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Solve quasi-linear PDE $uu_x+yu_y=x$ through the method of characteristics
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Related Question:
- PDE characteristic equation
I am doing a PDE problem from a course that I will take next semester. I watched this video in advance. It appears that I am doing something wrong with the initial data.
Consider the following first order PDE
begin{cases}
uu_x+yu_y=x, \
u(x,1)=2x.
end{cases}
- State the condition which guarantees that the initial surface $Gamma$ is not characteristic.
- Use the method of characteristics to find a solution of the PDE and discuss for which $(x,y)inmathbb R^2$ the solution exists.
For the second part, the general form of the method of characteristics is $au_x+bu_y=f$. Therefore, we have that
$$frac{dx}{a}=frac{dy}{b}=frac{du}{f}$$
or
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let
$$frac{dx}{u}=frac{du}{x}$$
Then,
$$ frac{du}{dx}=frac{x}{u}~~Rightarrow~~~ udu = xdx ~~Rightarrow~~~ frac{u^2}{2}=frac{x^2}{2}+C ~~Rightarrow~~~ C_1=x^2-u^2$$
Next, let
$$frac{dx}{u}=frac{dy}{y}$$
Then,
$$frac{dy}{dx}=frac{y}{u} ~~Rightarrow~~~ frac{1}{y}dy = frac{1}{u}dx ~~Rightarrow~~~ ln(y)=frac{x}{u}+C ~~Rightarrow~~~ C_2=y+e^{frac{x}{u}} $$
So, we have that $C_1=x^2-u^2$ and $C_2=y+e^{frac{x}{u}}$. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,
$$y+e^{frac{x}{u}}=F(x^2-u^2)$$
We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,
$$1+e^{frac{1}{2}}=F(-3x^2)$$
This doesn't appear to help us find the solution. If instead we set $C_1=F(C_2)$ then
$$x^2-u^2=F(y+e^{frac{x}{u}})$$
Applying the initial data produces
$$-3x^2=F(1+e^{frac{1}{2}})$$
That also appears to be incorrect. I must have made a mistake in applying the initial data. How can we apply the initial data to solve for $u$?
pde
asked Jan 7 at 23:24
Axion004Axion004
319312
$endgroup$
add a comment |
$begingroup$
Related Question:
- PDE characteristic equation
I am doing a PDE problem from a course that I will take next semester. I watched this video in advance. It appears that I am doing something wrong with the initial data.
Consider the following first order PDE
begin{cases}
uu_x+yu_y=x, \
u(x,1)=2x.
end{cases}
- State the condition which guarantees that the initial surface $Gamma$ is not characteristic.
- Use the method of characteristics to find a solution of the PDE and discuss for which $(x,y)inmathbb R^2$ the solution exists.
For the second part, the general form of the method of characteristics is $au_x+bu_y=f$. Therefore, we have that
$$frac{dx}{a}=frac{dy}{b}=frac{du}{f}$$
or
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let
$$frac{dx}{u}=frac{du}{x}$$
Then,
$$ frac{du}{dx}=frac{x}{u}~~Rightarrow~~~ udu = xdx ~~Rightarrow~~~ frac{u^2}{2}=frac{x^2}{2}+C ~~Rightarrow~~~ C_1=x^2-u^2$$
Next, let
$$frac{dx}{u}=frac{dy}{y}$$
Then,
$$frac{dy}{dx}=frac{y}{u} ~~Rightarrow~~~ frac{1}{y}dy = frac{1}{u}dx ~~Rightarrow~~~ ln(y)=frac{x}{u}+C ~~Rightarrow~~~ C_2=y+e^{frac{x}{u}} $$
So, we have that $C_1=x^2-u^2$ and $C_2=y+e^{frac{x}{u}}$. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,
$$y+e^{frac{x}{u}}=F(x^2-u^2)$$
We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,
$$1+e^{frac{1}{2}}=F(-3x^2)$$
This doesn't appear to help us find the solution. If instead we set $C_1=F(C_2)$ then
$$x^2-u^2=F(y+e^{frac{x}{u}})$$
Applying the initial data produces
$$-3x^2=F(1+e^{frac{1}{2}})$$
That also appears to be incorrect. I must have made a mistake in applying the initial data. How can we apply the initial data to solve for $u$?
pde
asked Jan 7 at 23:24
Axion004Axion004
319312
$endgroup$
add a comment |
$begingroup$
Related Question:
- PDE characteristic equation
I am doing a PDE problem from a course that I will take next semester. I watched this video in advance. It appears that I am doing something wrong with the initial data.
