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\Delta u=u_{{xx}}+u_{{yy}}=0\,u_{x}(0,y)=0\,
u_{x}(\pi ,y)=0\,
u(x,0)=K\cos x\,
u(x,1)=K\cos ^{2}x\,
t>0,\,\,0<x<\pi \,


{\frac  {X''}{X}}=-{\frac  {Y''}{Y}}=-\lambda ^{2}\,

New BC's:

X'(0)=0,X'(\pi )=0\,

Y(0)=K\cos x,Y(1)=K\cos ^{2}x\,

{\frac  {X''}{X}}=-{\frac  {Y''}{Y}}=-\lambda ^{2}\,

X(x)=c_{1}\cos \lambda x+c_{2}\sin \lambda x\,

X'(x)=-c_{1}\lambda \sin \lambda x+c_{2}\lambda \cos \lambda x\,

X'(0)=c_{2}\lambda =0\rightarrow c_{2}=0\,

X'(\pi )=-c_{2}\lambda \sin \lambda \pi =0\rightarrow \lambda =n,n=1,2,...\,

If n=0,\,\,X''=0,\,\,X_{0}(x)=c_{3}x+c_{4}\,,


X_{0}(x)=c_{4}\,. For mathematical ease, let c_{4}=1\,

So the solution for X(x)\, is

X_{n}(x)=d_{n}\cos nx,\,\,n=1,2,...\,


Y(y)=c_{5}\cosh ny+c_{6}\sinh ny\,

If n=0,\,\,Y''=0,\,\,Y_{0}(y)=c_{7}y+c_{8}\,

Using superposition to add all these solutions together,

u(x,t)=c_{7}y+c_{8}+\sum _{{n=1}}^{\infty }(A_{n}\cosh ny+B_{n}\sinh ny)\cos nx\,

u(x,0)=c_{8}+\sum _{{n=1}}^{\infty }A_{n}\cos nx=K\cos x\,

We are able to determine the coefficients just by comparing like terms.


u(x,y)=c_{7}y+K\cosh y\cos x+\sum _{{n=1}}^{\infty }B_{n}\sinh ny\cos nx\,

u(x,1)=c_{7}+K\cosh 1\cos x+\sum _{{n=1}}^{\infty }B_{n}\sinh n\cos nx=K\cos ^{2}x\,

Using \cos ^{2}x={\frac  {1+\cos 2x}{2}}\,

{\frac  {K}{2}}(1+\cos 2x)-K\cosh 1\cos x=c_{7}+\sum _{{n=1}}^{\infty }B_{n}\sinh n\cos nx\,

c_{7}={\frac  {K}{2}}\,

B_{1}={\frac  {-K\cosh 1}{\sinh 1}}\,

B_{2}={\frac  {K}{2\sinh 2}}\,


Now the solution is

u(x,y)={\frac  {Ky}{2}}+K\cosh y\cos x+{\frac  {-K\cosh 1}{\sinh 1}}\sinh y\cos x+{\frac  {K}{2\sinh 2}}\sinh 2y\cos 2x\,

Using the trig identity \sinh(A-B)=\sinh A\cosh B-\cosh A\sinh B\, the solution can be simplified to:

u(x,y)=K\left[{\frac  {1}{2}}y+{\frac  {\sinh(1-y)}{\sinh 1}}\cos x+{\frac  {\sinh 2y}{2\sinh 2}}\cos 2x\right]\,

Partial Differential Equations

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