Understanding an argument given in my textbook about a second order linear ODE with Dirichlet boundary conditions Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Homogeneous second-order differential equation with constant WronskianDetermining properties of solution of a second order ODEClarification about second order Linear ODE with constant coefficientsAbout zeroes of solutions of second order differential equationsProve that $y_1(x)=sin(x^2)$ can't be a solution for a linear homogeneous second order differential equation.Why do the Existence and Uniqueness Theorem and The Principle of Superposition not contradict each other?How to find how many solutions of a second order ODE satisfy given initial conditions?ODE Boundary Conditions Give Only Trivial SolutionWhat is a linearly independent solution for a differential equation?How Can the Following Couple of ODEs be solves at the Given Boundary Conditions?
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Understanding an argument given in my textbook about a second order linear ODE with Dirichlet boundary conditions
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Homogeneous second-order differential equation with constant WronskianDetermining properties of solution of a second order ODEClarification about second order Linear ODE with constant coefficientsAbout zeroes of solutions of second order differential equationsProve that $y_1(x)=sin(x^2)$ can't be a solution for a linear homogeneous second order differential equation.Why do the Existence and Uniqueness Theorem and The Principle of Superposition not contradict each other?How to find how many solutions of a second order ODE satisfy given initial conditions?ODE Boundary Conditions Give Only Trivial SolutionWhat is a linearly independent solution for a differential equation?How Can the Following Couple of ODEs be solves at the Given Boundary Conditions?
$begingroup$
Suppose $fracd^2ydx^2+omega y=f(x)$ is given for some well-defined function $f$, with $y(a)=alpha$ and $y(b)=beta$. To find the solutions, my textbook splits this into two cases, namely $omega=0$ and $omegane0$. For $omega=0$, the arguments are simple and clear. However, for $omegane0$, the following argument is given (note that this is a translation, with personal additions between [ ] brackets):
If $omegane0$, then the solutions to the DE with $f(x)=0$ which are equal to zero in $x=a$ are the multiples of $sin(omega(x-a))$. We notice [existence and] uniqueness of solutions [to the inhomogeneous case] if and only if $sin(omega(b-a))ne0$, in other words if $omega(b-a)$ is not an integer multiple of $pi$.
So as I see it, they try to prove the conditions for which the solutions to the homogeneous case are zero everywhere, since we know that only in this case the inhomogeneous solutions satisfy existence and uniqueness. When I try to piece together this argument in my own words, I can get to the conclusion in the first sentence. For $y_1(x)=sin(omega x)$ and $y_2(x)=cos(omega x)$ are linearly independent solutions to the homogeneous case, therefore the general solution is of the form $y(x)=c_1sin(omega x)+c_2cos(omega x)$, namely a linear combination of these two linearly independent solutions. Next, we notice that since the period of the two harmonic functions $y_1$ and $y_2$ is the same, $y$ is also harmonic and we can write $y(x)=psin(omega(x-q))$ for some $p,qinmathbb R$. For this equation to satisfy $y(a)=0$, we must have $q=a+n$ for some $ninmathbb Z$, so we can write $y(x)=psin(omega(x-a))$ without loss of generality. Indeed, by varying $c_1$ and $c_2$ we get all multiples of $sin(omega(x-a))$. However, the next sentence doesn't make sense to me at all. I expect them to try to get this result to be $0$ as well, since you want a function that is zero everywhere and satisfies $y(a)=y(b)=0$ to have existence and uniqueness of solutions to the inhomogeneous case.
I was wondering if my reasoning so far is correct and also if someone could explain that second sentence to me.
ordinary-differential-equations trigonometry initial-value-problems
asked Apr 1 at 13:57
TyronTyron
531213
$endgroup$
add a comment |
$begingroup$
Suppose $fracd^2ydx^2+omega y=f(x)$ is given for some well-defined function $f$, with $y(a)=alpha$ and $y(b)=beta$. To find the solutions, my textbook splits this into two cases, namely $omega=0$ and $omegane0$. For $omega=0$, the arguments are simple and clear. However, for $omegane0$, the following argument is given (note that this is a translation, with personal additions between [ ] brackets):
If $omegane0$, then the solutions to the DE with $f(x)=0$ which are equal to zero in $x=a$ are the multiples of $sin(omega(x-a))$. We notice [existence and] uniqueness of solutions [to the inhomogeneous case] if and only if $sin(omega(b-a))ne0$, in other words if $omega(b-a)$ is not an integer multiple of $pi$.
