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Grogs

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I've got a couple of BVP's I'm working on that are giving me some problems. First, I'm asked to solve the 1-D Heat Equation:

[tex]k \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}[/tex]

in a thin rod of length L subject to the following:

BC1) u(0, t) = 0

BC2) u(L, t) = 0

IC) 2 different problems. For the first, [itex]u(x,0) = 1 - 2 cos(\frac{9 \pi x}{L})[/itex]. For the second, [itex]u(x,0) = 3 cos(\frac{9 \pi x}{2 L})[/itex].

I can separate this problem into:

[tex]\frac{X''}{X} = \frac{T'}{k T} = - \lambda[/tex]

easily enough and apply the 2 BC's to get:

[tex]u_{n} (x,t) = A_{n} \sin(\frac{n \pi x}{L}) \exp(\frac{-k n^2 \pi^2 t}{L}) , n=1,2,3,...[/tex]

My problem is with applying the Initial condition and solving for u(x,t):

[tex]u(x,t) = \sum \limits_{n=1} ^ {\infty} A_{n} \sin(\frac{n \pi x}{L}) = IC[/tex]

My book shows me how to do a half-range expansion here, but according to my professor, I shouldn't need to. He even sent us an e-mail changing one of the IC's "so it would be an eigenfunction." How do I get there from here? Is there some easy trig identity to convert those IC's to a sin function I'm just missing?My 2nd question involves steady state heat conduction in a 2-D square plate of side L:

[tex]\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0[/tex]

Subject To:

BC1) [itex]\frac{\partial u}{\partial x} = 0[/itex] at x = 0

BC2) [itex]\frac{\partial u}{\partial x} = 0[/itex] at x = L

BC3) [itex]\frac{\partial u}{\partial y} = 0[/itex] at y = 0

BC4) u(x,L) = 5

I separate the Laplacian into:

[tex]\frac{X''}{X} = \frac{-Y'}{Y} = - \lambda[/tex]

and solve using the first 3 BC's, I get [itex]X = c_{1}[/itex] and [itex]Y = c_{4}[/itex] for [itex]\lambda = 0[/itex] and

[tex]

X(x) = c_{2} \sin(\frac{n \pi x}{L}) [/tex]

[tex]Y(y) = c_{4} \cosh(\frac{n \pi y}{L})[/tex]

for [itex]\lambda > 0[/itex].

Using superposition, I end up with:

[tex]u(x,y) = A_{0} + \sum \limits_{n=1} ^ {\infty} A_{n} \sin(\frac{n \pi x}{L}) \cosh(\frac{n \pi y}{L})[/tex]

When I apply the final BC, u(x,L) = 5, I end up with a solution of u(x,y) = 5. I guess I'm just looking for some sanity checking here. I can certainly see how u = 5 works here, but is it the

Thanks in advance,

Grogs

[tex]k \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}[/tex]

in a thin rod of length L subject to the following:

BC1) u(0, t) = 0

BC2) u(L, t) = 0

IC) 2 different problems. For the first, [itex]u(x,0) = 1 - 2 cos(\frac{9 \pi x}{L})[/itex]. For the second, [itex]u(x,0) = 3 cos(\frac{9 \pi x}{2 L})[/itex].

I can separate this problem into:

[tex]\frac{X''}{X} = \frac{T'}{k T} = - \lambda[/tex]

easily enough and apply the 2 BC's to get:

[tex]u_{n} (x,t) = A_{n} \sin(\frac{n \pi x}{L}) \exp(\frac{-k n^2 \pi^2 t}{L}) , n=1,2,3,...[/tex]

My problem is with applying the Initial condition and solving for u(x,t):

[tex]u(x,t) = \sum \limits_{n=1} ^ {\infty} A_{n} \sin(\frac{n \pi x}{L}) = IC[/tex]

My book shows me how to do a half-range expansion here, but according to my professor, I shouldn't need to. He even sent us an e-mail changing one of the IC's "so it would be an eigenfunction." How do I get there from here? Is there some easy trig identity to convert those IC's to a sin function I'm just missing?My 2nd question involves steady state heat conduction in a 2-D square plate of side L:

[tex]\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0[/tex]

Subject To:

BC1) [itex]\frac{\partial u}{\partial x} = 0[/itex] at x = 0

BC2) [itex]\frac{\partial u}{\partial x} = 0[/itex] at x = L

BC3) [itex]\frac{\partial u}{\partial y} = 0[/itex] at y = 0

BC4) u(x,L) = 5

I separate the Laplacian into:

[tex]\frac{X''}{X} = \frac{-Y'}{Y} = - \lambda[/tex]

and solve using the first 3 BC's, I get [itex]X = c_{1}[/itex] and [itex]Y = c_{4}[/itex] for [itex]\lambda = 0[/itex] and

[tex]

X(x) = c_{2} \sin(\frac{n \pi x}{L}) [/tex]

[tex]Y(y) = c_{4} \cosh(\frac{n \pi y}{L})[/tex]

for [itex]\lambda > 0[/itex].

Using superposition, I end up with:

[tex]u(x,y) = A_{0} + \sum \limits_{n=1} ^ {\infty} A_{n} \sin(\frac{n \pi x}{L}) \cosh(\frac{n \pi y}{L})[/tex]

When I apply the final BC, u(x,L) = 5, I end up with a solution of u(x,y) = 5. I guess I'm just looking for some sanity checking here. I can certainly see how u = 5 works here, but is it the

**only**solution?Thanks in advance,

Grogs

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