$left{begin{array}{lll} f_{t}+xf_{y}=0\ f|_{t=0}=f_{0}(x,y) end{array}right.$
Given $fin C^{1}({bf R}^{3})$ , then find the solution of the following pde :
$$left{begin{array}{lll}
f_{t}+xf_{y}=0\
f|_{t=0}=f_{0}(x,y)
end{array}right.$$
I tried to use the characteristic method to solve this question , but I failed . Any comment or advice I will be grateful .
differential-equations analysis pde characteristics
add a comment |
Given $fin C^{1}({bf R}^{3})$ , then find the solution of the following pde :
$$left{begin{array}{lll}
f_{t}+xf_{y}=0\
f|_{t=0}=f_{0}(x,y)
end{array}right.$$
I tried to use the characteristic method to solve this question , but I failed . Any comment or advice I will be grateful .
differential-equations analysis pde characteristics
add a comment |
Given $fin C^{1}({bf R}^{3})$ , then find the solution of the following pde :
$$left{begin{array}{lll}
f_{t}+xf_{y}=0\
f|_{t=0}=f_{0}(x,y)
end{array}right.$$
I tried to use the characteristic method to solve this question , but I failed . Any comment or advice I will be grateful .
differential-equations analysis pde characteristics
Given $fin C^{1}({bf R}^{3})$ , then find the solution of the following pde :
$$left{begin{array}{lll}
f_{t}+xf_{y}=0\
f|_{t=0}=f_{0}(x,y)
end{array}right.$$
I tried to use the characteristic method to solve this question , but I failed . Any comment or advice I will be grateful .
differential-equations analysis pde characteristics
differential-equations analysis pde characteristics
edited May 9 at 9:50
LutzL
55.8k42054
55.8k42054
asked Feb 3 at 7:18
user1992
6961510
6961510
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$$left{begin{array}{lll} f_{t}+xf_{y}=0\ f|_{t=0}=f_{0}(x,y) end{array}right.$$
The characteristic ODEs are : $quad frac{dt}{1}=frac{dy}{x}=frac{dx}{0}=frac{df}{0}$
A first family of characteristic curves comes from : $quad frac{dt}{1}=frac{dy}{x} quadimpliesquad y-xt=c_1$
A second family of characteristic curves comes from $quadfrac{dx}{0}=$finite$quadimpliesquad x=c_2$
A third family of characteristic curves comes from $quadfrac{df}{0}=$finite$quadimpliesquad f=c_3$
Thus, the general solution of the PDE, expressed on the form of implicite equation is :
$$Phileft(x:,: (t-xt) :,: f right)=0$$
where $Phi$ is any differentiable function of three variables.
( Doesn't matter the order of the variables since the function $Phi$ is arbitrary ).
Or alternatively, on explicit form :
$$f(x,y,t)=Fleft(x:,:(y-xt) right)$$
where $F$ is any differentiable function of two variables.
CONDITION :
$f(x,y,0)=f_0(x,y)=Fleft(x:,:(y-0) right) quadimpliesquad Fleft(x:,:y right)=f_0(x,y)$
Now, the function $F$ is determined. We put it into the above general solution where the second variable is $(y-xt)$. This leads to :
$$f(x,y,t)=f_0left(x:,:(y-xt) right)$$
Now , I understood . This explaining is clear and sound .
– user1992
Feb 3 at 9:50
You can solve the first equation in this way only after determining that $x$ is constant, as with $x$ a function the integration $y=int x,dt$ is not trivial.
– LutzL
Feb 3 at 10:46
On the characteristic we are looking for $x=c_2$. If one prefers, the above first characteristic can be introduced after the second. Doesn't matter the order.
– JJacquelin
Feb 3 at 11:38
add a comment |
The method of characteristics should tell you that the characteristic curves
$$
t(s),, x(s),,y(s),,z(s)=f(t(s),x(s),y(s))
$$
follow the ODE (up to re-parametrization)
$$
frac{dt}{ds}=1,, frac{dx}{dt}=0,, frac{dy}{dt}=x,, frac{dz}{ds}=0
$$
which has solutions
$$t=s+t_0,, x=x_0,, y=y_0+x_0s,, z=z_0$$ so that for all $s$ (discounting shocks etc., i.e., in a neighborhood of $s=0$)
begin{align}
f(t,x,y)&=z(s)=f(t_0+s,x_0,y_0+x_0s)\
&=z(0)=f(0,x_0,y_0)=f(0,x,y-xt)\
&=f_0(x,y-xt).
end{align}
where obviously the initial values where chosen to conform to the given boundary condition, that is, $t_0=0$ so that $t=s$, $x_0=x$, $y_0=y-x_0s=y-xt$.
