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General solution to Wave Equation (algebraic approach)
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Consider the Wave Equation
$$frac{partial^{2} u}{partial x^{2}} - c^{2}frac{partial^{2} u}{partial y^{2}} = 0$$
The algebraic approach from wikipedia is:
Take $alpha = x - cy$ and $beta = x + cy$. Thus, we have
$$frac{partial^{2} u}{partial alpha partial beta} = 0,$$
which leads to the general solution
$$u(x,y) = f(x-cy) + g(x+cy).$$
I didnt get the same conclusions with the initial variable change.
My attempt:
$$frac{partial^{2} u}{partial x^{2}} - c^{2}frac{partial^{2} u}{partial y^{2}} = left(frac{partial u}{partial x} - cfrac{partial u}{partial y}right)left(frac{partial u}{partial x} + cfrac{partial u}{partial y}right).$$
Now, take $alpha = x - cy$ and $beta = x + cy$. So,
$$frac{alpha + beta}{2} = xquadtext{e}quadfrac{beta - alpha}{2c} = y.$$
Thus,
$$frac{partial u}{partial alpha} = frac{partial u}{partial x}frac{partial x}{partial alpha} + frac{partial u}{partial y}frac{partial y}{partial alpha}$$
and
$$frac{partial u}{partial beta} = frac{partial u}{partial x}frac{partial x}{partial beta} + frac{partial u}{partial y}frac{partial y}{partial beta}.$$
Since,
$$frac{partial x}{partial alpha} = frac{1}{2}, frac{partial x}{partial beta} = frac{1}{2}, frac{partial y}{partial alpha} = -frac{1}{2c}, frac{partial y}{partial beta} = frac{1}{2c},$$
we have
$$frac{partial u}{partial alpha} = frac{1}{2}frac{partial u}{partial x} - frac{1}{2c}frac{partial u}{partial y}$$
and
$$frac{partial u}{partial beta} = frac{1}{2}frac{partial u}{partial x} + frac{1}{2c}frac{partial u}{partial y}.$$
Therefore,
$$frac{partial^{2} u}{partial alpha partial beta} = left(frac{1}{2}frac{partial u}{partial x} - frac{1}{2c}frac{partial u}{partial y}right)left(frac{1}{2}frac{partial u}{partial x} + frac{1}{2c}frac{partial u}{partial y}right) = frac{1}{4}frac{partial^{2} u}{partial x^{2}} - frac{1}{4c^{2}}frac{partial^{2} u}{partial y^{2}},$$
then
$$4frac{partial^{2} u}{partial alpha partial beta} = frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$$
I think I did something wrong. But I don't know exactly what or how to correct.
pde partial-derivative wave-equation
asked yesterday
Greg
758
add a comment |
up vote
1
down vote
favorite
Consider the Wave Equation
$$frac{partial^{2} u}{partial x^{2}} - c^{2}frac{partial^{2} u}{partial y^{2}} = 0$$
The algebraic approach from wikipedia is:
Take $alpha = x - cy$ and $beta = x + cy$. Thus, we have
$$frac{partial^{2} u}{partial alpha partial beta} = 0,$$
which leads to the general solution
$$u(x,y) = f(x-cy) + g(x+cy).$$
I didnt get the same conclusions with the initial variable change.
My attempt:
$$frac{partial^{2} u}{partial x^{2}} - c^{2}frac{partial^{2} u}{partial y^{2}} = left(frac{partial u}{partial x} - cfrac{partial u}{partial y}right)left(frac{partial u}{partial x} + cfrac{partial u}{partial y}right).$$
Now, take $alpha = x - cy$ and $beta = x + cy$. So,
$$frac{alpha + beta}{2} = xquadtext{e}quadfrac{beta - alpha}{2c} = y.$$
Thus,
$$frac{partial u}{partial alpha} = frac{partial u}{partial x}frac{partial x}{partial alpha} + frac{partial u}{partial y}frac{partial y}{partial alpha}$$
and
$$frac{partial u}{partial beta} = frac{partial u}{partial x}frac{partial x}{partial beta} + frac{partial u}{partial y}frac{partial y}{partial beta}.$$
Since,
$$frac{partial x}{partial alpha} = frac{1}{2}, frac{partial x}{partial beta} = frac{1}{2}, frac{partial y}{partial alpha} = -frac{1}{2c}, frac{partial y}{partial beta} = frac{1}{2c},$$
we have
$$frac{partial u}{partial alpha} = frac{1}{2}frac{partial u}{partial x} - frac{1}{2c}frac{partial u}{partial y}$$
and
$$frac{partial u}{partial beta} = frac{1}{2}frac{partial u}{partial x} + frac{1}{2c}frac{partial u}{partial y}.$$
Therefore,
$$frac{partial^{2} u}{partial alpha partial beta} = left(frac{1}{2}frac{partial u}{partial x} - frac{1}{2c}frac{partial u}{partial y}right)left(frac{1}{2}frac{partial u}{partial x} + frac{1}{2c}frac{partial u}{partial y}right) = frac{1}{4}frac{partial^{2} u}{partial x^{2}} - frac{1}{4c^{2}}frac{partial^{2} u}{partial y^{2}},$$
then
$$4frac{partial^{2} u}{partial alpha partial beta} = frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$$
I think I did something wrong. But I don't know exactly what or how to correct.
