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Method of Characteristics: Initial Conditions and the Unique Solution in This Linear PDE Example
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Find the general solution of the PDE $pu_x + u_y = u$, where $p$ is constant, in the domain $ (x, y) : x in mathbbR, 0 le y le Y $, subject to the initial condition $u(x, 0) = g(x), x in mathbbR$.
The parametric characteristic equations are
$$fracdxdt = p, fracdydt = 1, fracdudt = u$$
We can rewrite these in non-parametric form:
$$fracdxp = dt, fracdy1 = dt, fracduu = dt$$
$$rightarrow fracdxp = fracdy1 = fracduu$$
And this can be written as two new ODEs:
$$fracdxdy = p, fracdudy = u$$
Solving these ODEs using separation of variables, we get that $x = py + k$ and $u = Ae^y$ on any single characteristic, with $k$ and $A$ being two related constants.
Question 1: What is meant here by "related"? What does it mean for two constants to be "related"? Why are they "related"?
The values of $k$ and $A$ are constant along any one characteristic. As such, we can write the general solution as
$$x = py + k, u = A(k)e^y$$
$A$ is written here as a function of $k$. Is this because, again, $A$ is "related" to $k$?
Geometrically, the characters are straight lines, along which the solution varies exponentially.
The parameter $k$ can be eliminated:
$$u(x, y) = A(x - py)e^y$$
To use the initial condition, we set $y = 0$, and so $u(x, 0) = A(x) = g(x)$, and our unique solution is
$$u(x, y) = g(x - py)e^y$$
Question 2: Something about this doesn't seem right? Using the initial condition, we found that $u(x, 0) = A(x) = g(x)$. So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$. But in $u(x, y) = A(x - py)e^y$, we have that $u(x, y) = A(k) = A(x - py)e^y$, not $u(x, y) = A(x) = A(x - py)e^y$? So I don't see how it makes sense to replace $A$ with the function $g$ to get $u(x, y) = g(x - py)e^y$? Something doesn't seem right here.
Thank you for any guidance you could offer.
multivariable-calculus pde characteristics
asked Jul 30 at 10:02
handler's handle
848
add a comment |Â
up vote
3
down vote
favorite
Find the general solution of the PDE $pu_x + u_y = u$, where $p$ is constant, in the domain $ (x, y) : x in mathbbR, 0 le y le Y $, subject to the initial condition $u(x, 0) = g(x), x in mathbbR$.
The parametric characteristic equations are
$$fracdxdt = p, fracdydt = 1, fracdudt = u$$
We can rewrite these in non-parametric form:
$$fracdxp = dt, fracdy1 = dt, fracduu = dt$$
$$rightarrow fracdxp = fracdy1 = fracduu$$
And this can be written as two new ODEs:
$$fracdxdy = p, fracdudy = u$$
Solving these ODEs using separation of variables, we get that $x = py + k$ and $u = Ae^y$ on any single characteristic, with $k$ and $A$ being two related constants.
Question 1: What is meant here by "related"? What does it mean for two constants to be "related"? Why are they "related"?
The values of $k$ and $A$ are constant along any one characteristic. As such, we can write the general solution as
$$x = py + k, u = A(k)e^y$$
$A$ is written here as a function of $k$. Is this because, again, $A$ is "related" to $k$?
Geometrically, the characters are straight lines, along which the solution varies exponentially.
The parameter $k$ can be eliminated:
$$u(x, y) = A(x - py)e^y$$
To use the initial condition, we set $y = 0$, and so $u(x, 0) = A(x) = g(x)$, and our unique solution is
$$u(x, y) = g(x - py)e^y$$
Question 2: Something about this doesn't seem right? Using the initial condition, we found that $u(x, 0) = A(x) = g(x)$. So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$. But in $u(x, y) = A(x - py)e^y$, we have that $u(x, y) = A(k) = A(x - py)e^y$, not $u(x, y) = A(x) = A(x - py)e^y$? So I don't see how it makes sense to replace $A$ with the function $g$ to get $u(x, y) = g(x - py)e^y$? Something doesn't seem right here.
