$x^2 frac{partial^2{z} }{{partial{x^2}}}+2xy frac{partial^2{z} }{{partial{x}partial{y}}}+y^2...
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I want to solve this PDE. I seems like a classic problem but I don't know what to do when in second order PDE coefficients are polynomial themselves.
$$x^2 frac{partial^2{z} }{{partial{x^2}}}+2xy frac{partial^2{z} }{{partial{x}partial{y}}}+y^2 frac{partial^2{z} }{{partial{y^2}}}=0$$
calculus ordinary-differential-equations pde
asked Nov 28 '18 at 17:07
LeilaLeila
3,48553056
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closed as off-topic by amWhy, Leucippus, KReiser, Saad, Vidyanshu Mishra Nov 30 '18 at 8:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, KReiser, Saad, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 5 more comments
$begingroup$
I want to solve this PDE. I seems like a classic problem but I don't know what to do when in second order PDE coefficients are polynomial themselves.
$$x^2 frac{partial^2{z} }{{partial{x^2}}}+2xy frac{partial^2{z} }{{partial{x}partial{y}}}+y^2 frac{partial^2{z} }{{partial{y^2}}}=0$$
calculus ordinary-differential-equations pde
asked Nov 28 '18 at 17:07
LeilaLeila
3,48553056
$endgroup$
closed as off-topic by amWhy, Leucippus, KReiser, Saad, Vidyanshu Mishra Nov 30 '18 at 8:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, KReiser, Saad, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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It will be helpful to know that it's a parabolic PDE...
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– rafa11111
Nov 28 '18 at 17:26
1
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By guessing solutions of the form $x^{alpha} y^{beta}$ (by analogy with Euler ordinary differential equations) I found that either $x^{alpha} y^{-alpha}$ or $x^{alpha+1} y^{-alpha}$ are solutions. Based on that, it seems reasonable to guess that $f(x/y)$ and $x f(x/y)$ are both solutions - and by linearity, that would imply that $f(x/y) + x g(x/y)$ is a solution.
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– Daniel Schepler
Nov 28 '18 at 20:54
1
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$$x^{2} z_{xx} + 2xy z_{xy} + y^{2} z_{yy} = (x partial_{x} + y partial_{y})^{2} z$$ assuming $partial_{x} partial_{y} = partial_{y} partial_{x}$.
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– Mattos
Nov 28 '18 at 21:47
1
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Hmm, and that in turn suggests that it might be useful to translate the PDE into polar coordinates. If (purely hypothetically) the result were something like $r^2 cos theta sin theta frac{partial^2 z}{partial r^2} = 0$, then in the interiors of the quadrants that would imply $z = f(theta) + r g(theta)$ - and it shouldn't be too hard to massage this into $z = F(y/x) + x G(y/x)$.
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– Daniel Schepler
Nov 28 '18 at 21:50
1
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@RafaBudría Well I take it back then, what I originally wrote is incorrect. You are right. Thanks for catching that.
$endgroup$
– Mattos
Dec 3 '18 at 5:56
|
show 5 more comments
$begingroup$
I want to solve this PDE. I seems like a classic problem but I don't know what to do when in second order PDE coefficients are polynomial themselves.
$$x^2 frac{partial^2{z} }{{partial{x^2}}}+2xy frac{partial^2{z} }{{partial{x}partial{y}}}+y^2 frac{partial^2{z} }{{partial{y^2}}}=0$$
calculus ordinary-differential-equations pde
asked Nov 28 '18 at 17:07
LeilaLeila
3,48553056
$endgroup$
I want to solve this PDE. I seems like a classic problem but I don't know what to do when in second order PDE coefficients are polynomial themselves.
$$x^2 frac{partial^2{z} }{{partial{x^2}}}+2xy frac{partial^2{z} }{{partial{x}partial{y}}}+y^2 frac{partial^2{z} }{{partial{y^2}}}=0$$
calculus ordinary-differential-equations pde
calculus ordinary-differential-equations pde
asked Nov 28 '18 at 17:07
LeilaLeila
3,48553056
asked Nov 28 '18 at 17:07
LeilaLeila
3,48553056
asked Nov 28 '18 at 17:07
LeilaLeila
3,48553056
asked Nov 28 '18 at 17:07
LeilaLeila
3,48553056
asked Nov 28 '18 at 17:07
LeilaLeila
3,48553056
3,48553056
closed as off-topic by amWhy, Leucippus, KReiser, Saad, Vidyanshu Mishra Nov 30 '18 at 8:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, KReiser, Saad, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Leucippus, KReiser, Saad, Vidyanshu Mishra Nov 30 '18 at 8:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, KReiser, Saad, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
It will be helpful to know that it's a parabolic PDE...
$endgroup$
– rafa11111
Nov 28 '18 at 17:26
1
$begingroup$
By guessing solutions of the form $x^{alpha} y^{beta}$ (by analogy with Euler ordinary differential equations) I found that either $x^{alpha} y^{-alpha}$ or $x^{alpha+1} y^{-alpha}$ are solutions. Based on that, it seems reasonable to guess that $f(x/y)$ and $x f(x/y)$ are both solutions - and by linearity, that would imply that $f(x/y) + x g(x/y)$ is a solution.
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– Daniel Schepler
Nov 28 '18 at 20:54
1
$begingroup$
$$x^{2} z_{xx} + 2xy z_{xy} + y^{2} z_{yy} = (x partial_{x} + y partial_{y})^{2} z$$ assuming $partial_{x} partial_{y} = partial_{y} partial_{x}$.
$endgroup$
– Mattos
Nov 28 '18 at 21:47
1
$begingroup$
Hmm, and that in turn suggests that it might be useful to translate the PDE into polar coordinates. If (purely hypothetically) the result were something like $r^2 cos theta sin theta frac{partial^2 z}{partial r^2} = 0$, then in the interiors of the quadrants that would imply $z = f(theta) + r g(theta)$ - and it shouldn't be too hard to massage this into $z = F(y/x) + x G(y/x)$.
$endgroup$
– Daniel Schepler
Nov 28 '18 at 21:50
1
$begingroup$
@RafaBudría Well I take it back then, what I originally wrote is incorrect. You are right. Thanks for catching that.
$endgroup$
– Mattos
Dec 3 '18 at 5:56
|
show 5 more comments
2
$begingroup$
It will be helpful to know that it's a parabolic PDE...
$endgroup$
– rafa11111
Nov 28 '18 at 17:26
1
$begingroup$
By guessing solutions of the form $x^{alpha} y^{beta}$ (by analogy with Euler ordinary differential equations) I found that either $x^{alpha} y^{-alpha}$ or $x^{alpha+1} y^{-alpha}$ are solutions. Based on that, it seems reasonable to guess that $f(x/y)$ and $x f(x/y)$ are both solutions - and by linearity, that would imply that $f(x/y) + x g(x/y)$ is a solution.
