Solving Heat Equation

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• Sep 18th 2008, 03:20 PM
Avitar
Solving Heat Equation
Find a solution to the heat equation,

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

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

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

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

(Hint: find out the form that $\displaystyle f(s)$ must take by substituting $\displaystyle u(x,t)=f(k-xt)$ 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?
• Sep 18th 2008, 03:44 PM
shawsend
The heat equation is generally solved by separation of variables: Assume a solution of the form: $\displaystyle u(x,y)=X(x)T(t)$. That is, a product of a function of x and a function of t. Now, substitute the expression $\displaystyle X(x)T(t)$ 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:

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

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

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

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.
• Sep 18th 2008, 06:52 PM
ThePerfectHacker
Quote:

Originally Posted by Avitar
Find a solution to the heat equation,

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

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

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

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

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

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

Substitute that,
$\displaystyle k f'(x-kt) - kf''(x-kt) = 0 \implies f'(x-kt) = f''(x-kt)$.
Since $\displaystyle x-kt$ can be made any number is means,
$\displaystyle f ' ( z) = f'' (z)$ for any number $\displaystyle z$.
Thus, $\displaystyle f ' (z ) = ae^z \implies f(z) = ae^z + b$.
Thus, $\displaystyle f(x-kt) = ae^{x-kt} + b$ for some constants $\displaystyle a,b$.
• Sep 18th 2008, 07:48 PM
Avitar
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?