So as I see it, they try to prove the conditions for which the solutions to the homogeneous case are zero everywhere, since we know that only in this case the inhomogeneous solutions satisfy existence and uniqueness. When I try to piece together this argument in my own words, I can get to the conclusion in the first sentence. For $y_1(x)=sin(omega x)$ and $y_2(x)=cos(omega x)$ are linearly independent solutions to the homogeneous case, therefore the general solution is of the form $y(x)=c_1sin(omega x)+c_2cos(omega x)$, namely a linear combination of these two linearly independent solutions. Next, we notice that since the period of the two harmonic functions $y_1$ and $y_2$ is the same, $y$ is also harmonic and we can write $y(x)=psin(omega(x-q))$ for some $p,qinmathbb R$. For this equation to satisfy $y(a)=0$, we must have $q=a+n$ for some $ninmathbb Z$, so we can write $y(x)=psin(omega(x-a))$ without loss of generality. Indeed, by varying $c_1$ and $c_2$ we get all multiples of $sin(omega(x-a))$. However, the next sentence doesn't make sense to me at all. I expect them to try to get this result to be $0$ as well, since you want a function that is zero everywhere and satisfies $y(a)=y(b)=0$ to have existence and uniqueness of solutions to the inhomogeneous case.
I was wondering if my reasoning so far is correct and also if someone could explain that second sentence to me.
ordinary-differential-equations trigonometry initial-value-problems
asked Apr 1 at 13:57
TyronTyron
531213
$endgroup$
add a comment |
$begingroup$
Suppose $fracd^2ydx^2+omega y=f(x)$ is given for some well-defined function $f$, with $y(a)=alpha$ and $y(b)=beta$. To find the solutions, my textbook splits this into two cases, namely $omega=0$ and $omegane0$. For $omega=0$, the arguments are simple and clear. However, for $omegane0$, the following argument is given (note that this is a translation, with personal additions between [ ] brackets):
If $omegane0$, then the solutions to the DE with $f(x)=0$ which are equal to zero in $x=a$ are the multiples of $sin(omega(x-a))$. We notice [existence and] uniqueness of solutions [to the inhomogeneous case] if and only if $sin(omega(b-a))ne0$, in other words if $omega(b-a)$ is not an integer multiple of $pi$.
So as I see it, they try to prove the conditions for which the solutions to the homogeneous case are zero everywhere, since we know that only in this case the inhomogeneous solutions satisfy existence and uniqueness. When I try to piece together this argument in my own words, I can get to the conclusion in the first sentence. For $y_1(x)=sin(omega x)$ and $y_2(x)=cos(omega x)$ are linearly independent solutions to the homogeneous case, therefore the general solution is of the form $y(x)=c_1sin(omega x)+c_2cos(omega x)$, namely a linear combination of these two linearly independent solutions. Next, we notice that since the period of the two harmonic functions $y_1$ and $y_2$ is the same, $y$ is also harmonic and we can write $y(x)=psin(omega(x-q))$ for some $p,qinmathbb R$. For this equation to satisfy $y(a)=0$, we must have $q=a+n$ for some $ninmathbb Z$, so we can write $y(x)=psin(omega(x-a))$ without loss of generality. Indeed, by varying $c_1$ and $c_2$ we get all multiples of $sin(omega(x-a))$. However, the next sentence doesn't make sense to me at all. I expect them to try to get this result to be $0$ as well, since you want a function that is zero everywhere and satisfies $y(a)=y(b)=0$ to have existence and uniqueness of solutions to the inhomogeneous case.
I was wondering if my reasoning so far is correct and also if someone could explain that second sentence to me.
ordinary-differential-equations trigonometry initial-value-problems
asked Apr 1 at 13:57
TyronTyron
531213
$endgroup$
Suppose $fracd^2ydx^2+omega y=f(x)$ is given for some well-defined function $f$, with $y(a)=alpha$ and $y(b)=beta$. To find the solutions, my textbook splits this into two cases, namely $omega=0$ and $omegane0$. For $omega=0$, the arguments are simple and clear. However, for $omegane0$, the following argument is given (note that this is a translation, with personal additions between [ ] brackets):
If $omegane0$, then the solutions to the DE with $f(x)=0$ which are equal to zero in $x=a$ are the multiples of $sin(omega(x-a))$. We notice [existence and] uniqueness of solutions [to the inhomogeneous case] if and only if $sin(omega(b-a))ne0$, in other words if $omega(b-a)$ is not an integer multiple of $pi$.