$endgroup$
– Daniel Schepler
Nov 28 '18 at 20:54
1
$begingroup$
$$x^{2} z_{xx} + 2xy z_{xy} + y^{2} z_{yy} = (x partial_{x} + y partial_{y})^{2} z$$ assuming $partial_{x} partial_{y} = partial_{y} partial_{x}$.
$endgroup$
– Mattos
Nov 28 '18 at 21:47
1
$begingroup$
Hmm, and that in turn suggests that it might be useful to translate the PDE into polar coordinates. If (purely hypothetically) the result were something like $r^2 cos theta sin theta frac{partial^2 z}{partial r^2} = 0$, then in the interiors of the quadrants that would imply $z = f(theta) + r g(theta)$ - and it shouldn't be too hard to massage this into $z = F(y/x) + x G(y/x)$.
$endgroup$
– Daniel Schepler
Nov 28 '18 at 21:50
1
$begingroup$
@RafaBudría Well I take it back then, what I originally wrote is incorrect. You are right. Thanks for catching that.
$endgroup$
– Mattos
Dec 3 '18 at 5:56
2
2
$begingroup$
It will be helpful to know that it's a parabolic PDE...
$endgroup$
– rafa11111
Nov 28 '18 at 17:26
$begingroup$
It will be helpful to know that it's a parabolic PDE...
$endgroup$
– rafa11111
Nov 28 '18 at 17:26
1
1
$begingroup$
By guessing solutions of the form $x^{alpha} y^{beta}$ (by analogy with Euler ordinary differential equations) I found that either $x^{alpha} y^{-alpha}$ or $x^{alpha+1} y^{-alpha}$ are solutions. Based on that, it seems reasonable to guess that $f(x/y)$ and $x f(x/y)$ are both solutions - and by linearity, that would imply that $f(x/y) + x g(x/y)$ is a solution.
$endgroup$
– Daniel Schepler
Nov 28 '18 at 20:54
$begingroup$
By guessing solutions of the form $x^{alpha} y^{beta}$ (by analogy with Euler ordinary differential equations) I found that either $x^{alpha} y^{-alpha}$ or $x^{alpha+1} y^{-alpha}$ are solutions. Based on that, it seems reasonable to guess that $f(x/y)$ and $x f(x/y)$ are both solutions - and by linearity, that would imply that $f(x/y) + x g(x/y)$ is a solution.
$endgroup$
– Daniel Schepler
Nov 28 '18 at 20:54
1
1
$begingroup$
$$x^{2} z_{xx} + 2xy z_{xy} + y^{2} z_{yy} = (x partial_{x} + y partial_{y})^{2} z$$ assuming $partial_{x} partial_{y} = partial_{y} partial_{x}$.
$endgroup$
– Mattos
Nov 28 '18 at 21:47
$begingroup$
$$x^{2} z_{xx} + 2xy z_{xy} + y^{2} z_{yy} = (x partial_{x} + y partial_{y})^{2} z$$ assuming $partial_{x} partial_{y} = partial_{y} partial_{x}$.
$endgroup$
– Mattos
Nov 28 '18 at 21:47
1
1
$begingroup$
Hmm, and that in turn suggests that it might be useful to translate the PDE into polar coordinates. If (purely hypothetically) the result were something like $r^2 cos theta sin theta frac{partial^2 z}{partial r^2} = 0$, then in the interiors of the quadrants that would imply $z = f(theta) + r g(theta)$ - and it shouldn't be too hard to massage this into $z = F(y/x) + x G(y/x)$.
$endgroup$
– Daniel Schepler
Nov 28 '18 at 21:50
$begingroup$
Hmm, and that in turn suggests that it might be useful to translate the PDE into polar coordinates. If (purely hypothetically) the result were something like $r^2 cos theta sin theta frac{partial^2 z}{partial r^2} = 0$, then in the interiors of the quadrants that would imply $z = f(theta) + r g(theta)$ - and it shouldn't be too hard to massage this into $z = F(y/x) + x G(y/x)$.
$endgroup$
– Daniel Schepler
Nov 28 '18 at 21:50
1
1
$begingroup$
@RafaBudría Well I take it back then, what I originally wrote is incorrect. You are right. Thanks for catching that.
$endgroup$
– Mattos
Dec 3 '18 at 5:56
$begingroup$
@RafaBudría Well I take it back then, what I originally wrote is incorrect. You are right. Thanks for catching that.
$endgroup$
– Mattos
Dec 3 '18 at 5:56
|
show 5 more comments
3 Answers
3
active
oldest
votes
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Let us translate the given PDE into polar coordinates. Using $x = r cos theta$ and $y = r sin theta$, we get:
$$ frac{partial z}{partial r} = frac{partial z}{partial x} cdot frac{partial x}{partial r} + frac{partial z}{partial y} cdot frac{partial y}{partial r} = cos theta frac{partial z}{partial x} + sin theta frac{partial z}{partial y}.$$
Iterating this,
$$ frac{partial^2 z}{partial r^2} = cos theta left( cos theta frac{partial^2 z} {partial x^2} + sin theta frac{partial^2 z} {partial y partial x}right) + sin theta left( costheta frac{partial^2 z}{partial x partial y} + sintheta frac{partial^2 z}{partial y^2}right) = \
cos^2 theta frac{partial^2 z}{partial x^2} + 2 cos theta sin theta frac{partial^2 z}{partial x partial y} + sin^2 theta frac{partial^2 z}{partial y^2} = \
frac{1}{r^2} left( x^2 frac{partial^2 z}{partial x^2} + 2xy frac{partial^2 z}{partial x partial y} + y^2 frac{partial^2 z}{partial y^2} right).$$
Therefore, the given PDE is equivalent to $r^2 frac{partial^2 z}{partial r^2} = 0$ which has general solution of $z = f(theta) + r g(theta)$ (on any domain which is an annular sector not including the origin).
To relate this to the general solution from rafa11111's answer; or, to get a nice expression in terms of $x,y$: suppose our domain is contained either within the right half-plane or within the left half-plane. Then $theta = tan^{-1}(y/x)$, so we have:
$$f(theta) + r g(theta) = f(theta) + (r cos theta) (sec theta g(theta)) = f(tan^{-1}(y/x)) + x left(pmsqrt{1 + (y/x)^2} g(tan^{-1}(y/x)) right)$$
(with the sign depending on the specific half-plane).
Therefore, if $F(m) = f(tan^{-1}(m))$ and $G(m) = pm sqrt{1+m^2} g(tan^{-1}(m))$ (with the chosen branch of $tan^{-1}$ also depending on the half-plane) then $z = F(y/x) + x G(y/x)$.