pde partial-derivative wave-equation
asked yesterday
Greg
758
notice the end result looks something like the wave equation you started with originally..which is equal to zero. You either have a algebra mistake somewhere or need to show why your end result is ALSO zero.
– DaveNine
yesterday
@DaveNine, yeah! It seems that if I write the equation of where in the form: $displaystyle frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$, my attempt works. But, I've seen, in other references, the wave equation as in wikipedia, with the same change of variables and the same conclusion. So, I think maybe I should go a different way, but I don't know which one.
– Greg
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the Wave Equation
$$frac{partial^{2} u}{partial x^{2}} - c^{2}frac{partial^{2} u}{partial y^{2}} = 0$$
The algebraic approach from wikipedia is:
Take $alpha = x - cy$ and $beta = x + cy$. Thus, we have
$$frac{partial^{2} u}{partial alpha partial beta} = 0,$$
which leads to the general solution
$$u(x,y) = f(x-cy) + g(x+cy).$$
I didnt get the same conclusions with the initial variable change.
My attempt:
$$frac{partial^{2} u}{partial x^{2}} - c^{2}frac{partial^{2} u}{partial y^{2}} = left(frac{partial u}{partial x} - cfrac{partial u}{partial y}right)left(frac{partial u}{partial x} + cfrac{partial u}{partial y}right).$$
Now, take $alpha = x - cy$ and $beta = x + cy$. So,
$$frac{alpha + beta}{2} = xquadtext{e}quadfrac{beta - alpha}{2c} = y.$$
Thus,
$$frac{partial u}{partial alpha} = frac{partial u}{partial x}frac{partial x}{partial alpha} + frac{partial u}{partial y}frac{partial y}{partial alpha}$$
and
$$frac{partial u}{partial beta} = frac{partial u}{partial x}frac{partial x}{partial beta} + frac{partial u}{partial y}frac{partial y}{partial beta}.$$
Since,
$$frac{partial x}{partial alpha} = frac{1}{2}, frac{partial x}{partial beta} = frac{1}{2}, frac{partial y}{partial alpha} = -frac{1}{2c}, frac{partial y}{partial beta} = frac{1}{2c},$$
we have
$$frac{partial u}{partial alpha} = frac{1}{2}frac{partial u}{partial x} - frac{1}{2c}frac{partial u}{partial y}$$
and
$$frac{partial u}{partial beta} = frac{1}{2}frac{partial u}{partial x} + frac{1}{2c}frac{partial u}{partial y}.$$
Therefore,
$$frac{partial^{2} u}{partial alpha partial beta} = left(frac{1}{2}frac{partial u}{partial x} - frac{1}{2c}frac{partial u}{partial y}right)left(frac{1}{2}frac{partial u}{partial x} + frac{1}{2c}frac{partial u}{partial y}right) = frac{1}{4}frac{partial^{2} u}{partial x^{2}} - frac{1}{4c^{2}}frac{partial^{2} u}{partial y^{2}},$$
then
$$4frac{partial^{2} u}{partial alpha partial beta} = frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$$
I think I did something wrong. But I don't know exactly what or how to correct.
pde partial-derivative wave-equation
asked yesterday
Greg
758
Consider the Wave Equation
$$frac{partial^{2} u}{partial x^{2}} - c^{2}frac{partial^{2} u}{partial y^{2}} = 0$$
The algebraic approach from wikipedia is:
Take $alpha = x - cy$ and $beta = x + cy$. Thus, we have
$$frac{partial^{2} u}{partial alpha partial beta} = 0,$$
which leads to the general solution
$$u(x,y) = f(x-cy) + g(x+cy).$$
I didnt get the same conclusions with the initial variable change.
