# Thread: Solving Heat Equation

1. ## 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?

2. 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.

3. 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 $\displaystyle a,b$.

4. 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?