Consider the following first order PDE
begin{cases}
uu_x+yu_y=x, \
u(x,1)=2x.
end{cases}
- State the condition which guarantees that the initial surface $Gamma$ is not characteristic.
- Use the method of characteristics to find a solution of the PDE and discuss for which $(x,y)inmathbb R^2$ the solution exists.
For the second part, the general form of the method of characteristics is $au_x+bu_y=f$. Therefore, we have that
$$frac{dx}{a}=frac{dy}{b}=frac{du}{f}$$
or
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let
$$frac{dx}{u}=frac{du}{x}$$
Then,
$$ frac{du}{dx}=frac{x}{u}~~Rightarrow~~~ udu = xdx ~~Rightarrow~~~ frac{u^2}{2}=frac{x^2}{2}+C ~~Rightarrow~~~ C_1=x^2-u^2$$
Next, let
$$frac{dx}{u}=frac{dy}{y}$$
Then,
$$frac{dy}{dx}=frac{y}{u} ~~Rightarrow~~~ frac{1}{y}dy = frac{1}{u}dx ~~Rightarrow~~~ ln(y)=frac{x}{u}+C ~~Rightarrow~~~ C_2=y+e^{frac{x}{u}} $$
So, we have that $C_1=x^2-u^2$ and $C_2=y+e^{frac{x}{u}}$. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,
$$y+e^{frac{x}{u}}=F(x^2-u^2)$$
We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,
$$1+e^{frac{1}{2}}=F(-3x^2)$$
This doesn't appear to help us find the solution. If instead we set $C_1=F(C_2)$ then
$$x^2-u^2=F(y+e^{frac{x}{u}})$$
Applying the initial data produces
$$-3x^2=F(1+e^{frac{1}{2}})$$
That also appears to be incorrect. I must have made a mistake in applying the initial data. How can we apply the initial data to solve for $u$?
pde
asked Jan 7 at 23:24
Axion004Axion004
319312
$endgroup$
Related Question:
- PDE characteristic equation
I am doing a PDE problem from a course that I will take next semester. I watched this video in advance. It appears that I am doing something wrong with the initial data.
Consider the following first order PDE
begin{cases}
uu_x+yu_y=x, \
u(x,1)=2x.
end{cases}
- State the condition which guarantees that the initial surface $Gamma$ is not characteristic.
- Use the method of characteristics to find a solution of the PDE and discuss for which $(x,y)inmathbb R^2$ the solution exists.
For the second part, the general form of the method of characteristics is $au_x+bu_y=f$. Therefore, we have that
$$frac{dx}{a}=frac{dy}{b}=frac{du}{f}$$
or
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let
$$frac{dx}{u}=frac{du}{x}$$
Then,
$$ frac{du}{dx}=frac{x}{u}~~Rightarrow~~~ udu = xdx ~~Rightarrow~~~ frac{u^2}{2}=frac{x^2}{2}+C ~~Rightarrow~~~ C_1=x^2-u^2$$
Next, let
$$frac{dx}{u}=frac{dy}{y}$$
Then,
$$frac{dy}{dx}=frac{y}{u} ~~Rightarrow~~~ frac{1}{y}dy = frac{1}{u}dx ~~Rightarrow~~~ ln(y)=frac{x}{u}+C ~~Rightarrow~~~ C_2=y+e^{frac{x}{u}} $$
So, we have that $C_1=x^2-u^2$ and $C_2=y+e^{frac{x}{u}}$. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,
$$y+e^{frac{x}{u}}=F(x^2-u^2)$$
We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,
$$1+e^{frac{1}{2}}=F(-3x^2)$$
This doesn't appear to help us find the solution. If instead we set $C_1=F(C_2)$ then
$$x^2-u^2=F(y+e^{frac{x}{u}})$$
Applying the initial data produces
$$-3x^2=F(1+e^{frac{1}{2}})$$
That also appears to be incorrect. I must have made a mistake in applying the initial data. How can we apply the initial data to solve for $u$?
pde
pde
asked Jan 7 at 23:24
Axion004Axion004
319312
asked Jan 7 at 23:24
Axion004Axion004
319312
edited Jan 11 at 14:22
Axion004
asked Jan 7 at 23:24
Axion004Axion004
319312
asked Jan 7 at 23:24
Axion004Axion004
319312
asked Jan 7 at 23:24
Axion004Axion004
319312
319312
add a comment |
add a comment |
1 Answer
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For the first question, on $Gamma$ it is specified that $u(x,y)=2x$. Therefore, $u_x=2$ and $u_y=0$. Putting them into the PDE leads to $uu_x+yu_y=4x$ which is contradictory with $uu_x+yu_y=x$. So, the boundary condition is not on a characteristic.