As for how I came up with this approach: I first observed that the given PDE looks like an analogue of an Euler equation from ordinary differential equations. This suggested that some solution of the form $z = x^alpha y^beta$ might work. Plugging in this trial solution, we get:
$$alpha (alpha-1) x^alpha y^beta + 2 alpha beta x^alpha y^beta + beta (beta-1) x^alpha y^beta = [(alpha + beta)^2 - (alpha + beta)] x^alpha y^beta = 0.$$
Therefore, this does indeed give a solution whenever $alpha + beta in { 0, 1 }$. Thus, we find particular solutions of the form $z = x^{-beta} y^beta = (y/x)^beta$ and $z = x^{1-beta} y^beta = x (y/x)^beta$.
Once we see this, it is reasonable to conjecture that in general, $f(y/x)$ and $x g(y/x)$ might be solutions - which is then straightforward to check. Furthermore, by linearity of the equation, $z = f(y/x) + x g(y/x)$ is also a solution. At this point, we heuristically have "enough degrees of freedom" that we think it could possibly be the general solution. Also, from the appearance of $y/x$ in the solution, this suggests that the approach of converting the PDE into polar coordinates could be fruitful. (Another approach would be to observe that the hypothesized general solution is linear on any ray from the origin, which would suggest examining the behavior of a solution on such curves $x = t cos alpha$, $y = t sin alpha$ for fixed $alpha$. The calculation for this approach would end up looking much the same.)
answered Nov 28 '18 at 22:59
Daniel ScheplerDaniel Schepler
8,5141618
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WOW! Thanks for the answer and the insights you gave.
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– Leila
Dec 1 '18 at 8:39
add a comment |
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Let's introduce the variables $xi(x,y)$ and $eta(x,y)$. From sucessive application of chain rule one has
$$
z_{xx} = z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx},
$$
$$
z_{yy} = z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy},
$$
$$
z_{xy} = z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}.
$$
Substituting in the original PDE and colecting terms,
$$
x^2 left[z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx} right] + 2xy left[z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}right] + y^2 left[ z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy}right] = 0
$$
$$
left[ x^2 xi_x^2 + 2xy xi_xxi_y +y^2xi_y^2right] z_{xi xi} + 2left[x^2 xi_xeta_x+ xy(xi_xeta_y+xi_yeta_x)+y^2 xi_yeta_yright] z_{xi eta} + left[ x^2 eta_x^2+2xyeta_xeta_y+y^2 eta_y^2right] z_{eta eta} = phi,
$$
in which $phi$ represents the remaining terms. Setting the terms between the first pair of brackets to $0$ leads to
$$
left(frac{xxi_x}{yxi_y}right)^2 + 2 left(frac{xxi_x}{yxi_y}right) + 1 = 0,
$$
which is a quadratic equation on $xxi_x/yxi_y$, then
$$
frac{xi_x}{xi_y} = - frac{y}{x}.
$$
Since $dxi=xi_x dx + xi_y dy$, for $xi(x,y)=mathrm{const}$ we have $dxi=0$, leading to
$$
frac{dy}{dx} = - frac{xi_x}{xi_y} ,
$$
then
$$
frac{dy}{dx}=frac{y}{x},
$$
or $y=cx$. Therefore, $xi=mathrm{const}$ corresponds to $xi=y/x$.
Setting the terms between the second pair of brackets to $0$,
$$
x^2 frac{xi_x}{xi_y}eta_x+ xyleft(frac{xi_x}{xi_y}eta_y+eta_xright)+y^2 eta_y=0.
$$
Using what we found for $xi_x/xi_y$ we see that the expression vanishes identicaly to $0$, therefore the choice of $eta_y$ is arbitrary. Defining $eta=x$ leads to
$$
z_{eta eta} = 0,
$$
which is the canonical form of the parabolic equation. Integrating the equation,
$$
z(xi,eta) = f(xi) eta + g(xi),
$$
or
$$
z(x,y) = x fleft(frac{y}{x}right) + gleft(frac{y}{x}right).
$$
That is the same solution presented by Daniel Schepler in the comments.
This answer was based on this text about canonical forms.
answered Nov 28 '18 at 22:00
rafa11111rafa11111
1,1251417
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Do you have any additional conditions? I tried and ansatz of the form $z = log(Ax+By)$ and the left hand side yielded $-1$ so, since the equation is linear, any difference of logarithms of that form would solve the equation. Linear polynomials in $x$ and $y$ also make the left hand side vanish, so the following is a possible solution:
$$
z(x,y) = log(Ax+By) - log(Cx+Dy) + Ex +Fy + G
,,
$$
Where $A,dots,G$ are constants. Not sure if there are more general solutions, but I hope this helps.
answered Nov 28 '18 at 18:55
mlerma54mlerma54
1,177148
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Given the nature of the problem, I'm almost sure that a characteristic method-like solution is expected.
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– rafa11111
Nov 28 '18 at 19:09
1
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And this can be rewritten as $z = [log(A + B (y/x)) - log(C + D(y/x)) + G] + x [E + F (y/x)]$ which is a specialization of the general solution from rafa11111's answer.
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– Daniel Schepler
Nov 28 '18 at 22:10
1
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Good job! I was about to generalize my "ansatz"-based particular solution with a more general combination of the logarithms $$z(x,y) = sum_{i} lambda_i log(A_i x+B_i y) + Ex +Fy + G$$ where $sum_{i} lambda_i = 0$, but here is not need now.
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– mlerma54
Nov 28 '18 at 23:56
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us translate the given PDE into polar coordinates. Using $x = r cos theta$ and $y = r sin theta$, we get:
$$ frac{partial z}{partial r} = frac{partial z}{partial x} cdot frac{partial x}{partial r} + frac{partial z}{partial y} cdot frac{partial y}{partial r} = cos theta frac{partial z}{partial x} + sin theta frac{partial z}{partial y}.$$
Iterating this,
$$ frac{partial^2 z}{partial r^2} = cos theta left( cos theta frac{partial^2 z} {partial x^2} + sin theta frac{partial^2 z} {partial y partial x}right) + sin theta left( costheta frac{partial^2 z}{partial x partial y} + sintheta frac{partial^2 z}{partial y^2}right) = \
cos^2 theta frac{partial^2 z}{partial x^2} + 2 cos theta sin theta frac{partial^2 z}{partial x partial y} + sin^2 theta frac{partial^2 z}{partial y^2} = \
frac{1}{r^2} left( x^2 frac{partial^2 z}{partial x^2} + 2xy frac{partial^2 z}{partial x partial y} + y^2 frac{partial^2 z}{partial y^2} right).$$
Therefore, the given PDE is equivalent to $r^2 frac{partial^2 z}{partial r^2} = 0$ which has general solution of $z = f(theta) + r g(theta)$ (on any domain which is an annular sector not including the origin).
To relate this to the general solution from rafa11111's answer; or, to get a nice expression in terms of $x,y$: suppose our domain is contained either within the right half-plane or within the left half-plane. Then $theta = tan^{-1}(y/x)$, so we have:
$$f(theta) + r g(theta) = f(theta) + (r cos theta) (sec theta g(theta)) = f(tan^{-1}(y/x)) + x left(pmsqrt{1 + (y/x)^2} g(tan^{-1}(y/x)) right)$$
(with the sign depending on the specific half-plane).
