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Solve $\displaystyle \frac{\partial^2 z}{\partial x^2}+z=0$, given that when $x=0$, $z=e^y$ and $\displaystyle \frac{\partial z}{\partial x}=1$.

My Attempt: Integrating w.r.t x twice (keeping y constant)

$\displaystyle \frac{\partial z}{\partial x}+xz=f(y)$

$\displaystyle z+\frac{x^2}{2}z=xf(y)+g(y)$

The problem has a hint given : start with $\displaystyle z=f(y)\sin x+g(y)\cos x$. Uisng this obviously helps.

But if I go the standard way, I get stuck.. Am I missing something here or do I just memorise this hint and hope for the best.

user91500
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square_one
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2 Answers2

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Notice that you cannot integrate the way you did since $z$ is the dependent variable of your problem and therefore cannot be treated as a simple constant, just like you did in the $xz$ term.

You may want to try solutions of the form (since the PDE does not depend on $y$):

$$z(x,y) = C e^{qx},$$ which after substituting back into the PDE yields:

$$q^2 + 1 = 0, $$ and therefore we have $q = \pm i$, so the set of solutions is hence described by:

$$z(x,y) = C_1(y) \sin{x} + C_2(y) \cos{x},$$ where $C_{1,2}$ are arbitrary functions of $y$, which remains to be fixed by some boundary conditions. I'm sure you can now solve for $C_1$ and $C_2$ with the given information.

Hope this helps.

Edit: notice that I just treated $y$ as a constant, because the PDE is independent from $y$, and can be solved like a classical ODE with constant coefficients. I said nothing regarding $C$, but it naturally becomes $y$-dependent through the boundary conditions. Also notice that this is not the general approach to solve PDEs but it's heavily faster in this cases.

Cheers!

Dmoreno
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If you know how to solve higher order linear ODE, then you can just note that the PDE you have contains only derivatives with respect to $x$, so it is essentially a linear homogeneous ODE, with auxiliary equation $m^2 + 1 = 0 \Rightarrow m = \pm i$, for which the solution is $c_1\cos(x) + c_2\sin(x)$. As we actually have a PDE (with the only other independent variable being $y$), $c_1$ and $c_2$ should be functions of $y$ rather than pure constants.

If you want a "from scratch" solution, here goes...

Let $\partial_x = \dfrac{\partial}{\partial x}$. Then the given PDE is $\partial_x^2z + z = 0$, or $(\partial_x^2 + 1)z = 0$. This can be written as $(\partial_x + i)(\partial_x - i)z = 0$ (the validity of this factorization of the operator can be verified by actually carrying out the differentiations and verifying that this becomes the original equation - you will see that it is only a convenient notation). Thus we have to solve

$$(\partial_x + i)(\partial_x - i)z = 0$$

Let $(\partial_x - i)z = w$. Then the equation becomes

$(\partial_x + i)w = 0\Rightarrow\\ \partial_x w = -iw \Rightarrow\\ \dfrac{1}{w} \partial_xw = -i \Rightarrow\\ \partial_x (\log w) = -i \Rightarrow\\ \log w = -ix + C(y) \Rightarrow\\ w = c_1(y) e^{-ix} \qquad (\text{where we have written $e^{C(y)}$ as $c_1(y)$}) $

Now, as $(\partial_x - i)z = w$, we have

$ \partial_x z - iz = c_1(y)e^{-ix} = c_1 e^{-ix} \Rightarrow\\ \dfrac{\partial z}{\partial x} - iz = c_1 e^{-ix} $

This is a first order linear equation that can be solved using the integrating factor $e^{\int (-i)\, dx} = e^{-ix}$. Therefore:

$z\times \text{IF} = \displaystyle \int \text{RHS}\times \text{IF} \,dx \Rightarrow\\ z e^{-ix} = \displaystyle \int c_1 e^{-2ix} \,dx \Rightarrow\\ z e^{-ix} = c_1 \dfrac{e^{-2ix}}{-2i} + c_2 a = C_1 e^{-2ix} + C_2 \Rightarrow\\ z = C_1 e^{-ix} + C_2 e^{ix} \Rightarrow\\ z = C_1(\cos x - i \sin x) + C_2(\cos x + i \sin x) \Rightarrow\\ z = (C_1 + C_2) \cos x - i(C_1 - C_2) \sin x \Rightarrow\\ z = A \cos x + B \sin x $

Now as the equation is actually a PDE, $A$ and $B$ are functions of $y$, say $A = f(y)$ and $B = g(y)$. Thus

$$\boxed{z = f(y)\cos x + g(y) \sin x}$$

M. Vinay
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  • You should say here that $C_1$ and $C_2$ are complex conjugate (functions of $y$) and therefore $C_1 + C_2$ is real and $C_1 - C_2$ purely imaginary, so $A$ and $B$ are real (functions of $y$). – Dmoreno Jun 18 '14 at 14:27
  • Not necessarily. $A$ and $B$ might be complex constants too (depending on the given conditions). Or complex valued functions of $y$, as the case may be. The conditions given in the OP result in a real solution. – M. Vinay Jun 18 '14 at 15:12
  • Of course. I was confused with the solution of a real-coefficients ODE when the solutions of the characteristic equation are complex. Sorry! – Dmoreno Jun 18 '14 at 15:15