For the second question, we should follow the method of characteristics. Lets first write the general form as $au_x+bu_y=f$. Therefore, we have that
$$frac{dx}{a}=frac{dy}{b}=frac{du}{f}$$
or
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let
$$frac{dx}{u}=frac{du}{x}$$
Then,
$$ frac{du}{dx}=frac{x}{u}~~Rightarrow~~~ udu = xdx ~~Rightarrow~~~ frac{u^2}{2}=frac{x^2}{2}+C ~~Rightarrow~~~ C_1=u^2-x^2$$
Next, observe that $frac{a}{b}=frac{c}{d} iff frac{a+c}{b+d}=frac{c}{d}$. So,
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
can be written as
$$frac{dx+du}{u+x}=frac{dy}{y} ~~Rightarrow~~~ ln(x+u) = ln(y)+C ~~Rightarrow~~~ lnBig(frac{x+u}{y}Big)=C ~~Rightarrow~~~ C_2=frac{x+u}{y} $$
So, we have that $C_1=u^2-x^2$ and $C_2=dfrac{x+u}{y} $. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,
$$frac{x+u}{y} =F(u^2-x^2)$$
or
$$u=-x+yF(u^2-x^2)$$
We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,
$$2x=-x+f((2x)^2-x^2) ~~Rightarrow~~~ 3x=F(3x^2)$$
Letting $w=3x^2$, we see that
$$ x^2=frac{w}{3} ~~Rightarrow~~~ x=pmsqrt{frac{w}{3}}$$
So, $F(w)=pm3sqrt{frac{w}{3}}$. Therefore,
$$u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$$
Hence,
$$u(x,y)=-x+ypmsqrt{3(u^2-x^2)}$$
$$u_x=-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}}quadtext{and}quad u_y=1$$
Thus,
$$uu_x+yu_y=uBig(-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}})+y=...=x$$
No, it doesn't. I must have made a mistake somewhere.
answered Jan 8 at 21:39
Axion004Axion004
319312
$endgroup$
$begingroup$
It appears that a mistake is made after setting $3x=F(3x^2)$. This produces $u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$, which is not a function of x and y.
$endgroup$
– Axion004
Jan 12 at 19:58
add a comment |
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1 Answer
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$begingroup$
For the first question, on $Gamma$ it is specified that $u(x,y)=2x$. Therefore, $u_x=2$ and $u_y=0$. Putting them into the PDE leads to $uu_x+yu_y=4x$ which is contradictory with $uu_x+yu_y=x$. So, the boundary condition is not on a characteristic.
For the second question, we should follow the method of characteristics. Lets first write the general form as $au_x+bu_y=f$. Therefore, we have that
$$frac{dx}{a}=frac{dy}{b}=frac{du}{f}$$
or
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let
$$frac{dx}{u}=frac{du}{x}$$
Then,
$$ frac{du}{dx}=frac{x}{u}~~Rightarrow~~~ udu = xdx ~~Rightarrow~~~ frac{u^2}{2}=frac{x^2}{2}+C ~~Rightarrow~~~ C_1=u^2-x^2$$
Next, observe that $frac{a}{b}=frac{c}{d} iff frac{a+c}{b+d}=frac{c}{d}$. So,
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
can be written as
$$frac{dx+du}{u+x}=frac{dy}{y} ~~Rightarrow~~~ ln(x+u) = ln(y)+C ~~Rightarrow~~~ lnBig(frac{x+u}{y}Big)=C ~~Rightarrow~~~ C_2=frac{x+u}{y} $$
So, we have that $C_1=u^2-x^2$ and $C_2=dfrac{x+u}{y} $. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,
$$frac{x+u}{y} =F(u^2-x^2)$$
or
$$u=-x+yF(u^2-x^2)$$
We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,
$$2x=-x+f((2x)^2-x^2) ~~Rightarrow~~~ 3x=F(3x^2)$$
Letting $w=3x^2$, we see that
$$ x^2=frac{w}{3} ~~Rightarrow~~~ x=pmsqrt{frac{w}{3}}$$
So, $F(w)=pm3sqrt{frac{w}{3}}$. Therefore,
$$u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$$
Hence,
$$u(x,y)=-x+ypmsqrt{3(u^2-x^2)}$$
$$u_x=-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}}quadtext{and}quad u_y=1$$
Thus,
$$uu_x+yu_y=uBig(-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}})+y=...=x$$
No, it doesn't. I must have made a mistake somewhere.
answered Jan 8 at 21:39
Axion004Axion004
319312
$endgroup$
$begingroup$
It appears that a mistake is made after setting $3x=F(3x^2)$. This produces $u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$, which is not a function of x and y.