So as I see it, they try to prove the conditions for which the solutions to the homogeneous case are zero everywhere, since we know that only in this case the inhomogeneous solutions satisfy existence and uniqueness. When I try to piece together this argument in my own words, I can get to the conclusion in the first sentence. For $y_1(x)=sin(omega x)$ and $y_2(x)=cos(omega x)$ are linearly independent solutions to the homogeneous case, therefore the general solution is of the form $y(x)=c_1sin(omega x)+c_2cos(omega x)$, namely a linear combination of these two linearly independent solutions. Next, we notice that since the period of the two harmonic functions $y_1$ and $y_2$ is the same, $y$ is also harmonic and we can write $y(x)=psin(omega(x-q))$ for some $p,qinmathbb R$. For this equation to satisfy $y(a)=0$, we must have $q=a+n$ for some $ninmathbb Z$, so we can write $y(x)=psin(omega(x-a))$ without loss of generality. Indeed, by varying $c_1$ and $c_2$ we get all multiples of $sin(omega(x-a))$. However, the next sentence doesn't make sense to me at all. I expect them to try to get this result to be $0$ as well, since you want a function that is zero everywhere and satisfies $y(a)=y(b)=0$ to have existence and uniqueness of solutions to the inhomogeneous case.
I was wondering if my reasoning so far is correct and also if someone could explain that second sentence to me.
ordinary-differential-equations trigonometry initial-value-problems
ordinary-differential-equations trigonometry initial-value-problems
asked Apr 1 at 13:57
TyronTyron
531213
asked Apr 1 at 13:57
TyronTyron
531213
asked Apr 1 at 13:57
TyronTyron
531213
asked Apr 1 at 13:57
TyronTyron
531213
asked Apr 1 at 13:57
TyronTyron
531213
531213
add a comment |
add a comment |
1 Answer
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$begingroup$
The second condition is $y(b)=0$ which implies
$$ sin(omega(b-a))=0 $$
This occurs when $omega(b-a) = npi$ where $n$ is some integer.
answered Apr 1 at 14:08
DylanDylan
14.6k31127
$endgroup$
$begingroup$
Well... that's what I thought as well. But they require the expression to be NOT equal to zero, which is why I get stuck. I thought maybe it has something to do with the fact that we have -multiples- of $sin(omega(x-a))$ and somehow $sin(omega(b-a))ne0$ forces this expression to be zero everywhere. But I can't get anything from that.
$endgroup$
– Tyron
Apr 1 at 14:22
$begingroup$
We don't want the solution to be identically zero. If $omega(b-a)ne npi$, then the only way for $y(b)=0$ to hold is $y(x)equiv 0$ everywhere.
$endgroup$
– Dylan
Apr 1 at 14:45
$begingroup$
Okay yeah, I think I get it now. Thanks a bunch!
$endgroup$
– Tyron
Apr 1 at 14:59
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The second condition is $y(b)=0$ which implies
$$ sin(omega(b-a))=0 $$
This occurs when $omega(b-a) = npi$ where $n$ is some integer.
answered Apr 1 at 14:08
DylanDylan
14.6k31127
$endgroup$
$begingroup$
Well... that's what I thought as well. But they require the expression to be NOT equal to zero, which is why I get stuck. I thought maybe it has something to do with the fact that we have -multiples- of $sin(omega(x-a))$ and somehow $sin(omega(b-a))ne0$ forces this expression to be zero everywhere. But I can't get anything from that.
$endgroup$
– Tyron
Apr 1 at 14:22
$begingroup$
We don't want the solution to be identically zero. If $omega(b-a)ne npi$, then the only way for $y(b)=0$ to hold is $y(x)equiv 0$ everywhere.
$endgroup$
– Dylan
Apr 1 at 14:45
$begingroup$
Okay yeah, I think I get it now. Thanks a bunch!
$endgroup$
– Tyron
Apr 1 at 14:59
add a comment |
$begingroup$
The second condition is $y(b)=0$ which implies
$$ sin(omega(b-a))=0 $$
This occurs when $omega(b-a) = npi$ where $n$ is some integer.
answered Apr 1 at 14:08
DylanDylan
14.6k31127
$endgroup$
$begingroup$
Well... that's what I thought as well. But they require the expression to be NOT equal to zero, which is why I get stuck. I thought maybe it has something to do with the fact that we have -multiples- of $sin(omega(x-a))$ and somehow $sin(omega(b-a))ne0$ forces this expression to be zero everywhere. But I can't get anything from that.
