Solving this partial differential equation

[schniefen]

Homework Statement

Solve the differential equation $\frac{{\partial{f}}}{{\partial{x}}}-3\frac{{\partial{f}}}{{\partial{y}}}=0$ by introducing the variables $u=ax+y, v=x$ and choosing the constant $a$ appropriately.

Relevant Equations

The chain rule for multivariable functions.

Introducing the new variables $u$ and $v$, the chain rule gives

$\dfrac{{\partial{f}}}{{\partial{x}}}=\dfrac{{\partial{f}}}{{\partial{u}}} \dfrac{{\partial{u}}}{{\partial{x}}}+\dfrac{{\partial{f}}}{{\partial{v}}} \dfrac{{\partial{v}}}{{\partial{x}}}$

$-3\dfrac{{\partial{f}}}{{\partial{y}}}=-3(\dfrac{{\partial{f}}}{{\partial{u}}} \dfrac{{\partial{u}}}{{\partial{y}}}+\dfrac{{\partial{f}}}{{\partial{v}}} \dfrac{{\partial{v}}}{{\partial{y}}})$
​

$\dfrac{{\partial{u}}}{{\partial{x}}}$ equals $a$ and $\dfrac{{\partial{v}}}{{\partial{x}}}=1$. Also $\dfrac{{\partial{u}}}{{\partial{y}}}=1$ and $\dfrac{{\partial{v}}}{{\partial{y}}}=0$. So

$\dfrac{{\partial{f}}}{{\partial{x}}}=a\dfrac{{\partial{f}}}{{\partial{u}}}+\dfrac{{\partial{f}}}{{\partial{v}}}$

$-3\dfrac{{\partial{f}}}{{\partial{y}}}=-3\dfrac{{\partial{f}}}{{\partial{u}}}$

$\implies \dfrac{{\partial{f}}}{{\partial{x}}}-3\dfrac{{\partial{f}}}{{\partial{y}}}=\dfrac{{\partial{f}}}{{\partial{u}}}(a-3)+\dfrac{{\partial{f}}}{{\partial{v}}}$
​

The last equality should also equal $0$. How does one go from here to find the solution $f$?

 

[phyzguy]

Note the phrase that says, "choosing the constant a appropriately" Is there any choice of a that makes the problem easy?

 

[schniefen]

Bonus question. Solve for $x>0$ and $y>0$ the differential equation

$x\dfrac{{\partial{f}}}{{\partial{x}}}+y\dfrac{{\partial{f}}}{{\partial{y}}}=y \quad (1)$​

by introducing the new variables $x=u$ and $y=\frac{u}{v}$. The chain rule gives

$\dfrac{{\partial{f}}}{{\partial{x}}}=\dfrac{{\partial{f}}}{{\partial{u}}} \dfrac{{\partial{u}}}{{\partial{x}}}+\dfrac{{\partial{f}}}{{\partial{\frac{u}{v}}}} \dfrac{{\partial{\frac{u}{v}}}}{{\partial{x}}}$

$\dfrac{{\partial{f}}}{{\partial{y}}}=\dfrac{{\partial{f}}}{{\partial{u}}} \dfrac{{\partial{u}}}{{\partial{y}}}+\dfrac{{\partial{f}}}{{\partial{\frac{u}{v}}}} \dfrac{{\partial{\frac{u}{v}}}}{{\partial{y}}}$
​

From the change of variables it follows that $u(x)=x, v(x)=\frac{x}{y}, u(y)=vy$ and $v(y)=\frac{u}{y}$. Then

$\dfrac{{\partial{u}}}{{\partial{x}}}=1$,
$\dfrac{{\partial{\frac{u}{v}}}}{{\partial{x}}}=\dfrac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}=\dfrac{\frac{x}{y}-\frac{x}{y}}{\frac{x^2}{y^2}}=0$,
$\dfrac{{\partial{u}}}{{\partial{y}}}=v$,
$\dfrac{{\partial{\frac{u}{v}}}}{{\partial{y}}}=\dfrac{u'(y)v(y)-u(y)v'(y)}{v(y)^2}=\dfrac{\frac{uv}{y}+\frac{vyu}{y^2}}{\frac{u^2}{y^2}}=\dfrac{2vy}{u}=2$,
​

So $(1)$ is equivalent to

$u\dfrac{{\partial{f}}}{{\partial{u}}}+\dfrac{u}{v}(\dfrac{{\partial{f}}}{{\partial{u}}}v+\dfrac{{\partial{f}}}{{\partial{\frac{u}{v}}}}2)=\frac{u}{v}$

$\iff \quad 2(\dfrac{{\partial{f}}}{{\partial{u}}}+\dfrac{1}{v}\dfrac{{\partial{f}}}{{\partial{\frac{u}{v}}}})=\frac{1}{v}$​

How can one evaluate this expression further?

 

[phyzguy]

Before we move on, what was your solution to the first question?

 

[schniefen]

Before we move on, what was your solution to the first question?

Choosing $a=3$ yields $f(u,v)=C_2u^k+C_1$, where $C_2,C_1,k\in\mathbf{R}$, which can be reformulated as a function depending only on $u=3x+y$.

 

[phyzguy]

Choosing $a=3$ yields $f(u,v)=C_2u^k+C_1$, where $C_2,C_1,k\in\mathbf{R}$, which can be reformulated as a function depending only on $u=3x+y$.

Ignoring the constants, does the function f(u) have to be of the form f(u) = u^k?

 

[schniefen]

No, any function $f(3x+y)$ that is differentiable will do.

 

[Delta2]

I personally have problem understanding your work and how you apply the chain rule in #3 (at some point you calculate $\frac{\partial\frac{u}{v}}{\partial y}=2$ when it is obviously that $y=\frac{u}{v}$ hence that partial derivative is equal to 1).

I suggest you write down $u(x,y)$ and $v(x,y)$ as functions of x,y and then apply (in a standard way) the chain rule for partial derivatives as
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}$$ and
$$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}$$