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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

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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
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 $u(x,y)= \bigl(y-{3\over4}x\bigr)^3$.

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