Can I ask $t(x)~,x(s),~y(s),~z(s)=f(t(s),x(s),y(s))~$ ,what does it mean ?
– user1992
Feb 3 at 8:48
This is the parametrization of a characteristic curve of the problem. One needs to distinguish at least the function $z(s)$ from the function $f(t,x,y)$, to be perfectly correct one would also have to use different symbols for the free variables $t,x,y$ and the functions $t(s),x(s),y(s)$, but this kind of "abuse of notation" is usual in this situation.
– LutzL
Feb 3 at 10:36
add a comment |
If the PDE was just $f_{t} + f_{y} = 0$, then you would make a travelling wave ansatz $f(x,y,t) = f(y-t)$. As the derivative in $y$ has a non-constant coefficient $x$, it stands to reason that when differentiating our solution with respect to $t$, we should get an $x$ out the front. Hence, we can guess that $$f(x,y,t) = f_{0}(x, y - xt)$$ solves the problem, which you can check by differentiation it does.
Alternatively, make the guess $f(x,y,t) = g(alpha y + gamma t)$ which, after differentiation, yields
begin{align}
f_{t} &= gamma g' \
f_{y} &= alpha g' \
therefore f_{t} + x f_{y} &= 0 \
implies gamma g' + x alpha g' &= 0 \
implies gamma &= -x, quad alpha = 1 \
implies f(x,y,t) &= g(y - xt)
end{align}
Applying the initial condition yields
begin{align}
f(x,y,0) &= g(y) \
&= f_{0}(x,y) \
implies f(x,y,t) &= f_{0}(x, y - xt)
end{align}
yes , the function $f_0(y-xt)$ actually works .but I have no ideal that why you guess another function $g(alpha y+gamma t)$ to deduce the solution $f(x,y,t) = g(y-xt)$
– user1992
Feb 3 at 8:23
When I said alternatively, I meant that instead of using my first line of reasoning ('it stands to reason ...'), you can make an arbitrary wave ansatz and derive the values of $alpha, gamma$ that satisfy the PDE.
– Mattos
Feb 3 at 11:39
add a comment |
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$$left{begin{array}{lll} f_{t}+xf_{y}=0\ f|_{t=0}=f_{0}(x,y) end{array}right.$$
The characteristic ODEs are : $quad frac{dt}{1}=frac{dy}{x}=frac{dx}{0}=frac{df}{0}$
A first family of characteristic curves comes from : $quad frac{dt}{1}=frac{dy}{x} quadimpliesquad y-xt=c_1$
A second family of characteristic curves comes from $quadfrac{dx}{0}=$finite$quadimpliesquad x=c_2$
A third family of characteristic curves comes from $quadfrac{df}{0}=$finite$quadimpliesquad f=c_3$
Thus, the general solution of the PDE, expressed on the form of implicite equation is :
$$Phileft(x:,: (t-xt) :,: f right)=0$$
where $Phi$ is any differentiable function of three variables.
( Doesn't matter the order of the variables since the function $Phi$ is arbitrary ).
Or alternatively, on explicit form :
$$f(x,y,t)=Fleft(x:,:(y-xt) right)$$
where $F$ is any differentiable function of two variables.
CONDITION :
$f(x,y,0)=f_0(x,y)=Fleft(x:,:(y-0) right) quadimpliesquad Fleft(x:,:y right)=f_0(x,y)$
Now, the function $F$ is determined. We put it into the above general solution where the second variable is $(y-xt)$. This leads to :
$$f(x,y,t)=f_0left(x:,:(y-xt) right)$$
Now , I understood . This explaining is clear and sound .
– user1992
Feb 3 at 9:50
You can solve the first equation in this way only after determining that $x$ is constant, as with $x$ a function the integration $y=int x,dt$ is not trivial.
– LutzL
Feb 3 at 10:46
On the characteristic we are looking for $x=c_2$. If one prefers, the above first characteristic can be introduced after the second. Doesn't matter the order.
