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Kostya_I
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This is obviously not true if you allow general (i.e., varying along a sphere) boundary values. For example, let the boundary values be positive but zero on the hemispheres facing each other. Then $f(0,0,0)$ is positive but tends to zero both as the spheres get close to each other and as $d\to\infty$.

If the boundary value on each sphere is constant, then the result is true. First, you can assume that the constants are the same, say positive, by subtracting a solution with constants differing by sign, which vanishes at the origin. Then, denoting the solution with distance between the spheres $d$ and radii $r$ by $f_{d,r}$, we have for any $a>1,$ $$ f_{d,r}(0)=f_{ad,ar}(0)> f_{ad,r}(0), $$ because the values of $f_{ad,r}$ on the spheres of radius $ar$ are smaller than their constant boundary values on the spheres of radius $r$.

This is obviously not true if you allow general (i.e., varying along a sphere) boundary values. For example, let the boundary values be positive but zero on the hemispheres facing each other. Then $f(0,0,0)$ is positive but tends to zero both as the spheres get close to each other and as $d\to\infty$.

If the boundary value on each sphere is constant, then the result is true. First, you can assume that the constants are the same, by subtracting a solution with constants differing by sign, which vanishes at the origin. Then, denoting the solution with distance between the spheres $d$ and radii $r$ by $f_{d,r}$, we have for any $a>1,$ $$ f_{d,r}(0)=f_{ad,ar}(0)> f_{ad,r}(0), $$ because the values of $f_{ad,r}$ on the spheres of radius $ar$ are smaller than their constant boundary values on the spheres of radius $r$.

This is obviously not true if you allow general (i.e., varying along a sphere) boundary values. For example, let the boundary values be positive but zero on the hemispheres facing each other. Then $f(0,0,0)$ is positive but tends to zero both as the spheres get close to each other and as $d\to\infty$.

If the boundary value on each sphere is constant, then the result is true. First, you can assume that the constants are the same, say positive, by subtracting a solution with constants differing by sign, which vanishes at the origin. Then, denoting the solution with distance between the spheres $d$ and radii $r$ by $f_{d,r}$, we have for any $a>1,$ $$ f_{d,r}(0)=f_{ad,ar}(0)> f_{ad,r}(0), $$ because the values of $f_{ad,r}$ on the spheres of radius $ar$ are smaller than their constant boundary values on the spheres of radius $r$.

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Kostya_I
  • 9k
  • 1
  • 32
  • 51

This is obviously not true if you allow general (i.e., varying along a sphere) boundary values. For example, let the boundary values be positive but zero on the hemispheres facing each other. Then $f(0,0,0)$ is positive but tends to zero both as the spheres get close to each other and as $d\to\infty$.

If the boundary value on each sphere is constant, then the result is true. First, you can assume that the constants are the same, by subtracting a solution with constants differing by sign, which vanishes at the origin. Then, denoting the solution with distance between the spheres $d$ and radii $r$ by $f_{d,r}$, we have for any $a>1,$ $$ f_{d,r}(0)=f_{ad,ar}(0)> f_{ad,r}(0), $$ because the values of $f_{ad,r}$ on the spheres of radius $ar$ are smaller than their constant boundary values on the spheres of radius $r$.