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gerryliyana
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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
 one year ago
gerryliyana Group Title
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
 one year ago

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

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

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

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

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

gerryliyana Group TitleBest ResponseYou've already chosen the best response.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.., ??
 one year ago

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

mukushla Group TitleBest ResponseYou've already chosen the best response.3
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?
 one year ago

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

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
what's lambda ??
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

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

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

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
why \(\pm ??\)
 one year ago

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

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

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.3
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?
 one year ago

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

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

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

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

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
makes sense?
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
ah yea..,
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

mukushla Group TitleBest 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\[\lambda=\frac{n\pi}{10} \ \ \ \color\green{\text{eigenvalues}}\]
 one year ago

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

mukushla Group TitleBest ResponseYou've already chosen the best response.3
do u know why?
 one year ago

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

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

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

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