$endgroup$
– Tyron
Apr 1 at 14:22
$begingroup$
We don't want the solution to be identically zero. If $omega(b-a)ne npi$, then the only way for $y(b)=0$ to hold is $y(x)equiv 0$ everywhere.
$endgroup$
– Dylan
Apr 1 at 14:45
$begingroup$
Okay yeah, I think I get it now. Thanks a bunch!
$endgroup$
– Tyron
Apr 1 at 14:59
add a comment |
$begingroup$
The second condition is $y(b)=0$ which implies
$$ sin(omega(b-a))=0 $$
This occurs when $omega(b-a) = npi$ where $n$ is some integer.
answered Apr 1 at 14:08
DylanDylan
14.6k31127
$endgroup$
The second condition is $y(b)=0$ which implies
$$ sin(omega(b-a))=0 $$
This occurs when $omega(b-a) = npi$ where $n$ is some integer.
answered Apr 1 at 14:08
DylanDylan
14.6k31127
answered Apr 1 at 14:08
DylanDylan
14.6k31127
answered Apr 1 at 14:08
DylanDylan
14.6k31127
answered Apr 1 at 14:08
DylanDylan
14.6k31127
14.6k31127
$begingroup$
Well... that's what I thought as well. But they require the expression to be NOT equal to zero, which is why I get stuck. I thought maybe it has something to do with the fact that we have -multiples- of $sin(omega(x-a))$ and somehow $sin(omega(b-a))ne0$ forces this expression to be zero everywhere. But I can't get anything from that.
$endgroup$
– Tyron
Apr 1 at 14:22
$begingroup$
We don't want the solution to be identically zero. If $omega(b-a)ne npi$, then the only way for $y(b)=0$ to hold is $y(x)equiv 0$ everywhere.
$endgroup$
– Dylan
Apr 1 at 14:45
$begingroup$
Okay yeah, I think I get it now. Thanks a bunch!
$endgroup$
– Tyron
Apr 1 at 14:59
add a comment |
$begingroup$
Well... that's what I thought as well. But they require the expression to be NOT equal to zero, which is why I get stuck. I thought maybe it has something to do with the fact that we have -multiples- of $sin(omega(x-a))$ and somehow $sin(omega(b-a))ne0$ forces this expression to be zero everywhere. But I can't get anything from that.
$endgroup$
– Tyron
Apr 1 at 14:22
$begingroup$
We don't want the solution to be identically zero. If $omega(b-a)ne npi$, then the only way for $y(b)=0$ to hold is $y(x)equiv 0$ everywhere.
$endgroup$
– Dylan
Apr 1 at 14:45
$begingroup$
Okay yeah, I think I get it now. Thanks a bunch!
$endgroup$
– Tyron
Apr 1 at 14:59
$begingroup$
Well... that's what I thought as well. But they require the expression to be NOT equal to zero, which is why I get stuck. I thought maybe it has something to do with the fact that we have -multiples- of $sin(omega(x-a))$ and somehow $sin(omega(b-a))ne0$ forces this expression to be zero everywhere. But I can't get anything from that.
$endgroup$
– Tyron
Apr 1 at 14:22
$begingroup$
Well... that's what I thought as well. But they require the expression to be NOT equal to zero, which is why I get stuck. I thought maybe it has something to do with the fact that we have -multiples- of $sin(omega(x-a))$ and somehow $sin(omega(b-a))ne0$ forces this expression to be zero everywhere. But I can't get anything from that.
$endgroup$
– Tyron
Apr 1 at 14:22
$begingroup$
We don't want the solution to be identically zero. If $omega(b-a)ne npi$, then the only way for $y(b)=0$ to hold is $y(x)equiv 0$ everywhere.
$endgroup$
– Dylan
Apr 1 at 14:45
$begingroup$
We don't want the solution to be identically zero. If $omega(b-a)ne npi$, then the only way for $y(b)=0$ to hold is $y(x)equiv 0$ everywhere.
$endgroup$
– Dylan
Apr 1 at 14:45
$begingroup$
Okay yeah, I think I get it now. Thanks a bunch!
$endgroup$
– Tyron
Apr 1 at 14:59
$begingroup$
Okay yeah, I think I get it now. Thanks a bunch!
$endgroup$
– Tyron
Apr 1 at 14:59
add a comment |
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