Therefore, if $F(m) = f(tan^{-1}(m))$ and $G(m) = pm sqrt{1+m^2} g(tan^{-1}(m))$ (with the chosen branch of $tan^{-1}$ also depending on the half-plane) then $z = F(y/x) + x G(y/x)$.
As for how I came up with this approach: I first observed that the given PDE looks like an analogue of an Euler equation from ordinary differential equations. This suggested that some solution of the form $z = x^alpha y^beta$ might work. Plugging in this trial solution, we get:
$$alpha (alpha-1) x^alpha y^beta + 2 alpha beta x^alpha y^beta + beta (beta-1) x^alpha y^beta = [(alpha + beta)^2 - (alpha + beta)] x^alpha y^beta = 0.$$
Therefore, this does indeed give a solution whenever $alpha + beta in { 0, 1 }$. Thus, we find particular solutions of the form $z = x^{-beta} y^beta = (y/x)^beta$ and $z = x^{1-beta} y^beta = x (y/x)^beta$.
Once we see this, it is reasonable to conjecture that in general, $f(y/x)$ and $x g(y/x)$ might be solutions - which is then straightforward to check. Furthermore, by linearity of the equation, $z = f(y/x) + x g(y/x)$ is also a solution. At this point, we heuristically have "enough degrees of freedom" that we think it could possibly be the general solution. Also, from the appearance of $y/x$ in the solution, this suggests that the approach of converting the PDE into polar coordinates could be fruitful. (Another approach would be to observe that the hypothesized general solution is linear on any ray from the origin, which would suggest examining the behavior of a solution on such curves $x = t cos alpha$, $y = t sin alpha$ for fixed $alpha$. The calculation for this approach would end up looking much the same.)
answered Nov 28 '18 at 22:59
Daniel ScheplerDaniel Schepler
8,5141618
$endgroup$
$begingroup$
WOW! Thanks for the answer and the insights you gave.
$endgroup$
– Leila
Dec 1 '18 at 8:39
add a comment |
$begingroup$
Let us translate the given PDE into polar coordinates. Using $x = r cos theta$ and $y = r sin theta$, we get:
$$ frac{partial z}{partial r} = frac{partial z}{partial x} cdot frac{partial x}{partial r} + frac{partial z}{partial y} cdot frac{partial y}{partial r} = cos theta frac{partial z}{partial x} + sin theta frac{partial z}{partial y}.$$
Iterating this,
$$ frac{partial^2 z}{partial r^2} = cos theta left( cos theta frac{partial^2 z} {partial x^2} + sin theta frac{partial^2 z} {partial y partial x}right) + sin theta left( costheta frac{partial^2 z}{partial x partial y} + sintheta frac{partial^2 z}{partial y^2}right) = \
cos^2 theta frac{partial^2 z}{partial x^2} + 2 cos theta sin theta frac{partial^2 z}{partial x partial y} + sin^2 theta frac{partial^2 z}{partial y^2} = \
frac{1}{r^2} left( x^2 frac{partial^2 z}{partial x^2} + 2xy frac{partial^2 z}{partial x partial y} + y^2 frac{partial^2 z}{partial y^2} right).$$
Therefore, the given PDE is equivalent to $r^2 frac{partial^2 z}{partial r^2} = 0$ which has general solution of $z = f(theta) + r g(theta)$ (on any domain which is an annular sector not including the origin).
To relate this to the general solution from rafa11111's answer; or, to get a nice expression in terms of $x,y$: suppose our domain is contained either within the right half-plane or within the left half-plane. Then $theta = tan^{-1}(y/x)$, so we have:
$$f(theta) + r g(theta) = f(theta) + (r cos theta) (sec theta g(theta)) = f(tan^{-1}(y/x)) + x left(pmsqrt{1 + (y/x)^2} g(tan^{-1}(y/x)) right)$$
(with the sign depending on the specific half-plane).
Therefore, if $F(m) = f(tan^{-1}(m))$ and $G(m) = pm sqrt{1+m^2} g(tan^{-1}(m))$ (with the chosen branch of $tan^{-1}$ also depending on the half-plane) then $z = F(y/x) + x G(y/x)$.
As for how I came up with this approach: I first observed that the given PDE looks like an analogue of an Euler equation from ordinary differential equations. This suggested that some solution of the form $z = x^alpha y^beta$ might work. Plugging in this trial solution, we get:
$$alpha (alpha-1) x^alpha y^beta + 2 alpha beta x^alpha y^beta + beta (beta-1) x^alpha y^beta = [(alpha + beta)^2 - (alpha + beta)] x^alpha y^beta = 0.$$
Therefore, this does indeed give a solution whenever $alpha + beta in { 0, 1 }$. Thus, we find particular solutions of the form $z = x^{-beta} y^beta = (y/x)^beta$ and $z = x^{1-beta} y^beta = x (y/x)^beta$.
Once we see this, it is reasonable to conjecture that in general, $f(y/x)$ and $x g(y/x)$ might be solutions - which is then straightforward to check. Furthermore, by linearity of the equation, $z = f(y/x) + x g(y/x)$ is also a solution. At this point, we heuristically have "enough degrees of freedom" that we think it could possibly be the general solution. Also, from the appearance of $y/x$ in the solution, this suggests that the approach of converting the PDE into polar coordinates could be fruitful. (Another approach would be to observe that the hypothesized general solution is linear on any ray from the origin, which would suggest examining the behavior of a solution on such curves $x = t cos alpha$, $y = t sin alpha$ for fixed $alpha$. The calculation for this approach would end up looking much the same.)
answered Nov 28 '18 at 22:59
Daniel ScheplerDaniel Schepler
8,5141618
$endgroup$
$begingroup$
WOW! Thanks for the answer and the insights you gave.
$endgroup$
– Leila
Dec 1 '18 at 8:39
add a comment |
$begingroup$
Let us translate the given PDE into polar coordinates. Using $x = r cos theta$ and $y = r sin theta$, we get:
$$ frac{partial z}{partial r} = frac{partial z}{partial x} cdot frac{partial x}{partial r} + frac{partial z}{partial y} cdot frac{partial y}{partial r} = cos theta frac{partial z}{partial x} + sin theta frac{partial z}{partial y}.$$
Iterating this,
$$ frac{partial^2 z}{partial r^2} = cos theta left( cos theta frac{partial^2 z} {partial x^2} + sin theta frac{partial^2 z} {partial y partial x}right) + sin theta left( costheta frac{partial^2 z}{partial x partial y} + sintheta frac{partial^2 z}{partial y^2}right) = \
cos^2 theta frac{partial^2 z}{partial x^2} + 2 cos theta sin theta frac{partial^2 z}{partial x partial y} + sin^2 theta frac{partial^2 z}{partial y^2} = \
frac{1}{r^2} left( x^2 frac{partial^2 z}{partial x^2} + 2xy frac{partial^2 z}{partial x partial y} + y^2 frac{partial^2 z}{partial y^2} right).$$
Therefore, the given PDE is equivalent to $r^2 frac{partial^2 z}{partial r^2} = 0$ which has general solution of $z = f(theta) + r g(theta)$ (on any domain which is an annular sector not including the origin).
