## anonymous 3 years ago Find the steady-state temperature distribution for the semi-infinite plate problem if the temperature of the bottom edge is T = f(x) = x (in degrees; that is, the temperature at x cm is x degrees), the temperature of the other sides is 0o, and the width of the plate is 10 cm.

1. anonymous

@UnkleRhaukus @oldrin.bataku @sirm3d @niksva @ParthKohli and others, would you kindly help me?

2. anonymous

would u give us a geometry of problem? :)

3. anonymous

i would give a partial differential equation...,

4. anonymous

ok, and boundary conditions

5. anonymous

$\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0$???

6. anonymous

Hint: Answer:$T = \frac{ 20 }{ \pi } \sum_{n=1}^{\infty} \frac{ (-1)^{n+1} }{ n } e^{-n \pi y/10} \sin (n \pi x/10)$ How to get that.., ??

7. anonymous

ok, u must use "separation of variables"

8. anonymous

suppose that steady state solution can be written as$T(x,y)=X(x).Y(y)$plug in the original equation, u will get$X''Y+XY''=0$$\frac{X''}{X}=-\frac{Y''}{Y}=\color\red{\pm}\lambda^2$now what do u think positive sign or positive?

9. anonymous

sorry positive or negative?

10. anonymous

what's lambda ??

11. anonymous

lambda is a positive number, an arbitrary constant...but why we put it there? see this equation$\frac{X''}{X}=-\frac{Y''}{Y}$LHS is a function of only $$x$$ and RHS is a function of only $$y$$ its not possible unless both sides are equal to a constant number

12. anonymous

sorry i've no idea about positive and negative, but i just know that it's separation constant

13. anonymous

what do u think? what are boundary conditions?

14. anonymous

why $$\pm ??$$

15. anonymous

there are several methods to verify the sign of constant of separation

16. anonymous

the boundary conditions is l from 0 up to 10, right ?

17. anonymous

what math u are in? Advanced Engineering Mathematics?

18. anonymous

mathematical physics ..,

19. anonymous

ok, boundary conditions$T=f(x) \ \ @ \ \ y=0$$T=0 \ \ @ \ \ y \rightarrow \infty$$T=0 \ \ @ \ \ x=0$$T=0 \ \ @ \ \ x=10$what do u think?

20. anonymous

i think |dw:1366283650608:dw|

21. anonymous

|dw:1366283758632:dw|

22. anonymous

exactly, and $$f(x)=x$$ of course

23. anonymous

so lets think what will be the sign of $$\lambda^2$$

24. anonymous

because of solution may be either positive or negative ??

25. anonymous

ah..., i don't know

26. anonymous

first suppose positive$\frac{X''}{X}=\lambda^2$$X''-\lambda^2 X=0$$X=A\sinh \lambda x+B\cosh \lambda x$set boundary conditions$T=0 \ \ @ \ \ x=0$will result in $$X(0)=0$$ and it gives $$B=0$$$T=0 \ \ @ \ \ x=10$will result in $$X(10)=0$$ and it gives $$A=0$$ so $$X$$ will be equal to zero which is not acceptable so we must choose negative sign

27. anonymous

makes sense?

28. anonymous

ah yea..,

29. anonymous

ok, so we have$\frac{X''}{X}=-\frac{Y''}{Y}=\color\green{-}\lambda^2$$X''+\lambda^2 X=0$$Y''-\lambda^2 Y=0$$X=A\sin \lambda x+B\cos \lambda x$$Y=A\sinh \lambda y+B\cosh \lambda y$and finally setting boundary conditions

30. anonymous

$T=0 \ \ @ \ \ x=0$will result in $$X(0)=0$$ and it gives $$B=0$$$T=0 \ \ @ \ \ x=0$will result in $$X(10)=0$$ and it gives $$10\lambda=n\pi$$ and$\lambda=\frac{n\pi}{10} \ \ \ \color\green{\text{eigenvalues}}$

31. anonymous

or if u want get the exact form wich is given in solution u can write$Y=Ce^{\lambda y} +De^{-\lambda y}$

32. anonymous

do u know why?

33. anonymous

general solution of$Y''-\lambda^2 Y=0$can be written as$Y=C\sinh \lambda y+D\cosh \lambda y$or$Y=Ce^{\lambda y} +De^{-\lambda y}$thats for u...find out why :)

34. anonymous

now we know as $$y \rightarrow \infty$$ temperature approches zero so $$C=0$$

35. anonymous

its a simple limit calculation, right?

36. anonymous

so summing the answers for $$n=1,2,3,...$$ we will have$T = \sum_{n=1}^{\infty} a_ne^{-\frac{n \pi}{10} y} \sin (\frac{n \pi}{10} x)$for calculating $$a_n$$ final boundary condition will be useful$T=f(x)=x \ \ @ \ \ y=0$so$x = \sum_{n=1}^{\infty} a_n\sin (\frac{n \pi}{10} x)$from fourier series$a_n=\frac{2}{10} \int_{0}^{10} x\sin (\frac{n \pi}{10} x) \ \text{d}x=\frac{20}{n \pi} (-1)^{n+1}$We are done,I hope its helpful

37. anonymous

let me know if it is unclear.

38. anonymous

ah thnak you so much :)