Nonlinear first order ODE with quadratic in the derivative
$begingroup$
This equation shouldn't be so hard, and yet I'm stymied.
$$
left( frac{dw}{dz} right )^2 + alpha frac{dw}{dz} + w beta = 0
$$
with $w(0) = w_0>0$ $w(L) = 0$ for some known L and $alpha(z)>0$ and $beta(z)>0$ known. $alpha$ and $beta$ actually begin life as functions of known $w$, so I dont worry about existence of solution, what I want to prove is that I can invert back from $alpha$ and $beta$ to give unique $w$. $d alpha / dz = beta + C$ for some constant $C$ if it helps.
I can solve for some special cases but I'm interested in a general expression, or at least to prove uniqueness. Interestingly, those special cases are not unique if I just put constraint at $z=0$ but become so with the $z=L$ constraint.
ordinary-differential-equations quadratics nonlinear-system
asked Jun 20 '14 at 3:35
TunnellerTunneller
111
$endgroup$
add a comment |
$begingroup$
This equation shouldn't be so hard, and yet I'm stymied.
$$
left( frac{dw}{dz} right )^2 + alpha frac{dw}{dz} + w beta = 0
$$
with $w(0) = w_0>0$ $w(L) = 0$ for some known L and $alpha(z)>0$ and $beta(z)>0$ known. $alpha$ and $beta$ actually begin life as functions of known $w$, so I dont worry about existence of solution, what I want to prove is that I can invert back from $alpha$ and $beta$ to give unique $w$. $d alpha / dz = beta + C$ for some constant $C$ if it helps.
I can solve for some special cases but I'm interested in a general expression, or at least to prove uniqueness. Interestingly, those special cases are not unique if I just put constraint at $z=0$ but become so with the $z=L$ constraint.
ordinary-differential-equations quadratics nonlinear-system
asked Jun 20 '14 at 3:35
TunnellerTunneller
111
$endgroup$
$begingroup$
You may check a book "odinary ODE" by Ince.
$endgroup$
– mike
Jun 20 '14 at 4:16
$begingroup$
Hey, Ince is online... I would not have expected that. Ploughing through now.... archive.org/details/ordinarydifferen029666mbp
$endgroup$
– Tunneller
Jun 20 '14 at 12:41
$begingroup$
Do you really mean that "α and β actually begin life as known functions of w"?
$endgroup$
– mike
Jun 20 '14 at 13:15
$begingroup$
Yes..... Actually that is correct. In the real world, w is the velocity of fluid and $alpha = theta - Cz$ and $beta = theta^prime$ are measurable parameters from the flow. So then in the real world, we can consider measuring $theta$ to get $alpha$ and $beta$ and that as to give a way of computing $w$.
$endgroup$
– Tunneller
Jun 20 '14 at 14:46
$begingroup$
Ince seems to be proposing a Taylor expansion of those square roots and writing the answer as infinite sum. Not the most satisfying solution...
$endgroup$
– Tunneller
Jun 20 '14 at 14:48
add a comment |
$begingroup$
This equation shouldn't be so hard, and yet I'm stymied.
$$
left( frac{dw}{dz} right )^2 + alpha frac{dw}{dz} + w beta = 0
$$
with $w(0) = w_0>0$ $w(L) = 0$ for some known L and $alpha(z)>0$ and $beta(z)>0$ known. $alpha$ and $beta$ actually begin life as functions of known $w$, so I dont worry about existence of solution, what I want to prove is that I can invert back from $alpha$ and $beta$ to give unique $w$. $d alpha / dz = beta + C$ for some constant $C$ if it helps.
