erok81 said:
Homework Statement
I am not even sure if the title is correct - it's day two of the class and I am already lost beyond belief. Anyway...here is the question.
Consider the equation (1) \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} = 0
Where u=u(x,t) is the unknown function. If f is any differentiable function of a single variable and we set u(x,t)=f(x-t) then u is a solution of (1). I'm not sure this part is even needed.
Yes, that's true and very important. For any differentiable function, f, the derivative of f(x- t) with respect to x, by the chain rule, is the usual derivative of f, evaluated at x-t, times the derivative of x- t with respect to x, which is 1: f_x(x- t)= f'(x- t).
Similarly, the derivative of f(x- t) with respect to t, is the usual derivative of f, evaluated at x- t, times the derivative of x- t with respect to t, which is -1: f_t(x-t)= -f'(x- t).
That is, u_t+ u_x= -f'(x-t)+ f'(x- t)= 0.
Because this is a
first degree, in both x and t, equation, just as the general solution of a first degree ordinary differential equation depends on a single arbitrary
constant, the general solution to this partial differential equation depends on a single arbitrary
function.
That is, the general solution to this differential equation
is u(x,y)= f(x- t) where f can be any differentiable function of a single variable.
Which solution of (1) is equal to xe^(-x^2) on the x-axis?
On the x-axis, t= 0 so \
u(x,0)= f(x- 0)= f(x)= xe^{-x^2}
That immediately tells you that
u(x,t)= f(x- t)= (x- t)e^{-(x- t)^2}.
Homework Equations
n/a
The Attempt at a Solution
I have zero idea of where to start. Well, I could probably guess a solution from my ODE knowledge, but that isn't why I am taking this course. :)
Any tips, pointers, etc?