My attempt:
$$frac{partial^{2} u}{partial x^{2}} - c^{2}frac{partial^{2} u}{partial y^{2}} = left(frac{partial u}{partial x} - cfrac{partial u}{partial y}right)left(frac{partial u}{partial x} + cfrac{partial u}{partial y}right).$$
Now, take $alpha = x - cy$ and $beta = x + cy$. So,
$$frac{alpha + beta}{2} = xquadtext{e}quadfrac{beta - alpha}{2c} = y.$$
Thus,
$$frac{partial u}{partial alpha} = frac{partial u}{partial x}frac{partial x}{partial alpha} + frac{partial u}{partial y}frac{partial y}{partial alpha}$$
and
$$frac{partial u}{partial beta} = frac{partial u}{partial x}frac{partial x}{partial beta} + frac{partial u}{partial y}frac{partial y}{partial beta}.$$
Since,
$$frac{partial x}{partial alpha} = frac{1}{2}, frac{partial x}{partial beta} = frac{1}{2}, frac{partial y}{partial alpha} = -frac{1}{2c}, frac{partial y}{partial beta} = frac{1}{2c},$$
we have
$$frac{partial u}{partial alpha} = frac{1}{2}frac{partial u}{partial x} - frac{1}{2c}frac{partial u}{partial y}$$
and
$$frac{partial u}{partial beta} = frac{1}{2}frac{partial u}{partial x} + frac{1}{2c}frac{partial u}{partial y}.$$
Therefore,
$$frac{partial^{2} u}{partial alpha partial beta} = left(frac{1}{2}frac{partial u}{partial x} - frac{1}{2c}frac{partial u}{partial y}right)left(frac{1}{2}frac{partial u}{partial x} + frac{1}{2c}frac{partial u}{partial y}right) = frac{1}{4}frac{partial^{2} u}{partial x^{2}} - frac{1}{4c^{2}}frac{partial^{2} u}{partial y^{2}},$$
then
$$4frac{partial^{2} u}{partial alpha partial beta} = frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$$
I think I did something wrong. But I don't know exactly what or how to correct.
pde partial-derivative wave-equation
pde partial-derivative wave-equation
asked yesterday
Greg
758
asked yesterday
Greg
758
edited yesterday
asked yesterday
Greg
758
asked yesterday
Greg
758
asked yesterday
Greg
758
758
notice the end result looks something like the wave equation you started with originally..which is equal to zero. You either have a algebra mistake somewhere or need to show why your end result is ALSO zero.
– DaveNine
yesterday
@DaveNine, yeah! It seems that if I write the equation of where in the form: $displaystyle frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$, my attempt works. But, I've seen, in other references, the wave equation as in wikipedia, with the same change of variables and the same conclusion. So, I think maybe I should go a different way, but I don't know which one.
– Greg
yesterday
add a comment |
notice the end result looks something like the wave equation you started with originally..which is equal to zero. You either have a algebra mistake somewhere or need to show why your end result is ALSO zero.
– DaveNine
yesterday
@DaveNine, yeah! It seems that if I write the equation of where in the form: $displaystyle frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$, my attempt works. But, I've seen, in other references, the wave equation as in wikipedia, with the same change of variables and the same conclusion. So, I think maybe I should go a different way, but I don't know which one.
– Greg
yesterday
notice the end result looks something like the wave equation you started with originally..which is equal to zero. You either have a algebra mistake somewhere or need to show why your end result is ALSO zero.
– DaveNine
yesterday
notice the end result looks something like the wave equation you started with originally..which is equal to zero. You either have a algebra mistake somewhere or need to show why your end result is ALSO zero.
– DaveNine
yesterday
@DaveNine, yeah! It seems that if I write the equation of where in the form: $displaystyle frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$, my attempt works. But, I've seen, in other references, the wave equation as in wikipedia, with the same change of variables and the same conclusion. So, I think maybe I should go a different way, but I don't know which one.