I can solve for some special cases but I'm interested in a general expression, or at least to prove uniqueness. Interestingly, those special cases are not unique if I just put constraint at $z=0$ but become so with the $z=L$ constraint.
ordinary-differential-equations quadratics nonlinear-system
asked Jun 20 '14 at 3:35
TunnellerTunneller
111
$endgroup$
This equation shouldn't be so hard, and yet I'm stymied.
$$
left( frac{dw}{dz} right )^2 + alpha frac{dw}{dz} + w beta = 0
$$
with $w(0) = w_0>0$ $w(L) = 0$ for some known L and $alpha(z)>0$ and $beta(z)>0$ known. $alpha$ and $beta$ actually begin life as functions of known $w$, so I dont worry about existence of solution, what I want to prove is that I can invert back from $alpha$ and $beta$ to give unique $w$. $d alpha / dz = beta + C$ for some constant $C$ if it helps.
I can solve for some special cases but I'm interested in a general expression, or at least to prove uniqueness. Interestingly, those special cases are not unique if I just put constraint at $z=0$ but become so with the $z=L$ constraint.
ordinary-differential-equations quadratics nonlinear-system
ordinary-differential-equations quadratics nonlinear-system
asked Jun 20 '14 at 3:35
TunnellerTunneller
111
asked Jun 20 '14 at 3:35
TunnellerTunneller
111
asked Jun 20 '14 at 3:35
TunnellerTunneller
111
asked Jun 20 '14 at 3:35
TunnellerTunneller
111
asked Jun 20 '14 at 3:35
TunnellerTunneller
111
111
$begingroup$
You may check a book "odinary ODE" by Ince.
$endgroup$
– mike
Jun 20 '14 at 4:16
$begingroup$
Hey, Ince is online... I would not have expected that. Ploughing through now.... archive.org/details/ordinarydifferen029666mbp
$endgroup$
– Tunneller
Jun 20 '14 at 12:41
$begingroup$
Do you really mean that "α and β actually begin life as known functions of w"?
$endgroup$
– mike
Jun 20 '14 at 13:15
$begingroup$
Yes..... Actually that is correct. In the real world, w is the velocity of fluid and $alpha = theta - Cz$ and $beta = theta^prime$ are measurable parameters from the flow. So then in the real world, we can consider measuring $theta$ to get $alpha$ and $beta$ and that as to give a way of computing $w$.
$endgroup$
– Tunneller
Jun 20 '14 at 14:46
$begingroup$
Ince seems to be proposing a Taylor expansion of those square roots and writing the answer as infinite sum. Not the most satisfying solution...
$endgroup$
– Tunneller
Jun 20 '14 at 14:48
add a comment |
$begingroup$
You may check a book "odinary ODE" by Ince.
$endgroup$
– mike
Jun 20 '14 at 4:16
$begingroup$
Hey, Ince is online... I would not have expected that. Ploughing through now.... archive.org/details/ordinarydifferen029666mbp
$endgroup$
– Tunneller
Jun 20 '14 at 12:41
$begingroup$
Do you really mean that "α and β actually begin life as known functions of w"?
$endgroup$
– mike
Jun 20 '14 at 13:15
$begingroup$
Yes..... Actually that is correct. In the real world, w is the velocity of fluid and $alpha = theta - Cz$ and $beta = theta^prime$ are measurable parameters from the flow. So then in the real world, we can consider measuring $theta$ to get $alpha$ and $beta$ and that as to give a way of computing $w$.
$endgroup$
– Tunneller
Jun 20 '14 at 14:46
$begingroup$
Ince seems to be proposing a Taylor expansion of those square roots and writing the answer as infinite sum. Not the most satisfying solution...
$endgroup$
– Tunneller
Jun 20 '14 at 14:48
$begingroup$
You may check a book "odinary ODE" by Ince.
$endgroup$
– mike
Jun 20 '14 at 4:16
$begingroup$
You may check a book "odinary ODE" by Ince.
$endgroup$
– mike
Jun 20 '14 at 4:16
$begingroup$
Hey, Ince is online... I would not have expected that. Ploughing through now.... archive.org/details/ordinarydifferen029666mbp
$endgroup$
– Tunneller
Jun 20 '14 at 12:41
$begingroup$
Hey, Ince is online... I would not have expected that. Ploughing through now.... archive.org/details/ordinarydifferen029666mbp
$endgroup$
– Tunneller
Jun 20 '14 at 12:41
$begingroup$
Do you really mean that "α and β actually begin life as known functions of w"?