$endgroup$
– Axion004
Jan 12 at 19:58
add a comment |
$begingroup$
For the first question, on $Gamma$ it is specified that $u(x,y)=2x$. Therefore, $u_x=2$ and $u_y=0$. Putting them into the PDE leads to $uu_x+yu_y=4x$ which is contradictory with $uu_x+yu_y=x$. So, the boundary condition is not on a characteristic.
For the second question, we should follow the method of characteristics. Lets first write the general form as $au_x+bu_y=f$. Therefore, we have that
$$frac{dx}{a}=frac{dy}{b}=frac{du}{f}$$
or
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let
$$frac{dx}{u}=frac{du}{x}$$
Then,
$$ frac{du}{dx}=frac{x}{u}~~Rightarrow~~~ udu = xdx ~~Rightarrow~~~ frac{u^2}{2}=frac{x^2}{2}+C ~~Rightarrow~~~ C_1=u^2-x^2$$
Next, observe that $frac{a}{b}=frac{c}{d} iff frac{a+c}{b+d}=frac{c}{d}$. So,
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
can be written as
$$frac{dx+du}{u+x}=frac{dy}{y} ~~Rightarrow~~~ ln(x+u) = ln(y)+C ~~Rightarrow~~~ lnBig(frac{x+u}{y}Big)=C ~~Rightarrow~~~ C_2=frac{x+u}{y} $$
So, we have that $C_1=u^2-x^2$ and $C_2=dfrac{x+u}{y} $. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,
$$frac{x+u}{y} =F(u^2-x^2)$$
or
$$u=-x+yF(u^2-x^2)$$
We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,
$$2x=-x+f((2x)^2-x^2) ~~Rightarrow~~~ 3x=F(3x^2)$$
Letting $w=3x^2$, we see that
$$ x^2=frac{w}{3} ~~Rightarrow~~~ x=pmsqrt{frac{w}{3}}$$
So, $F(w)=pm3sqrt{frac{w}{3}}$. Therefore,
$$u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$$
Hence,
$$u(x,y)=-x+ypmsqrt{3(u^2-x^2)}$$
$$u_x=-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}}quadtext{and}quad u_y=1$$
Thus,
$$uu_x+yu_y=uBig(-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}})+y=...=x$$
No, it doesn't. I must have made a mistake somewhere.
answered Jan 8 at 21:39
Axion004Axion004
319312
$endgroup$
$begingroup$
It appears that a mistake is made after setting $3x=F(3x^2)$. This produces $u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$, which is not a function of x and y.
$endgroup$
– Axion004
Jan 12 at 19:58
add a comment |
$begingroup$
For the first question, on $Gamma$ it is specified that $u(x,y)=2x$. Therefore, $u_x=2$ and $u_y=0$. Putting them into the PDE leads to $uu_x+yu_y=4x$ which is contradictory with $uu_x+yu_y=x$. So, the boundary condition is not on a characteristic.
For the second question, we should follow the method of characteristics. Lets first write the general form as $au_x+bu_y=f$. Therefore, we have that
$$frac{dx}{a}=frac{dy}{b}=frac{du}{f}$$
or
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let
$$frac{dx}{u}=frac{du}{x}$$
Then,
$$ frac{du}{dx}=frac{x}{u}~~Rightarrow~~~ udu = xdx ~~Rightarrow~~~ frac{u^2}{2}=frac{x^2}{2}+C ~~Rightarrow~~~ C_1=u^2-x^2$$
Next, observe that $frac{a}{b}=frac{c}{d} iff frac{a+c}{b+d}=frac{c}{d}$. So,
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
can be written as
$$frac{dx+du}{u+x}=frac{dy}{y} ~~Rightarrow~~~ ln(x+u) = ln(y)+C ~~Rightarrow~~~ lnBig(frac{x+u}{y}Big)=C ~~Rightarrow~~~ C_2=frac{x+u}{y} $$
So, we have that $C_1=u^2-x^2$ and $C_2=dfrac{x+u}{y} $. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,
$$frac{x+u}{y} =F(u^2-x^2)$$
or
$$u=-x+yF(u^2-x^2)$$
We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,
$$2x=-x+f((2x)^2-x^2) ~~Rightarrow~~~ 3x=F(3x^2)$$
Letting $w=3x^2$, we see that
$$ x^2=frac{w}{3} ~~Rightarrow~~~ x=pmsqrt{frac{w}{3}}$$
So, $F(w)=pm3sqrt{frac{w}{3}}$. Therefore,
$$u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$$
Hence,
$$u(x,y)=-x+ypmsqrt{3(u^2-x^2)}$$
$$u_x=-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}}quadtext{and}quad u_y=1$$
Thus,
$$uu_x+yu_y=uBig(-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}})+y=...=x$$
No, it doesn't. I must have made a mistake somewhere.
