gerryliyana Group Title 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. one year ago one year ago

1. gerryliyana Group Title

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

2. mukushla Group Title

would u give us a geometry of problem? :)

3. gerryliyana Group Title

i would give a partial differential equation...,

4. mukushla Group Title

ok, and boundary conditions

5. mukushla Group Title

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

6. gerryliyana Group Title

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. mukushla Group Title

ok, u must use "separation of variables"

8. mukushla Group Title

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. mukushla Group Title

sorry positive or negative?

10. gerryliyana Group Title

what's lambda ??

11. mukushla Group Title

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. gerryliyana Group Title

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

13. mukushla Group Title

what do u think? what are boundary conditions?

14. gerryliyana Group Title

why $$\pm ??$$

15. mukushla Group Title

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

16. gerryliyana Group Title

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

17. mukushla Group Title

what math u are in? Advanced Engineering Mathematics?

18. gerryliyana Group Title

mathematical physics ..,

19. mukushla Group Title

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. gerryliyana Group Title

i think |dw:1366283650608:dw|

21. gerryliyana Group Title

|dw:1366283758632:dw|

22. mukushla Group Title

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

23. mukushla Group Title

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

24. gerryliyana Group Title

because of solution may be either positive or negative ??

25. gerryliyana Group Title

ah..., i don't know

26. mukushla Group Title

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. mukushla Group Title

makes sense?

28. gerryliyana Group Title

ah yea..,

29. mukushla Group Title

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. mukushla Group Title

$T=0 \ \ @ \ \ x=0$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. mukushla Group Title

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

32. mukushla Group Title

do u know why?

33. mukushla Group Title

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. mukushla Group Title

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

35. mukushla Group Title

its a simple limit calculation, right?

36. mukushla Group Title

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. mukushla Group Title

let me know if it is unclear.

38. gerryliyana Group Title

ah thnak you so much :)