Confused about Burgers' Equation with an inhomogeneous RHS
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I am having trouble solving a question and would appreciate some help. The question is:
Solve the PDE:
$$u_t+ucdot u_x=1, >>>u(x,0)=x$$
for $tgeq0$.
Now I know that the Characteristic Equations are given by:
begin{align*}
frac{partial x}{partial t} &= u(t)\
frac{partial z}{partial t} &= 1
end{align*}
I also made auxiliary conditions:
begin{align*}
x(0) &= x_0\
z(0) &= z_0 = g(x_0)
end{align*}
Thus integrating and using auxiliary conditions, I solved that:
$$z(t)=t+z_0, >> x(t) = g(x_0)cdot t+frac{t^2}{2}+x_0$$
From there I solved for $g(x_0)$:
$$g(x_0) = frac{x(t)}{t}-frac{x_0}{t}-frac{t}{2}$$
By the auxiliary condition $u(x,0)=x$, $g(x_0)=x_0$, and with this in mind I solved for $x_0$:
$$x_0 = frac{x}{t+1}-frac{t^2}{2(t+1)}$$
So I deduced that $$u(x,t) = frac{2x-t^2}{2(t+1)}$$
Did I go about this correctly? Or did I make a mistake somewhere?
pde
edited Dec 5 at 17:03
Harry49
5,90421030
asked Sep 15 '17 at 1:31
Felicio Grande
499619
add a comment |
up vote
1
down vote
favorite
I am having trouble solving a question and would appreciate some help. The question is:
Solve the PDE:
$$u_t+ucdot u_x=1, >>>u(x,0)=x$$
for $tgeq0$.
Now I know that the Characteristic Equations are given by:
begin{align*}
frac{partial x}{partial t} &= u(t)\
frac{partial z}{partial t} &= 1
end{align*}
I also made auxiliary conditions:
begin{align*}
x(0) &= x_0\
z(0) &= z_0 = g(x_0)
end{align*}
Thus integrating and using auxiliary conditions, I solved that:
$$z(t)=t+z_0, >> x(t) = g(x_0)cdot t+frac{t^2}{2}+x_0$$
From there I solved for $g(x_0)$:
$$g(x_0) = frac{x(t)}{t}-frac{x_0}{t}-frac{t}{2}$$
By the auxiliary condition $u(x,0)=x$, $g(x_0)=x_0$, and with this in mind I solved for $x_0$:
$$x_0 = frac{x}{t+1}-frac{t^2}{2(t+1)}$$
So I deduced that $$u(x,t) = frac{2x-t^2}{2(t+1)}$$
Did I go about this correctly? Or did I make a mistake somewhere?
pde
edited Dec 5 at 17:03
Harry49
5,90421030
asked Sep 15 '17 at 1:31
Felicio Grande
499619
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am having trouble solving a question and would appreciate some help. The question is:
Solve the PDE:
$$u_t+ucdot u_x=1, >>>u(x,0)=x$$
for $tgeq0$.
Now I know that the Characteristic Equations are given by:
begin{align*}
frac{partial x}{partial t} &= u(t)\
frac{partial z}{partial t} &= 1
end{align*}
I also made auxiliary conditions:
begin{align*}
x(0) &= x_0\
z(0) &= z_0 = g(x_0)
end{align*}
Thus integrating and using auxiliary conditions, I solved that:
$$z(t)=t+z_0, >> x(t) = g(x_0)cdot t+frac{t^2}{2}+x_0$$
From there I solved for $g(x_0)$:
$$g(x_0) = frac{x(t)}{t}-frac{x_0}{t}-frac{t}{2}$$
By the auxiliary condition $u(x,0)=x$, $g(x_0)=x_0$, and with this in mind I solved for $x_0$:
$$x_0 = frac{x}{t+1}-frac{t^2}{2(t+1)}$$
So I deduced that $$u(x,t) = frac{2x-t^2}{2(t+1)}$$
Did I go about this correctly? Or did I make a mistake somewhere?
pde
edited Dec 5 at 17:03
Harry49
5,90421030
asked Sep 15 '17 at 1:31
Felicio Grande
499619
I am having trouble solving a question and would appreciate some help. The question is:
Solve the PDE:
$$u_t+ucdot u_x=1, >>>u(x,0)=x$$
for $tgeq0$.
