# Change of variables in Heat Equation (and Fourier Series)

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1. Feb 5, 2017

Q: Suppose $u(x,t)$ satisfies the heat equation for $0<x<a$ with the usual initial condition $u(x,0)=f(x)$, and the temperature given to be a non-zero constant C on the surfaces $x=0$ and $x=a$.
We have BCs $u(0,t) = u(a,t) = C.$ Our standard method for finding u doesn't work here, since $e^{-k(\frac{n\pi}a)^2t}sin(\frac{n\pi}a)$ does not satisfy these BCs.
Make a change of variable from $u$ to $v=u-C.$ Show that $v$ satisfies the heat equation with BCs $v=0$ at $x=0$ and $x=a.$
Write down the solution for $v(x,t).$Deduce an expression for $u(x,t)$ in terms of constants $c_1,c_2,\ldots,$ and write down a formula for $c_n.$
[Harder] Now suppose the BCs are $u(0,t) = C$, $u(a,t)=D$ for constants $C,D.$ How could you solve the case?

My question: These are extensions to homework which I'd like try to attempt, but I don't know where to start with the change of variable

2. Feb 5, 2017

### pasmith

Substitute $u(x,t) = v(x,t) + C$ into the PDE and boundary and initial conditions you are given for $u$ to obtain a PDE and boundary and initial conditions satisfied by $v$.

3. Feb 5, 2017

From the given BCs for $u$, am I right in saying that BCs for $v$ is $v(0,t)=v(a,t)=u(0,t)+C=2C$? Also by substituting $u(x,t)=v(x,t)+C$ into the PDE do you mean partially differentiate it then substitute in like $$v=u+C$$ $$\frac{\partial{v}}{\partial{t}}=\frac{\partial{u}}{\partial{t}}$$ $$\frac{\partial^2{v}}{\partial^2{x}}=\frac{\partial^2{u}}{\partial^2{x}}$$ so $$\frac{\partial{u}}{\partial{t}}=K\frac{\partial^2{u}}{\partial{x}^2}$$ becomes $$\frac{\partial{v}}{\partial{t}}=K\frac{\partial^2{v}}{\partial{x}^2}$$