– Greg
yesterday
@DaveNine, yeah! It seems that if I write the equation of where in the form: $displaystyle frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$, my attempt works. But, I've seen, in other references, the wave equation as in wikipedia, with the same change of variables and the same conclusion. So, I think maybe I should go a different way, but I don't know which one.
– Greg
yesterday
add a comment |
1 Answer
1
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The infinite domain wave equation looks like the following
$$ begin{align}begin{cases}u_{tt} + c^{2} u_{xx} = 0 & x in mathbb{R} , t > 0 \ u(x,0) = g(x) , u_{t}(x,0) = h(x) & x in mathbb{R} end{cases} end{align} tag{1}$$
this is called the d'Alembert solution to the global Cauchy problem.
It says you should have the following change of variables
$$ r = x+ct , s = x-ct tag{2}$$
$$ x = frac{1}{2}(r+s) , t=frac{1}{2c}(r-s) tag{3} $$
$$ frac{partial }{partial r} = frac{partial }{partial x}frac{partial x}{partial r}+frac{partial }{partial t}frac{partial t}{partial r} = frac{1}{2c}bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg) tag{4}$$
$$ frac{partial }{partial s} = frac{partial }{partial x}frac{partial x}{partial s}+frac{partial }{partial t}frac{partial t}{partial s} = frac{1}{2c}bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)tag{5}$$
this becomes
$$ -4c^{2} frac{partial^{2} u}{partial r partial s} = bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg)u = frac{partial^{2} u}{partial t^{2}} - c^{2} frac{partial^{2} u}{partial x^{2}} tag{6} $$
I can simply substitute in symbols here. You use $alpha, beta$ and $x,y$
$$ alpha = x+cy , beta = x-cy tag{7}$$
$$ x = frac{1}{2}(alpha+beta) , y=frac{1}{2c}(beta-alpha) tag{8} $$
answered yesterday
Ryan Howe
2,3841323
It was with this method that I solved (not as well written as your answer is +1). But now I want to solve using only variable change and simple techniques of Real Analysis: partial derivatives, Schwartz Theorem for commutativity of partial derivatives (I'll suppose $u in C^{1}$) and things like that.
– Greg
yesterday
1
math.ubc.ca/~peirce/M257_316_2012_Lecture_21.pdf I think it looks like this on page $4$.
– Ryan Howe
yesterday
This seems very helpful!
– Greg
yesterday
1
i'll edit it in a moment so the answer is just that.
– Ryan Howe
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The infinite domain wave equation looks like the following
$$ begin{align}begin{cases}u_{tt} + c^{2} u_{xx} = 0 & x in mathbb{R} , t > 0 \ u(x,0) = g(x) , u_{t}(x,0) = h(x) & x in mathbb{R} end{cases} end{align} tag{1}$$
this is called the d'Alembert solution to the global Cauchy problem.
It says you should have the following change of variables
$$ r = x+ct , s = x-ct tag{2}$$
$$ x = frac{1}{2}(r+s) , t=frac{1}{2c}(r-s) tag{3} $$
$$ frac{partial }{partial r} = frac{partial }{partial x}frac{partial x}{partial r}+frac{partial }{partial t}frac{partial t}{partial r} = frac{1}{2c}bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg) tag{4}$$
$$ frac{partial }{partial s} = frac{partial }{partial x}frac{partial x}{partial s}+frac{partial }{partial t}frac{partial t}{partial s} = frac{1}{2c}bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)tag{5}$$
this becomes
$$ -4c^{2} frac{partial^{2} u}{partial r partial s} = bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg)u = frac{partial^{2} u}{partial t^{2}} - c^{2} frac{partial^{2} u}{partial x^{2}} tag{6} $$
I can simply substitute in symbols here. You use $alpha, beta$ and $x,y$
$$ alpha = x+cy , beta = x-cy tag{7}$$
$$ x = frac{1}{2}(alpha+beta) , y=frac{1}{2c}(beta-alpha) tag{8} $$
answered yesterday
Ryan Howe
2,3841323
It was with this method that I solved (not as well written as your answer is +1). But now I want to solve using only variable change and simple techniques of Real Analysis: partial derivatives, Schwartz Theorem for commutativity of partial derivatives (I'll suppose $u in C^{1}$) and things like that.
– Greg
yesterday
1
math.ubc.ca/~peirce/M257_316_2012_Lecture_21.pdf I think it looks like this on page $4$.
