Solve $u_x + 4xu_y = 1 + u^2$ for $u(0,y)=y$
$begingroup$
I got a weird result so I'm not sure I did this right
Let the initial condition be $u(0, y_0) = y_0 $ for some $y_0$
By the method of characteristics let
$$frac{dx}{ds} = 1 to x = s + A$$
$$x(s=0) = A = 0 to x(s) = s$$
$$frac{dy}{ds} = 4x = 4s to y = 2s^2 + B$$
$$y(s=0) = B = y_0 to y(s) = 2s^2+y_0$$
$$frac{du}{ds} = 1 + u^2 to arctan u = s + C to u(s)=tan(s+C)$$
$$u(0) = tan(C) = y_0 to C=arctan y_0$$
Now we have that
$$u(s) = tan(s+C) = tan(x + arctan y_0)$$
Using $$y(s) = 2s^2+y_0 to y_0 = y - 2s^2$$
We can substitute in $u$ to get
$$u(s) = tan(x + arctan(y-2s^2)) = tan(x + arctan(y-2x^2)) = u(x,y)$$
proof-verification pde characteristics
asked Dec 14 '18 at 7:59
The BoscoThe Bosco
600212
$endgroup$
add a comment |
$begingroup$
I got a weird result so I'm not sure I did this right
Let the initial condition be $u(0, y_0) = y_0 $ for some $y_0$
By the method of characteristics let
$$frac{dx}{ds} = 1 to x = s + A$$
$$x(s=0) = A = 0 to x(s) = s$$
$$frac{dy}{ds} = 4x = 4s to y = 2s^2 + B$$
$$y(s=0) = B = y_0 to y(s) = 2s^2+y_0$$
$$frac{du}{ds} = 1 + u^2 to arctan u = s + C to u(s)=tan(s+C)$$
$$u(0) = tan(C) = y_0 to C=arctan y_0$$
Now we have that
$$u(s) = tan(s+C) = tan(x + arctan y_0)$$
Using $$y(s) = 2s^2+y_0 to y_0 = y - 2s^2$$
We can substitute in $u$ to get
$$u(s) = tan(x + arctan(y-2s^2)) = tan(x + arctan(y-2x^2)) = u(x,y)$$
proof-verification pde characteristics
asked Dec 14 '18 at 7:59
The BoscoThe Bosco
600212
$endgroup$
add a comment |
$begingroup$
I got a weird result so I'm not sure I did this right
Let the initial condition be $u(0, y_0) = y_0 $ for some $y_0$
By the method of characteristics let
$$frac{dx}{ds} = 1 to x = s + A$$
$$x(s=0) = A = 0 to x(s) = s$$
$$frac{dy}{ds} = 4x = 4s to y = 2s^2 + B$$
$$y(s=0) = B = y_0 to y(s) = 2s^2+y_0$$
$$frac{du}{ds} = 1 + u^2 to arctan u = s + C to u(s)=tan(s+C)$$
$$u(0) = tan(C) = y_0 to C=arctan y_0$$
Now we have that
$$u(s) = tan(s+C) = tan(x + arctan y_0)$$
Using $$y(s) = 2s^2+y_0 to y_0 = y - 2s^2$$
We can substitute in $u$ to get
$$u(s) = tan(x + arctan(y-2s^2)) = tan(x + arctan(y-2x^2)) = u(x,y)$$
proof-verification pde characteristics
asked Dec 14 '18 at 7:59
The BoscoThe Bosco
600212
$endgroup$
I got a weird result so I'm not sure I did this right
Let the initial condition be $u(0, y_0) = y_0 $ for some $y_0$
By the method of characteristics let
$$frac{dx}{ds} = 1 to x = s + A$$
$$x(s=0) = A = 0 to x(s) = s$$
$$frac{dy}{ds} = 4x = 4s to y = 2s^2 + B$$
$$y(s=0) = B = y_0 to y(s) = 2s^2+y_0$$
$$frac{du}{ds} = 1 + u^2 to arctan u = s + C to u(s)=tan(s+C)$$
$$u(0) = tan(C) = y_0 to C=arctan y_0$$
Now we have that
$$u(s) = tan(s+C) = tan(x + arctan y_0)$$
Using $$y(s) = 2s^2+y_0 to y_0 = y - 2s^2$$
We can substitute in $u$ to get
$$u(s) = tan(x + arctan(y-2s^2)) = tan(x + arctan(y-2x^2)) = u(x,y)$$
proof-verification pde characteristics
proof-verification pde characteristics
asked Dec 14 '18 at 7:59
The BoscoThe Bosco
600212
asked Dec 14 '18 at 7:59
The BoscoThe Bosco
600212
edited Dec 14 '18 at 8:06
The Bosco
asked Dec 14 '18 at 7:59
The BoscoThe Bosco
600212
asked Dec 14 '18 at 7:59
The BoscoThe Bosco
600212
asked Dec 14 '18 at 7:59
The BoscoThe Bosco
600212
600212
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, your answer is correct.
