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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.

Mathematics
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@UnkleRhaukus @oldrin.bataku @sirm3d @niksva @ParthKohli and others, would you kindly help me?
would u give us a geometry of problem? :)
i would give a partial differential equation...,

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Other answers:

ok, and boundary conditions
\[\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0\]???
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.., ??
ok, u must use "separation of variables"
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?
sorry positive or negative?
what's lambda ??
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
sorry i've no idea about positive and negative, but i just know that it's separation constant
what do u think? what are boundary conditions?
why \(\pm ??\)
there are several methods to verify the sign of constant of separation
the boundary conditions is l from 0 up to 10, right ?
what math u are in? Advanced Engineering Mathematics?
mathematical physics ..,
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?
i think |dw:1366283650608:dw|
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exactly, and \(f(x)=x\) of course
so lets think what will be the sign of \(\lambda^2\)
because of solution may be either positive or negative ??
ah..., i don't know
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
makes sense?
ah yea..,
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
\[T=0 \ \ @ \ \ x=0\]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\[\lambda=\frac{n\pi}{10} \ \ \ \color\green{\text{eigenvalues}}\]
or if u want get the exact form wich is given in solution u can write\[Y=Ce^{\lambda y} +De^{-\lambda y} \]
do u know why?
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 :)
now we know as \(y \rightarrow \infty\) temperature approches zero so \(C=0\)
its a simple limit calculation, right?
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
let me know if it is unclear.
ah thnak you so much :)

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