Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is my solution to the following problem correct?

Solve $4u_x+3u_y=0$ subject to $u(0,y)=y^3$

Changing coordinates so that we have: $\displaystyle \frac{\partial x}{\partial \alpha}=4$ and $\displaystyle \frac{\partial y}{\partial \beta}=3$, so take $x=4\alpha$ and $y=3\beta+\gamma$, we then get:

$\displaystyle \frac{\partial u}{\partial \alpha}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \alpha} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial \beta}=0$

so we have that $u=f(\gamma)$ and as $\gamma=y-\frac{3}{4}x$ we get $f(y-\frac{3}{4}\gamma)$, letting $x=0$, $f(y)=y^3$ which gives $u(x,y)=f(y-\frac{3}{4}x)=(y-\frac{3}{4}x)^3$

Thanks very much for any help

share|improve this question
1  
With $\alpha=\beta$? –  Did Apr 12 '12 at 11:42
    
Yeah, got a bit confused there but I see now-thanks –  hmmmm Apr 12 '12 at 17:26
add comment

1 Answer 1

up vote 1 down vote accepted

The $\beta$ in your argument makes no sense: You want to replace the variables $x$ and $y$ by two new variables $\alpha$ and $\gamma$ such that $${\partial \tilde u\over\partial\alpha}\equiv0\ .\qquad(1)$$ Inspecting the given equation you propose $x:=4\alpha$, $\ y:=3\alpha +\gamma$, and then you easily prove that in this way $(1)$ is satisfied. It follows that any solution $u$, expressed in terms of $\alpha$ and $\gamma$, is a function of $\gamma$ alone: $$\tilde u(\alpha,\gamma)=f(\gamma)$$ with an arbitrary function $f$ of one variable. In terms of $x$ and $y$ this means that $$u(x,y)= f\bigl(y-{3\over4}x\bigr)\ ,$$ and plugging in the initial condition you indeed get $u(x,y)= \bigl(y-{3\over4}x\bigr)^3$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.