Thank you for any guidance you could offer.
multivariable-calculus pde characteristics
asked Jul 30 at 10:02
handler's handle
848
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find the general solution of the PDE $pu_x + u_y = u$, where $p$ is constant, in the domain $ (x, y) : x in mathbbR, 0 le y le Y $, subject to the initial condition $u(x, 0) = g(x), x in mathbbR$.
The parametric characteristic equations are
$$fracdxdt = p, fracdydt = 1, fracdudt = u$$
We can rewrite these in non-parametric form:
$$fracdxp = dt, fracdy1 = dt, fracduu = dt$$
$$rightarrow fracdxp = fracdy1 = fracduu$$
And this can be written as two new ODEs:
$$fracdxdy = p, fracdudy = u$$
Solving these ODEs using separation of variables, we get that $x = py + k$ and $u = Ae^y$ on any single characteristic, with $k$ and $A$ being two related constants.
Question 1: What is meant here by "related"? What does it mean for two constants to be "related"? Why are they "related"?
The values of $k$ and $A$ are constant along any one characteristic. As such, we can write the general solution as
$$x = py + k, u = A(k)e^y$$
$A$ is written here as a function of $k$. Is this because, again, $A$ is "related" to $k$?
Geometrically, the characters are straight lines, along which the solution varies exponentially.
The parameter $k$ can be eliminated:
$$u(x, y) = A(x - py)e^y$$
To use the initial condition, we set $y = 0$, and so $u(x, 0) = A(x) = g(x)$, and our unique solution is
$$u(x, y) = g(x - py)e^y$$
Question 2: Something about this doesn't seem right? Using the initial condition, we found that $u(x, 0) = A(x) = g(x)$. So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$. But in $u(x, y) = A(x - py)e^y$, we have that $u(x, y) = A(k) = A(x - py)e^y$, not $u(x, y) = A(x) = A(x - py)e^y$? So I don't see how it makes sense to replace $A$ with the function $g$ to get $u(x, y) = g(x - py)e^y$? Something doesn't seem right here.
Thank you for any guidance you could offer.
multivariable-calculus pde characteristics
asked Jul 30 at 10:02
handler's handle
848
Find the general solution of the PDE $pu_x + u_y = u$, where $p$ is constant, in the domain $ (x, y) : x in mathbbR, 0 le y le Y $, subject to the initial condition $u(x, 0) = g(x), x in mathbbR$.
The parametric characteristic equations are
$$fracdxdt = p, fracdydt = 1, fracdudt = u$$
We can rewrite these in non-parametric form:
$$fracdxp = dt, fracdy1 = dt, fracduu = dt$$
$$rightarrow fracdxp = fracdy1 = fracduu$$
And this can be written as two new ODEs:
$$fracdxdy = p, fracdudy = u$$
Solving these ODEs using separation of variables, we get that $x = py + k$ and $u = Ae^y$ on any single characteristic, with $k$ and $A$ being two related constants.
Question 1: What is meant here by "related"? What does it mean for two constants to be "related"? Why are they "related"?
The values of $k$ and $A$ are constant along any one characteristic. As such, we can write the general solution as
$$x = py + k, u = A(k)e^y$$
$A$ is written here as a function of $k$. Is this because, again, $A$ is "related" to $k$?
Geometrically, the characters are straight lines, along which the solution varies exponentially.
The parameter $k$ can be eliminated:
$$u(x, y) = A(x - py)e^y$$
To use the initial condition, we set $y = 0$, and so $u(x, 0) = A(x) = g(x)$, and our unique solution is
$$u(x, y) = g(x - py)e^y$$
Question 2: Something about this doesn't seem right? Using the initial condition, we found that $u(x, 0) = A(x) = g(x)$. So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$. But in $u(x, y) = A(x - py)e^y$, we have that $u(x, y) = A(k) = A(x - py)e^y$, not $u(x, y) = A(x) = A(x - py)e^y$? So I don't see how it makes sense to replace $A$ with the function $g$ to get $u(x, y) = g(x - py)e^y$? Something doesn't seem right here.