$endgroup$
– mike
Jun 20 '14 at 13:15
$begingroup$
Do you really mean that "α and β actually begin life as known functions of w"?
$endgroup$
– mike
Jun 20 '14 at 13:15
$begingroup$
Yes..... Actually that is correct. In the real world, w is the velocity of fluid and $alpha = theta - Cz$ and $beta = theta^prime$ are measurable parameters from the flow. So then in the real world, we can consider measuring $theta$ to get $alpha$ and $beta$ and that as to give a way of computing $w$.
$endgroup$
– Tunneller
Jun 20 '14 at 14:46
$begingroup$
Yes..... Actually that is correct. In the real world, w is the velocity of fluid and $alpha = theta - Cz$ and $beta = theta^prime$ are measurable parameters from the flow. So then in the real world, we can consider measuring $theta$ to get $alpha$ and $beta$ and that as to give a way of computing $w$.
$endgroup$
– Tunneller
Jun 20 '14 at 14:46
$begingroup$
Ince seems to be proposing a Taylor expansion of those square roots and writing the answer as infinite sum. Not the most satisfying solution...
$endgroup$
– Tunneller
Jun 20 '14 at 14:48
$begingroup$
Ince seems to be proposing a Taylor expansion of those square roots and writing the answer as infinite sum. Not the most satisfying solution...
$endgroup$
– Tunneller
Jun 20 '14 at 14:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Start with
$$left( frac{dw}{dz} right )^2 + alpha(z) frac{dw}{dz} + w beta(z) = 0$$
We obtain:
$$frac{dw}{dz}=-frac{1}{2}alpha(z)+frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(1)$$
$$frac{dw}{dz}=-frac{1}{2}alpha(z)-frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(2)$$
You might be integrate (1) and (2) to find $w(z)$.
EDIT:
Not quite yet.
$w(z)$ appeared in the square root.
answered Jun 20 '14 at 3:51
mikemike
4,36421019
$endgroup$
$begingroup$
Yes, exactly. For the application at hand, I can prove that the term in the integrand is positive, but not strictly positive. It will touch zero. This means (I think) that I dont get even a simple proof that each of these ODE's has a unique solution as the square root is not Lipschitz at the zero.
$endgroup$
– Tunneller
Jun 20 '14 at 9:01
$begingroup$
Touching zero is fine. If $alpha(z)=A(w(z))$ and $beta(z)=B(w(z))$, then it will help.
$endgroup$
– mike
Jun 20 '14 at 10:58
$begingroup$
No obvious inversion to get back from w(z). In the formula, $theta$ will be measured data and $C$ a known constant. $$left( frac{dw}{dz} right )^2 + (theta(z) - Cz) frac{dw}{dz} + w frac{d theta}{dz} =0 $$
$endgroup$
– Tunneller
Jun 20 '14 at 12:44
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Start with
$$left( frac{dw}{dz} right )^2 + alpha(z) frac{dw}{dz} + w beta(z) = 0$$
We obtain:
$$frac{dw}{dz}=-frac{1}{2}alpha(z)+frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(1)$$
$$frac{dw}{dz}=-frac{1}{2}alpha(z)-frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(2)$$
You might be integrate (1) and (2) to find $w(z)$.
EDIT:
Not quite yet.
$w(z)$ appeared in the square root.
answered Jun 20 '14 at 3:51
mikemike
4,36421019
$endgroup$
$begingroup$
Yes, exactly. For the application at hand, I can prove that the term in the integrand is positive, but not strictly positive. It will touch zero. This means (I think) that I dont get even a simple proof that each of these ODE's has a unique solution as the square root is not Lipschitz at the zero.
$endgroup$
– Tunneller
Jun 20 '14 at 9:01
$begingroup$
Touching zero is fine. If $alpha(z)=A(w(z))$ and $beta(z)=B(w(z))$, then it will help.
