Mixing
Order 2 differential equations and linearly dependent solutions
$begingroup$
I have difficulties to resolve the following exercice. I try with Wronski, but i haven't obtain good result.
Let $y_1$ and $y_2$ two linearly independent solutions of the equation
$$
(P(x)y')'+q(x)y=0
$$
on the interval $[a,b]$ with $P(x) >0$.
Prouve that $y_1$ and $y_2$ do not vanish together in the same time.
Prouve that if $y_1$ and $y_2$ are not equal to zero, and if they are linearly dependent, then $y_1$ and $y_2$ vanish together in the same time.
Thank you in advance for the help.
real-analysis functional-analysis ordinary-differential-equations pde
$endgroup$
add a comment |
$begingroup$
I have difficulties to resolve the following exercice. I try with Wronski, but i haven't obtain good result.
Let $y_1$ and $y_2$ two linearly independent solutions of the equation
$$
(P(x)y')'+q(x)y=0
$$
on the interval $[a,b]$ with $P(x) >0$.
Prouve that $y_1$ and $y_2$ do not vanish together in the same time.
Prouve that if $y_1$ and $y_2$ are not equal to zero, and if they are linearly dependent, then $y_1$ and $y_2$ vanish together in the same time.
Thank you in advance for the help.
real-analysis functional-analysis ordinary-differential-equations pde
$endgroup$
add a comment |
$begingroup$
I have difficulties to resolve the following exercice. I try with Wronski, but i haven't obtain good result.
Let $y_1$ and $y_2$ two linearly independent solutions of the equation
$$
(P(x)y')'+q(x)y=0
$$
on the interval $[a,b]$ with $P(x) >0$.
Prouve that $y_1$ and $y_2$ do not vanish together in the same time.
Prouve that if $y_1$ and $y_2$ are not equal to zero, and if they are linearly dependent, then $y_1$ and $y_2$ vanish together in the same time.
Thank you in advance for the help.
real-analysis functional-analysis ordinary-differential-equations pde
$endgroup$
I have difficulties to resolve the following exercice. I try with Wronski, but i haven't obtain good result.
Let $y_1$ and $y_2$ two linearly independent solutions of the equation
$$
(P(x)y')'+q(x)y=0
$$
on the interval $[a,b]$ with $P(x) >0$.
Prouve that $y_1$ and $y_2$ do not vanish together in the same time.
Prouve that if $y_1$ and $y_2$ are not equal to zero, and if they are linearly dependent, then $y_1$ and $y_2$ vanish together in the same time.
Thank you in advance for the help.
real-analysis functional-analysis ordinary-differential-equations pde
real-analysis functional-analysis ordinary-differential-equations pde
asked May 19 '17 at 10:27
user415040
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
1.) If $y_1(x^*)=0=y_2(x^*)$, then
$$
y(x)=αy_1(x)+βy_2(x) ~~ text{ with } ~ α=y_2'(x^*),~ β=-y_1'(x^*),
$$
is a non-trivial linear combination with $y(x^*)=y'(x^*)=0$. As $α=0$ would imply $y_2(x^*)=0=y_2'(x^*)$, both coefficients are non-zero.
2.) is trivial as $y_2=cy_1$.
answered May 19 '17 at 11:35
LutzLLutzL
58k42054
$endgroup$
$begingroup$
Thank you for the answer. Please, i don't understant 1. Can you more explain please. Why you choose $y_1'(t*)$ and $y_2'(t*)$ whith $y_1(t*)=0=y_2(t*)$. And another question, why we suppose that $P(x)>0$? And if $P(x) <0$?
$endgroup$
– user415040
May 19 '17 at 12:04
$begingroup$
1.) to get $y'(t^*)=0$. Check again what linear dependence means, use the general definition in the context of more than 2 vectors. -- You have to exclude that $P(x)=0$ at some point in the interval, i.e., the sign stays constant, positive is the default.
$endgroup$
– LutzL
May 19 '17 at 12:26
$begingroup$
$y_1$ et $y_2$ are linearly dependent means that $$forall alpha, beta in mathbb{R}: alpha y_1(x)+ beta y_2(x)=0 => alpha=beta=0$$ you put $alpha=y'_1(x*)$ and $beta=y'_2(x*)$ with $y_1(x*)=0=y_2(x*)$. I don't understand
$endgroup$
– user415040
May 19 '17 at 12:33
$begingroup$
For a homogeneous second order linear DE, what does it mean for a solution $y$ if $y(x^*)=0$ and $y'(x^*)=0$? And if you look closer you will find that $α=y_2(x^*)$ and $β=-y_1'(x^*)$. Follow-up: Why would $α=0$ or $β=0$ be impossible?