Thank you for any guidance you could offer.
multivariable-calculus pde characteristics
asked Jul 30 at 10:02
handler's handle
848
asked Jul 30 at 10:02
handler's handle
848
asked Jul 30 at 10:02
handler's handle
848
asked Jul 30 at 10:02
handler's handle
848
848
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
"So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$"
Yes, it is exactly this way, so we can say too that $A$ as a function of $k$ is equal to $g$ as a function of $k$. In $u(x, y) = A(x - py)e^y$ we have $u(x, y) = A(k)e^y=g(k)e^y=g(x-py)e^y$ because $k=x-py$, as stated the ODEs.
"Related" means that there must be a functional relation between these constants $k$ and $A$, so is, "each $k$ has its $A$" determining a curve (as intersection of two surfaces $x-py=k$ and $u=Ae^y$) into a particular solution (and another value of $k$ has another correct value of $A$ to fit the curve so determined into the surface solution): remember that the solutions are surfaces constructed has a family of curves (the characteristics).
answered Aug 1 at 8:30
Rafa BudrÃa
4,7661525
Thank you for this. I think I understand now.
– handler's handle
Aug 1 at 13:48
add a comment |Â
up vote
1
down vote
Let take the problem from the system of characteristic ODEs that you correctly got :
$$rightarrow fracdxp = fracdy1 = fracduu$$
A first characteristic equation comes from $quadfracdxp = fracdy1$ :
$$x-py=c_1$$
A second characteristic equation comes from $quadfracdy1 = fracduu$ :
$$ue^-y=c_2$$
The general solution of the PDE expressed on the form of an implicit equation is :
$$Phileft(x-py:,:ue^-y right)=0$$
$Phi$ is an arbitrary function of two variables.
Equivalently, on explicit form :
$$ue^-y=F(x-py)$$
$F$ is an arbitrary function.
So, the general solution is :
$$u(x,y)=e^yF(x-py)$$
Boundary condition :
$u(X,0)=g(X)=e^0F(X-0)=F(X)quad$ any $X$, doesn't matter the notation of the variable.
Now the function $F(X)$ is known : $$F(X)=g(X)$$
We put it into the above general solution where $X=x-py$.
The particular solution which satisfies the boundary condition is :
$$u(x,y)=e^y g(x-py)$$
So, your result was correct. The variant to get it is equivalent. May be less confusing.
answered Aug 1 at 9:07
JJacquelin
39.7k21649
Yes, I found this clearer! Thank you for this.
– handler's handle
Aug 1 at 13:43
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
"So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$"
Yes, it is exactly this way, so we can say too that $A$ as a function of $k$ is equal to $g$ as a function of $k$. In $u(x, y) = A(x - py)e^y$ we have $u(x, y) = A(k)e^y=g(k)e^y=g(x-py)e^y$ because $k=x-py$, as stated the ODEs.
"Related" means that there must be a functional relation between these constants $k$ and $A$, so is, "each $k$ has its $A$" determining a curve (as intersection of two surfaces $x-py=k$ and $u=Ae^y$) into a particular solution (and another value of $k$ has another correct value of $A$ to fit the curve so determined into the surface solution): remember that the solutions are surfaces constructed has a family of curves (the characteristics).
answered Aug 1 at 8:30
Rafa BudrÃa
4,7661525
Thank you for this. I think I understand now.
– handler's handle
Aug 1 at 13:48
add a comment |Â
up vote
1
down vote
"So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$"
Yes, it is exactly this way, so we can say too that $A$ as a function of $k$ is equal to $g$ as a function of $k$. In $u(x, y) = A(x - py)e^y$ we have $u(x, y) = A(k)e^y=g(k)e^y=g(x-py)e^y$ because $k=x-py$, as stated the ODEs.
"Related" means that there must be a functional relation between these constants $k$ and $A$, so is, "each $k$ has its $A$" determining a curve (as intersection of two surfaces $x-py=k$ and $u=Ae^y$) into a particular solution (and another value of $k$ has another correct value of $A$ to fit the curve so determined into the surface solution): remember that the solutions are surfaces constructed has a family of curves (the characteristics).
answered Aug 1 at 8:30
Rafa BudrÃa
4,7661525
Thank you for this. I think I understand now.
