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 one year ago
Find the steadystate temperature distribution for the semiinfinite 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
Find the steadystate temperature distribution for the semiinfinite 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.

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gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus @oldrin.bataku @sirm3d @niksva @ParthKohli and others, would you kindly help me?

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3would u give us a geometry of problem? :)

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0i would give a partial differential equation...,

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3ok, and boundary conditions

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0\]???

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0Hint: 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.., ??

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3ok, u must use "separation of variables"

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3suppose 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?

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3sorry positive or negative?

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3lambda 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

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0sorry i've no idea about positive and negative, but i just know that it's separation constant

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3what do u think? what are boundary conditions?

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3there are several methods to verify the sign of constant of separation

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0the boundary conditions is l from 0 up to 10, right ?

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3what math u are in? Advanced Engineering Mathematics?

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0mathematical physics ..,

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3ok, boundary conditions\[T=f(x) \ \ @ \ \ y=0\]\[T=0 \ \ @ \ \ y \rightarrow \infty\]\[T=0 \ \ @ \ \ x=0\]\[T=0 \ \ @ \ \ x=10\]what do u think?

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0i think dw:1366283650608:dw

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0dw:1366283758632:dw

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3exactly, and \(f(x)=x\) of course

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3so lets think what will be the sign of \(\lambda^2\)

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0because of solution may be either positive or negative ??

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0ah..., i don't know

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3first 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

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3ok, 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

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3\[T=0 \ \ @ \ \ x=0\]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\[\lambda=\frac{n\pi}{10} \ \ \ \color\green{\text{eigenvalues}}\]

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3or if u want get the exact form wich is given in solution u can write\[Y=Ce^{\lambda y} +De^{\lambda y} \]

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3general 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 :)

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3now we know as \(y \rightarrow \infty\) temperature approches zero so \(C=0\)

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3its a simple limit calculation, right?

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3so 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

mukushla
 one year ago
Best ResponseYou've already chosen the best response.3let me know if it is unclear.

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0ah thnak you so much :)
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