Abel's THM and the Wronskian - W=0?
Hi guys,
Questions about Wronskians. So consider a general second order DE: y" + py' + qy = 0 (*)
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Q: By evaluating the Wronskian, or otherwise, find functions p(x) and q(x) such that has solutions y1(x) = 1 + cos x and y2(x) = sin x. What is the value of W(pi)? Is there a unique solution to the differential equation for 0 < x < infinity with initial conditions y(0) = 0, y′(0) = 1? Why or why not?
---
I have found p(x) and q(x) - that's fine. Now, evaluating W(pi), I get W(pi) = 0.
But W cannot be 0 for all values in the interval we are considering, namely 0 < x < infinity; indeed, by Abel's theorem, either W is not 0 for all x, or W = 0 for all x. So in the context of this question, does that mean we cannot find a unique solution since W(pi) = 0??
If anyone could clear this up it would be much appreciated.
Questions about Wronskians. So consider a general second order DE: y" + py' + qy = 0 (*)
---
Q: By evaluating the Wronskian, or otherwise, find functions p(x) and q(x) such that has solutions y1(x) = 1 + cos x and y2(x) = sin x. What is the value of W(pi)? Is there a unique solution to the differential equation for 0 < x < infinity with initial conditions y(0) = 0, y′(0) = 1? Why or why not?
---
I have found p(x) and q(x) - that's fine. Now, evaluating W(pi), I get W(pi) = 0.
But W cannot be 0 for all values in the interval we are considering, namely 0 < x < infinity; indeed, by Abel's theorem, either W is not 0 for all x, or W = 0 for all x. So in the context of this question, does that mean we cannot find a unique solution since W(pi) = 0??
If anyone could clear this up it would be much appreciated.
Original post by Peter8837
Hi guys,
Questions about Wronskians. So consider a general second order DE: y" + py' + qy = 0 (*)
---
Q: By evaluating the Wronskian, or otherwise, find functions p(x) and q(x) such that has solutions y1(x) = 1 + cos x and y2(x) = sin x. What is the value of W(pi)? Is there a unique solution to the differential equation for 0 < x < infinity with initial conditions y(0) = 0, y′(0) = 1? Why or why not?
---
I have found p(x) and q(x) - that's fine. Now, evaluating W(pi), I get W(pi) = 0.
But W cannot be 0 for all values in the interval we are considering, namely 0 < x < infinity; indeed, by Abel's theorem, either W is not 0 for all x, or W = 0 for all x. So in the context of this question, does that mean we cannot find a unique solution since W(pi) = 0??
If anyone could clear this up it would be much appreciated.
Questions about Wronskians. So consider a general second order DE: y" + py' + qy = 0 (*)
---
Q: By evaluating the Wronskian, or otherwise, find functions p(x) and q(x) such that has solutions y1(x) = 1 + cos x and y2(x) = sin x. What is the value of W(pi)? Is there a unique solution to the differential equation for 0 < x < infinity with initial conditions y(0) = 0, y′(0) = 1? Why or why not?
---
I have found p(x) and q(x) - that's fine. Now, evaluating W(pi), I get W(pi) = 0.
But W cannot be 0 for all values in the interval we are considering, namely 0 < x < infinity; indeed, by Abel's theorem, either W is not 0 for all x, or W = 0 for all x. So in the context of this question, does that mean we cannot find a unique solution since W(pi) = 0??
If anyone could clear this up it would be much appreciated.
It's been about 3 years since I last did this kind of thing so this might not be the right approach, but one way of looking at this is to note that all solutions to the DE must be of the form ay1 + by2, and since you know y1(0), y1'(0), y2(0), y2'(0), you can get simultaneous equations for a and b using the initial conditions you've got. If these simultaneous equations uniquely determine a and b, then there's a unique solution, otherwise there isn't. This approach doesn't use the Wronskian so it might be that this is wrong, or that this is right but the examiners are looking for something else.
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