Solve IVP of $(u_t )^2 + (u_x )^2 − u^2 = 0$ using method of characteristics
$begingroup$
Consider the nonlinear first-order initial-value problem:
$$(u_t )^2 + (u_x )^2 − u^2 = 0$$
with initial condition $u(x, 0) = Ae^{−sqrt{1+x^2}}$.
(a) Find its solution for all $t > 0$ using the method of
characteristics.
(b) Describe its behavior for all $t > 0$. Does it remain a continuous
function of $x$ for all $t$?
(c) Find $lim u(x, t)$.
My attempt:
Let $x(s)=langle x(s),t(s) rangle, p= langle p_1(s),p_2(s) rangle = langle u_x,u_t rangle , z = u(x,t)$
$$F(p,z,x)=p_1^2+p_2^2-z^2, frac{dF}{dp}=langle 2p_1,2p_2 rangle, frac{dF}{dz}=2z, frac{dF}{dx}=langle 0,0 rangle $$
Characteristics:
begin{align}
p_s &= frac{dF}{dx}-frac{dF}{dz}p=-langle 0,0 rangle -2zlangle p_1,p_2 rangle \
z_s &= frac{dF}{dp}p= langle 2p_1,2p_2rangle cdot langle p_1,p_2 rangle =2p_1^2+2p_2^2\
x_s &= frac{dF}{dp}= langle 2p_1,2p_2 rangle
end{align}
IVP:
$$ x_0 =r quad t_0 =0, quad u_0=z_0=Ae^{−sqrt{1+r^2}} $$
$$ u_x(x,0) = -frac{Ae^{−sqrt{1+x^2}}x}{{sqrt{1+x^2}}}implies p_{1_0}= -frac{Ae^{−sqrt{1+r^2}}r}{{sqrt{1+r^2}}} $$
$$ p_{1_0}^2+p_{2_0}^2-u_0^2 = 0 implies p_{2_0}^2=A^2e^{-2sqrt{1+r^2}}-frac{A^2e^{−2sqrt{1+r^2}}r^2}{1+r^2}=frac{Ae^{−sqrt{1+r^2}}}{{sqrt{1+r^2}}}
$$
What's the next step? Solving $p_s,z_s,x_s$ seems a little bit painful...
pde characteristics
edited Dec 10 '18 at 17:06
Harry49
6,18331132
asked Dec 9 '18 at 19:16
dxdydzdxdydz
3009
$endgroup$
add a comment |
$begingroup$
Consider the nonlinear first-order initial-value problem:
$$(u_t )^2 + (u_x )^2 − u^2 = 0$$
with initial condition $u(x, 0) = Ae^{−sqrt{1+x^2}}$.
(a) Find its solution for all $t > 0$ using the method of
characteristics.
(b) Describe its behavior for all $t > 0$. Does it remain a continuous
function of $x$ for all $t$?
(c) Find $lim u(x, t)$.
My attempt:
Let $x(s)=langle x(s),t(s) rangle, p= langle p_1(s),p_2(s) rangle = langle u_x,u_t rangle , z = u(x,t)$
$$F(p,z,x)=p_1^2+p_2^2-z^2, frac{dF}{dp}=langle 2p_1,2p_2 rangle, frac{dF}{dz}=2z, frac{dF}{dx}=langle 0,0 rangle $$
Characteristics:
begin{align}
p_s &= frac{dF}{dx}-frac{dF}{dz}p=-langle 0,0 rangle -2zlangle p_1,p_2 rangle \
z_s &= frac{dF}{dp}p= langle 2p_1,2p_2rangle cdot langle p_1,p_2 rangle =2p_1^2+2p_2^2\
x_s &= frac{dF}{dp}= langle 2p_1,2p_2 rangle
end{align}
IVP:
$$ x_0 =r quad t_0 =0, quad u_0=z_0=Ae^{−sqrt{1+r^2}} $$
$$ u_x(x,0) = -frac{Ae^{−sqrt{1+x^2}}x}{{sqrt{1+x^2}}}implies p_{1_0}= -frac{Ae^{−sqrt{1+r^2}}r}{{sqrt{1+r^2}}} $$
$$ p_{1_0}^2+p_{2_0}^2-u_0^2 = 0 implies p_{2_0}^2=A^2e^{-2sqrt{1+r^2}}-frac{A^2e^{−2sqrt{1+r^2}}r^2}{1+r^2}=frac{Ae^{−sqrt{1+r^2}}}{{sqrt{1+r^2}}}
$$
What's the next step? Solving $p_s,z_s,x_s$ seems a little bit painful...
pde characteristics
edited Dec 10 '18 at 17:06
Harry49
6,18331132
asked Dec 9 '18 at 19:16
dxdydzdxdydz
3009
$endgroup$
add a comment |
$begingroup$
Consider the nonlinear first-order initial-value problem:
$$(u_t )^2 + (u_x )^2 − u^2 = 0$$
with initial condition $u(x, 0) = Ae^{−sqrt{1+x^2}}$.
(a) Find its solution for all $t > 0$ using the method of
characteristics.