$endgroup$
– LutzL
May 19 '17 at 13:07
$begingroup$
it means that $y$ is equal to zero. How you remark that $alpha= y_2(x*)$ and $beta=-y'_1(x*)$. You use Wronski? I am very lost
$endgroup$
– user415040
May 19 '17 at 13:10
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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$begingroup$
1.) If $y_1(x^*)=0=y_2(x^*)$, then
$$
y(x)=αy_1(x)+βy_2(x) ~~ text{ with } ~ α=y_2'(x^*),~ β=-y_1'(x^*),
$$
is a non-trivial linear combination with $y(x^*)=y'(x^*)=0$. As $α=0$ would imply $y_2(x^*)=0=y_2'(x^*)$, both coefficients are non-zero.
2.) is trivial as $y_2=cy_1$.
answered May 19 '17 at 11:35
LutzLLutzL
58k42054
$endgroup$
$begingroup$
Thank you for the answer. Please, i don't understant 1. Can you more explain please. Why you choose $y_1'(t*)$ and $y_2'(t*)$ whith $y_1(t*)=0=y_2(t*)$. And another question, why we suppose that $P(x)>0$? And if $P(x) <0$?
$endgroup$
– user415040
May 19 '17 at 12:04
$begingroup$
1.) to get $y'(t^*)=0$. Check again what linear dependence means, use the general definition in the context of more than 2 vectors. -- You have to exclude that $P(x)=0$ at some point in the interval, i.e., the sign stays constant, positive is the default.
$endgroup$
– LutzL
May 19 '17 at 12:26
$begingroup$
$y_1$ et $y_2$ are linearly dependent means that $$forall alpha, beta in mathbb{R}: alpha y_1(x)+ beta y_2(x)=0 => alpha=beta=0$$ you put $alpha=y'_1(x*)$ and $beta=y'_2(x*)$ with $y_1(x*)=0=y_2(x*)$. I don't understand
$endgroup$
– user415040
May 19 '17 at 12:33
$begingroup$
For a homogeneous second order linear DE, what does it mean for a solution $y$ if $y(x^*)=0$ and $y'(x^*)=0$? And if you look closer you will find that $α=y_2(x^*)$ and $β=-y_1'(x^*)$. Follow-up: Why would $α=0$ or $β=0$ be impossible?
$endgroup$
– LutzL
May 19 '17 at 13:07
$begingroup$
it means that $y$ is equal to zero. How you remark that $alpha= y_2(x*)$ and $beta=-y'_1(x*)$. You use Wronski? I am very lost
$endgroup$
– user415040
May 19 '17 at 13:10
|
show 1 more comment
$begingroup$
1.) If $y_1(x^*)=0=y_2(x^*)$, then
$$
y(x)=αy_1(x)+βy_2(x) ~~ text{ with } ~ α=y_2'(x^*),~ β=-y_1'(x^*),
$$
is a non-trivial linear combination with $y(x^*)=y'(x^*)=0$. As $α=0$ would imply $y_2(x^*)=0=y_2'(x^*)$, both coefficients are non-zero.
2.) is trivial as $y_2=cy_1$.
answered May 19 '17 at 11:35
LutzLLutzL
58k42054
$endgroup$
$begingroup$
Thank you for the answer. Please, i don't understant 1. Can you more explain please. Why you choose $y_1'(t*)$ and $y_2'(t*)$ whith $y_1(t*)=0=y_2(t*)$. And another question, why we suppose that $P(x)>0$? And if $P(x) <0$?
$endgroup$
– user415040
May 19 '17 at 12:04
$begingroup$
1.) to get $y'(t^*)=0$. Check again what linear dependence means, use the general definition in the context of more than 2 vectors. -- You have to exclude that $P(x)=0$ at some point in the interval, i.e., the sign stays constant, positive is the default.
$endgroup$
– LutzL
May 19 '17 at 12:26
$begingroup$
$y_1$ et $y_2$ are linearly dependent means that $$forall alpha, beta in mathbb{R}: alpha y_1(x)+ beta y_2(x)=0 => alpha=beta=0$$ you put $alpha=y'_1(x*)$ and $beta=y'_2(x*)$ with $y_1(x*)=0=y_2(x*)$. I don't understand
$endgroup$
– user415040
May 19 '17 at 12:33
$begingroup$
For a homogeneous second order linear DE, what does it mean for a solution $y$ if $y(x^*)=0$ and $y'(x^*)=0$? And if you look closer you will find that $α=y_2(x^*)$ and $β=-y_1'(x^*)$. Follow-up: Why would $α=0$ or $β=0$ be impossible?
