f(x)$
Parameterizing your variable as $\displaystyle x=r,\; t=s,\; u=z$ and assuming that you start on a characteristic curve with initial conditions $\displaystyle \Gamma: x = r,\; t = 0,\; u = f(s)$ then the solution to the IVP can be found by the ODE
$\displaystyle \dfrac{dx}{ds} = b(r,s,z)$
$\displaystyle \dfrac{dt}{ds} = a(r,s,z)$
$\displaystyle \dfrac{dz}{ds} = c(r,s,z)$