To relate this to the general solution from rafa11111's answer; or, to get a nice expression in terms of $x,y$: suppose our domain is contained either within the right half-plane or within the left half-plane. Then $theta = tan^{-1}(y/x)$, so we have:
$$f(theta) + r g(theta) = f(theta) + (r cos theta) (sec theta g(theta)) = f(tan^{-1}(y/x)) + x left(pmsqrt{1 + (y/x)^2} g(tan^{-1}(y/x)) right)$$
(with the sign depending on the specific half-plane).
Therefore, if $F(m) = f(tan^{-1}(m))$ and $G(m) = pm sqrt{1+m^2} g(tan^{-1}(m))$ (with the chosen branch of $tan^{-1}$ also depending on the half-plane) then $z = F(y/x) + x G(y/x)$.
As for how I came up with this approach: I first observed that the given PDE looks like an analogue of an Euler equation from ordinary differential equations. This suggested that some solution of the form $z = x^alpha y^beta$ might work. Plugging in this trial solution, we get:
$$alpha (alpha-1) x^alpha y^beta + 2 alpha beta x^alpha y^beta + beta (beta-1) x^alpha y^beta = [(alpha + beta)^2 - (alpha + beta)] x^alpha y^beta = 0.$$
Therefore, this does indeed give a solution whenever $alpha + beta in { 0, 1 }$. Thus, we find particular solutions of the form $z = x^{-beta} y^beta = (y/x)^beta$ and $z = x^{1-beta} y^beta = x (y/x)^beta$.
Once we see this, it is reasonable to conjecture that in general, $f(y/x)$ and $x g(y/x)$ might be solutions - which is then straightforward to check. Furthermore, by linearity of the equation, $z = f(y/x) + x g(y/x)$ is also a solution. At this point, we heuristically have "enough degrees of freedom" that we think it could possibly be the general solution. Also, from the appearance of $y/x$ in the solution, this suggests that the approach of converting the PDE into polar coordinates could be fruitful. (Another approach would be to observe that the hypothesized general solution is linear on any ray from the origin, which would suggest examining the behavior of a solution on such curves $x = t cos alpha$, $y = t sin alpha$ for fixed $alpha$. The calculation for this approach would end up looking much the same.)
answered Nov 28 '18 at 22:59
Daniel ScheplerDaniel Schepler
8,5141618
$endgroup$
Let us translate the given PDE into polar coordinates. Using $x = r cos theta$ and $y = r sin theta$, we get:
$$ frac{partial z}{partial r} = frac{partial z}{partial x} cdot frac{partial x}{partial r} + frac{partial z}{partial y} cdot frac{partial y}{partial r} = cos theta frac{partial z}{partial x} + sin theta frac{partial z}{partial y}.$$
Iterating this,
$$ frac{partial^2 z}{partial r^2} = cos theta left( cos theta frac{partial^2 z} {partial x^2} + sin theta frac{partial^2 z} {partial y partial x}right) + sin theta left( costheta frac{partial^2 z}{partial x partial y} + sintheta frac{partial^2 z}{partial y^2}right) = \
cos^2 theta frac{partial^2 z}{partial x^2} + 2 cos theta sin theta frac{partial^2 z}{partial x partial y} + sin^2 theta frac{partial^2 z}{partial y^2} = \
frac{1}{r^2} left( x^2 frac{partial^2 z}{partial x^2} + 2xy frac{partial^2 z}{partial x partial y} + y^2 frac{partial^2 z}{partial y^2} right).$$
Therefore, the given PDE is equivalent to $r^2 frac{partial^2 z}{partial r^2} = 0$ which has general solution of $z = f(theta) + r g(theta)$ (on any domain which is an annular sector not including the origin).
To relate this to the general solution from rafa11111's answer; or, to get a nice expression in terms of $x,y$: suppose our domain is contained either within the right half-plane or within the left half-plane. Then $theta = tan^{-1}(y/x)$, so we have:
$$f(theta) + r g(theta) = f(theta) + (r cos theta) (sec theta g(theta)) = f(tan^{-1}(y/x)) + x left(pmsqrt{1 + (y/x)^2} g(tan^{-1}(y/x)) right)$$
(with the sign depending on the specific half-plane).
Therefore, if $F(m) = f(tan^{-1}(m))$ and $G(m) = pm sqrt{1+m^2} g(tan^{-1}(m))$ (with the chosen branch of $tan^{-1}$ also depending on the half-plane) then $z = F(y/x) + x G(y/x)$.
As for how I came up with this approach: I first observed that the given PDE looks like an analogue of an Euler equation from ordinary differential equations. This suggested that some solution of the form $z = x^alpha y^beta$ might work. Plugging in this trial solution, we get:
$$alpha (alpha-1) x^alpha y^beta + 2 alpha beta x^alpha y^beta + beta (beta-1) x^alpha y^beta = [(alpha + beta)^2 - (alpha + beta)] x^alpha y^beta = 0.$$
Therefore, this does indeed give a solution whenever $alpha + beta in { 0, 1 }$. Thus, we find particular solutions of the form $z = x^{-beta} y^beta = (y/x)^beta$ and $z = x^{1-beta} y^beta = x (y/x)^beta$.
Once we see this, it is reasonable to conjecture that in general, $f(y/x)$ and $x g(y/x)$ might be solutions - which is then straightforward to check. Furthermore, by linearity of the equation, $z = f(y/x) + x g(y/x)$ is also a solution. At this point, we heuristically have "enough degrees of freedom" that we think it could possibly be the general solution. Also, from the appearance of $y/x$ in the solution, this suggests that the approach of converting the PDE into polar coordinates could be fruitful. (Another approach would be to observe that the hypothesized general solution is linear on any ray from the origin, which would suggest examining the behavior of a solution on such curves $x = t cos alpha$, $y = t sin alpha$ for fixed $alpha$. The calculation for this approach would end up looking much the same.)
answered Nov 28 '18 at 22:59
Daniel ScheplerDaniel Schepler
8,5141618
answered Nov 28 '18 at 22:59
Daniel ScheplerDaniel Schepler
8,5141618
answered Nov 28 '18 at 22:59
Daniel ScheplerDaniel Schepler
8,5141618
answered Nov 28 '18 at 22:59
Daniel ScheplerDaniel Schepler
8,5141618
8,5141618
$begingroup$
WOW! Thanks for the answer and the insights you gave.
