1
$\begingroup$

I encountered this problem in class, and I did not understand how to solve it:

The problem:

Find all functions $ f \in C²$ such that $$ \dfrac{\partial²f}{\partial x²} + \dfrac{\partial²f}{\partial x \partial y} - 6\frac{\partial² f}{\partial v²} = 1, $$ using the variable substitution $$ \begin{cases} u = x + ay \\ v = x + by. \end{cases} $$

My attempt at solving it

First I used the chain rule to get $$ \dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial u} + \dfrac{\partial f}{\partial v} $$ $$\dfrac{\partial f}{\partial y} = a\cdot\dfrac{\partial f}{\partial u} + b\cdot \dfrac{\partial f}{\partial v} $$ Then I used the chain rule to derive $\dfrac{\partial f}{\partial x²}$, $\dfrac{\partial²f}{\partial x\partial y}$ and $\dfrac{\partial²f}{\partial y²}$, which yielded $$ \dfrac{\partial²f}{\partial x²} = \dfrac{\partial²f}{\partial u²} + 2\dfrac{\partial²f}{\partial u \partial v} + \dfrac{\partial²f}{\partial v²}, $$ $$ \dfrac{\partial²f}{\partial x \partial y} = a \dfrac{\partial²f}{\partial u \partial v} +(a + b)\dfrac{\partial²f}{\partial } +b\dfrac{\partial²f}{\partial v²}, $$ $$ \dfrac{\partial²f}{\partial y²} = a^2\dfrac{\partial²f}{\partial u²} +2ab\dfrac{\partial²f}{\partial u \partial v} + b^2\dfrac{\partial²f}{\partial v²}. $$ This can be inserted in the original equation, giving us $$ (1+a-6a²)\dfrac{\partial²f}{\partial u²} + (2+b+a-12ab)\dfrac{\partial²f}{\partial u \partial v} + (1+b-6b²)\dfrac{\partial²f}{\partial v²} $$ but I don't know where to go from here. When I attempt solving for $a$ and $b$ I end up with much more complicated expressions than I expect to get in this exercise (like $a=\frac{1}{12} \pm \sqrt{1/12² + 1/6}$) when I try to solve for 0 as the coefficients. The answer to the question says that they chose $a=1/2$ and $b=1/3$, but when I insert these values I only get the first coefficient to become 0, and I only know how to solve this if I am left with a single term.

How can this be solved?

edit (solution):

I didn't see the common denominator $\frac{1}{12}$ under the square-root when solving for $a$ in $a = \frac{1}{12} \pm \sqrt{\frac{1}{12²} + \frac{1}{6}}$. It could be rearranged into $a = \frac{1 \pm \sqrt{1 + 24}}{12} = \frac{1 \pm 5}{12}$. After that, one can solve for $b$ to make the second term go to $0$ as the answer suggested, and then integrate the last term to solve the differential equation (remember to add the functions of the other variable when integrating).

The answer from @tarkovsky123 helped me know that the rest of the solution was correct.

$\endgroup$

1 Answer 1

0
$\begingroup$

After completing your transformation you want to choose a and b such that the coefficient in front of the mixed derivative and the one homogenous in either u or v equals zero. You can (for example) choose to solve for a in the first quadratic equation. The solutions are a=1/2 and a=-1/3. Choose one of these, e.g. a =1/2 and solve for b in the second equation. Then you can continue solving the differential equation with known one-dimensional tools.

$\endgroup$
3
  • $\begingroup$ How do you see that it should be $a=1/2$? When I solve it like a quadratic equation, I get $a = (1/12² - 1/6)^{1/2} + 1/12$. Am I solving the quadratic equation for $a$ incorrectly? 0.o $\endgroup$
    – dekuShrub
    Apr 25, 2018 at 13:39
  • $\begingroup$ Try put a= -1/3 or a=1/2 into the first equation. Clearly it equals zero. If you complete the square you get the expression (a-1/2)(a+1/3). So choose one of these and solve for b in the second equation. Then continue solving the pde with these a and b. $\endgroup$ Apr 25, 2018 at 13:49
  • $\begingroup$ This helped me find where I went wrong, so I mark this as the answer. See my edited question for solution. $\endgroup$
    – dekuShrub
    Apr 26, 2018 at 9:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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