– JJacquelin
Feb 3 at 11:38
add a comment |
$$left{begin{array}{lll} f_{t}+xf_{y}=0\ f|_{t=0}=f_{0}(x,y) end{array}right.$$
The characteristic ODEs are : $quad frac{dt}{1}=frac{dy}{x}=frac{dx}{0}=frac{df}{0}$
A first family of characteristic curves comes from : $quad frac{dt}{1}=frac{dy}{x} quadimpliesquad y-xt=c_1$
A second family of characteristic curves comes from $quadfrac{dx}{0}=$finite$quadimpliesquad x=c_2$
A third family of characteristic curves comes from $quadfrac{df}{0}=$finite$quadimpliesquad f=c_3$
Thus, the general solution of the PDE, expressed on the form of implicite equation is :
$$Phileft(x:,: (t-xt) :,: f right)=0$$
where $Phi$ is any differentiable function of three variables.
( Doesn't matter the order of the variables since the function $Phi$ is arbitrary ).
Or alternatively, on explicit form :
$$f(x,y,t)=Fleft(x:,:(y-xt) right)$$
where $F$ is any differentiable function of two variables.
CONDITION :
$f(x,y,0)=f_0(x,y)=Fleft(x:,:(y-0) right) quadimpliesquad Fleft(x:,:y right)=f_0(x,y)$
Now, the function $F$ is determined. We put it into the above general solution where the second variable is $(y-xt)$. This leads to :
$$f(x,y,t)=f_0left(x:,:(y-xt) right)$$
Now , I understood . This explaining is clear and sound .
– user1992
Feb 3 at 9:50
You can solve the first equation in this way only after determining that $x$ is constant, as with $x$ a function the integration $y=int x,dt$ is not trivial.
– LutzL
Feb 3 at 10:46
On the characteristic we are looking for $x=c_2$. If one prefers, the above first characteristic can be introduced after the second. Doesn't matter the order.
– JJacquelin
Feb 3 at 11:38
add a comment |
$$left{begin{array}{lll} f_{t}+xf_{y}=0\ f|_{t=0}=f_{0}(x,y) end{array}right.$$
The characteristic ODEs are : $quad frac{dt}{1}=frac{dy}{x}=frac{dx}{0}=frac{df}{0}$
A first family of characteristic curves comes from : $quad frac{dt}{1}=frac{dy}{x} quadimpliesquad y-xt=c_1$
A second family of characteristic curves comes from $quadfrac{dx}{0}=$finite$quadimpliesquad x=c_2$
A third family of characteristic curves comes from $quadfrac{df}{0}=$finite$quadimpliesquad f=c_3$
Thus, the general solution of the PDE, expressed on the form of implicite equation is :
$$Phileft(x:,: (t-xt) :,: f right)=0$$
where $Phi$ is any differentiable function of three variables.
( Doesn't matter the order of the variables since the function $Phi$ is arbitrary ).
Or alternatively, on explicit form :
$$f(x,y,t)=Fleft(x:,:(y-xt) right)$$
where $F$ is any differentiable function of two variables.
CONDITION :
$f(x,y,0)=f_0(x,y)=Fleft(x:,:(y-0) right) quadimpliesquad Fleft(x:,:y right)=f_0(x,y)$
Now, the function $F$ is determined. We put it into the above general solution where the second variable is $(y-xt)$. This leads to :
$$f(x,y,t)=f_0left(x:,:(y-xt) right)$$
$$left{begin{array}{lll} f_{t}+xf_{y}=0\ f|_{t=0}=f_{0}(x,y) end{array}right.$$
The characteristic ODEs are : $quad frac{dt}{1}=frac{dy}{x}=frac{dx}{0}=frac{df}{0}$
A first family of characteristic curves comes from : $quad frac{dt}{1}=frac{dy}{x} quadimpliesquad y-xt=c_1$
A second family of characteristic curves comes from $quadfrac{dx}{0}=$finite$quadimpliesquad x=c_2$
A third family of characteristic curves comes from $quadfrac{df}{0}=$finite$quadimpliesquad f=c_3$
Thus, the general solution of the PDE, expressed on the form of implicite equation is :
$$Phileft(x:,: (t-xt) :,: f right)=0$$
where $Phi$ is any differentiable function of three variables.