$endgroup$
– Leila
Dec 1 '18 at 8:39
add a comment |
$begingroup$
WOW! Thanks for the answer and the insights you gave.
$endgroup$
– Leila
Dec 1 '18 at 8:39
$begingroup$
WOW! Thanks for the answer and the insights you gave.
$endgroup$
– Leila
Dec 1 '18 at 8:39
$begingroup$
WOW! Thanks for the answer and the insights you gave.
$endgroup$
– Leila
Dec 1 '18 at 8:39
add a comment |
$begingroup$
Let's introduce the variables $xi(x,y)$ and $eta(x,y)$. From sucessive application of chain rule one has
$$
z_{xx} = z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx},
$$
$$
z_{yy} = z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy},
$$
$$
z_{xy} = z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}.
$$
Substituting in the original PDE and colecting terms,
$$
x^2 left[z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx} right] + 2xy left[z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}right] + y^2 left[ z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy}right] = 0
$$
$$
left[ x^2 xi_x^2 + 2xy xi_xxi_y +y^2xi_y^2right] z_{xi xi} + 2left[x^2 xi_xeta_x+ xy(xi_xeta_y+xi_yeta_x)+y^2 xi_yeta_yright] z_{xi eta} + left[ x^2 eta_x^2+2xyeta_xeta_y+y^2 eta_y^2right] z_{eta eta} = phi,
$$
in which $phi$ represents the remaining terms. Setting the terms between the first pair of brackets to $0$ leads to
$$
left(frac{xxi_x}{yxi_y}right)^2 + 2 left(frac{xxi_x}{yxi_y}right) + 1 = 0,
$$
which is a quadratic equation on $xxi_x/yxi_y$, then
$$
frac{xi_x}{xi_y} = - frac{y}{x}.
$$
Since $dxi=xi_x dx + xi_y dy$, for $xi(x,y)=mathrm{const}$ we have $dxi=0$, leading to
$$
frac{dy}{dx} = - frac{xi_x}{xi_y} ,
$$
then
$$
frac{dy}{dx}=frac{y}{x},
$$
or $y=cx$. Therefore, $xi=mathrm{const}$ corresponds to $xi=y/x$.
Setting the terms between the second pair of brackets to $0$,
$$
x^2 frac{xi_x}{xi_y}eta_x+ xyleft(frac{xi_x}{xi_y}eta_y+eta_xright)+y^2 eta_y=0.
$$
Using what we found for $xi_x/xi_y$ we see that the expression vanishes identicaly to $0$, therefore the choice of $eta_y$ is arbitrary. Defining $eta=x$ leads to
$$
z_{eta eta} = 0,
$$
which is the canonical form of the parabolic equation. Integrating the equation,
$$
z(xi,eta) = f(xi) eta + g(xi),
$$
or
$$
z(x,y) = x fleft(frac{y}{x}right) + gleft(frac{y}{x}right).
$$
That is the same solution presented by Daniel Schepler in the comments.
This answer was based on this text about canonical forms.
answered Nov 28 '18 at 22:00
rafa11111rafa11111
1,1251417
$endgroup$
add a comment |
$begingroup$
Let's introduce the variables $xi(x,y)$ and $eta(x,y)$. From sucessive application of chain rule one has
$$
z_{xx} = z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx},
$$
$$
z_{yy} = z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy},
$$
$$
z_{xy} = z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}.
$$
Substituting in the original PDE and colecting terms,
$$
x^2 left[z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx} right] + 2xy left[z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}right] + y^2 left[ z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy}right] = 0
$$
$$
left[ x^2 xi_x^2 + 2xy xi_xxi_y +y^2xi_y^2right] z_{xi xi} + 2left[x^2 xi_xeta_x+ xy(xi_xeta_y+xi_yeta_x)+y^2 xi_yeta_yright] z_{xi eta} + left[ x^2 eta_x^2+2xyeta_xeta_y+y^2 eta_y^2right] z_{eta eta} = phi,
$$
in which $phi$ represents the remaining terms. Setting the terms between the first pair of brackets to $0$ leads to
$$
left(frac{xxi_x}{yxi_y}right)^2 + 2 left(frac{xxi_x}{yxi_y}right) + 1 = 0,
$$
which is a quadratic equation on $xxi_x/yxi_y$, then
$$
frac{xi_x}{xi_y} = - frac{y}{x}.
$$
Since $dxi=xi_x dx + xi_y dy$, for $xi(x,y)=mathrm{const}$ we have $dxi=0$, leading to
$$
frac{dy}{dx} = - frac{xi_x}{xi_y} ,
$$
then
$$
frac{dy}{dx}=frac{y}{x},
$$
or $y=cx$. Therefore, $xi=mathrm{const}$ corresponds to $xi=y/x$.
Setting the terms between the second pair of brackets to $0$,
$$
x^2 frac{xi_x}{xi_y}eta_x+ xyleft(frac{xi_x}{xi_y}eta_y+eta_xright)+y^2 eta_y=0.
$$
Using what we found for $xi_x/xi_y$ we see that the expression vanishes identicaly to $0$, therefore the choice of $eta_y$ is arbitrary. Defining $eta=x$ leads to
$$
z_{eta eta} = 0,
$$
which is the canonical form of the parabolic equation. Integrating the equation,
$$
z(xi,eta) = f(xi) eta + g(xi),
$$
or
$$
z(x,y) = x fleft(frac{y}{x}right) + gleft(frac{y}{x}right).
$$
That is the same solution presented by Daniel Schepler in the comments.
This answer was based on this text about canonical forms.
answered Nov 28 '18 at 22:00
rafa11111rafa11111
1,1251417
$endgroup$
add a comment |
$begingroup$
Let's introduce the variables $xi(x,y)$ and $eta(x,y)$. From sucessive application of chain rule one has
$$
z_{xx} = z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx},
$$
$$
z_{yy} = z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy},
$$
$$
z_{xy} = z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}.
$$
Substituting in the original PDE and colecting terms,
$$
x^2 left[z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx} right] + 2xy left[z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}right] + y^2 left[ z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy}right] = 0
$$
$$
left[ x^2 xi_x^2 + 2xy xi_xxi_y +y^2xi_y^2right] z_{xi xi} + 2left[x^2 xi_xeta_x+ xy(xi_xeta_y+xi_yeta_x)+y^2 xi_yeta_yright] z_{xi eta} + left[ x^2 eta_x^2+2xyeta_xeta_y+y^2 eta_y^2right] z_{eta eta} = phi,
$$
in which $phi$ represents the remaining terms. Setting the terms between the first pair of brackets to $0$ leads to
$$
left(frac{xxi_x}{yxi_y}right)^2 + 2 left(frac{xxi_x}{yxi_y}right) + 1 = 0,
$$
which is a quadratic equation on $xxi_x/yxi_y$, then
$$
frac{xi_x}{xi_y} = - frac{y}{x}.
$$
Since $dxi=xi_x dx + xi_y dy$, for $xi(x,y)=mathrm{const}$ we have $dxi=0$, leading to
$$
frac{dy}{dx} = - frac{xi_x}{xi_y} ,
$$
then
$$
frac{dy}{dx}=frac{y}{x},
$$
or $y=cx$. Therefore, $xi=mathrm{const}$ corresponds to $xi=y/x$.
