Solving Heat Equation

Avitar · #1 $Old$ September 18th, 2008, 04:20 PM

Find a solution to the heat equation,

$\frac{{\partial u}}{{\partial t}}-k\frac{{\partial^2\ u}}{{\partial x^2}}=0$

(k>0 is a constant)
which takes the form

$u(x,t)=f(x-kt)$

for some $f(s)$ which is twice differentiable in s.

(Hint: find out the form that $f(s)$ must take by substituting $u(x,t)=f(k-xt)$ into the equation.)

I...dono what i'm supposed to do. We went over deriving the heat equation in class and proving a solution is unique, but not so much the actual solving part. How do I go about solving such kinds of problems?

shawsend · #2 $Old$ September 18th, 2008, 04:44 PM

The heat equation is generally solved by separation of variables: Assume a solution of the form: $u(x,y)=X(x)T(t)$ . That is, a product of a function of x and a function of t. Now, substitute the expression $X(x)T(t)$ into the PDE and obtain separate ODEs in terms of t and x. Solve those ODEs under suitable "well-posed" conditions (boundary and initial conditions) to arrive at a solution. I'd recommend "Basic Partial Differential Equations" by D. Bleecker and G. CSordas. Here's a well-posed problem:

$\textbf{D.E.}\quad u_t=ku_{xx}\quad 0\leq x\leq L,\; t\geq 0$

$\textbf{B.C}\quad u(0,t)=A(t)\quad u(L,t)=B(t)$

$\textbf{I.C}\quad u(x,0)=f(x)$

If you get into PDEs, always try to "pose the problem well" (correctly define the boundary and initial conditions). The style of these will change with the PDE.

ThePerfectHacker · #3 $Old$ September 18th, 2008, 07:52 PM

Quote:

Originally Posted by Avitar $View Post$

Find a solution to the heat equation,

$\frac{{\partial u}}{{\partial t}}-k\frac{{\partial^2\ u}}{{\partial x^2}}=0$

(k>0 is a constant)
which takes the form

$u(x,t)=f(x-kt)$

for some $f(s)$ which is twice differentiable in s.

(Hint: find out the form that $f(s)$ must take by substituting $u(x,t)=f(k-xt)$ into the equation.)

$u_t = -kf'(x-kt)$ and $u_x = f'(x-kt)$ and $u_{xx} = f''(x-kt)$ .

Substitute that,
$k f'(x-kt) - kf''(x-kt) = 0 \implies f'(x-kt) = f''(x-kt)$ .
Since $x-kt$ can be made any number is means,
$f ' ( z) = f'' (z)$ for any number $z$ .
Thus, $f ' (z ) = ae^z \implies f(z) = ae^z + b$ .
Thus, $f(x-kt) = ae^{x-kt} + b$ for some constants $a,b$ .

Avitar · #4 $Old$ September 18th, 2008, 08:48 PM

Alright, I see what you did there. That helps enormously, thanks. Although, I have to ask what happened to the negative sign from finding the partial with respect to u during the substitution? Is there something I don't know about that lets us disregard it, or should it still be there?