– handler's handle
Aug 1 at 13:48
add a comment |Â
up vote
1
down vote
up vote
1
down vote
"So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$"
Yes, it is exactly this way, so we can say too that $A$ as a function of $k$ is equal to $g$ as a function of $k$. In $u(x, y) = A(x - py)e^y$ we have $u(x, y) = A(k)e^y=g(k)e^y=g(x-py)e^y$ because $k=x-py$, as stated the ODEs.
"Related" means that there must be a functional relation between these constants $k$ and $A$, so is, "each $k$ has its $A$" determining a curve (as intersection of two surfaces $x-py=k$ and $u=Ae^y$) into a particular solution (and another value of $k$ has another correct value of $A$ to fit the curve so determined into the surface solution): remember that the solutions are surfaces constructed has a family of curves (the characteristics).
answered Aug 1 at 8:30
Rafa BudrÃa
4,7661525
"So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$"
Yes, it is exactly this way, so we can say too that $A$ as a function of $k$ is equal to $g$ as a function of $k$. In $u(x, y) = A(x - py)e^y$ we have $u(x, y) = A(k)e^y=g(k)e^y=g(x-py)e^y$ because $k=x-py$, as stated the ODEs.
"Related" means that there must be a functional relation between these constants $k$ and $A$, so is, "each $k$ has its $A$" determining a curve (as intersection of two surfaces $x-py=k$ and $u=Ae^y$) into a particular solution (and another value of $k$ has another correct value of $A$ to fit the curve so determined into the surface solution): remember that the solutions are surfaces constructed has a family of curves (the characteristics).
answered Aug 1 at 8:30
Rafa BudrÃa
4,7661525
answered Aug 1 at 8:30
Rafa BudrÃa
4,7661525
answered Aug 1 at 8:30
Rafa BudrÃa
4,7661525
answered Aug 1 at 8:30
Rafa BudrÃa
4,7661525
4,7661525
Thank you for this. I think I understand now.
– handler's handle
Aug 1 at 13:48
add a comment |Â
Thank you for this. I think I understand now.
– handler's handle
Aug 1 at 13:48
Thank you for this. I think I understand now.
– handler's handle
Aug 1 at 13:48
Thank you for this. I think I understand now.
– handler's handle
Aug 1 at 13:48
add a comment |Â
up vote
1
down vote
Let take the problem from the system of characteristic ODEs that you correctly got :
$$rightarrow fracdxp = fracdy1 = fracduu$$
A first characteristic equation comes from $quadfracdxp = fracdy1$ :
$$x-py=c_1$$
A second characteristic equation comes from $quadfracdy1 = fracduu$ :
$$ue^-y=c_2$$
The general solution of the PDE expressed on the form of an implicit equation is :
$$Phileft(x-py:,:ue^-y right)=0$$
$Phi$ is an arbitrary function of two variables.
Equivalently, on explicit form :
$$ue^-y=F(x-py)$$
$F$ is an arbitrary function.
So, the general solution is :
$$u(x,y)=e^yF(x-py)$$
Boundary condition :
$u(X,0)=g(X)=e^0F(X-0)=F(X)quad$ any $X$, doesn't matter the notation of the variable.
Now the function $F(X)$ is known : $$F(X)=g(X)$$
We put it into the above general solution where $X=x-py$.
The particular solution which satisfies the boundary condition is :
$$u(x,y)=e^y g(x-py)$$
So, your result was correct. The variant to get it is equivalent. May be less confusing.
answered Aug 1 at 9:07
JJacquelin
39.7k21649
Yes, I found this clearer! Thank you for this.
– handler's handle
Aug 1 at 13:43
add a comment |Â
up vote
1
down vote
Let take the problem from the system of characteristic ODEs that you correctly got :
$$rightarrow fracdxp = fracdy1 = fracduu$$
A first characteristic equation comes from $quadfracdxp = fracdy1$ :
$$x-py=c_1$$
A second characteristic equation comes from $quadfracdy1 = fracduu$ :
$$ue^-y=c_2$$
The general solution of the PDE expressed on the form of an implicit equation is :
$$Phileft(x-py:,:ue^-y right)=0$$
$Phi$ is an arbitrary function of two variables.
