Let z(xi,eta) have continuous partial derivatives

Let $z(\xi,\eta)$ have continuous partial derivatives, and let: $$ u(x,y) = \frac12\int_0^x d\xi\int_{\xi-x+y}^{x+y-\xi} z(\xi,\eta)\,d\eta. $$ Show that: $$ \frac{\partial^2u}{\partial x^2} - \frac{\partial^2u}{\partial y^2} = z(x,y). $$

Step 1 · First derivative with respect to $x$

Write $u(x,y)=\frac12\displaystyle\int_0^x\varphi(\xi,x,y)\,d\xi$, where $\varphi(\xi,x,y)=\displaystyle\int_{\xi-x+y}^{x+y-\xi}z(\xi,\eta)\,d\eta$. By Leibniz's rule (the upper limit $x$ contributes a boundary term, and $x$ also sits inside $\varphi$):

$$ \frac{\partial u}{\partial x} = \frac12\left[\varphi(x,x,y) + \int_0^x\frac{\partial\varphi}{\partial x}\,d\xi\right]. $$

The boundary term vanishes, since setting $\xi=x$ collapses the limits of $\varphi$ to a single point:

$$ \varphi(x,x,y) = \int_{x-x+y}^{x+y-x} z(x,\eta)\,d\eta = \int_y^y z(x,\eta)\,d\eta = 0. $$

Differentiating $\varphi$ under its own integral sign (Leibniz's rule again, on the inner integral's variable limits):

$$ \frac{\partial\varphi}{\partial x} = z(\xi,x+y-\xi)\cdot1 - z(\xi,\xi-x+y)\cdot(-1) = z(\xi,x+y-\xi)+z(\xi,\xi-x+y). $$

Hence:

$$ \frac{\partial u}{\partial x} = \frac12\int_0^x\big[z(\xi,x+y-\xi)+z(\xi,\xi-x+y)\big]\,d\xi. $$

Step 2 · First derivative with respect to $y$

Here $x$ (the outer limit) does not depend on $y$, so only the inner limits contribute, via the same rule:

$$ \frac{\partial\varphi}{\partial y} = z(\xi,x+y-\xi)\cdot1 - z(\xi,\xi-x+y)\cdot1 = z(\xi,x+y-\xi)-z(\xi,\xi-x+y), $$

so:

$$ \frac{\partial u}{\partial y} = \frac12\int_0^x\big[z(\xi,x+y-\xi)-z(\xi,\xi-x+y)\big]\,d\xi. $$

Step 3 · Second derivative with respect to $x$

Apply Leibniz's rule once more to $u_x=\frac12\displaystyle\int_0^x\psi(\xi,x,y)\,d\xi$ with $\psi=z(\xi,x+y-\xi)+z(\xi,\xi-x+y)$:

$$ \frac{\partial^2u}{\partial x^2} = \frac12\left[\psi(x,x,y)+\int_0^x\frac{\partial\psi}{\partial x}\,d\xi\right]. $$

The boundary term no longer vanishes — both pieces of $\psi(x,x,y)$ evaluate at $\eta=y$:

$$ \psi(x,x,y) = z(x,x+y-x)+z(x,x-x+y) = z(x,y)+z(x,y) = 2z(x,y). $$

Differentiating the integrand, using $z_\eta$ for the partial derivative of $z$ with respect to its second argument:

$$ \frac{\partial\psi}{\partial x} = z_\eta(\xi,x+y-\xi)\cdot1 - z_\eta(\xi,\xi-x+y)\cdot(-1)\cdot(-1) = z_\eta(\xi,x+y-\xi)-z_\eta(\xi,\xi-x+y). $$

So:

$$ \frac{\partial^2u}{\partial x^2} = \frac12\left[2z(x,y)+\int_0^x\big(z_\eta(\xi,x+y-\xi)-z_\eta(\xi,\xi-x+y)\big)\,d\xi\right] = z(x,y)+\frac12\int_0^x\big[z_\eta(\xi,x+y-\xi)-z_\eta(\xi,\xi-x+y)\big]\,d\xi. $$

Step 4 · Second derivative with respect to $y$

Differentiate $u_y=\frac12\displaystyle\int_0^x\big[z(\xi,x+y-\xi)-z(\xi,\xi-x+y)\big]\,d\xi$ with respect to $y$; the outer limit $x$ is independent of $y$, so no boundary term arises:

$$ \frac{\partial^2u}{\partial y^2} = \frac12\int_0^x\big[z_\eta(\xi,x+y-\xi)-z_\eta(\xi,\xi-x+y)\big]\,d\xi. $$

This is exactly the same integral that appeared as the correction term in Step 3.

Step 5 · Subtract

$$ \frac{\partial^2u}{\partial x^2}-\frac{\partial^2u}{\partial y^2} = \left[z(x,y)+\frac12\int_0^x(\cdots)\,d\xi\right] - \frac12\int_0^x(\cdots)\,d\xi = z(x,y). $$

Additional Notes

• The region of integration $\{(\xi,\eta):0\leqslant\xi\leqslant x,\ \xi-x+y\leqslant\eta\leqslant x+y-\xi\}$ is precisely the characteristic triangle for the one-dimensional wave operator $\partial_x^2-\partial_y^2$, with apex $(x,y)$ and base on the line $\xi=0$ — this formula is the double-integral (Riemann/d'Alembert-type) representation of the solution to $u_{xx}-u_{yy}=z$ with zero Cauchy data on $x=0$.
• The vanishing of the boundary term $\varphi(x,x,y)$ in Step 1, but not of $\psi(x,x,y)$ in Step 3, is the crux of the computation: it is exactly why one derivative in $x$ produces no source term while two derivatives produce precisely $z(x,y)$, matching the coefficient of $z$ in the PDE.