( Doesn't matter the order of the variables since the function $Phi$ is arbitrary ).
Or alternatively, on explicit form :
$$f(x,y,t)=Fleft(x:,:(y-xt) right)$$
where $F$ is any differentiable function of two variables.
CONDITION :
$f(x,y,0)=f_0(x,y)=Fleft(x:,:(y-0) right) quadimpliesquad Fleft(x:,:y right)=f_0(x,y)$
Now, the function $F$ is determined. We put it into the above general solution where the second variable is $(y-xt)$. This leads to :
$$f(x,y,t)=f_0left(x:,:(y-xt) right)$$
answered Feb 3 at 9:41
JJacquelin
42.5k21750
42.5k21750
Now , I understood . This explaining is clear and sound .
– user1992
Feb 3 at 9:50
You can solve the first equation in this way only after determining that $x$ is constant, as with $x$ a function the integration $y=int x,dt$ is not trivial.
– LutzL
Feb 3 at 10:46
On the characteristic we are looking for $x=c_2$. If one prefers, the above first characteristic can be introduced after the second. Doesn't matter the order.
– JJacquelin
Feb 3 at 11:38
add a comment |
Now , I understood . This explaining is clear and sound .
– user1992
Feb 3 at 9:50
You can solve the first equation in this way only after determining that $x$ is constant, as with $x$ a function the integration $y=int x,dt$ is not trivial.
– LutzL
Feb 3 at 10:46
On the characteristic we are looking for $x=c_2$. If one prefers, the above first characteristic can be introduced after the second. Doesn't matter the order.
– JJacquelin
Feb 3 at 11:38
Now , I understood . This explaining is clear and sound .
– user1992
Feb 3 at 9:50
Now , I understood . This explaining is clear and sound .
– user1992
Feb 3 at 9:50
You can solve the first equation in this way only after determining that $x$ is constant, as with $x$ a function the integration $y=int x,dt$ is not trivial.
– LutzL
Feb 3 at 10:46
You can solve the first equation in this way only after determining that $x$ is constant, as with $x$ a function the integration $y=int x,dt$ is not trivial.
– LutzL
Feb 3 at 10:46
On the characteristic we are looking for $x=c_2$. If one prefers, the above first characteristic can be introduced after the second. Doesn't matter the order.
– JJacquelin
Feb 3 at 11:38
On the characteristic we are looking for $x=c_2$. If one prefers, the above first characteristic can be introduced after the second. Doesn't matter the order.
– JJacquelin
Feb 3 at 11:38
add a comment |
The method of characteristics should tell you that the characteristic curves
$$
t(s),, x(s),,y(s),,z(s)=f(t(s),x(s),y(s))
$$
follow the ODE (up to re-parametrization)
$$
frac{dt}{ds}=1,, frac{dx}{dt}=0,, frac{dy}{dt}=x,, frac{dz}{ds}=0
$$
which has solutions
$$t=s+t_0,, x=x_0,, y=y_0+x_0s,, z=z_0$$ so that for all $s$ (discounting shocks etc., i.e., in a neighborhood of $s=0$)
begin{align}
f(t,x,y)&=z(s)=f(t_0+s,x_0,y_0+x_0s)\
&=z(0)=f(0,x_0,y_0)=f(0,x,y-xt)\
&=f_0(x,y-xt).
end{align}
where obviously the initial values where chosen to conform to the given boundary condition, that is, $t_0=0$ so that $t=s$, $x_0=x$, $y_0=y-x_0s=y-xt$.
Can I ask $t(x)~,x(s),~y(s),~z(s)=f(t(s),x(s),y(s))~$ ,what does it mean ?
– user1992
Feb 3 at 8:48
This is the parametrization of a characteristic curve of the problem. One needs to distinguish at least the function $z(s)$ from the function $f(t,x,y)$, to be perfectly correct one would also have to use different symbols for the free variables $t,x,y$ and the functions $t(s),x(s),y(s)$, but this kind of "abuse of notation" is usual in this situation.