Equivalently, on explicit form :
$$ue^-y=F(x-py)$$
$F$ is an arbitrary function.
So, the general solution is :
$$u(x,y)=e^yF(x-py)$$
Boundary condition :
$u(X,0)=g(X)=e^0F(X-0)=F(X)quad$ any $X$, doesn't matter the notation of the variable.
Now the function $F(X)$ is known : $$F(X)=g(X)$$
We put it into the above general solution where $X=x-py$.
The particular solution which satisfies the boundary condition is :
$$u(x,y)=e^y g(x-py)$$
So, your result was correct. The variant to get it is equivalent. May be less confusing.
answered Aug 1 at 9:07
JJacquelin
39.7k21649
Yes, I found this clearer! Thank you for this.
– handler's handle
Aug 1 at 13:43
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let take the problem from the system of characteristic ODEs that you correctly got :
$$rightarrow fracdxp = fracdy1 = fracduu$$
A first characteristic equation comes from $quadfracdxp = fracdy1$ :
$$x-py=c_1$$
A second characteristic equation comes from $quadfracdy1 = fracduu$ :
$$ue^-y=c_2$$
The general solution of the PDE expressed on the form of an implicit equation is :
$$Phileft(x-py:,:ue^-y right)=0$$
$Phi$ is an arbitrary function of two variables.
Equivalently, on explicit form :
$$ue^-y=F(x-py)$$
$F$ is an arbitrary function.
So, the general solution is :
$$u(x,y)=e^yF(x-py)$$
Boundary condition :
$u(X,0)=g(X)=e^0F(X-0)=F(X)quad$ any $X$, doesn't matter the notation of the variable.
Now the function $F(X)$ is known : $$F(X)=g(X)$$
We put it into the above general solution where $X=x-py$.
The particular solution which satisfies the boundary condition is :
$$u(x,y)=e^y g(x-py)$$
So, your result was correct. The variant to get it is equivalent. May be less confusing.
answered Aug 1 at 9:07
JJacquelin
39.7k21649
Let take the problem from the system of characteristic ODEs that you correctly got :
$$rightarrow fracdxp = fracdy1 = fracduu$$
A first characteristic equation comes from $quadfracdxp = fracdy1$ :
$$x-py=c_1$$
A second characteristic equation comes from $quadfracdy1 = fracduu$ :
$$ue^-y=c_2$$
The general solution of the PDE expressed on the form of an implicit equation is :
$$Phileft(x-py:,:ue^-y right)=0$$
$Phi$ is an arbitrary function of two variables.
Equivalently, on explicit form :
$$ue^-y=F(x-py)$$
$F$ is an arbitrary function.
So, the general solution is :
$$u(x,y)=e^yF(x-py)$$
Boundary condition :
$u(X,0)=g(X)=e^0F(X-0)=F(X)quad$ any $X$, doesn't matter the notation of the variable.
Now the function $F(X)$ is known : $$F(X)=g(X)$$
We put it into the above general solution where $X=x-py$.
The particular solution which satisfies the boundary condition is :
$$u(x,y)=e^y g(x-py)$$
So, your result was correct. The variant to get it is equivalent. May be less confusing.
answered Aug 1 at 9:07
JJacquelin
39.7k21649
answered Aug 1 at 9:07
JJacquelin
39.7k21649
answered Aug 1 at 9:07
JJacquelin
39.7k21649
answered Aug 1 at 9:07
JJacquelin
39.7k21649
39.7k21649
Yes, I found this clearer! Thank you for this.
– handler's handle
Aug 1 at 13:43
add a comment |Â
Yes, I found this clearer! Thank you for this.
– handler's handle
Aug 1 at 13:43
Yes, I found this clearer! Thank you for this.
– handler's handle
Aug 1 at 13:43
Yes, I found this clearer! Thank you for this.
– handler's handle
Aug 1 at 13:43
add a comment |Â
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