Now I know that the Characteristic Equations are given by:
begin{align*}
frac{partial x}{partial t} &= u(t)\
frac{partial z}{partial t} &= 1
end{align*}
I also made auxiliary conditions:
begin{align*}
x(0) &= x_0\
z(0) &= z_0 = g(x_0)
end{align*}
Thus integrating and using auxiliary conditions, I solved that:
$$z(t)=t+z_0, >> x(t) = g(x_0)cdot t+frac{t^2}{2}+x_0$$
From there I solved for $g(x_0)$:
$$g(x_0) = frac{x(t)}{t}-frac{x_0}{t}-frac{t}{2}$$
By the auxiliary condition $u(x,0)=x$, $g(x_0)=x_0$, and with this in mind I solved for $x_0$:
$$x_0 = frac{x}{t+1}-frac{t^2}{2(t+1)}$$
So I deduced that $$u(x,t) = frac{2x-t^2}{2(t+1)}$$
Did I go about this correctly? Or did I make a mistake somewhere?
pde
pde
edited Dec 5 at 17:03
Harry49
5,90421030
asked Sep 15 '17 at 1:31
Felicio Grande
499619
edited Dec 5 at 17:03
Harry49
5,90421030
asked Sep 15 '17 at 1:31
Felicio Grande
499619
edited Dec 5 at 17:03
Harry49
5,90421030
edited Dec 5 at 17:03
Harry49
5,90421030
edited Dec 5 at 17:03
Harry49
5,90421030
5,90421030
asked Sep 15 '17 at 1:31
Felicio Grande
499619
asked Sep 15 '17 at 1:31
Felicio Grande
499619
asked Sep 15 '17 at 1:31
Felicio Grande
499619
499619
add a comment |
add a comment |
1 Answer
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1
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You found the characteristic curves correctly: they are the parabolas
$$
x = x_0+x_0t + frac{t^2}{2}
$$
So, the characteristic through $(x,t)$ originates at $x_0 = dfrac{2x-t^2}{2(t+1)}$, as you also found. The mistake is in concluding that $u(x,t) = x_0$. This would be true for homogeneous PDE. But the nonhomogeneity means that $u$ (represented by $z$ in the characteristic equation) changes along the characteristic. Specifically, $z=x_0+t$. So the final answer is
$$
u(x,t) = x_0 + t = dfrac{2x-t^2}{2(t+1)} + t = dfrac{2x+t+t^2}{2(t+1)}
$$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You found the characteristic curves correctly: they are the parabolas
$$
x = x_0+x_0t + frac{t^2}{2}
$$
So, the characteristic through $(x,t)$ originates at $x_0 = dfrac{2x-t^2}{2(t+1)}$, as you also found. The mistake is in concluding that $u(x,t) = x_0$. This would be true for homogeneous PDE. But the nonhomogeneity means that $u$ (represented by $z$ in the characteristic equation) changes along the characteristic. Specifically, $z=x_0+t$. So the final answer is
$$
u(x,t) = x_0 + t = dfrac{2x-t^2}{2(t+1)} + t = dfrac{2x+t+t^2}{2(t+1)}
$$
add a comment |
up vote
1
down vote
accepted
You found the characteristic curves correctly: they are the parabolas
$$
x = x_0+x_0t + frac{t^2}{2}
$$
So, the characteristic through $(x,t)$ originates at $x_0 = dfrac{2x-t^2}{2(t+1)}$, as you also found. The mistake is in concluding that $u(x,t) = x_0$. This would be true for homogeneous PDE. But the nonhomogeneity means that $u$ (represented by $z$ in the characteristic equation) changes along the characteristic. Specifically, $z=x_0+t$. So the final answer is
$$
u(x,t) = x_0 + t = dfrac{2x-t^2}{2(t+1)} + t = dfrac{2x+t+t^2}{2(t+1)}
$$
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You found the characteristic curves correctly: they are the parabolas
$$
x = x_0+x_0t + frac{t^2}{2}
$$
So, the characteristic through $(x,t)$ originates at $x_0 = dfrac{2x-t^2}{2(t+1)}$, as you also found. The mistake is in concluding that $u(x,t) = x_0$. This would be true for homogeneous PDE. But the nonhomogeneity means that $u$ (represented by $z$ in the characteristic equation) changes along the characteristic. Specifically, $z=x_0+t$. So the final answer is
$$
u(x,t) = x_0 + t = dfrac{2x-t^2}{2(t+1)} + t = dfrac{2x+t+t^2}{2(t+1)}
$$
You found the characteristic curves correctly: they are the parabolas
$$
x = x_0+x_0t + frac{t^2}{2}
$$
So, the characteristic through $(x,t)$ originates at $x_0 = dfrac{2x-t^2}{2(t+1)}$, as you also found. The mistake is in concluding that $u(x,t) = x_0$. This would be true for homogeneous PDE. But the nonhomogeneity means that $u$ (represented by $z$ in the characteristic equation) changes along the characteristic. Specifically, $z=x_0+t$. So the final answer is
$$
u(x,t) = x_0 + t = dfrac{2x-t^2}{2(t+1)} + t = dfrac{2x+t+t^2}{2(t+1)}
$$
answered Sep 15 '17 at 16:09
user357151
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