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Trying to solve an ODE (nonlinear, first-order)

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Re: Trying to solve an ODE (nonlinear, first-order)

The part that I'm struggling with at the moment is the translation from

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to

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I've followed the equations as stated from the last e-mail & at the mon=ment I get numbers >10^307 so obviously something wrong.

regards

Andy

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Re: Trying to solve an ODE (nonlinear, first-order)

I had not looked at the mail in detail, but I gues Sungyeop would have to explain his substitution in more detail. I am not sure if his intitial second order ODe1 is correct.

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Re: Trying to solve an ODE (nonlinear, first-order)

After a long absent, I would like to make some further notes. Thank you very much in advace.

1) Negative boundary condition V(d) = -0.6 V

As Werner mentioned, mathematically, V(d) cannot take a negative value. And, I have noticed that a negative V(d) is also physically non-sense. Sorry for my carelessness. For it to be physically and mathematically meaningful, V(d) should always be 0.6 V. In fact, in the problem, V(0) is the electrical potential imposed by Au(gold) electrode and V(d) is that of Al(Aluminium).

2) The slow change of V(d) in response to the change of C(the electric field, which is dV/dx).

As mentioned in the last part of 1), we are dealing with a problem where the electrical potential at two electrodes V(0) and V(d) are fixed at 0 and 0.6V, respectively. Thus, in principle, V(0) and V(C) should not be changing in response to the change of C(the electric field).

The ODE solver was written with one boundary condition. It would be better if we can impose two boundary conditions V(d) = 0.6 V as well as V(0) = 0 V. But I have no clue, for now, how to do that.

Anyway, the proposed solution was very useful.

The variable that I am trying to sweep is 'sigma' from 0.05 to a value where V(d) restes on 0.6 V, while struggling to extract data from the plot using excel component. (recently uploaded an other thread on that problem. If time allows, please have a look.)

I hope this response would help you understand the problem better. Thank you.

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