MHB Finding Temperature in a Rectangular Plate: $\nabla^2u=0$

Dustinsfl
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The steady state temperature $u(x,y)$ in a rectangular plate $0\leq x\leq L$, $0\leq y\leq M$, is sought, under the condition that the edge $x = 0$ is maintained at zero degrees, $x = L$ is kept at $u(L,y) = y$ degrees, and the edges $y = 0$ and $y = M$ are insulated. The appropriate differential equation $\nabla^2u = 0$.

Since the vertical boundary conditions are insulated, wouldn't this be the same as just dealing with $u_t=u_{xx}$ since Fourier coefficients for those boundaries will be 0?
 
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Saying that the boundaries are insulated does NOT mean that the Fourier coefficients are 0. It means that the derivatives there are 0 and so the coefficients of the sine terms are 0.
 
Wasn't thinking figured out this post.
 
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I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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