(b) Describe its behavior for all $t > 0$. Does it remain a continuous
function of $x$ for all $t$?
(c) Find $lim u(x, t)$.
My attempt:
Let $x(s)=langle x(s),t(s) rangle, p= langle p_1(s),p_2(s) rangle = langle u_x,u_t rangle , z = u(x,t)$
$$F(p,z,x)=p_1^2+p_2^2-z^2, frac{dF}{dp}=langle 2p_1,2p_2 rangle, frac{dF}{dz}=2z, frac{dF}{dx}=langle 0,0 rangle $$
Characteristics:
begin{align}
p_s &= frac{dF}{dx}-frac{dF}{dz}p=-langle 0,0 rangle -2zlangle p_1,p_2 rangle \
z_s &= frac{dF}{dp}p= langle 2p_1,2p_2rangle cdot langle p_1,p_2 rangle =2p_1^2+2p_2^2\
x_s &= frac{dF}{dp}= langle 2p_1,2p_2 rangle
end{align}
IVP:
$$ x_0 =r quad t_0 =0, quad u_0=z_0=Ae^{−sqrt{1+r^2}} $$
$$ u_x(x,0) = -frac{Ae^{−sqrt{1+x^2}}x}{{sqrt{1+x^2}}}implies p_{1_0}= -frac{Ae^{−sqrt{1+r^2}}r}{{sqrt{1+r^2}}} $$
$$ p_{1_0}^2+p_{2_0}^2-u_0^2 = 0 implies p_{2_0}^2=A^2e^{-2sqrt{1+r^2}}-frac{A^2e^{−2sqrt{1+r^2}}r^2}{1+r^2}=frac{Ae^{−sqrt{1+r^2}}}{{sqrt{1+r^2}}}
$$
What's the next step? Solving $p_s,z_s,x_s$ seems a little bit painful...
pde characteristics
edited Dec 10 '18 at 17:06
Harry49
6,18331132
asked Dec 9 '18 at 19:16
dxdydzdxdydz
3009
$endgroup$
Consider the nonlinear first-order initial-value problem:
$$(u_t )^2 + (u_x )^2 − u^2 = 0$$
with initial condition $u(x, 0) = Ae^{−sqrt{1+x^2}}$.
(a) Find its solution for all $t > 0$ using the method of
characteristics.
(b) Describe its behavior for all $t > 0$. Does it remain a continuous
function of $x$ for all $t$?
(c) Find $lim u(x, t)$.
My attempt:
Let $x(s)=langle x(s),t(s) rangle, p= langle p_1(s),p_2(s) rangle = langle u_x,u_t rangle , z = u(x,t)$
$$F(p,z,x)=p_1^2+p_2^2-z^2, frac{dF}{dp}=langle 2p_1,2p_2 rangle, frac{dF}{dz}=2z, frac{dF}{dx}=langle 0,0 rangle $$
Characteristics:
begin{align}
p_s &= frac{dF}{dx}-frac{dF}{dz}p=-langle 0,0 rangle -2zlangle p_1,p_2 rangle \
z_s &= frac{dF}{dp}p= langle 2p_1,2p_2rangle cdot langle p_1,p_2 rangle =2p_1^2+2p_2^2\
x_s &= frac{dF}{dp}= langle 2p_1,2p_2 rangle
end{align}
IVP:
$$ x_0 =r quad t_0 =0, quad u_0=z_0=Ae^{−sqrt{1+r^2}} $$
$$ u_x(x,0) = -frac{Ae^{−sqrt{1+x^2}}x}{{sqrt{1+x^2}}}implies p_{1_0}= -frac{Ae^{−sqrt{1+r^2}}r}{{sqrt{1+r^2}}} $$
$$ p_{1_0}^2+p_{2_0}^2-u_0^2 = 0 implies p_{2_0}^2=A^2e^{-2sqrt{1+r^2}}-frac{A^2e^{−2sqrt{1+r^2}}r^2}{1+r^2}=frac{Ae^{−sqrt{1+r^2}}}{{sqrt{1+r^2}}}
$$
What's the next step? Solving $p_s,z_s,x_s$ seems a little bit painful...
pde characteristics
pde characteristics
edited Dec 10 '18 at 17:06
Harry49
6,18331132
asked Dec 9 '18 at 19:16
dxdydzdxdydz
3009
edited Dec 10 '18 at 17:06
Harry49
6,18331132
asked Dec 9 '18 at 19:16
dxdydzdxdydz
3009
edited Dec 10 '18 at 17:06
Harry49
6,18331132
edited Dec 10 '18 at 17:06
Harry49
6,18331132
edited Dec 10 '18 at 17:06
Harry49
6,18331132
6,18331132
asked Dec 9 '18 at 19:16
dxdydzdxdydz
3009
asked Dec 9 '18 at 19:16
dxdydzdxdydz
3009
asked Dec 9 '18 at 19:16
dxdydzdxdydz
3009
3009
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$(u_t )^2 + (u_x )^2 = u^2 $$
$$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
HINT:
$u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
$$(v_t )^2 + (v_x )^2 =1$$
$$v(x,0)=-sqrt{1+x^2}$$
This kind of PDE is well known (Eikonal PDE) :
https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).
https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.
https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.
