Solve the Indicial Equation to Show c_1\,=\,c_2\,=\,0

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The first term of the equation is xr= a0x0= a0. The second term is (r+1) a1x1= (r+1) a1x. In order that the sum of those be 0, you must choose r so that a0= a1= 0. That's the indicial equation.
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VinnyCee
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Homework Statement



Apply the method of Frobenius to find the roots of the indicial equation to show that [itex]c_1\,=\,c_2\,=\,0[/itex].

The equation in question is a 2nd order DE that was https://www.physicsforums.com/showthread.php?t=177492".

[tex]z\,\frac{d^2T}{dx^2}\,+\,\frac{z}{x}\,\frac{dT}{dx}\,-\,z\,y\,=\,0[/tex]

If you look at the thread where this equation was derived, we assume that [itex]z\,=\,x^2[/itex].

[tex]x^2\,\frac{d^2T}{dx^2}\,+\,\frac{x^2}{x}\,\frac{dT}{dx}\,-\,x^2\,y\,=\,0[/tex]

Homework Equations



http://en.wikipedia.org/wiki/Frobenius_method" [Broken]

The Attempt at a Solution



There is a regular, singular point at [itex]x_0\,=\,0[/itex]. We seek a solution of the form

[tex]\sum_0^\infty\,a_n\,x^{n\,+\,r}[/tex]

Differentiating that sum twice and substituting into the steady-state heat balance equation above and bringing the x's into the sums

[tex]\sum_0^\infty\,(n\,+\,r)(n\,+\,r\,+1)\,a_n\,x^{n\,+\,r}\,+\,\sum_0^\infty\,(n\,+\,r)\,a_n\,x^{n\,+\,r\,-1}\,-\,\sum_0^\infty\,a_n\,x^{n\,+\,r\,+\,2}\,+\,T_a\,x^2\,=\,0[/tex]

Note the last term, it arises from using [itex]y\,=\,T\,-\,T_a[/itex] from the https://www.physicsforums.com/showthread.php?t=177492". I really don't know what to do with it, I don't think it is constant, but how do I combine into sums and how do I deal with the constant [itex]T_a[/itex]? I am going to eliminate it, but don't know why!

Now, changing the indicies to combine the summations

[tex]\sum_0^\infty\,\left[(n\,+\,r)(n\,+\,r\,+\,1)\,a_n\,+\,(n\,+\,r\,+1)\,a_{n\,+\,1}\,-\,a_{n\,-\,2}\right]\,x^{n\,+\,r}\,=\,0[/tex]

Now, I get the indicial equation

[tex](r^2\,-\,r)\,a_0\,+\,(r\,+\,1)\,a_1\,=\,0[/tex]

[tex]r^2\,-\,r\,=\,0\,\longrightarrow\,r^2\,-\,r\,=\,0\,\longrightarrow\,r\,=\,0,\,1[/tex]

But the roots are both supposed to be zero, what did I do wrong?
 
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  • #2
VinnyCee said:

Homework Statement



Apply the method of Frobenius to find the roots of the indicial equation to show that [itex]c_1\,=\,c_2\,=\,0[/itex].

The equation in question is a 2nd order DE that was https://www.physicsforums.com/showthread.php?t=177492".

[tex]z\,\frac{d^2T}{dx^2}\,+\,\frac{z}{x}\,\frac{dT}{dx}\,-\,z\,y\,=\,0[/tex]

If you look at the thread where this equation was derived, we assume that [itex]z\,=\,x^2[/itex].