$endgroup$
– mike
Jun 20 '14 at 10:58
$begingroup$
No obvious inversion to get back from w(z). In the formula, $theta$ will be measured data and $C$ a known constant. $$left( frac{dw}{dz} right )^2 + (theta(z) - Cz) frac{dw}{dz} + w frac{d theta}{dz} =0 $$
$endgroup$
– Tunneller
Jun 20 '14 at 12:44
add a comment |
$begingroup$
Start with
$$left( frac{dw}{dz} right )^2 + alpha(z) frac{dw}{dz} + w beta(z) = 0$$
We obtain:
$$frac{dw}{dz}=-frac{1}{2}alpha(z)+frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(1)$$
$$frac{dw}{dz}=-frac{1}{2}alpha(z)-frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(2)$$
You might be integrate (1) and (2) to find $w(z)$.
EDIT:
Not quite yet.
$w(z)$ appeared in the square root.
answered Jun 20 '14 at 3:51
mikemike
4,36421019
$endgroup$
$begingroup$
Yes, exactly. For the application at hand, I can prove that the term in the integrand is positive, but not strictly positive. It will touch zero. This means (I think) that I dont get even a simple proof that each of these ODE's has a unique solution as the square root is not Lipschitz at the zero.
$endgroup$
– Tunneller
Jun 20 '14 at 9:01
$begingroup$
Touching zero is fine. If $alpha(z)=A(w(z))$ and $beta(z)=B(w(z))$, then it will help.
$endgroup$
– mike
Jun 20 '14 at 10:58
$begingroup$
No obvious inversion to get back from w(z). In the formula, $theta$ will be measured data and $C$ a known constant. $$left( frac{dw}{dz} right )^2 + (theta(z) - Cz) frac{dw}{dz} + w frac{d theta}{dz} =0 $$
$endgroup$
– Tunneller
Jun 20 '14 at 12:44
add a comment |
$begingroup$
Start with
$$left( frac{dw}{dz} right )^2 + alpha(z) frac{dw}{dz} + w beta(z) = 0$$
We obtain:
$$frac{dw}{dz}=-frac{1}{2}alpha(z)+frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(1)$$
$$frac{dw}{dz}=-frac{1}{2}alpha(z)-frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(2)$$
You might be integrate (1) and (2) to find $w(z)$.
EDIT:
Not quite yet.
$w(z)$ appeared in the square root.
answered Jun 20 '14 at 3:51
mikemike
4,36421019
$endgroup$
Start with
$$left( frac{dw}{dz} right )^2 + alpha(z) frac{dw}{dz} + w beta(z) = 0$$
We obtain:
$$frac{dw}{dz}=-frac{1}{2}alpha(z)+frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(1)$$
$$frac{dw}{dz}=-frac{1}{2}alpha(z)-frac{1}{2}left(alpha^2(z)-4w beta(z)right)^{1/2}......(2)$$
You might be integrate (1) and (2) to find $w(z)$.
EDIT:
Not quite yet.
$w(z)$ appeared in the square root.
answered Jun 20 '14 at 3:51
mikemike
4,36421019
answered Jun 20 '14 at 3:51
mikemike
4,36421019
answered Jun 20 '14 at 3:51
mikemike
4,36421019
answered Jun 20 '14 at 3:51
mikemike
4,36421019
4,36421019
$begingroup$
Yes, exactly. For the application at hand, I can prove that the term in the integrand is positive, but not strictly positive. It will touch zero. This means (I think) that I dont get even a simple proof that each of these ODE's has a unique solution as the square root is not Lipschitz at the zero.
$endgroup$
– Tunneller
Jun 20 '14 at 9:01
$begingroup$
Touching zero is fine. If $alpha(z)=A(w(z))$ and $beta(z)=B(w(z))$, then it will help.