$endgroup$
– LutzL
May 19 '17 at 13:07
$begingroup$
it means that $y$ is equal to zero. How you remark that $alpha= y_2(x*)$ and $beta=-y'_1(x*)$. You use Wronski? I am very lost
$endgroup$
– user415040
May 19 '17 at 13:10
|
show 1 more comment
$begingroup$
1.) If $y_1(x^*)=0=y_2(x^*)$, then
$$
y(x)=αy_1(x)+βy_2(x) ~~ text{ with } ~ α=y_2'(x^*),~ β=-y_1'(x^*),
$$
is a non-trivial linear combination with $y(x^*)=y'(x^*)=0$. As $α=0$ would imply $y_2(x^*)=0=y_2'(x^*)$, both coefficients are non-zero.
2.) is trivial as $y_2=cy_1$.
answered May 19 '17 at 11:35
LutzLLutzL
58k42054
$endgroup$
1.) If $y_1(x^*)=0=y_2(x^*)$, then
$$
y(x)=αy_1(x)+βy_2(x) ~~ text{ with } ~ α=y_2'(x^*),~ β=-y_1'(x^*),
$$
is a non-trivial linear combination with $y(x^*)=y'(x^*)=0$. As $α=0$ would imply $y_2(x^*)=0=y_2'(x^*)$, both coefficients are non-zero.
2.) is trivial as $y_2=cy_1$.
answered May 19 '17 at 11:35
LutzLLutzL
58k42054
edited Jan 10 at 23:27
answered May 19 '17 at 11:35
LutzLLutzL
58k42054
answered May 19 '17 at 11:35
LutzLLutzL
58k42054
answered May 19 '17 at 11:35
LutzLLutzL
58k42054
58k42054
$begingroup$
Thank you for the answer. Please, i don't understant 1. Can you more explain please. Why you choose $y_1'(t*)$ and $y_2'(t*)$ whith $y_1(t*)=0=y_2(t*)$. And another question, why we suppose that $P(x)>0$? And if $P(x) <0$?
$endgroup$
– user415040
May 19 '17 at 12:04
$begingroup$
1.) to get $y'(t^*)=0$. Check again what linear dependence means, use the general definition in the context of more than 2 vectors. -- You have to exclude that $P(x)=0$ at some point in the interval, i.e., the sign stays constant, positive is the default.
$endgroup$
– LutzL
May 19 '17 at 12:26
$begingroup$
$y_1$ et $y_2$ are linearly dependent means that $$forall alpha, beta in mathbb{R}: alpha y_1(x)+ beta y_2(x)=0 => alpha=beta=0$$ you put $alpha=y'_1(x*)$ and $beta=y'_2(x*)$ with $y_1(x*)=0=y_2(x*)$. I don't understand
$endgroup$
– user415040
May 19 '17 at 12:33
$begingroup$
For a homogeneous second order linear DE, what does it mean for a solution $y$ if $y(x^*)=0$ and $y'(x^*)=0$? And if you look closer you will find that $α=y_2(x^*)$ and $β=-y_1'(x^*)$. Follow-up: Why would $α=0$ or $β=0$ be impossible?
$endgroup$
– LutzL
May 19 '17 at 13:07
$begingroup$
it means that $y$ is equal to zero. How you remark that $alpha= y_2(x*)$ and $beta=-y'_1(x*)$. You use Wronski? I am very lost
$endgroup$
– user415040
May 19 '17 at 13:10
|
show 1 more comment
$begingroup$
Thank you for the answer. Please, i don't understant 1. Can you more explain please. Why you choose $y_1'(t*)$ and $y_2'(t*)$ whith $y_1(t*)=0=y_2(t*)$. And another question, why we suppose that $P(x)>0$? And if $P(x) <0$?
$endgroup$
– user415040
May 19 '17 at 12:04
$begingroup$
1.) to get $y'(t^*)=0$. Check again what linear dependence means, use the general definition in the context of more than 2 vectors. -- You have to exclude that $P(x)=0$ at some point in the interval, i.e., the sign stays constant, positive is the default.