– LutzL
Feb 3 at 10:36
add a comment |
The method of characteristics should tell you that the characteristic curves
$$
t(s),, x(s),,y(s),,z(s)=f(t(s),x(s),y(s))
$$
follow the ODE (up to re-parametrization)
$$
frac{dt}{ds}=1,, frac{dx}{dt}=0,, frac{dy}{dt}=x,, frac{dz}{ds}=0
$$
which has solutions
$$t=s+t_0,, x=x_0,, y=y_0+x_0s,, z=z_0$$ so that for all $s$ (discounting shocks etc., i.e., in a neighborhood of $s=0$)
begin{align}
f(t,x,y)&=z(s)=f(t_0+s,x_0,y_0+x_0s)\
&=z(0)=f(0,x_0,y_0)=f(0,x,y-xt)\
&=f_0(x,y-xt).
end{align}
where obviously the initial values where chosen to conform to the given boundary condition, that is, $t_0=0$ so that $t=s$, $x_0=x$, $y_0=y-x_0s=y-xt$.
Can I ask $t(x)~,x(s),~y(s),~z(s)=f(t(s),x(s),y(s))~$ ,what does it mean ?
– user1992
Feb 3 at 8:48
This is the parametrization of a characteristic curve of the problem. One needs to distinguish at least the function $z(s)$ from the function $f(t,x,y)$, to be perfectly correct one would also have to use different symbols for the free variables $t,x,y$ and the functions $t(s),x(s),y(s)$, but this kind of "abuse of notation" is usual in this situation.
– LutzL
Feb 3 at 10:36
add a comment |
The method of characteristics should tell you that the characteristic curves
$$
t(s),, x(s),,y(s),,z(s)=f(t(s),x(s),y(s))
$$
follow the ODE (up to re-parametrization)
$$
frac{dt}{ds}=1,, frac{dx}{dt}=0,, frac{dy}{dt}=x,, frac{dz}{ds}=0
$$
which has solutions
$$t=s+t_0,, x=x_0,, y=y_0+x_0s,, z=z_0$$ so that for all $s$ (discounting shocks etc., i.e., in a neighborhood of $s=0$)
begin{align}
f(t,x,y)&=z(s)=f(t_0+s,x_0,y_0+x_0s)\
&=z(0)=f(0,x_0,y_0)=f(0,x,y-xt)\
&=f_0(x,y-xt).
end{align}
where obviously the initial values where chosen to conform to the given boundary condition, that is, $t_0=0$ so that $t=s$, $x_0=x$, $y_0=y-x_0s=y-xt$.
The method of characteristics should tell you that the characteristic curves
$$
t(s),, x(s),,y(s),,z(s)=f(t(s),x(s),y(s))
$$
follow the ODE (up to re-parametrization)
$$
frac{dt}{ds}=1,, frac{dx}{dt}=0,, frac{dy}{dt}=x,, frac{dz}{ds}=0
$$
which has solutions
$$t=s+t_0,, x=x_0,, y=y_0+x_0s,, z=z_0$$ so that for all $s$ (discounting shocks etc., i.e., in a neighborhood of $s=0$)
begin{align}
f(t,x,y)&=z(s)=f(t_0+s,x_0,y_0+x_0s)\
&=z(0)=f(0,x_0,y_0)=f(0,x,y-xt)\
&=f_0(x,y-xt).
end{align}
where obviously the initial values where chosen to conform to the given boundary condition, that is, $t_0=0$ so that $t=s$, $x_0=x$, $y_0=y-x_0s=y-xt$.
edited Feb 3 at 10:43
answered Feb 3 at 8:42
LutzL
55.8k42054
55.8k42054
Can I ask $t(x)~,x(s),~y(s),~z(s)=f(t(s),x(s),y(s))~$ ,what does it mean ?
– user1992
Feb 3 at 8:48
This is the parametrization of a characteristic curve of the problem. One needs to distinguish at least the function $z(s)$ from the function $f(t,x,y)$, to be perfectly correct one would also have to use different symbols for the free variables $t,x,y$ and the functions $t(s),x(s),y(s)$, but this kind of "abuse of notation" is usual in this situation.
– LutzL
Feb 3 at 10:36
add a comment |
Can I ask $t(x)~,x(s),~y(s),~z(s)=f(t(s),x(s),y(s))~$ ,what does it mean ?