Setting the terms between the second pair of brackets to $0$,
$$
x^2 frac{xi_x}{xi_y}eta_x+ xyleft(frac{xi_x}{xi_y}eta_y+eta_xright)+y^2 eta_y=0.
$$
Using what we found for $xi_x/xi_y$ we see that the expression vanishes identicaly to $0$, therefore the choice of $eta_y$ is arbitrary. Defining $eta=x$ leads to
$$
z_{eta eta} = 0,
$$
which is the canonical form of the parabolic equation. Integrating the equation,
$$
z(xi,eta) = f(xi) eta + g(xi),
$$
or
$$
z(x,y) = x fleft(frac{y}{x}right) + gleft(frac{y}{x}right).
$$
That is the same solution presented by Daniel Schepler in the comments.
This answer was based on this text about canonical forms.
answered Nov 28 '18 at 22:00
rafa11111rafa11111
1,1251417
$endgroup$
Let's introduce the variables $xi(x,y)$ and $eta(x,y)$. From sucessive application of chain rule one has
$$
z_{xx} = z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx},
$$
$$
z_{yy} = z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy},
$$
$$
z_{xy} = z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}.
$$
Substituting in the original PDE and colecting terms,
$$
x^2 left[z_{xi xi} xi_x^2 + 2z_{xi eta} xi_x eta_x + z_{eta eta} eta_x^2 + z_xi xi_{xx} + z_eta eta_{xx} right] + 2xy left[z_{xi xi} xi_x xi_y + z_{xi eta} (xi_x eta_y+xi_y eta_x) + z_{eta eta} eta_x eta_y + z_xi xi_{xy} + z_eta eta_{xy}right] + y^2 left[ z_{xi xi} xi_y^2 + 2z_{xi eta} xi_y eta_y + z_{eta eta} eta_y^2 + z_xi xi_{yy} + z_eta eta_{yy}right] = 0
$$
$$
left[ x^2 xi_x^2 + 2xy xi_xxi_y +y^2xi_y^2right] z_{xi xi} + 2left[x^2 xi_xeta_x+ xy(xi_xeta_y+xi_yeta_x)+y^2 xi_yeta_yright] z_{xi eta} + left[ x^2 eta_x^2+2xyeta_xeta_y+y^2 eta_y^2right] z_{eta eta} = phi,
$$
in which $phi$ represents the remaining terms. Setting the terms between the first pair of brackets to $0$ leads to
$$
left(frac{xxi_x}{yxi_y}right)^2 + 2 left(frac{xxi_x}{yxi_y}right) + 1 = 0,
$$
which is a quadratic equation on $xxi_x/yxi_y$, then
$$
frac{xi_x}{xi_y} = - frac{y}{x}.
$$
Since $dxi=xi_x dx + xi_y dy$, for $xi(x,y)=mathrm{const}$ we have $dxi=0$, leading to
$$
frac{dy}{dx} = - frac{xi_x}{xi_y} ,
$$
then
$$
frac{dy}{dx}=frac{y}{x},
$$
or $y=cx$. Therefore, $xi=mathrm{const}$ corresponds to $xi=y/x$.
Setting the terms between the second pair of brackets to $0$,
$$
x^2 frac{xi_x}{xi_y}eta_x+ xyleft(frac{xi_x}{xi_y}eta_y+eta_xright)+y^2 eta_y=0.
$$
Using what we found for $xi_x/xi_y$ we see that the expression vanishes identicaly to $0$, therefore the choice of $eta_y$ is arbitrary. Defining $eta=x$ leads to
$$
z_{eta eta} = 0,
$$
which is the canonical form of the parabolic equation. Integrating the equation,
$$
z(xi,eta) = f(xi) eta + g(xi),
$$
or
$$
z(x,y) = x fleft(frac{y}{x}right) + gleft(frac{y}{x}right).
$$
That is the same solution presented by Daniel Schepler in the comments.
This answer was based on this text about canonical forms.
answered Nov 28 '18 at 22:00
rafa11111rafa11111
1,1251417
edited Nov 28 '18 at 22:14
answered Nov 28 '18 at 22:00
rafa11111rafa11111
1,1251417
answered Nov 28 '18 at 22:00
rafa11111rafa11111
1,1251417
answered Nov 28 '18 at 22:00
rafa11111rafa11111
1,1251417
1,1251417
add a comment |
add a comment |
$begingroup$
Do you have any additional conditions? I tried and ansatz of the form $z = log(Ax+By)$ and the left hand side yielded $-1$ so, since the equation is linear, any difference of logarithms of that form would solve the equation. Linear polynomials in $x$ and $y$ also make the left hand side vanish, so the following is a possible solution:
$$
z(x,y) = log(Ax+By) - log(Cx+Dy) + Ex +Fy + G
,,
$$
Where $A,dots,G$ are constants. Not sure if there are more general solutions, but I hope this helps.
answered Nov 28 '18 at 18:55
mlerma54mlerma54
1,177148
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Given the nature of the problem, I'm almost sure that a characteristic method-like solution is expected.
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– rafa11111
Nov 28 '18 at 19:09
1
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And this can be rewritten as $z = [log(A + B (y/x)) - log(C + D(y/x)) + G] + x [E + F (y/x)]$ which is a specialization of the general solution from rafa11111's answer.
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– Daniel Schepler
Nov 28 '18 at 22:10
1
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Good job! I was about to generalize my "ansatz"-based particular solution with a more general combination of the logarithms $$z(x,y) = sum_{i} lambda_i log(A_i x+B_i y) + Ex +Fy + G$$ where $sum_{i} lambda_i = 0$, but here is not need now.
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– mlerma54
Nov 28 '18 at 23:56
add a comment |
$begingroup$
Do you have any additional conditions? I tried and ansatz of the form $z = log(Ax+By)$ and the left hand side yielded $-1$ so, since the equation is linear, any difference of logarithms of that form would solve the equation. Linear polynomials in $x$ and $y$ also make the left hand side vanish, so the following is a possible solution:
$$
z(x,y) = log(Ax+By) - log(Cx+Dy) + Ex +Fy + G
,,
$$
Where $A,dots,G$ are constants. Not sure if there are more general solutions, but I hope this helps.
answered Nov 28 '18 at 18:55
mlerma54mlerma54
1,177148
$endgroup$
$begingroup$
Given the nature of the problem, I'm almost sure that a characteristic method-like solution is expected.