$endgroup$
– mike
Jun 20 '14 at 10:58
$begingroup$
No obvious inversion to get back from w(z). In the formula, $theta$ will be measured data and $C$ a known constant. $$left( frac{dw}{dz} right )^2 + (theta(z) - Cz) frac{dw}{dz} + w frac{d theta}{dz} =0 $$
$endgroup$
– Tunneller
Jun 20 '14 at 12:44
add a comment |
$begingroup$
Yes, exactly. For the application at hand, I can prove that the term in the integrand is positive, but not strictly positive. It will touch zero. This means (I think) that I dont get even a simple proof that each of these ODE's has a unique solution as the square root is not Lipschitz at the zero.
$endgroup$
– Tunneller
Jun 20 '14 at 9:01
$begingroup$
Touching zero is fine. If $alpha(z)=A(w(z))$ and $beta(z)=B(w(z))$, then it will help.
$endgroup$
– mike
Jun 20 '14 at 10:58
$begingroup$
No obvious inversion to get back from w(z). In the formula, $theta$ will be measured data and $C$ a known constant. $$left( frac{dw}{dz} right )^2 + (theta(z) - Cz) frac{dw}{dz} + w frac{d theta}{dz} =0 $$
$endgroup$
– Tunneller
Jun 20 '14 at 12:44
$begingroup$
Yes, exactly. For the application at hand, I can prove that the term in the integrand is positive, but not strictly positive. It will touch zero. This means (I think) that I dont get even a simple proof that each of these ODE's has a unique solution as the square root is not Lipschitz at the zero.
$endgroup$
– Tunneller
Jun 20 '14 at 9:01
$begingroup$
Yes, exactly. For the application at hand, I can prove that the term in the integrand is positive, but not strictly positive. It will touch zero. This means (I think) that I dont get even a simple proof that each of these ODE's has a unique solution as the square root is not Lipschitz at the zero.
$endgroup$
– Tunneller
Jun 20 '14 at 9:01
$begingroup$
Touching zero is fine. If $alpha(z)=A(w(z))$ and $beta(z)=B(w(z))$, then it will help.
$endgroup$
– mike
Jun 20 '14 at 10:58
$begingroup$
Touching zero is fine. If $alpha(z)=A(w(z))$ and $beta(z)=B(w(z))$, then it will help.
$endgroup$
– mike
Jun 20 '14 at 10:58
$begingroup$
No obvious inversion to get back from w(z). In the formula, $theta$ will be measured data and $C$ a known constant. $$left( frac{dw}{dz} right )^2 + (theta(z) - Cz) frac{dw}{dz} + w frac{d theta}{dz} =0 $$
$endgroup$
– Tunneller
Jun 20 '14 at 12:44
$begingroup$
No obvious inversion to get back from w(z). In the formula, $theta$ will be measured data and $C$ a known constant. $$left( frac{dw}{dz} right )^2 + (theta(z) - Cz) frac{dw}{dz} + w frac{d theta}{dz} =0 $$
$endgroup$
– Tunneller
Jun 20 '14 at 12:44
add a comment |
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$begingroup$
You may check a book "odinary ODE" by Ince.
$endgroup$
– mike
Jun 20 '14 at 4:16
$begingroup$
Hey, Ince is online... I would not have expected that. Ploughing through now.... archive.org/details/ordinarydifferen029666mbp
$endgroup$
– Tunneller
Jun 20 '14 at 12:41
$begingroup$
Do you really mean that "α and β actually begin life as known functions of w"?
$endgroup$
– mike
Jun 20 '14 at 13:15
$begingroup$
Yes..... Actually that is correct. In the real world, w is the velocity of fluid and $alpha = theta - Cz$ and $beta = theta^prime$ are measurable parameters from the flow. So then in the real world, we can consider measuring $theta$ to get $alpha$ and $beta$ and that as to give a way of computing $w$.
$endgroup$
– Tunneller
Jun 20 '14 at 14:46
$begingroup$
Ince seems to be proposing a Taylor expansion of those square roots and writing the answer as infinite sum. Not the most satisfying solution...
$endgroup$
– Tunneller
Jun 20 '14 at 14:48