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.


1 Answer 1


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.

  • $\begingroup$ 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

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