– Ryan Howe
yesterday
This seems very helpful!
– Greg
yesterday
1
i'll edit it in a moment so the answer is just that.
– Ryan Howe
yesterday
add a comment |
up vote
1
down vote
accepted
The infinite domain wave equation looks like the following
$$ begin{align}begin{cases}u_{tt} + c^{2} u_{xx} = 0 & x in mathbb{R} , t > 0 \ u(x,0) = g(x) , u_{t}(x,0) = h(x) & x in mathbb{R} end{cases} end{align} tag{1}$$
this is called the d'Alembert solution to the global Cauchy problem.
It says you should have the following change of variables
$$ r = x+ct , s = x-ct tag{2}$$
$$ x = frac{1}{2}(r+s) , t=frac{1}{2c}(r-s) tag{3} $$
$$ frac{partial }{partial r} = frac{partial }{partial x}frac{partial x}{partial r}+frac{partial }{partial t}frac{partial t}{partial r} = frac{1}{2c}bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg) tag{4}$$
$$ frac{partial }{partial s} = frac{partial }{partial x}frac{partial x}{partial s}+frac{partial }{partial t}frac{partial t}{partial s} = frac{1}{2c}bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)tag{5}$$
this becomes
$$ -4c^{2} frac{partial^{2} u}{partial r partial s} = bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg)u = frac{partial^{2} u}{partial t^{2}} - c^{2} frac{partial^{2} u}{partial x^{2}} tag{6} $$
I can simply substitute in symbols here. You use $alpha, beta$ and $x,y$
$$ alpha = x+cy , beta = x-cy tag{7}$$
$$ x = frac{1}{2}(alpha+beta) , y=frac{1}{2c}(beta-alpha) tag{8} $$
answered yesterday
Ryan Howe
2,3841323
It was with this method that I solved (not as well written as your answer is +1). But now I want to solve using only variable change and simple techniques of Real Analysis: partial derivatives, Schwartz Theorem for commutativity of partial derivatives (I'll suppose $u in C^{1}$) and things like that.
– Greg
yesterday
1
math.ubc.ca/~peirce/M257_316_2012_Lecture_21.pdf I think it looks like this on page $4$.
– Ryan Howe
yesterday
This seems very helpful!
– Greg
yesterday
1
i'll edit it in a moment so the answer is just that.
– Ryan Howe
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The infinite domain wave equation looks like the following
$$ begin{align}begin{cases}u_{tt} + c^{2} u_{xx} = 0 & x in mathbb{R} , t > 0 \ u(x,0) = g(x) , u_{t}(x,0) = h(x) & x in mathbb{R} end{cases} end{align} tag{1}$$
this is called the d'Alembert solution to the global Cauchy problem.
It says you should have the following change of variables
$$ r = x+ct , s = x-ct tag{2}$$
$$ x = frac{1}{2}(r+s) , t=frac{1}{2c}(r-s) tag{3} $$
$$ frac{partial }{partial r} = frac{partial }{partial x}frac{partial x}{partial r}+frac{partial }{partial t}frac{partial t}{partial r} = frac{1}{2c}bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg) tag{4}$$
$$ frac{partial }{partial s} = frac{partial }{partial x}frac{partial x}{partial s}+frac{partial }{partial t}frac{partial t}{partial s} = frac{1}{2c}bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)tag{5}$$
this becomes
$$ -4c^{2} frac{partial^{2} u}{partial r partial s} = bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg)u = frac{partial^{2} u}{partial t^{2}} - c^{2} frac{partial^{2} u}{partial x^{2}} tag{6} $$
I can simply substitute in symbols here. You use $alpha, beta$ and $x,y$
$$ alpha = x+cy , beta = x-cy tag{7}$$
$$ x = frac{1}{2}(alpha+beta) , y=frac{1}{2c}(beta-alpha) tag{8} $$
answered yesterday
Ryan Howe
2,3841323
The infinite domain wave equation looks like the following
$$ begin{align}begin{cases}u_{tt} + c^{2} u_{xx} = 0 & x in mathbb{R} , t > 0 \ u(x,0) = g(x) , u_{t}(x,0) = h(x) & x in mathbb{R} end{cases} end{align} tag{1}$$
this is called the d'Alembert solution to the global Cauchy problem.
