Calculating partial derivative.

Question

Let $f=f(x,y)$ a function of two times continuously differentiable.

Given :

$\frac{\partial^2{f}}{\partial{x^2}}-\frac{\partial^2{f}}{\partial{y^2}}=0$

$f(x,2x)=x$

$\frac{\partial{f}}{\partial{x}}(x,2x)=x^2$

Question:

Find $\frac{\partial^2{f}}{\partial{x^2}}(x,2x)$ and $\frac{\partial^2{f}}{\partial{x}\partial{y}}(x,2x)$.

I've tried determining the function $f(x,y)$ out of the given info but in vain and I'm not even sure if it's possible to determine the function $f(x,y)$ as there could be multiple functions with the same partial derivatives? (correct me if I'm wrong).

So I'm assuming there's a simple way to solve this problem.

If anyone could give me hints or point me in the right direction I'd appreciate it!

I'd also be grateful if anyone knows somewhere where i could find similar problems to this one.

Thanks in advance.

Answer 1

Let $g(x) := f(x,2x)$. Then we know that $g(x) =x$. Thus by multidimensional chain-rule $$\tag{1} 1= g'(x) = \partial_x f(x,2x)+2\partial_y f(x,2x).$$ This shows that $$\partial_y f(x,2x) = \frac{1-x^2}{2}.$$ Now, we apply the same argument to $h(x) = \partial_x f(x,2x) =x^2$. This gives $$\tag{2}2x = h'(x)= \partial_x^2 f(x,2x) + 2 \partial_{xy} f(x,2x).$$ And for $l(x):=\partial_y f(x,2x)$, we have $$\tag{3}-x = l'(x) = \partial_{xy} f(x,2x)+2 \partial_{y}^2 f(x,2x).$$ Subtracting $2 \times(2)-(3)$ gives $$5x=3\partial_{xy} f(x,2x)+2\partial_x^2 f(x,2x)-2\partial_{y}^2 f(x,2x)=3\partial_{xy} f(x,2x),$$ because $\partial_x^2 f = \partial_y^2 f$. So we find $$\partial_{xy} f(x,2x) =\frac{5}{3}x.$$ Inserting in $(3)$ gives now $$\partial_{x}^2 f(x,2x)=\partial_{y}^2 f(x,2x) = - \frac{4}{3} x.$$ Additonal Comment: However, the question is if such a function exists and if it is unique.

Second Comment: By solving a linear equation I have cooked up following example: $$f(x,y)=- \frac{13}{18} x^2 y+\frac{7}{9}y^2 x+\frac{7}{27}x^3-\frac{13}{54} y^3+\frac{1}{2}y$$ As Andrew commented, the equation in question is a Cauchy problem for the wave equation and can be treated accordingly. Thus, the solution is unique.

Answer 2

There's a very direct way of solving this. Because we are always interested in $y$ around $2x$, let's just represent the function as a series:

$$f(x,y)=\sum_{k=0}^\infty a_k(x) (y-2x)^k$$

Immediately we can see from the last two conditions:

$$a_0(x)=x$$

$$a_0'(x)-2a_1(x)=x^2 \quad \Rightarrow a_1(x)=\frac{1-x^2}{2}$$

And from the first condition after some work:

$$a_k''-4(k+1)a_{k+1}'+3(k+1)(k+2)a_{k+2}=0$$

And now we can find every coefficient:

$$a_k=\frac{4(k-1) a_{k-1}'-a_{k-2}''}{3k(k-1)}$$

We find:

$$a_2=\frac{-4x}{6}=-\frac{2}{3} x$$

Now to get the answers we need:

$$f_{xx} (x,2x)=a_0''-4a_1'+8a_2=4x-\frac{16}{3} x=-\frac{4}{3} x$$

$$f_{xy} (x,2x)=a_1'-4a_2=-x+\frac{8}{3} x=\frac{5}{3} x$$

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Using the recursion formula further, we obtain:

$$a_3=-\frac{13}{54}$$

$$a_{k >3}=0$$

Which makes the function to be:

$$f(x,y)=x+\frac{1}{2}(1-x^2)(y-2x)-\frac{2}{3} x (y-2x)^2-\frac{13}{54} (y-2x)^3=$$

$$=\frac{1}{54} (14 x^3 - 39 x^2 y + 42 x y^2 - 13 y^3 + 27 y)$$

Answer 3

To elaborate on various comments:

Claim. A function $f:\>{\mathbb R}^2\to{\mathbb R}$ satisfying the wave equation $f_{xx}-f_{yy}\equiv0$ can be written in the form $$f(x,y)=a(x-y)+b(x+y)\tag{1}$$ for certain functions $a$, $b$ of one variable.

This is very well known. Here is a sketch of proof: Consider the auxiliary function $$g(u,v):=f\left({u+v\over2},{-u+v\over2}\right)\ .$$ Using $f_{xx}-f_{yy}\equiv0$ one easily verifies that $g_{uv}\equiv0$, i.e., $(g_u)_v\equiv0$. This implies that $g_u$ depends only on $u$, and one further integration gives $g(u,v)=a(u)+b(v)$, hence $$f(x,y)=g(x-y,x+y)=a(x-y)+b(x+y)\ .\qquad\square$$ We are given that $$f(x,2x)=a(-x)+b(3x)=x\ ,$$ which implies $f(0,0)=0$ and $$-a'(-x)+3b'(3x)=1\ .\tag{2}$$ Furthermore $f_x(x,y)=a'(x-y)+b'(x+y)$ implies $f_x(x,2x)=a'(-x)+b'(3x)$, so that we obtain the condition $$a'(-x)+b'(3x)=x^2\ .\tag{3}$$ Solving $(2)\wedge(3)$ gives $$a'(-x)={3x^2-1\over 4},\qquad b'(3x)={x^2+1\over4}$$ and therefore $$a'(x)={3x^2-1\over4},\qquad b'(x)={x^2+9\over36}\ .$$ Since we may take $a(0)=b(0)=0$ we then obtain $$a(x)={x^3-x\over4},\qquad b(x)={x^3+27x\over108}\ .$$ Plugging this into $(1)$ gives the uniquely determined $f$ appearing as "example" in p4sch's answer.