– user1992
Feb 3 at 8:48
This is the parametrization of a characteristic curve of the problem. One needs to distinguish at least the function $z(s)$ from the function $f(t,x,y)$, to be perfectly correct one would also have to use different symbols for the free variables $t,x,y$ and the functions $t(s),x(s),y(s)$, but this kind of "abuse of notation" is usual in this situation.
– LutzL
Feb 3 at 10:36
Can I ask $t(x)~,x(s),~y(s),~z(s)=f(t(s),x(s),y(s))~$ ,what does it mean ?
– user1992
Feb 3 at 8:48
Can I ask $t(x)~,x(s),~y(s),~z(s)=f(t(s),x(s),y(s))~$ ,what does it mean ?
– user1992
Feb 3 at 8:48
This is the parametrization of a characteristic curve of the problem. One needs to distinguish at least the function $z(s)$ from the function $f(t,x,y)$, to be perfectly correct one would also have to use different symbols for the free variables $t,x,y$ and the functions $t(s),x(s),y(s)$, but this kind of "abuse of notation" is usual in this situation.
– LutzL
Feb 3 at 10:36
This is the parametrization of a characteristic curve of the problem. One needs to distinguish at least the function $z(s)$ from the function $f(t,x,y)$, to be perfectly correct one would also have to use different symbols for the free variables $t,x,y$ and the functions $t(s),x(s),y(s)$, but this kind of "abuse of notation" is usual in this situation.
– LutzL
Feb 3 at 10:36
add a comment |
If the PDE was just $f_{t} + f_{y} = 0$, then you would make a travelling wave ansatz $f(x,y,t) = f(y-t)$. As the derivative in $y$ has a non-constant coefficient $x$, it stands to reason that when differentiating our solution with respect to $t$, we should get an $x$ out the front. Hence, we can guess that $$f(x,y,t) = f_{0}(x, y - xt)$$ solves the problem, which you can check by differentiation it does.
Alternatively, make the guess $f(x,y,t) = g(alpha y + gamma t)$ which, after differentiation, yields
begin{align}
f_{t} &= gamma g' \
f_{y} &= alpha g' \
therefore f_{t} + x f_{y} &= 0 \
implies gamma g' + x alpha g' &= 0 \
implies gamma &= -x, quad alpha = 1 \
implies f(x,y,t) &= g(y - xt)
end{align}
Applying the initial condition yields
begin{align}
f(x,y,0) &= g(y) \
&= f_{0}(x,y) \
implies f(x,y,t) &= f_{0}(x, y - xt)
end{align}
yes , the function $f_0(y-xt)$ actually works .but I have no ideal that why you guess another function $g(alpha y+gamma t)$ to deduce the solution $f(x,y,t) = g(y-xt)$
– user1992
Feb 3 at 8:23
When I said alternatively, I meant that instead of using my first line of reasoning ('it stands to reason ...'), you can make an arbitrary wave ansatz and derive the values of $alpha, gamma$ that satisfy the PDE.
– Mattos
Feb 3 at 11:39
add a comment |
If the PDE was just $f_{t} + f_{y} = 0$, then you would make a travelling wave ansatz $f(x,y,t) = f(y-t)$. As the derivative in $y$ has a non-constant coefficient $x$, it stands to reason that when differentiating our solution with respect to $t$, we should get an $x$ out the front. Hence, we can guess that $$f(x,y,t) = f_{0}(x, y - xt)$$ solves the problem, which you can check by differentiation it does.
Alternatively, make the guess $f(x,y,t) = g(alpha y + gamma t)$ which, after differentiation, yields
begin{align}
f_{t} &= gamma g' \
f_{y} &= alpha g' \
therefore f_{t} + x f_{y} &= 0 \
implies gamma g' + x alpha g' &= 0 \
implies gamma &= -x, quad alpha = 1 \
implies f(x,y,t) &= g(y - xt)
end{align}
Applying the initial condition yields
begin{align}
f(x,y,0) &= g(y) \
&= f_{0}(x,y) \
implies f(x,y,t) &= f_{0}(x, y - xt)
end{align}
yes , the function $f_0(y-xt)$ actually works .but I have no ideal that why you guess another function $g(alpha y+gamma t)$ to deduce the solution $f(x,y,t) = g(y-xt)$
– user1992
Feb 3 at 8:23
When I said alternatively, I meant that instead of using my first line of reasoning ('it stands to reason ...'), you can make an arbitrary wave ansatz and derive the values of $alpha, gamma$ that satisfy the PDE.