It says you should have the following change of variables
$$ r = x+ct , s = x-ct tag{2}$$
$$ x = frac{1}{2}(r+s) , t=frac{1}{2c}(r-s) tag{3} $$
$$ frac{partial }{partial r} = frac{partial }{partial x}frac{partial x}{partial r}+frac{partial }{partial t}frac{partial t}{partial r} = frac{1}{2c}bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg) tag{4}$$
$$ frac{partial }{partial s} = frac{partial }{partial x}frac{partial x}{partial s}+frac{partial }{partial t}frac{partial t}{partial s} = frac{1}{2c}bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)tag{5}$$
this becomes
$$ -4c^{2} frac{partial^{2} u}{partial r partial s} = bigg( frac{partial }{partial t} - cfrac{partial }{partial x}bigg)bigg( frac{partial }{partial t} + cfrac{partial }{partial x}bigg)u = frac{partial^{2} u}{partial t^{2}} - c^{2} frac{partial^{2} u}{partial x^{2}} tag{6} $$
I can simply substitute in symbols here. You use $alpha, beta$ and $x,y$
$$ alpha = x+cy , beta = x-cy tag{7}$$
$$ x = frac{1}{2}(alpha+beta) , y=frac{1}{2c}(beta-alpha) tag{8} $$
answered yesterday
Ryan Howe
2,3841323
edited yesterday
answered yesterday
Ryan Howe
2,3841323
answered yesterday
Ryan Howe
2,3841323
answered yesterday
Ryan Howe
2,3841323
2,3841323
It was with this method that I solved (not as well written as your answer is +1). But now I want to solve using only variable change and simple techniques of Real Analysis: partial derivatives, Schwartz Theorem for commutativity of partial derivatives (I'll suppose $u in C^{1}$) and things like that.
– Greg
yesterday
1
math.ubc.ca/~peirce/M257_316_2012_Lecture_21.pdf I think it looks like this on page $4$.
– Ryan Howe
yesterday
This seems very helpful!
– Greg
yesterday
1
i'll edit it in a moment so the answer is just that.
– Ryan Howe
yesterday
add a comment |
It was with this method that I solved (not as well written as your answer is +1). But now I want to solve using only variable change and simple techniques of Real Analysis: partial derivatives, Schwartz Theorem for commutativity of partial derivatives (I'll suppose $u in C^{1}$) and things like that.
– Greg
yesterday
1
math.ubc.ca/~peirce/M257_316_2012_Lecture_21.pdf I think it looks like this on page $4$.
– Ryan Howe
yesterday
This seems very helpful!
– Greg
yesterday
1
i'll edit it in a moment so the answer is just that.
– Ryan Howe
yesterday
It was with this method that I solved (not as well written as your answer is +1). But now I want to solve using only variable change and simple techniques of Real Analysis: partial derivatives, Schwartz Theorem for commutativity of partial derivatives (I'll suppose $u in C^{1}$) and things like that.
– Greg
yesterday
It was with this method that I solved (not as well written as your answer is +1). But now I want to solve using only variable change and simple techniques of Real Analysis: partial derivatives, Schwartz Theorem for commutativity of partial derivatives (I'll suppose $u in C^{1}$) and things like that.
– Greg
yesterday
1
1
math.ubc.ca/~peirce/M257_316_2012_Lecture_21.pdf I think it looks like this on page $4$.
– Ryan Howe
yesterday
math.ubc.ca/~peirce/M257_316_2012_Lecture_21.pdf I think it looks like this on page $4$.
– Ryan Howe
yesterday
This seems very helpful!
– Greg
yesterday
This seems very helpful!
– Greg
yesterday
1
1
i'll edit it in a moment so the answer is just that.
– Ryan Howe
yesterday
i'll edit it in a moment so the answer is just that.
– Ryan Howe
yesterday
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notice the end result looks something like the wave equation you started with originally..which is equal to zero. You either have a algebra mistake somewhere or need to show why your end result is ALSO zero.
– DaveNine
yesterday
@DaveNine, yeah! It seems that if I write the equation of where in the form: $displaystyle frac{partial^{2} u}{partial x^{2}} - frac{1}{c^{2}}frac{partial^{2} u}{partial y^{2}}$, my attempt works. But, I've seen, in other references, the wave equation as in wikipedia, with the same change of variables and the same conclusion. So, I think maybe I should go a different way, but I don't know which one.
– Greg
yesterday