Help Two fused semi-infinite rods - hard integral

[ab959]

Homework Statement

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 ($$K_1$$, $$K_2$$).

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

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 doesn't 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.

 

[mighty2000]

Thank you for posting your problem. It seems like you are on the right track, but there are a few things that could be improved in your approach.

Firstly, when setting u_1=u_2 at x=0, you need to also ensure that the derivatives of u_1 and u_2 are equal at x=0. This means that you need to set K_1 u_{1x}=K_2 u_{2x} at x=0 as well. This should help you to get a non-zero solution.

Secondly, when evaluating the integrals, it may be helpful to use variable substitutions to simplify the expressions. For example, for the first integral, you could use the substitution u=x+y and v=x-y, which would simplify the integrand to exp(-u^2/4piK_1t) and exp(-v^2/4piK_1t). This could help to simplify the expression and make it easier to evaluate.

Lastly, it may also be helpful to check your work and make sure that the final solution satisfies all of the given boundary conditions and equations. If it does not, then there may be a mistake somewhere in your calculations.

Overall, using the technique of images may also be a valid approach to solving this problem. It may be worth exploring both approaches to see which one yields a simpler solution.

I hope this helps and good luck with your problem!