Of course, the boundary conditions are diferent, but with the same method you will find the solution :
$$v(x,t)=-sqrt{(t+1)^2+x^2}$$
$$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$
answered Dec 10 '18 at 12:47
JJacquelinJJacquelin
43.4k21853
$endgroup$
$begingroup$
So u is a continuous function of x for all t, can we say anything else about the behavior of u?
$endgroup$
– dxdydz
Dec 11 '18 at 0:34
$begingroup$
$u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
$endgroup$
– JJacquelin
Dec 11 '18 at 13:54
add a comment |
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$begingroup$
$$(u_t )^2 + (u_x )^2 = u^2 $$
$$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
HINT:
$u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
$$(v_t )^2 + (v_x )^2 =1$$
$$v(x,0)=-sqrt{1+x^2}$$
This kind of PDE is well known (Eikonal PDE) :
https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).
https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.
https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.
Of course, the boundary conditions are diferent, but with the same method you will find the solution :
$$v(x,t)=-sqrt{(t+1)^2+x^2}$$
$$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$
answered Dec 10 '18 at 12:47
JJacquelinJJacquelin
43.4k21853
$endgroup$
$begingroup$
So u is a continuous function of x for all t, can we say anything else about the behavior of u?
$endgroup$
– dxdydz
Dec 11 '18 at 0:34
$begingroup$
$u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
$endgroup$
– JJacquelin
Dec 11 '18 at 13:54
add a comment |
$begingroup$
$$(u_t )^2 + (u_x )^2 = u^2 $$
$$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
HINT:
$u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
$$(v_t )^2 + (v_x )^2 =1$$
$$v(x,0)=-sqrt{1+x^2}$$
This kind of PDE is well known (Eikonal PDE) :
https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).
https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.
https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.
Of course, the boundary conditions are diferent, but with the same method you will find the solution :
$$v(x,t)=-sqrt{(t+1)^2+x^2}$$
$$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$
answered Dec 10 '18 at 12:47
JJacquelinJJacquelin
43.4k21853
$endgroup$
$begingroup$
So u is a continuous function of x for all t, can we say anything else about the behavior of u?
$endgroup$
– dxdydz
Dec 11 '18 at 0:34
$begingroup$
$u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
$endgroup$
– JJacquelin
Dec 11 '18 at 13:54
add a comment |
$begingroup$
$$(u_t )^2 + (u_x )^2 = u^2 $$
$$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
HINT:
$u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
$$(v_t )^2 + (v_x )^2 =1$$
$$v(x,0)=-sqrt{1+x^2}$$
This kind of PDE is well known (Eikonal PDE) :
https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).
https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.
https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.
Of course, the boundary conditions are diferent, but with the same method you will find the solution :
$$v(x,t)=-sqrt{(t+1)^2+x^2}$$
$$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$
answered Dec 10 '18 at 12:47
JJacquelinJJacquelin
43.4k21853
$endgroup$
$$(u_t )^2 + (u_x )^2 = u^2 $$
$$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
HINT:
$u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
$$(v_t )^2 + (v_x )^2 =1$$
$$v(x,0)=-sqrt{1+x^2}$$
This kind of PDE is well known (Eikonal PDE) :
https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).
https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.
https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.
Of course, the boundary conditions are diferent, but with the same method you will find the solution :
$$v(x,t)=-sqrt{(t+1)^2+x^2}$$
$$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$
answered Dec 10 '18 at 12:47
JJacquelinJJacquelin
43.4k21853
edited Dec 10 '18 at 12:58
answered Dec 10 '18 at 12:47
JJacquelinJJacquelin
43.4k21853
answered Dec 10 '18 at 12:47
JJacquelinJJacquelin
43.4k21853
answered Dec 10 '18 at 12:47
JJacquelinJJacquelin
43.4k21853
43.4k21853
$begingroup$
So u is a continuous function of x for all t, can we say anything else about the behavior of u?
$endgroup$
– dxdydz
Dec 11 '18 at 0:34
$begingroup$
$u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
$endgroup$
– JJacquelin
Dec 11 '18 at 13:54
add a comment |
$begingroup$
So u is a continuous function of x for all t, can we say anything else about the behavior of u?
$endgroup$
– dxdydz
Dec 11 '18 at 0:34
$begingroup$
$u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
$endgroup$
– JJacquelin
Dec 11 '18 at 13:54
$begingroup$
So u is a continuous function of x for all t, can we say anything else about the behavior of u?
$endgroup$
– dxdydz
Dec 11 '18 at 0:34
$begingroup$
So u is a continuous function of x for all t, can we say anything else about the behavior of u?
$endgroup$
– dxdydz
Dec 11 '18 at 0:34
$begingroup$
$u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
$endgroup$
– JJacquelin
Dec 11 '18 at 13:54
$begingroup$
$u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
$endgroup$
– JJacquelin
Dec 11 '18 at 13:54
add a comment |
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