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| CL470 Quote | Reply | | Please help to integrate hyperbolic function posted on: 11/9/2006 9:05:57 PM Hi! Can anyone show me step-by-step how to integrate this function (Poisson-Boltzmann Equation)? dψ/dx = [-2κkT/(ze)]*sinh (zeψ/2kT) ψ is a function of x, ψ(x) with boundary conditions: x=0,ψ(x)=ψ0, x=∞,ψ(x)=0 The answer is tanh(zeψ/4kT)=tanh(zeψ0/4kT)exp(-κx) Thank you! |
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herbert
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Please help to integrate hyperbolic function
replied on: 11/10/2006 6:22:09 AM before expecting advice I strongly recommend to translate this gibberish into something readable... |
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CL470
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Please help to integrate hyperbolic function
replied on: 11/10/2006 7:41:43 AM Sorry! I did not realise this. dp/dx = [-2KkT/(ze)]*sinh (zep/2kT) K,k,T,z,e are all constants p is a function of x, p(x) with boundary conditions: x=0,p(x)=p0, x=infinity;,p(x)=0 The answer is tanh(zep/4kT)=tanh(zep0/4kT)exp(-Kx) |
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jehovah0121
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Please help to integrate hyperbolic function
replied on: 11/10/2006 11:07:06 AM Let me have a try, if you don't mind. Too many constants take place in this equation, so we let A = 2kT/ze [1] and then dp/dx = -AK sinh(p/A) Let f = p/A [2] and we get df/dx = -K sinh(f) that is df / sinh(f) = -K dx Note that sinh(f) = [exp(f) - exp(-f)] / 2 so 2 df / [exp(f) - exp(-f)] = -K dx Multiply the upper and lower part of the left side of the equation (I don't know how to say it in English) by exp(f) together and we get 2 d exp(f) / [exp(2f) - 1] = -K dx Let u = exp(f) [3] and thus there is 2 du / [u^2 - 1] = -K dx which can be transformed into du / (u + 1) - du / (u - 1) = K dx Integrate both side ln {|u + 1| / |u - 1|} = K x + C where C represents a constant. And use a 'exp' to both side you get |u + 1| / |u - 1| = D exp(Kx) Here D = exp(C) > 0. In order to get rid of the absolute value symbols, you can write (u + 1)/(u - 1) = E exp(Kx) [4] E = D or -D, and this is not important. We just need to know that E is an arbitary constant. Using [3] and [4] we know that [exp(f) + 1] / [exp(f) - 1] = E exp(Kx) that is [exp(f/2) + exp(-f/2)] / [exp(f/2) - exp(-f/2)] = E exp(Kx) Using the definition of the functions sinh and cosh we obtain the following cosh(f/2) / sinh(f/2) = E exp(Kx) or tanh(f/2) = G exp(-Kx) where G = 1/E. No we are quite close to the answer. According to [2] we can write tanh(p/2A) = G exp(-Kx) [5] Consider the boundary conditions. When x = 0, p = p0, so from the above we know G = tanh(p0/2A) With reference to [1] you'll easily get the answer. The second boundary condition can be used to check the answer. When x -> +infinity, the right side of [5] -> zero (assuming that K > 0 is right). And now the left side. when p -> 0, tanh(p/2A) -> zero, too. So the solution we've just aquired meets the requirements of the problem. Wish my solution is good enough for you. 
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CL470
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Please help to integrate hyperbolic function
replied on: 11/10/2006 7:56:43 PM Thanks very much! You really help me to understand better. I have one question here: 2 df / [exp(f) - exp(-f)] = -K dx Multiply the upper and lower part of the left side of the equation (I don't know how to say it in English) by exp(f) together and we get 2 d exp(f) / [exp(2f) - 1] = -K dx I know when we multiply with exp(f), it will become exp(f)df (numerator part), can we assume exp(f)df = d exp(f)? I am not very sure about this 'transformation'. Probably you can help to further explain. Thanks. |
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jehovah0121
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Please help to integrate hyperbolic function
replied on: 11/12/2006 6:29:19 AM OK, let's change the variable first. f(x) = exp(x) we know that its derivative(hope I spell it correctly) is f'(x) = exp(x) and besides, f'(x) is equivalent to df/dx, which is d(exp(x))/dx = exp(x) and then d(exp(x)) = exp(x) dx Pay more attention to the relation between the f'(x) and the df(x). Understanding this will solve a lot of problems in your calculus learning.
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CL470
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Please help to integrate hyperbolic function
replied on: 11/13/2006 7:25:36 PM I got it. Thanks very much. 
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