# Partial Differential eqn - method of characteristics

• May 8th 2010, 11:57 PM
monster
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?
• May 9th 2010, 07:27 AM
Danny
Quote:

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 $r = x + t$ and $s = x- t$ so $Q = F(r) + G(s)$ so

$
\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}
$
• May 9th 2010, 08:01 AM
lvleph
Consider the IVP
$a(x,t,u) u_t + b(x,t,u) u_x = c(x,t,u),\; u(x,0) = f(x)$
Parameterizing your variable as $x=r,\; t=s,\; u=z$ and assuming that you start on a characteristic curve with initial conditions $\Gamma: x = r,\; t = 0,\; u = f(s)$ then the solution to the IVP can be found by the ODE
$
\dfrac{dx}{ds} = b(r,s,z)
$

$
\dfrac{dt}{ds} = a(r,s,z)
$

$
\dfrac{dz}{ds} = c(r,s,z)
$