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The general solution of PDE $u_{xx} +u_{yy}=0$. There are four options given (correct option is given as d):

a) $ u=f(x+iy)-g(x-iy)$

b) $ u=f(x-iy)-g(x-iy)$

c) $ u=f(x-iy)+g(x+iy)$

d) $ u=f(x+iy)+g(x-iy)$

My attempt:

This is a homogeneous linear equation with constant coefficients:

So, its auxiliary equation is $m^2+1=0$. $\implies m=\pm i $

The textbook I am referring to (which also has this question) gives the following general solution for complex roots :

$ u=f_1(y+ix)+f_1(y-ix)+i\left[f_2(y+ix)+f_2(y-ix) \right]$

This question says general solution is $ u=C_1(x+iy)+C_2(x-iy)$

Which one is correct ? Moreover, I don't see any difference between options (a),(c) and (d).

Please advise.

square_one
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1 Answers1

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This is a ridiculously bad multiple-choice question. You are right that (a), (c), and (d) are all correct answers. The only way to have (d) as the unique correct answer is to interpret "correct" as "identical in appearance to what was presented in class".

The form $$u=C_1(x+iy)+C_2(x-iy)\tag{1}$$ with complex-valued functions $C_1,C_2$, is also correct.

The textbook I am referring to (which also has this question) gives the following general solution for complex roots : $$u=f_1(y+ix)+f_1(y-ix)+i\left[f_2(y+ix)+f_2(y-ix) \right] \tag{2}$$

As written, this is not the general solution. Indeed, any $u$ of the form (2) satisfies $u(-x,y)=u(x,y)$. So, the formula (2) does not include $u(x,y)=x$, which is obviously a solution of the PDE.

It would be correct to write $$u=f_1(y+ix)+g_1(y-ix)+i\left[f_2(y+ix)+g_2(y-ix) \right] \tag{3}$$ with real-valued $f_1,f_2,g_1,g_2$. This is just a result of separating the real and imaginary parts in $u=f (y+ix)+g (y-ix) $.