$endgroup$
– rafa11111
Nov 28 '18 at 19:09
1
$begingroup$
And this can be rewritten as $z = [log(A + B (y/x)) - log(C + D(y/x)) + G] + x [E + F (y/x)]$ which is a specialization of the general solution from rafa11111's answer.
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– Daniel Schepler
Nov 28 '18 at 22:10
1
$begingroup$
Good job! I was about to generalize my "ansatz"-based particular solution with a more general combination of the logarithms $$z(x,y) = sum_{i} lambda_i log(A_i x+B_i y) + Ex +Fy + G$$ where $sum_{i} lambda_i = 0$, but here is not need now.
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– mlerma54
Nov 28 '18 at 23:56
add a comment |
$begingroup$
Do you have any additional conditions? I tried and ansatz of the form $z = log(Ax+By)$ and the left hand side yielded $-1$ so, since the equation is linear, any difference of logarithms of that form would solve the equation. Linear polynomials in $x$ and $y$ also make the left hand side vanish, so the following is a possible solution:
$$
z(x,y) = log(Ax+By) - log(Cx+Dy) + Ex +Fy + G
,,
$$
Where $A,dots,G$ are constants. Not sure if there are more general solutions, but I hope this helps.
answered Nov 28 '18 at 18:55
mlerma54mlerma54
1,177148
$endgroup$
Do you have any additional conditions? I tried and ansatz of the form $z = log(Ax+By)$ and the left hand side yielded $-1$ so, since the equation is linear, any difference of logarithms of that form would solve the equation. Linear polynomials in $x$ and $y$ also make the left hand side vanish, so the following is a possible solution:
$$
z(x,y) = log(Ax+By) - log(Cx+Dy) + Ex +Fy + G
,,
$$
Where $A,dots,G$ are constants. Not sure if there are more general solutions, but I hope this helps.
answered Nov 28 '18 at 18:55
mlerma54mlerma54
1,177148
answered Nov 28 '18 at 18:55
mlerma54mlerma54
1,177148
answered Nov 28 '18 at 18:55
mlerma54mlerma54
1,177148
answered Nov 28 '18 at 18:55
mlerma54mlerma54
1,177148
1,177148
$begingroup$
Given the nature of the problem, I'm almost sure that a characteristic method-like solution is expected.
$endgroup$
– rafa11111
Nov 28 '18 at 19:09
1
$begingroup$
And this can be rewritten as $z = [log(A + B (y/x)) - log(C + D(y/x)) + G] + x [E + F (y/x)]$ which is a specialization of the general solution from rafa11111's answer.
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– Daniel Schepler
Nov 28 '18 at 22:10
1
$begingroup$
Good job! I was about to generalize my "ansatz"-based particular solution with a more general combination of the logarithms $$z(x,y) = sum_{i} lambda_i log(A_i x+B_i y) + Ex +Fy + G$$ where $sum_{i} lambda_i = 0$, but here is not need now.
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– mlerma54
Nov 28 '18 at 23:56
add a comment |
$begingroup$
Given the nature of the problem, I'm almost sure that a characteristic method-like solution is expected.
$endgroup$
– rafa11111
Nov 28 '18 at 19:09
1
$begingroup$
And this can be rewritten as $z = [log(A + B (y/x)) - log(C + D(y/x)) + G] + x [E + F (y/x)]$ which is a specialization of the general solution from rafa11111's answer.
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– Daniel Schepler
Nov 28 '18 at 22:10
1
$begingroup$
Good job! I was about to generalize my "ansatz"-based particular solution with a more general combination of the logarithms $$z(x,y) = sum_{i} lambda_i log(A_i x+B_i y) + Ex +Fy + G$$ where $sum_{i} lambda_i = 0$, but here is not need now.
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– mlerma54
Nov 28 '18 at 23:56
$begingroup$
Given the nature of the problem, I'm almost sure that a characteristic method-like solution is expected.
$endgroup$
– rafa11111
Nov 28 '18 at 19:09
$begingroup$
Given the nature of the problem, I'm almost sure that a characteristic method-like solution is expected.
$endgroup$
– rafa11111
Nov 28 '18 at 19:09
1
1
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And this can be rewritten as $z = [log(A + B (y/x)) - log(C + D(y/x)) + G] + x [E + F (y/x)]$ which is a specialization of the general solution from rafa11111's answer.
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– Daniel Schepler
Nov 28 '18 at 22:10
$begingroup$
And this can be rewritten as $z = [log(A + B (y/x)) - log(C + D(y/x)) + G] + x [E + F (y/x)]$ which is a specialization of the general solution from rafa11111's answer.
$endgroup$
– Daniel Schepler
Nov 28 '18 at 22:10
1
1
$begingroup$
Good job! I was about to generalize my "ansatz"-based particular solution with a more general combination of the logarithms $$z(x,y) = sum_{i} lambda_i log(A_i x+B_i y) + Ex +Fy + G$$ where $sum_{i} lambda_i = 0$, but here is not need now.
$endgroup$
– mlerma54
Nov 28 '18 at 23:56
$begingroup$
Good job! I was about to generalize my "ansatz"-based particular solution with a more general combination of the logarithms $$z(x,y) = sum_{i} lambda_i log(A_i x+B_i y) + Ex +Fy + G$$ where $sum_{i} lambda_i = 0$, but here is not need now.
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– mlerma54
Nov 28 '18 at 23:56
add a comment |
2
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It will be helpful to know that it's a parabolic PDE...
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– rafa11111
Nov 28 '18 at 17:26
1
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By guessing solutions of the form $x^{alpha} y^{beta}$ (by analogy with Euler ordinary differential equations) I found that either $x^{alpha} y^{-alpha}$ or $x^{alpha+1} y^{-alpha}$ are solutions. Based on that, it seems reasonable to guess that $f(x/y)$ and $x f(x/y)$ are both solutions - and by linearity, that would imply that $f(x/y) + x g(x/y)$ is a solution.
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– Daniel Schepler
Nov 28 '18 at 20:54
1
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$$x^{2} z_{xx} + 2xy z_{xy} + y^{2} z_{yy} = (x partial_{x} + y partial_{y})^{2} z$$ assuming $partial_{x} partial_{y} = partial_{y} partial_{x}$.
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– Mattos
Nov 28 '18 at 21:47
1
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Hmm, and that in turn suggests that it might be useful to translate the PDE into polar coordinates. If (purely hypothetically) the result were something like $r^2 cos theta sin theta frac{partial^2 z}{partial r^2} = 0$, then in the interiors of the quadrants that would imply $z = f(theta) + r g(theta)$ - and it shouldn't be too hard to massage this into $z = F(y/x) + x G(y/x)$.
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– Daniel Schepler
Nov 28 '18 at 21:50
1
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@RafaBudría Well I take it back then, what I originally wrote is incorrect. You are right. Thanks for catching that.
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– Mattos
Dec 3 '18 at 5:56