answered Jan 8 at 21:39
Axion004Axion004
319312
$endgroup$
For the first question, on $Gamma$ it is specified that $u(x,y)=2x$. Therefore, $u_x=2$ and $u_y=0$. Putting them into the PDE leads to $uu_x+yu_y=4x$ which is contradictory with $uu_x+yu_y=x$. So, the boundary condition is not on a characteristic.
For the second question, we should follow the method of characteristics. Lets first write the general form as $au_x+bu_y=f$. Therefore, we have that
$$frac{dx}{a}=frac{dy}{b}=frac{du}{f}$$
or
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let
$$frac{dx}{u}=frac{du}{x}$$
Then,
$$ frac{du}{dx}=frac{x}{u}~~Rightarrow~~~ udu = xdx ~~Rightarrow~~~ frac{u^2}{2}=frac{x^2}{2}+C ~~Rightarrow~~~ C_1=u^2-x^2$$
Next, observe that $frac{a}{b}=frac{c}{d} iff frac{a+c}{b+d}=frac{c}{d}$. So,
$$frac{dx}{u}=frac{dy}{y}=frac{du}{x}$$
can be written as
$$frac{dx+du}{u+x}=frac{dy}{y} ~~Rightarrow~~~ ln(x+u) = ln(y)+C ~~Rightarrow~~~ lnBig(frac{x+u}{y}Big)=C ~~Rightarrow~~~ C_2=frac{x+u}{y} $$
So, we have that $C_1=u^2-x^2$ and $C_2=dfrac{x+u}{y} $. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,
$$frac{x+u}{y} =F(u^2-x^2)$$
or
$$u=-x+yF(u^2-x^2)$$
We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,
$$2x=-x+f((2x)^2-x^2) ~~Rightarrow~~~ 3x=F(3x^2)$$
Letting $w=3x^2$, we see that
$$ x^2=frac{w}{3} ~~Rightarrow~~~ x=pmsqrt{frac{w}{3}}$$
So, $F(w)=pm3sqrt{frac{w}{3}}$. Therefore,
$$u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$$
Hence,
$$u(x,y)=-x+ypmsqrt{3(u^2-x^2)}$$
$$u_x=-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}}quadtext{and}quad u_y=1$$
Thus,
$$uu_x+yu_y=uBig(-1pmfrac{sqrt{3}{x}}{sqrt{u^2-x^2}})+y=...=x$$
No, it doesn't. I must have made a mistake somewhere.
answered Jan 8 at 21:39
Axion004Axion004
319312
edited Jan 11 at 22:52
answered Jan 8 at 21:39
Axion004Axion004
319312
answered Jan 8 at 21:39
Axion004Axion004
319312
answered Jan 8 at 21:39
Axion004Axion004
319312
319312
$begingroup$
It appears that a mistake is made after setting $3x=F(3x^2)$. This produces $u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$, which is not a function of x and y.
$endgroup$
– Axion004
Jan 12 at 19:58
add a comment |
$begingroup$
It appears that a mistake is made after setting $3x=F(3x^2)$. This produces $u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$, which is not a function of x and y.
$endgroup$
– Axion004
Jan 12 at 19:58
$begingroup$
It appears that a mistake is made after setting $3x=F(3x^2)$. This produces $u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$, which is not a function of x and y.
$endgroup$
– Axion004
Jan 12 at 19:58
$begingroup$
It appears that a mistake is made after setting $3x=F(3x^2)$. This produces $u=-x+yF(u^2-x^2) = -x+ypm3sqrt{frac{u^2-x^2}{3}} = -x+ypmsqrt{3(u^2-x^2)}$, which is not a function of x and y.
$endgroup$
– Axion004
Jan 12 at 19:58
add a comment |
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