[tex]x^2\,\frac{d^2T}{dx^2}\,+\,\frac{x^2}{x}\,\frac{dT}{dx}\,-\,x^2\,y\,=\,0[/tex]
So the equation is really
[tex]x^2\frac{d^2T}{dx^2}+ x\frac{dT}{dx}- x^y= 0[/tex]
and with y= T-Ta, that's
[tex]x^2\frac{d^2T}{dx^2}+ x\frac{dT}{dx}- x^2T- x^2Ta= 0[/tex]


Homework Equations



http://en.wikipedia.org/wiki/Frobenius_method" [Broken]



The Attempt at a Solution



There is a regular, singular point at [itex]x_0\,=\,0[/itex]. We seek a solution of the form

[tex]\sum_0^\infty\,a_n\,x^{n\,+\,r}[/tex]

Differentiating that sum twice and substituting into the steady-state heat balance equation above and bringing the x's into the sums

[tex]\sum_0^\infty\,(n\,+\,r)(n\,+\,r\,+1)\,a_n\,x^{n\,+\,r}\,+\,\sum_0^\infty\,(n\,+\,r)\,a_n\,x^{n\,+\,r\,-1}\,-\,\sum_0^\infty\,a_n\,x^{n\,+\,r\,+\,2}\,+\,T_a\,x^2\,=\,0[/tex]

Note the last term, it arises from using [itex]y\,=\,T\,-\,T_a[/itex] from the https://www.physicsforums.com/showthread.php?t=177492". I really don't know what to do with it, I don't think it is constant, but how do I combine into sums and how do I deal with the constant [itex]T_a[/itex]? I am going to eliminate it, but don't know why!

Now, changing the indicies to combine the summations

[tex]\sum_0^\infty\,\left[(n\,+\,r)(n\,+\,r\,+\,1)\,a_n\,+\,(n\,+\,r\,+1)\,a_{n\,+\,1}\,-\,a_{n\,-\,2}\right]\,x^{n\,+\,r}\,=\,0[/tex]

Now, I get the indicial equation

[tex](r^2\,-\,r)\,a_0\,+\,(r\,+\,1)\,a_1\,=\,0[/tex]
No, don't change the indicies yet. You should never have any "a" except a0 in the indicial equation. Going back to the equation before you changed the indicies, observe that the lowest power occurs when n= 0 (that's always true) in the middle sum. Then you get [itex]a_0 r x^r[/itex] and that is the ONLY way you get xr in the entire sum. In order that that be 0 (so the entire sum will be identically 0) is to have either a0= 0 or r= 0. The whole point of Frobenious' method is to choose r so that the first term, a0 is NOT 0. We must have r= 0.

[tex]r^2\,-\,r\,=\,0\,\longrightarrow\,r^2\,-\,r\,=\,0\,\longrightarrow\,r\,=\,0,\,1[/tex]

But the roots are both supposed to be zero, what did I do wrong?

As far as the "Tax2" term is concerned, include Ta in the equation for the coefficient of x2 but no other power.
 
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1. What is an indicial equation?

An indicial equation is a type of differential equation that involves the variable x raised to a power, known as an index. The equation is typically solved by using the method of Frobenius, which involves finding a series solution.

2. How do you solve an indicial equation?

To solve an indicial equation, you can use the method of Frobenius, which involves finding a series solution. The first step is to assume a solution in the form of a power series and then substitute it into the indicial equation. This will result in a recurrence relation, which can be used to find the coefficients of the series and ultimately solve the equation.

3. What does it mean for c_1 and c_2 to equal 0 in the solution?

If c_1 and c_2 are both equal to 0 in the solution of an indicial equation, it means that the series solution reduces to a polynomial solution. This is a special case and occurs when the roots of the indicial equation are equal or differ by an integer.

4. Can an indicial equation have multiple solutions?

Yes, an indicial equation can have multiple solutions. This typically occurs when the indicial equation has distinct roots, resulting in different series solutions. In some cases, the indicial equation may also have repeated roots, leading to a single solution with a different form.

5. What are some real-world applications of solving indicial equations?

Solving indicial equations is essential in many fields of science and engineering, such as physics, chemistry, and biology. For example, in quantum mechanics, the Schrödinger equation can be solved using the method of Frobenius, which involves solving an indicial equation. Indicial equations also arise in the study of differential equations in fluid mechanics, electromagnetism, and other areas of physics.

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