# Inhomogeneous heat equation Neumann boundary conditions with f(x,t)=cos(2x).

Consider the following inhomogeneous heat equation with Neumann boundary conditions. \begin{cases} u_t-u_{xx}=\cos(2x)\\ u(x,0)=x \\ u_x(0,t)=u_x(\pi,t)=0 \end{cases}\text{ for }\: \begin{align} 0 By separation of variables I arrive at the homogeneous solution $$u_h(x,t)=\sum_{i=0}^\infty A_ne^{-n^2t}\cos(nx)$$ $$A_n= \begin{cases} \pi /2 & n=0\\ -4/(\pi n^2)& \text{if n is odd}\\ 0& \text{if n is even} \end{cases}$$ On trying to find the inhomogeneous solution by applying Duhamel's principle i.e. using $$u_p(x,t)=\int_0^t\sum_{i=0}^\infty B_n(s)\cos(nx)e^{-n^2(t-s)}ds$$ where $$B_n(s)=2/\pi \int_0^\pi \cos(nx)\cos(2x) dx$$ I get $$B_n(s)=0$$ What went wrong? Any assistance much appreciated.

• $n$ isn't fixed, it can be any integer. Can you think of an integer where $B_n(s)$ is nonzero? – Ninad Munshi Nov 19 '19 at 20:22
• I can write cos(nx)cos(2x) as 1/2(cos((n+2)x)+cos((n-2)x)) which integrates to terms in sin which vanish for multiples of pi. So I am struggling to find an integral n that produces anything other than zero. I am clearly missing something? A pi/2 would be nice – OEB Nov 20 '19 at 16:37
• Thank you. I missed the obvious!!! – OEB Nov 21 '19 at 0:19

Using the Laplace transform

$${\cal L}\left(u_t-u_{xx}-\cos(2x)\right) = sU(s,x)-u(0,x)-U_{xx}(s,x)-\frac 1s\cos(2x)$$

and now solving

$$sU(s,x)-x-U_{xx}(s,x)-\frac 1s\cos(2x)=0,\ \ U_x(s,0)=U_x(s,\pi)$$

we have

$$U(s,x) = \frac{1}{s(s+4)}\left((s+4)x+\cos(2x)-(s(s+4))(e^{\sqrt s x}-e^{\sqrt s(s-x)})C(s)\right)$$

now assuming that $$U(s,x)$$ remains limited as $$x\to\infty$$ we have $$C(s) = 0$$ and then

$$U(s,x) = \frac xs+\frac{\cos(2x)}{s(s+4)}$$

with inverse

$$u(t,x) = \frac 14\left(1-e^{-4 t}\right) \cos (2 x)+x$$