这是一份ed.ac爱丁堡格大学MATH10086作业代写的成功案

$$
\sigma_{x}=\frac{\partial^{2} U}{\partial x^{2}}, \quad \tau_{x y}=-\frac{\partial^{2} U}{\partial x \partial y}, \quad \sigma_{y}=\frac{\partial^{2} U}{\partial y^{2}}=0
$$
Keeping them in mind, we are able to combin into a single biharmonic equation defined by:
$$
\frac{\partial^{4} U}{\partial x^{4}}+2 \frac{\partial^{4} U}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} U}{\partial y^{4}}=0
$$
The solution to the above equation is that biharmonic function $U$ must be of the following form
$$
U(x, y)=\operatorname{Re}[\bar{z} \varphi(z)+\chi(z)]
$$

MATH10086 COURSE NOTES ：

$$
\alpha(z) \varphi(z)+\beta(z) z \overline{\varphi^{\prime}(z)}+\beta(z) \overline{\psi(z)}=F \text { for } z \in \Gamma
$$
where $\alpha, \beta$ and $F$ are associated with the boundary conditions. For force boundary conditions, we have
$$
\alpha=1, \quad \beta=1, \quad F=i \int_{z_{0}}^{z}\left(X_{n}+i Y_{n}\right) d s
$$
whereas for displacement boundary conditions,
$$
\alpha=\kappa, \quad \beta=-1, \quad F=2 \mu(\tilde{u}+i \tilde{v}) .
$$