Find a solution of $u_{xx}+u_{yy} = Ax+By+C$ for given real constants $A,B$ and $C$. How many more solutions can be produced?

Question

1. Observe that $u_1(x,y)=x^2$ solves $u_{xx}+u{yy} =2$ and $u_2(x,y) =cx^3+dy^3$ solves $u_{xx}+u_{yy} =6cx+6dy$ for real constants $c$ and $d$.

Question: a. Find a solution of $u_{xx}+u_{yy} = Ax+By+C$ for given real constants $A,B$ and $C$.

My attempt:

We need to find a solution that satisfies $u_{xx}+u_{yy} = Ax+By+C$.

First, we notice that we are working with two variables: x and y. Since x only appears next to A, we assume that the y variable must be next to B and C.

Suppose we have the following equation: $\frac{1}{6}Ax^3+\frac{1}{6}By^3+\frac{1}{2}Cy^2$

Taking the first partial derivatives in terms of x and y, we have,

$u_x = \frac{1}{2}Ax^2$ and $u_y = \frac{1}{2}By^2+Cy$

Adding $u_x$ and $u_y$ produces the following result:

$u_x+u_y=\frac{1}{2}Ax^2+\frac{1}{2}By^2+Cy$

Taking the derivatives of $u_x$ and $u_y$, we have

$u_{xx} =Ax$ and $u_{yy} = By+C$

Adding $u_{xx}$ and $u_{yy}$ gives us $Ax+By+C$, so we have proven that $\frac{1}{6}Ax^3+\frac{1}{6}By^3+\frac{1}{2}Cy^2$ satisfies $u_{xx}+u_{yy} = Ax+By+C$.

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Question: b. How many more solutions of the problem in $(a)$ be produced?

Ok. This is where I get stuck... well not exactly stuck more like unsure. This is what I think. Suppose we have fractions with bigger denominators, then it would be impossible to achieve $Ax+By+C$.

For example, suppose we have an equation $\frac{1}{12}Ax^3+\frac{1}{12}By^3+\frac{1}{6}y^2C$

Taking the $u_x$ and $u_y$ we have

$\frac{1}{4}Ax^2+\frac{1}{4}By^2+\frac{1}{3}yC$

and the $u_{xx}$ and $u_{yy}$

would be $\frac{1}{2}Ax+\frac{1}{2}By+\frac{1}{3}C$ Unless I am able to multiply by 2 and manipulate the C part, then the equation won't match.

Also, the question did say whether or not we are able to generate more solutions that will satisfy $u_{xx}+u_{yy}=Ax+By+C$ which is a very strict condition. It doesn't say that we can have an integer that would allow us to multiply through the left side to equal the right side.

If we have fractional exponents then we have to consider that the exponent will change when we take the derivative and if we have a $e^x$ then taking the derivative will keep the $e^x$ and that doesn't satisfy the equation either. Therefore, no additional solutions can be produced.

Answer 1

Using the Laplace-operator $\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$, your equation writes $$ \Delta u = Ax + By + C. $$ Now, if both $u$ and $v$ are solutions of this equation, we obtain by considering the difference $\Delta(u-v)=0$. Functions $h\in C^2$ with $\Delta h=0$ are called harmonic. Conversely, if $u$ is a solution and $h$ is harmonic, then $u+h$ is again a solution. This shows the following: Whenever you have found one solution, then every other solution can be generated by adding a harmonic function!

Answer 2

You have found one solution to the equation. Now you want to find other solutions. Suppose you add $Dx$ to your solution. What happens. Well, if you differentiate $Dx$ (where $D$ is just a constant) twice wrt $x$ or twice wrt $y$ you get nil. So your solution $+Dx$ is also a solution. Similarly you can add $Ey$ and still have a solution.

The usual way of approaching this to say that if $u=f(x,y)$ is another solution, then we can put $f(x,y)=\frac{1}{6}Ax^3+\frac{1}{6}By^3+\frac{1}{2}Cy^2+g(x,y)$. Then $g$ must satisfy the equation $g_{xx}+g_{yy}=0$, so we just have to solve that equation.