$displaystyle dx=frac{dy}{4x}=frac{du}{1+u^2}$
$4x dx=dyimplies 2x^2-y=c_1$
$displaystyle dx=frac{du}{1+u^2}implies x-tan^{-1}u=c_2$
The general solution is given by $f(2x^2-y,x-tan^{-1}u)=0$
$implies x-tan^{-1}u=g(2x^2-y)$
$u(0,y)=yimplies g(-y)=-tan^{-1}yimplies g(y)=tan^{-1}y$
The answer is $u=tan[x-tan^{-1}(2x^2-y)]$
answered Dec 14 '18 at 8:37
Shubham JohriShubham Johri
5,192717
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1 Answer
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1 Answer
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$begingroup$
Yes, your answer is correct.
$displaystyle dx=frac{dy}{4x}=frac{du}{1+u^2}$
$4x dx=dyimplies 2x^2-y=c_1$
$displaystyle dx=frac{du}{1+u^2}implies x-tan^{-1}u=c_2$
The general solution is given by $f(2x^2-y,x-tan^{-1}u)=0$
$implies x-tan^{-1}u=g(2x^2-y)$
$u(0,y)=yimplies g(-y)=-tan^{-1}yimplies g(y)=tan^{-1}y$
The answer is $u=tan[x-tan^{-1}(2x^2-y)]$
answered Dec 14 '18 at 8:37
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
Yes, your answer is correct.
$displaystyle dx=frac{dy}{4x}=frac{du}{1+u^2}$
$4x dx=dyimplies 2x^2-y=c_1$
$displaystyle dx=frac{du}{1+u^2}implies x-tan^{-1}u=c_2$
The general solution is given by $f(2x^2-y,x-tan^{-1}u)=0$
$implies x-tan^{-1}u=g(2x^2-y)$
$u(0,y)=yimplies g(-y)=-tan^{-1}yimplies g(y)=tan^{-1}y$
The answer is $u=tan[x-tan^{-1}(2x^2-y)]$
answered Dec 14 '18 at 8:37
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
Yes, your answer is correct.
$displaystyle dx=frac{dy}{4x}=frac{du}{1+u^2}$
$4x dx=dyimplies 2x^2-y=c_1$
$displaystyle dx=frac{du}{1+u^2}implies x-tan^{-1}u=c_2$
The general solution is given by $f(2x^2-y,x-tan^{-1}u)=0$
$implies x-tan^{-1}u=g(2x^2-y)$
$u(0,y)=yimplies g(-y)=-tan^{-1}yimplies g(y)=tan^{-1}y$
The answer is $u=tan[x-tan^{-1}(2x^2-y)]$
answered Dec 14 '18 at 8:37
Shubham JohriShubham Johri
5,192717
$endgroup$
Yes, your answer is correct.
$displaystyle dx=frac{dy}{4x}=frac{du}{1+u^2}$
$4x dx=dyimplies 2x^2-y=c_1$
$displaystyle dx=frac{du}{1+u^2}implies x-tan^{-1}u=c_2$
The general solution is given by $f(2x^2-y,x-tan^{-1}u)=0$
$implies x-tan^{-1}u=g(2x^2-y)$
$u(0,y)=yimplies g(-y)=-tan^{-1}yimplies g(y)=tan^{-1}y$
The answer is $u=tan[x-tan^{-1}(2x^2-y)]$
answered Dec 14 '18 at 8:37
Shubham JohriShubham Johri
5,192717
answered Dec 14 '18 at 8:37
Shubham JohriShubham Johri
5,192717
answered Dec 14 '18 at 8:37
Shubham JohriShubham Johri
5,192717
answered Dec 14 '18 at 8:37
Shubham JohriShubham Johri
5,192717
5,192717
add a comment |
add a comment |
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