$endgroup$
– LutzL
May 19 '17 at 12:26
$begingroup$
$y_1$ et $y_2$ are linearly dependent means that $$forall alpha, beta in mathbb{R}: alpha y_1(x)+ beta y_2(x)=0 => alpha=beta=0$$ you put $alpha=y'_1(x*)$ and $beta=y'_2(x*)$ with $y_1(x*)=0=y_2(x*)$. I don't understand
$endgroup$
– user415040
May 19 '17 at 12:33
$begingroup$
For a homogeneous second order linear DE, what does it mean for a solution $y$ if $y(x^*)=0$ and $y'(x^*)=0$? And if you look closer you will find that $α=y_2(x^*)$ and $β=-y_1'(x^*)$. Follow-up: Why would $α=0$ or $β=0$ be impossible?
$endgroup$
– LutzL
May 19 '17 at 13:07
$begingroup$
it means that $y$ is equal to zero. How you remark that $alpha= y_2(x*)$ and $beta=-y'_1(x*)$. You use Wronski? I am very lost
$endgroup$
– user415040
May 19 '17 at 13:10
$begingroup$
Thank you for the answer. Please, i don't understant 1. Can you more explain please. Why you choose $y_1'(t*)$ and $y_2'(t*)$ whith $y_1(t*)=0=y_2(t*)$. And another question, why we suppose that $P(x)>0$? And if $P(x) <0$?
$endgroup$
– user415040
May 19 '17 at 12:04
$begingroup$
Thank you for the answer. Please, i don't understant 1. Can you more explain please. Why you choose $y_1'(t*)$ and $y_2'(t*)$ whith $y_1(t*)=0=y_2(t*)$. And another question, why we suppose that $P(x)>0$? And if $P(x) <0$?
$endgroup$
– user415040
May 19 '17 at 12:04
$begingroup$
1.) to get $y'(t^*)=0$. Check again what linear dependence means, use the general definition in the context of more than 2 vectors. -- You have to exclude that $P(x)=0$ at some point in the interval, i.e., the sign stays constant, positive is the default.
$endgroup$
– LutzL
May 19 '17 at 12:26
$begingroup$
1.) to get $y'(t^*)=0$. Check again what linear dependence means, use the general definition in the context of more than 2 vectors. -- You have to exclude that $P(x)=0$ at some point in the interval, i.e., the sign stays constant, positive is the default.
$endgroup$
– LutzL
May 19 '17 at 12:26
$begingroup$
$y_1$ et $y_2$ are linearly dependent means that $$forall alpha, beta in mathbb{R}: alpha y_1(x)+ beta y_2(x)=0 => alpha=beta=0$$ you put $alpha=y'_1(x*)$ and $beta=y'_2(x*)$ with $y_1(x*)=0=y_2(x*)$. I don't understand
$endgroup$
– user415040
May 19 '17 at 12:33
$begingroup$
$y_1$ et $y_2$ are linearly dependent means that $$forall alpha, beta in mathbb{R}: alpha y_1(x)+ beta y_2(x)=0 => alpha=beta=0$$ you put $alpha=y'_1(x*)$ and $beta=y'_2(x*)$ with $y_1(x*)=0=y_2(x*)$. I don't understand
$endgroup$
– user415040
May 19 '17 at 12:33
$begingroup$
For a homogeneous second order linear DE, what does it mean for a solution $y$ if $y(x^*)=0$ and $y'(x^*)=0$? And if you look closer you will find that $α=y_2(x^*)$ and $β=-y_1'(x^*)$. Follow-up: Why would $α=0$ or $β=0$ be impossible?
$endgroup$
– LutzL
May 19 '17 at 13:07
$begingroup$
For a homogeneous second order linear DE, what does it mean for a solution $y$ if $y(x^*)=0$ and $y'(x^*)=0$? And if you look closer you will find that $α=y_2(x^*)$ and $β=-y_1'(x^*)$. Follow-up: Why would $α=0$ or $β=0$ be impossible?
$endgroup$
– LutzL
May 19 '17 at 13:07
$begingroup$
it means that $y$ is equal to zero. How you remark that $alpha= y_2(x*)$ and $beta=-y'_1(x*)$. You use Wronski? I am very lost
$endgroup$
– user415040
May 19 '17 at 13:10
$begingroup$
it means that $y$ is equal to zero. How you remark that $alpha= y_2(x*)$ and $beta=-y'_1(x*)$. You use Wronski? I am very lost
$endgroup$
– user415040
May 19 '17 at 13:10
|
show 1 more comment
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