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Author Topic: Problem Set 1: Problem 4  (Read 6534 times)
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เกียรติศักดิ์
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« on: November 11, 2006, 04:24:17 PM »

Here is the questioning part.

Let  u(x; \lambda) and  v(x; \lambda) be the fundamental solutions of the Liouville equation, i.e.  u and  v are two linearly-independent solutions in terms of which all other solutions may be expressed (for a give value \lambda). Then there are constants  A and  B which allow any solution  y to be expressed as a linear combination of this fundamental set:

 y(x; \lambda) = Au(x; \lambda) + Bv(x; \lambda)

These constants are determined by requiring  y(x; \lambda) to satisfy the boundary conditions:

 \begin{array}{rcl} y(a; \lambda) &=& Au(a; \lambda) + Bv(a; \lambda) \\ y(b; \lambda) &=& Au(b; \lambda) + Bv(b; \lambda) \end{array}

Use this to show that the solution  y(x; \lambda) is unique, i.e., that there is one and only one solution corresponding to an eigenvalue of the Liouville equation.

I've been thinking for hours how to show... in vain. It'd be very nice if you give some hints. Smiley I guess there has something to do with the Wronskian, but that doesn't carry me quite far... Cry
« Last Edit: November 11, 2006, 04:47:12 PM by ปิยพงษ์ - Head Admin » Logged

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« Reply #1 on: November 11, 2006, 05:20:53 PM »

...
These constants are determined by requiring  y(x; \lambda) to satisfy the boundary conditions:

 \begin{array}{rcl} y(a; \lambda) &=& Au(a; \lambda) + Bv(a; \lambda) \\ y(b; \lambda) &=& Au(b; \lambda) + Bv(b; \lambda) \end{array}
...

If you had read the question carefully you should have noticed that the boundary conditions are:

 \begin{array}{rcl} y(a; \lambda) &=& Au(a; \lambda) + Bv(a; \lambda) = 0 \\ y(b; \lambda) &=& Au(b; \lambda) + Bv(b; \lambda) = 0\end{array}

 coolsmiley
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« Reply #2 on: November 11, 2006, 05:27:26 PM »

Oh, sorry for that redfaced.

But I really have noticed them, and used them in the calculation of the Wronskian. buck2
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« Reply #3 on: November 11, 2006, 05:50:12 PM »

...
But I really have noticed them, and used them in the calculation of the Wronskian. buck2

If you write the two equations as a matrix acting on a column vector with A and B as its elements. You will get nontrivial solutions only if the determinant of ....   Grin
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