# Help! Two fused semi-infinite rods - hard integral

1. Oct 11, 2007

### ab959

1. The problem statement, all variables and given/known data

The problem is to find an function that describes the tempurate variation with time and position of an infinitely long heated rod that is made by fusing together two semi infinite rods at x=0. They have perfect thermal contact but are made of two different materials and have different conductivity coefficients ($$K_1$$, $$K_2$$).

2. Relevant equations

$$u_t=K u_{xx}$$
u(x,0)=f(x)

for x>=0:
$$u=u_1$$.
for x<0:
$$u=u_2$$.

at x=0:
$$u_1=u_2=g(t)$$
$$K_1 u_{1x}=K_2 u_{2x}$$

3. The attempt at a solution

Once I created the equations above I tried to solve each case separately for two semi infinite rods and then impose the boundary conditions to so I can give the function g(t) a value that will properly govern the join.

I derive:
$$u_1(x,t)=\frac{1}{\sqrt{4\pi K_1 t}} \int_0^{inf}exp(\frac{-(x+y)^2}{4\pi K_1 t}-exp\frac{-(x-y)^2}{4\pi K_1 t}) f(s)ds + \int_0^t\frac{x}{\sqrt{4\pi K\1 (t-s)^3}}exp\frac{-x^2}{4\pi K_1 (t-s)} g(s)ds$$

$$u_2(x,t)=\frac{1}{\sqrt{4\pi K_2 t}} \int_0^{inf}exp(\frac{-(x+y)^2}{4\pi K_2 t}-exp\frac{-(x-y)^2}{4\pi K_2 t}) f(s)ds - \int_0^t\frac{x}{\sqrt{4\pi K_2(t-s)^3}}exp\frac{-x^2}{4\pi K_2 (t-s)} g(s)ds$$

but get stuck when I set $$u_1=u_2$$ at x=0 it doesnt work out to give anything (i.e. u=0). So I was thinking of trying to evaluate the integral before setting x=0 but I get stuck!

I end up reducing the problem to something like this:

$$\int_0^t\frac{x}{\sqrt{4\pi K_2(t-s)^3}}exp\frac{-x^2}{4\pi K_2 (t-s)} g(s)ds+\int_0^t\frac{x}{\sqrt{4\pi K_2(t-s)^3}}exp\frac{-x^2}{4\pi K_2 (t-s)} g(s)ds =0$$

Any suggestions on how to simplify this further and climb this hurdle would be appreciated. Do you think this is the correct approach? I was also considering using the technique of images which is used to derive the semi infinite rod equation but I thought this would be easier.