Heat Equation with Zero-Temperature Boundaries

Homogeneous Linear ODEs with Constant Coefficients


As I was finalizing my study of the method of separation of variables, I ran into the inconsistency that I had forgotten how to solve homogeneous linear ODEs with constant coefficients, that is, equations of the formad2ydx2+bdydx+cy=0Basically, we perform the following lambda substitution:aλ2+bλ+c=0and solve the resulting quadratic equation for λ:λ=b±b24ac2aThere are three cases:

  • If b24ac>0, then we have a solution of the formy=c1eλ1x+c2eλ2x
  • If  b24ac=0, then we have a solution of the formy=c1eλx+c2xeλx
  • If  b24ac<0, then we have a solution of the formy=eαx(c1cosβx+c2sinβx)

The last one is derived from Euler's formula eiθ=cosθ+isinθ, and the solution is taken to have the form λ=α±iβ.

Exampley+3y+2y=6To find the complementary function yc, we perform a lambda substitution and solve:λ2+3λ+2=0λ=2 or 1yc=c1ex+c2e2xThe particular solution has the form yp=A. Differentiating and substituting accordingly gives2A=6A=3Therefore, by the superposition principle, the general solution to the ODE isy=yc+yp=c1ex+c2e2x+3

The Method of Separation of Variables


We want to solve the PDEut=k2ux2with initial and boundary conditionsu(x,0)=f(x)u(0,t)=0u(L,t)=0To do that, let's assume the solution has the formu(x,t)=φ(x)G(t)Don't ask, just accept it. There's a reason behind picking such form, but it wouldn't help to explain it now. With that said, let's plug this solution into our PDE to obtainφ(x)dGdt=kG(t)d2φdx2Now, let's separate variables!1kG(t)dGdt=1φ(x)d2φdx2If you look at what we just obtained, you're eventually going to ask yourself: "how's it possible for a function in terms of t to be equal to a function in terms of x? This makes no sense at all!" Well, of course it doesn't, unless they're both equal to the same constant! Let's call this constant λ:1kG(t)dGdt=1φ(x)d2φdx2=λAgain, the reason we separated our equation this way and the reason we put a negative sign before λ are for the purpose of convenience only. It'll all make sense down the road, so stop scratching your head and keep reading! By the way, notice that our partial derivatives turned into normal ones because we're dealing with functions with one variable.

Now, let's solve the first ODEdGdt=λkG(t)Our famous toolkit methods lead to the solutionG(t)=ceλktWith that out of the way, we can now solve the second ODE (it's a BVP)d2φdx2=λφφ(0)=0φ(L)=0Performing a lambda substitution would be confusing, so let's call it an m substitution. It'll yieldm=±λAgain we have three cases:

  • If λ>0, thenφ=c1cosλx+c2sinλxFrom the boundary conditions,φ(0)=0c1=0φ(L)=0c2sinλL=0To avoid the trivial solution, we must have thatsinλL=0λ=(nπL)2where nN. As a result, we're left withφ(x)=c2sinnπxL
  • If λ=0, thenφ=c1+c2xFrom the boundary conditions,φ(0)=0c1=0φ(L)=0c2=0Therefore, λ=0 is not an eigenvalue for this problem.

  • If λ<0, thenφ=c1esx+c2esxwhere λ=s. Let's use the hyperbolic functions to represent thisφ=c3coshsx+c4sinhsxFrom the boundary conditions,φ(0)=0c3=0φ(L)=0c4sinhsL=0c4=0because sL>0 and sinh is never zero for a positive argument. Therefore, λ<0 is not an eigenvalue for this problem, either.

In summary, we obtained product solutions of the heat equationu(x,t)=BsinnπxLek(nπ/L)2tn=1,2,3,Finally, by the principle of superposition, we claim that the following infinite series is the solution of our heat conduction problem:u(x,t)=n=1BnsinnπxLek(nπ/L)2tPhew! I guess I could now do the same thing for the multidimensional case, lol.