dRic2
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Understanding the Eigenvalue Concept in Diffusion Equations
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I don't know about others but I have no interest in getting a kink in my neck turning my head sideways just because you can't be bothered to show your image right-side-updRic2 said:
dRic2
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Neither one. I was making a statement of fact.dRic2 said:
dRic2
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I think "please re-upload the image" would be enough. I'm not upset about it, I just want to tell you that (I think) it sounds mean: everyone can make a mistake. Words are powerful.
BTW I'm not offended, I was just saying.
BTW I'm not offended, I was just saying.
Voq
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I found it funny because i was just in pose with my head at 90 degree.
dRic2
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DrClaude said:
"Combustion" (Physical and chemical fundamentals, modeling and simulation) J. Warnatz, U. Maas, R.W. Dibble
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I've put a fair amount of time into this problem today, but haven't gotten very far. Maybe this can help you think about what's going on here, though. The business about eigenvalues refers to the linear algebra concept of linear transformations and function spaces, which are a slight variation of the concept of a vector space.
For a linear transformation L, ##\lambda## is an eigenvalue with associated eigenvector ##\vec x## provided that ##L\vec x = \lambda \vec x##. IOW, for an eigenvector, ##L\vec x## produces a scaled version of the same vector.
In this problem, we're talking about eigenvalues and eigenfunctions, so the equation above would be ##Lf = \lambda f##. As above, for an eigenfunction, ##Lf## results in a scalar multiple of the same function. The differential equation could be written as ##L f = 0##, where the tranformation is the operator ##L = a\frac{d^2}{dz^2} - v \frac{d}{dz} - \delta \cdot A \exp \left[-\frac E {R(T_b - \delta)}\right]##. Applying the linear operator L to a function f would entail working with this linear combination of f, f', and f''.
To find the eigenfunction basis (which would consist of two functions, being that the DE is second order), one would need to solve the DE ##L f = \lambda f##, which as the text states, would be quite complicated. Even for much simpler DEs there is a lot of work.
An eigenfunction would be some exponential function for which the exponent would involve the associated eigenvalue, ##\lambda##. I don't have any advice on the "Nevertheless, it can be shown easily that ... " part.
For a linear transformation L, ##\lambda## is an eigenvalue with associated eigenvector ##\vec x## provided that ##L\vec x = \lambda \vec x##. IOW, for an eigenvector, ##L\vec x## produces a scaled version of the same vector.
In this problem, we're talking about eigenvalues and eigenfunctions, so the equation above would be ##Lf = \lambda f##. As above, for an eigenfunction, ##Lf## results in a scalar multiple of the same function. The differential equation could be written as ##L f = 0##, where the tranformation is the operator ##L = a\frac{d^2}{dz^2} - v \frac{d}{dz} - \delta \cdot A \exp \left[-\frac E {R(T_b - \delta)}\right]##. Applying the linear operator L to a function f would entail working with this linear combination of f, f', and f''.
To find the eigenfunction basis (which would consist of two functions, being that the DE is second order), one would need to solve the DE ##L f = \lambda f##, which as the text states, would be quite complicated. Even for much simpler DEs there is a lot of work.
An eigenfunction would be some exponential function for which the exponent would involve the associated eigenvalue, ##\lambda##. I don't have any advice on the "Nevertheless, it can be shown easily that ... " part.
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dRic2
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First of all thank you very much for the effort.Mark44 said:
I'm familiar with eigenvalues/eigenfunctions thanks to quantum mechanics. But I can't even understand the meaning of this. The operator ##L## refers to the function ##\delta(z)## so what does "if v has the eigenvalue..." mean? How can v have an eigenvalue if the differential equation is written for ##\delta##?
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The best answer I can come up with is that the above means "if v is replaced by the eigenvalue ##v_L##..." Other than that, no idea.dRic2 said:
It seems to be slightly idiosyncratic (although in some cases the relation with more "traditional" eigenvalue problems can be more evident), but in combustion you often get such differential equations that depend on some parameters that have non-trivial solutions only for certain relations between the parameters.
In the case at hand, ##\delta(z) = 0## is the trivial solution. Given the diffusivity ##\alpha## and the Arrhenius parameters ##A## and ##E##, then the equation has a non-trivial solution ##\delta(z)## only for a particular value of ##v##, which is then called an eigenvalue.
In the case at hand, ##\delta(z) = 0## is the trivial solution. Given the diffusivity ##\alpha## and the Arrhenius parameters ##A## and ##E##, then the equation has a non-trivial solution ##\delta(z)## only for a particular value of ##v##, which is then called an eigenvalue.
dRic2
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Thank you for the replies. It's far more clear now!
I just have one more question out of curiosity:
any ideas to show the "eigenvalue" is ##v_L = \sqrt {\alpha k}##?PS: Am I correct to assume that "eigenvalue" is just used in a misleading/improper way here ?
I just have one more question out of curiosity:
DrClaude said:
any ideas to show the "eigenvalue" is ##v_L = \sqrt {\alpha k}##?PS: Am I correct to assume that "eigenvalue" is just used in a misleading/improper way here ?