– Mattos
Feb 3 at 11:39
add a comment |
If the PDE was just $f_{t} + f_{y} = 0$, then you would make a travelling wave ansatz $f(x,y,t) = f(y-t)$. As the derivative in $y$ has a non-constant coefficient $x$, it stands to reason that when differentiating our solution with respect to $t$, we should get an $x$ out the front. Hence, we can guess that $$f(x,y,t) = f_{0}(x, y - xt)$$ solves the problem, which you can check by differentiation it does.
Alternatively, make the guess $f(x,y,t) = g(alpha y + gamma t)$ which, after differentiation, yields
begin{align}
f_{t} &= gamma g' \
f_{y} &= alpha g' \
therefore f_{t} + x f_{y} &= 0 \
implies gamma g' + x alpha g' &= 0 \
implies gamma &= -x, quad alpha = 1 \
implies f(x,y,t) &= g(y - xt)
end{align}
Applying the initial condition yields
begin{align}
f(x,y,0) &= g(y) \
&= f_{0}(x,y) \
implies f(x,y,t) &= f_{0}(x, y - xt)
end{align}
If the PDE was just $f_{t} + f_{y} = 0$, then you would make a travelling wave ansatz $f(x,y,t) = f(y-t)$. As the derivative in $y$ has a non-constant coefficient $x$, it stands to reason that when differentiating our solution with respect to $t$, we should get an $x$ out the front. Hence, we can guess that $$f(x,y,t) = f_{0}(x, y - xt)$$ solves the problem, which you can check by differentiation it does.
Alternatively, make the guess $f(x,y,t) = g(alpha y + gamma t)$ which, after differentiation, yields
begin{align}
f_{t} &= gamma g' \
f_{y} &= alpha g' \
therefore f_{t} + x f_{y} &= 0 \
implies gamma g' + x alpha g' &= 0 \
implies gamma &= -x, quad alpha = 1 \
implies f(x,y,t) &= g(y - xt)
end{align}
Applying the initial condition yields
begin{align}
f(x,y,0) &= g(y) \
&= f_{0}(x,y) \
implies f(x,y,t) &= f_{0}(x, y - xt)
end{align}
edited Nov 26 at 3:17
answered Feb 3 at 7:54
Mattos
2,73021321
2,73021321
yes , the function $f_0(y-xt)$ actually works .but I have no ideal that why you guess another function $g(alpha y+gamma t)$ to deduce the solution $f(x,y,t) = g(y-xt)$
– user1992
Feb 3 at 8:23
When I said alternatively, I meant that instead of using my first line of reasoning ('it stands to reason ...'), you can make an arbitrary wave ansatz and derive the values of $alpha, gamma$ that satisfy the PDE.
– Mattos
Feb 3 at 11:39
add a comment |
yes , the function $f_0(y-xt)$ actually works .but I have no ideal that why you guess another function $g(alpha y+gamma t)$ to deduce the solution $f(x,y,t) = g(y-xt)$
– user1992
Feb 3 at 8:23
When I said alternatively, I meant that instead of using my first line of reasoning ('it stands to reason ...'), you can make an arbitrary wave ansatz and derive the values of $alpha, gamma$ that satisfy the PDE.
– Mattos
Feb 3 at 11:39
yes , the function $f_0(y-xt)$ actually works .but I have no ideal that why you guess another function $g(alpha y+gamma t)$ to deduce the solution $f(x,y,t) = g(y-xt)$
– user1992
Feb 3 at 8:23
yes , the function $f_0(y-xt)$ actually works .but I have no ideal that why you guess another function $g(alpha y+gamma t)$ to deduce the solution $f(x,y,t) = g(y-xt)$
– user1992
Feb 3 at 8:23
When I said alternatively, I meant that instead of using my first line of reasoning ('it stands to reason ...'), you can make an arbitrary wave ansatz and derive the values of $alpha, gamma$ that satisfy the PDE.
– Mattos
Feb 3 at 11:39
When I said alternatively, I meant that instead of using my first line of reasoning ('it stands to reason ...'), you can make an arbitrary wave ansatz and derive the values of $alpha, gamma$ that satisfy the PDE.
– Mattos
Feb 3 at 11:39
add a comment |
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